Chapter 5 Continuity And Differentiability

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}\frac{1− \cos 4x}{8x^2},& x≠0\\k,&x=0\end{cases}$

To Find:

The value of the constant $k$ such that the function $f(x)$ is continuous at $x=0$.

Solution:

For the function $f(x)$ to be continuous at $x=0$, the limit of the function as $x$ approaches $0$ must be equal to the value of the function at $x=0$. That is:

From the definition of the function $f(x)$, we can directly find the value of $f(0)$ by using the second case ($x=0$):

$f(0) = k$

Next, we need to evaluate the limit of $f(x)$ as $x$ approaches $0$. For $x \neq 0$, the function is defined as $\frac{1 - \cos 4x}{8x^2}$. So, we consider the limit:

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{1− \cos 4x}{8x^2}$

This limit is in the indeterminate form $\frac{0}{0}$ as $x \to 0$. We can use the trigonometric identity $1 - \cos 2\theta = 2\sin^2\theta$. Let $\theta = 2x$, so $2\theta = 4x$. Then, $1 - \cos 4x = 2\sin^2(2x)$.

Substitute this into the limit expression:

$\lim\limits_{x \to 0} \frac{2\sin^2(2x)}{8x^2} = \lim\limits_{x \to 0} \frac{\sin^2(2x)}{4x^2}$

We can rewrite the denominator to match the argument of the sine function:

$= \lim\limits_{x \to 0} \frac{(\sin(2x))^2}{(2x)^2}$

$= \lim\limits_{x \to 0} \left(\frac{\sin(2x)}{2x}\right)^2$

Let $y = 2x$. As $x \to 0$, $y \to 0$. Using the standard fundamental trigonometric limit $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$, we get:

$= \left(\lim\limits_{y \to 0} \frac{\sin y}{y}\right)^2$

$= (1)^2$

$= 1$

So, the limit of the function as $x$ approaches $0$ is:

$\lim\limits_{x \to 0} f(x) = 1$

For the function to be continuous at $x=0$, we apply the condition from equation (i):

Substituting the values we found for the limit and $f(0)$:

$1 = k$

Therefore, the value of the constant $k$ for which the function $f(x)$ is continuous at $x=0$ is 1.

Alternate Method for Limit Calculation

Alternatively, we can evaluate the limit $\lim\limits_{x \to 0} \frac{1− \cos 4x}{8x^2}$ using L'Hopital's Rule, since it is in the indeterminate form $\frac{0}{0}$.

Apply L'Hopital's Rule by differentiating the numerator and the denominator with respect to $x$:

$\lim\limits_{x \to 0} \frac{1− \cos 4x}{8x^2} = \lim\limits_{x \to 0} \frac{\frac{d}{dx}(1− \cos 4x)}{\frac{d}{dx}(8x^2)}$

$= \lim\limits_{x \to 0} \frac{-(-4\sin 4x)}{16x}$

$= \lim\limits_{x \to 0} \frac{4\sin 4x}{16x}$

$= \lim\limits_{x \to 0} \frac{\sin 4x}{4x}$

Let $y = 4x$. As $x \to 0$, $y \to 0$. Using the standard limit $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$, we get:

$= \lim\limits_{y \to 0} \frac{\sin y}{y} = 1$

This confirms that $\lim\limits_{x \to 0} f(x) = 1$. Equating this limit to $f(0) = k$ (from the continuity condition $\lim\limits_{x \to 0} f(x) = f(0)$) gives $k=1$.

Given:

The function is $f(x) = \sin x \cdot \cos x$.

To Discuss:

The continuity of the function $f(x)$.

Solution:

A function $f(x)$ is said to be continuous on an interval if it is continuous at every point in that interval.

We know that the function $\sin x$ is continuous for all real numbers. Its domain is $\mathbb{R}$.

We also know that the function $\cos x$ is continuous for all real numbers. Its domain is $\mathbb{R}$.

The given function $f(x) = \sin x \cdot \cos x$ is the product of two functions, $g(x) = \sin x$ and $h(x) = \cos x$.

A key property of continuous functions states that if two functions are continuous at a point (or on an interval), their product is also continuous at that point (or on that interval).

Since both $g(x) = \sin x$ and $h(x) = \cos x$ are continuous for all real numbers, their product $f(x) = g(x) \cdot h(x) = \sin x \cdot \cos x$ must also be continuous for all real numbers.

Therefore, the function $f(x) = \sin x \cdot \cos x$ is continuous on its entire domain, which is $\mathbb{R}$ (all real numbers).

Alternate Method (Using Limit Definition)

To check the continuity of $f(x)$ at an arbitrary point $c \in \mathbb{R}$, we need to verify if $\lim\limits_{x \to c} f(x) = f(c)$.

First, find $f(c)$:

$f(c) = \sin c \cos c$

Next, find the limit of $f(x)$ as $x$ approaches $c$:

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (\sin x \cos x)$

Since $\sin x$ and $\cos x$ are continuous functions, we know that $\lim\limits_{x \to c} \sin x = \sin c$ and $\lim\limits_{x \to c} \cos x = \cos c$.

Using the property of limits that the limit of a product is the product of the limits:

$\lim\limits_{x \to c} (\sin x \cos x) = (\lim\limits_{x \to c} \sin x) \cdot (\lim\limits_{x \to c} \cos x)$

$= (\sin c) \cdot (\cos c)$

$= \sin c \cos c$

So, we have $\lim\limits_{x \to c} f(x) = \sin c \cos c$ and $f(c) = \sin c \cos c$.

Since $\lim\limits_{x \to c} f(x) = f(c)$ for any arbitrary real number $c$, the function $f(x) = \sin x \cos x$ is continuous for all real numbers.

Alternate Method (Using Trigonometric Identity)

We can also rewrite the function using the identity $\sin 2x = 2 \sin x \cos x$.

$f(x) = \sin x \cos x = \frac{1}{2} \sin 2x$

Let $g(x) = \sin x$. We know $g(x)$ is continuous for all real numbers. The function $\sin 2x$ is a composition of the function $g(y) = \sin y$ and $y = 2x$. Since $y = 2x$ is a polynomial function, it is continuous for all real numbers.

The composition of two continuous functions is continuous. Therefore, $\sin 2x$ is continuous for all real numbers.

Multiplying a continuous function by a constant ($\frac{1}{2}$) results in a continuous function.

Thus, $f(x) = \frac{1}{2} \sin 2x$ is continuous for all real numbers.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}\frac{x^3+x^2 − 16x + 20}{(x − 2)^2}, & x≠2\\k,& x=2\end{cases}$

The function is continuous at $x = 2$.

To Find:

The value of the constant $k$.

Solution:

For the function $f(x)$ to be continuous at $x=2$, the limit of the function as $x$ approaches $2$ must be equal to the value of the function at $x=2$. That is:

From the definition of the function $f(x)$, we can directly find the value of $f(2)$ by using the second case ($x=2$):

$f(2) = k$

Next, we need to evaluate the limit of $f(x)$ as $x$ approaches $2$. For $x \neq 2$, the function is defined as $\frac{x^3+x^2 − 16x + 20}{(x − 2)^2}$. So, we consider the limit:

$\lim\limits_{x \to 2} f(x) = \lim\limits_{x \to 2} \frac{x^3+x^2 − 16x + 20}{(x − 2)^2}$

Let's evaluate the numerator and denominator at $x=2$:

Numerator at $x=2$: $2^3 + 2^2 - 16(2) + 20 = 8 + 4 - 32 + 20 = 12 - 32 + 20 = 32 - 32 = 0$.

Denominator at $x=2$: $(2 - 2)^2 = 0^2 = 0$.

The limit is in the indeterminate form $\frac{0}{0}$. Since the denominator has a factor of $(x-2)^2$, the numerator must also have $(x-2)^2$ as a factor for the limit to exist and be finite (which it must be for continuity).

We can factor the numerator $x^3+x^2 − 16x + 20$. Since $x=2$ is a root (because plugging in 2 gives 0), $(x-2)$ is a factor. We can perform polynomial division or synthetic division.

Using synthetic division with root 2:

$\begin{array}{c|cccc} 2 & 1 & 1 & -16 & 20 \\ & & 2 & 6 & -20 \\ \hline & 1 & 3 & -10 & 0 \end{array}$

The quotient is $x^2 + 3x - 10$. So, $x^3+x^2 − 16x + 20 = (x-2)(x^2 + 3x - 10)$.

Now, factor the quadratic $x^2 + 3x - 10$. We look for two numbers that multiply to -10 and add to 3. These numbers are 5 and -2.

$x^2 + 3x - 10 = (x+5)(x-2)$

So, the numerator is $(x-2)(x+5)(x-2) = (x-2)^2(x+5)$.

Substitute the factored numerator back into the limit expression:

$\lim\limits_{x \to 2} \frac{(x-2)^2(x+5)}{(x − 2)^2}$

For $x \neq 2$, $(x-2)^2 \neq 0$, so we can cancel the common factor $(x-2)^2$:

$\lim\limits_{x \to 2} (x+5)$

Now, substitute $x=2$ into the simplified expression:

$2 + 5 = 7$

So, the limit of the function as $x$ approaches $2$ is:

$\lim\limits_{x \to 2} f(x) = 7$

For the function to be continuous at $x=2$, we apply the condition from equation (i):

Substituting the values we found for the limit and $f(2)$:

$7 = k$

Therefore, the value of the constant $k$ for which the function $f(x)$ is continuous at $x=2$ is 7.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}x \sin \frac{1}{x},& x≠0\\0,& x=0\end{cases}$

To Show:

The function $f(x)$ is continuous at $x=0$.

Solution:

For the function $f(x)$ to be continuous at $x=0$, we need to show that the limit of the function as $x$ approaches $0$ exists and is equal to the value of the function at $x=0$. That is, we must show:

First, let's find the value of $f(0)$. From the definition of the function:

$f(0) = 0$

Next, we need to evaluate the limit of $f(x)$ as $x$ approaches $0$. For $x \neq 0$, the function is given by $f(x) = x \sin \frac{1}{x}$. So we consider:

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} x \sin \frac{1}{x}$

As $x \to 0$, the term $\frac{1}{x}$ approaches infinity (or minus infinity), and the function $\sin \frac{1}{x}$ oscillates between $-1$ and $1$. The limit of $\sin \frac{1}{x}$ as $x \to 0$ does not exist. However, we know that the sine function is bounded. For any non-zero value of $x$:

$-1 \leq \sin \frac{1}{x} \leq 1$

Now, multiply the inequality by $|x|$. Since $|x| \geq 0$, the direction of the inequality remains the same:

$-|x| \leq |x| \sin \frac{1}{x} \leq |x|$

We know that $|x| \sin \frac{1}{x}$ is either equal to $x \sin \frac{1}{x}$ (if $x>0$) or $-x \sin \frac{1}{x}$ (if $x<0$). However, we can write $-|x| \leq x \sin \frac{1}{x} \leq |x|$ directly because if $x>0$, we multiply by $x$, getting $-x \leq x \sin \frac{1}{x} \leq x$, which is $-|x| \leq x \sin \frac{1}{x} \leq |x|$. If $x<0$, we multiply by $x$ (a negative number) and reverse the inequality, $-x \geq x \sin \frac{1}{x} \geq x$, which is equivalent to $x \leq x \sin \frac{1}{x} \leq -x$. Since $x<0$, $|x| = -x$, so this is $x \leq x \sin \frac{1}{x} \leq |x|$. The previous form $-|x| \leq x \sin \frac{1}{x} \leq |x|$ covers both cases correctly (for $x<0$, $-|x|=x$).

Consider the limits of the bounding functions as $x \to 0$:

$\lim\limits_{x \to 0} (-|x|) = -|0| = 0$

$\lim\limits_{x \to 0} |x| = |0| = 0$

Since $\lim\limits_{x \to 0} (-|x|) = 0$ and $\lim\limits_{x \to 0} |x| = 0$, by the Squeeze Theorem, the limit of the function in between must also be $0$.

Therefore:

$\lim\limits_{x \to 0} x \sin \frac{1}{x} = 0$

So, $\lim\limits_{x \to 0} f(x) = 0$.

Now we compare the limit with the function value at $x=0$:

$\lim\limits_{x \to 0} f(x) = 0$

$f(0) = 0$

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the condition for continuity at $x=0$ is satisfied.

Thus, the function $f(x)$ is continuous at $x = 0$.

Given:

The function $f(x) = \frac{1}{x − 1}$.

The composite function $y = f[f(x)]$.

To Find:

The points of discontinuity of the composite function $y = f[f(x)]$.

Solution:

A composite function $f[g(x)]$ is discontinuous at a point $a$ if:

1. $a$ is a point of discontinuity for the inner function $g(x)$.

2. $g(a)$ is a point of discontinuity for the outer function $f(y)$, where $y = g(x)$.

In this problem, the inner function is $g(x) = f(x) = \frac{1}{x-1}$, and the outer function is $f(y) = \frac{1}{y-1}$.

First, let's find the points of discontinuity for the inner function $f(x) = \frac{1}{x-1}$.

The function $f(x)$ is a rational function. It is defined for all real numbers except where the denominator is zero.

The denominator is zero when $x - 1 = 0$, which implies $x = 1$.

So, the inner function $f(x)$ is discontinuous at $x = 1$. This is one point of discontinuity for $f[f(x)]$ based on condition 1.

Next, let's find the points where the outer function $f(y) = \frac{1}{y-1}$ is discontinuous.

The outer function $f(y)$ is discontinuous when its input $y$ is equal to $1$. That is, when $y = 1$.

Now, we need to find the values of $x$ for which the output of the inner function, $f(x)$, is equal to the discontinuity point of the outer function. Based on condition 2, $f[f(x)]$ is discontinuous when $f(x) = 1$.

Set the inner function equal to 1:

$f(x) = 1$

$\frac{1}{x-1} = 1$

Assuming $x-1 \neq 0$ (i.e., $x \neq 1$), we can multiply both sides by $(x-1)$:

$1 = 1 \cdot (x-1)$

$1 = x - 1$

Add 1 to both sides:

$x = 1 + 1$

$x = 2$

So, when $x = 2$, the inner function gives $f(2) = \frac{1}{2-1} = \frac{1}{1} = 1$. Since the outer function $f(y)$ is discontinuous at $y=1$, the composite function $f[f(x)]$ is discontinuous at $x = 2$. This is another point of discontinuity for $f[f(x)]$ based on condition 2.

The points of discontinuity of the composite function $f[f(x)]$ are the union of the points found in the two steps above.

Points from condition 1 (discontinuity of inner function): $x=1$.

Points from condition 2 (where inner function's output causes outer function discontinuity): $x=2$.

Therefore, the points of discontinuity for $y = f[f(x)]$ are $x=1$ and $x=2$.

Alternate Method (By finding the explicit formula for $f[f(x)]$)

First, find the explicit formula for the composite function $f[f(x)]$.

$f(x) = \frac{1}{x-1}$

$f[f(x)] = f\left(\frac{1}{x-1}\right)$

Substitute $\frac{1}{x-1}$ into $f(y) = \frac{1}{y-1}$ wherever $y$ appears:

$f[f(x)] = \frac{1}{\left(\frac{1}{x-1}\right) - 1}$

Simplify the denominator:

$\frac{1}{x-1} - 1 = \frac{1}{x-1} - \frac{x-1}{x-1} = \frac{1 - (x-1)}{x-1} = \frac{1 - x + 1}{x-1} = \frac{2-x}{x-1}$

So, $f[f(x)] = \frac{1}{\frac{2-x}{x-1}}$.

Invert and multiply:

$f[f(x)] = 1 \cdot \frac{x-1}{2-x} = \frac{x-1}{2-x}$.

Thus, the composite function is $f[f(x)] = \frac{x-1}{2-x}$.

Now, we need to find the points of discontinuity of this composite function. A rational function is discontinuous where its denominator is zero.

The denominator of $\frac{x-1}{2-x}$ is zero when $2 - x = 0$, which implies $x = 2$.

However, we must also consider the domain of the inner function $f(x)$. The composite function $f[f(x)]$ is only defined if the inner function $f(x)$ is defined. The inner function $f(x) = \frac{1}{x-1}$ is undefined at $x=1$. If $x=1$, $f(1)$ does not exist, and therefore $f[f(1)]$ is also undefined.

So, the points where $f[f(x)]$ is undefined (and thus discontinuous) are where the simplified expression's denominator is zero OR where the inner function is undefined.

From the simplified expression $\frac{x-1}{2-x}$, discontinuity occurs at $x=2$.

From the inner function $f(x)=\frac{1}{x-1}$, discontinuity occurs at $x=1$.

Combining these points, the points of discontinuity are $x=1$ and $x=2$.

Given:

The function $f(x) = x|x|$ for all $x \in \mathbb{R}$.

We can write the function $f(x)$ in a piecewise form based on the definition of $|x|$:

• If $x \geq 0$, then $|x| = x$. So, $f(x) = x \cdot x = x^2$.
• If $x < 0$, then $|x| = -x$. So, $f(x) = x \cdot (-x) = -x^2$.

Thus, the function can be written as:

$f(x) = \begin{cases} x^2, & x \geq 0 \\ -x^2, & x < 0 \end{cases}$

To Discuss:

The derivability of the function $f(x)$ at the point $x = 0$.

Solution:

A function $f(x)$ is derivable at a point $x=c$ if and only if the limit of the difference quotient exists at $x=c$. This means the left-hand derivative (LHD) at $x=c$ must be equal to the right-hand derivative (RHD) at $x=c$.

The derivative of $f(x)$ at $x=c$, denoted by $f'(c)$, is defined as:

$f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$

For derivability at $x=0$, we need to evaluate $f'(0) = \lim\limits_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim\limits_{h \to 0} \frac{f(h) - f(0)}{h}$.

First, let's find the value of the function at $x=0$. Since $0 \geq 0$, we use the first part of the piecewise definition:

$f(0) = 0^2 = 0$

Now, we calculate the Left-Hand Derivative (LHD) and the Right-Hand Derivative (RHD) at $x=0$.

Left-Hand Derivative (LHD) at $x=0$:

LHD at $x=0$ is denoted by $f'(0^-)$ and is defined as:

$f'(0^-) = \lim\limits_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim\limits_{h \to 0^-} \frac{f(h) - 0}{h} = \lim\limits_{h \to 0^-} \frac{f(h)}{h}$

Since $h \to 0^-$, this means $h$ is a small negative number ($h < 0$). For $x < 0$, $f(x) = -x^2$. Therefore, for $h < 0$, $f(h) = -h^2$.

Substitute $f(h) = -h^2$ into the LHD expression:

$f'(0^-) = \lim\limits_{h \to 0^-} \frac{-h^2}{h}$

For $h \neq 0$, we can cancel $h$ from the numerator and the denominator:

$f'(0^-) = \lim\limits_{h \to 0^-} (-h)$

As $h$ approaches $0$ from the left side (through negative values), $-h$ approaches $0$ from the right side (through positive values).

$f'(0^-) = 0$

Right-Hand Derivative (RHD) at $x=0$:

RHD at $x=0$ is denoted by $f'(0^+)$ and is defined as:

$f'(0^+) = \lim\limits_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim\limits_{h \to 0^+} \frac{f(h) - 0}{h} = \lim\limits_{h \to 0^+} \frac{f(h)}{h}$

Since $h \to 0^+$, this means $h$ is a small positive number ($h > 0$). For $x \geq 0$, $f(x) = x^2$. Therefore, for $h > 0$, $f(h) = h^2$.

Substitute $f(h) = h^2$ into the RHD expression:

$f'(0^+) = \lim\limits_{h \to 0^+} \frac{h^2}{h}$

For $h \neq 0$, we can cancel $h$ from the numerator and the denominator:

$f'(0^+) = \lim\limits_{h \to 0^+} (h)$

As $h$ approaches $0$ from the right side (through positive values), $h$ approaches $0$.

$f'(0^+) = 0$

Conclusion:

We have calculated the Left-Hand Derivative and the Right-Hand Derivative at $x=0$:

LHD at $x=0$ = $0$

RHD at $x=0$ = $0$

Since the LHD at $x=0$ is equal to the RHD at $x=0$:

The limit of the difference quotient exists at $x=0$, and its value is $0$.

Therefore, the function $f(x) = x|x|$ is derivable at $x = 0$, and its derivative at $x=0$ is $f'(0) = 0$.

Given:

The function $y = \sqrt{\tan \sqrt{x}}$.

To Find:

The derivative of the function with respect to $x$, i.e., $\frac{dy}{dx}$.

Solution:

We will differentiate the given function using the Chain Rule. The chain rule states that if $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. For nested functions, we apply it iteratively.

Let the given function be $y = \sqrt{\tan \sqrt{x}}$.

This can be viewed as a composition of functions:

$y = \sqrt{u}$, where $u = \tan v$, and $v = \sqrt{x}$.

We differentiate each part with respect to its variable:

1. Differentiate $y$ with respect to $u$:

$y = \sqrt{u} = u^{1/2}$

$\frac{dy}{du} = \frac{d}{du}(u^{1/2}) = \frac{1}{2} u^{1/2 - 1} = \frac{1}{2} u^{-1/2} = \frac{1}{2\sqrt{u}}$

Substituting $u = \tan \sqrt{x}$ back, we get:

$\frac{dy}{du} = \frac{1}{2\sqrt{\tan \sqrt{x}}}$

2. Differentiate $u$ with respect to $v$:

$u = \tan v$

$\frac{du}{dv} = \frac{d}{dv}(\tan v) = \sec^2 v$

Substituting $v = \sqrt{x}$ back, we get:

$\frac{du}{dv} = \sec^2 \sqrt{x}$

3. Differentiate $v$ with respect to $x$:

$v = \sqrt{x} = x^{1/2}$

$\frac{dv}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{1/2 - 1} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$

Now, apply the Chain Rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$

$\frac{dy}{dx} = \left(\frac{1}{2\sqrt{\tan \sqrt{x}}}\right) \cdot (\sec^2 \sqrt{x}) \cdot \left(\frac{1}{2\sqrt{x}}\right)$

Multiply the terms:

$\frac{dy}{dx} = \frac{1 \cdot \sec^2 \sqrt{x} \cdot 1}{2\sqrt{\tan \sqrt{x}} \cdot 1 \cdot 2\sqrt{x}}$

$\frac{dy}{dx} = \frac{\sec^2 \sqrt{x}}{4\sqrt{x}\sqrt{\tan \sqrt{x}}}$

This can also be written as:

$\frac{dy}{dx} = \frac{\sec^2 \sqrt{x}}{4\sqrt{x \tan \sqrt{x}}}$

Therefore, the derivative of $\sqrt{\tan \sqrt{x}}$ with respect to $x$ is $\frac{\sec^2 \sqrt{x}}{4\sqrt{x \tan \sqrt{x}}}$.

Given:

The equation is $y = \tan(x + y)$.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We are given an implicit equation relating $y$ and $x$. To find $\frac{dy}{dx}$, we use implicit differentiation. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and using the chain rule where necessary.

Differentiate both sides of the equation $y = \tan(x+y)$ with respect to $x$:

$\frac{d}{dx}(y) = \frac{d}{dx}(\tan(x+y))$

The left side is straightforward:

$\frac{d}{dx}(y) = \frac{dy}{dx}$

For the right side, we differentiate $\tan(x+y)$ using the chain rule. The derivative of $\tan u$ with respect to $u$ is $\sec^2 u$. Here, $u = x+y$. So, we have:

$\frac{d}{dx}(\tan(x+y)) = \sec^2(x+y) \cdot \frac{d}{dx}(x+y)$

Now, differentiate the term $(x+y)$ with respect to $x$:

$\frac{d}{dx}(x+y) = \frac{d}{dx}(x) + \frac{d}{dx}(y)$

$\frac{d}{dx}(x+y) = 1 + \frac{dy}{dx}$

Substitute this back into the derivative of the right side:

$\frac{d}{dx}(\tan(x+y)) = \sec^2(x+y) \cdot (1 + \frac{dy}{dx})$

Now, equate the derivatives of the left and right sides:

$\frac{dy}{dx} = \sec^2(x+y) \cdot (1 + \frac{dy}{dx})$

Expand the right side:

$\frac{dy}{dx} = \sec^2(x+y) + \sec^2(x+y) \frac{dy}{dx}$

Collect all terms containing $\frac{dy}{dx}$ on one side of the equation. Subtract $\sec^2(x+y) \frac{dy}{dx}$ from both sides:

$\frac{dy}{dx} - \sec^2(x+y) \frac{dy}{dx} = \sec^2(x+y)$

Factor out $\frac{dy}{dx}$ from the terms on the left side:

$\frac{dy}{dx} (1 - \sec^2(x+y)) = \sec^2(x+y)$

Solve for $\frac{dy}{dx}$ by dividing both sides by $(1 - \sec^2(x+y))$:

$\frac{dy}{dx} = \frac{\sec^2(x+y)}{1 - \sec^2(x+y)}$

We can simplify the denominator using the trigonometric identity $\sec^2 \theta - \tan^2 \theta = 1$, which implies $1 - \sec^2 \theta = -\tan^2 \theta$. Using $\theta = x+y$:

$1 - \sec^2(x+y) = -\tan^2(x+y)$

Substitute this into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{\sec^2(x+y)}{-\tan^2(x+y)}$

Alternatively, we can express this in terms of $\sin$ and $\cos$, or using the original equation $y = \tan(x+y)$.

Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$, the numerator becomes $1 + \tan^2(x+y)$. Since $y = \tan(x+y)$, $\tan^2(x+y) = y^2$.

So, the numerator is $1+y^2$. The denominator is $-\tan^2(x+y) = -y^2$.

$\frac{dy}{dx} = \frac{1+y^2}{-y^2} = -\frac{1+y^2}{y^2}$

or

$\frac{dy}{dx} = -\left(\frac{1}{y^2} + 1\right)$

Using $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$\frac{\sec^2(x+y)}{-\tan^2(x+y)} = \frac{1/\cos^2(x+y)}{-\sin^2(x+y)/\cos^2(x+y)} = \frac{1}{\cos^2(x+y)} \cdot \frac{\cos^2(x+y)}{-\sin^2(x+y)} = \frac{1}{-\sin^2(x+y)}$

$\frac{dy}{dx} = -\frac{1}{\sin^2(x+y)} = -\text{cosec}^2(x+y)$

All these forms are equivalent. The simplest forms are usually considered to be those without the trigonometric functions of $(x+y)$.

The derivative is:

$\frac{dy}{dx} = -\text{cosec}^2(x+y)$

or

$\frac{dy}{dx} = -\frac{1+y^2}{y^2}$

Given:

The equation $e^x + e^y = e^{x+y}$.

To Prove:

$\frac{dy}{dx} = -e^{y-x}$.

Proof:

We are given the equation $e^x + e^y = e^{x+y}$. We need to find $\frac{dy}{dx}$ by differentiating this equation implicitly with respect to $x$. Remember that $y$ is a function of $x$, so we use the chain rule for terms involving $y$.

Differentiate both sides of the equation with respect to $x$:

$\frac{d}{dx}(e^x + e^y) = \frac{d}{dx}(e^{x+y})$

Differentiate the terms on the left side:

$\frac{d}{dx}(e^x) + \frac{d}{dx}(e^y)$

The derivative of $e^x$ with respect to $x$ is $e^x$.

The derivative of $e^y$ with respect to $x$ requires the chain rule. We differentiate $e^y$ with respect to $y$, and then multiply by $\frac{dy}{dx}$. The derivative of $e^y$ with respect to $y$ is $e^y$. So, $\frac{d}{dx}(e^y) = e^y \frac{dy}{dx}$.

Thus, the left side becomes:

$e^x + e^y \frac{dy}{dx}$

Now, differentiate the term on the right side, $e^{x+y}$, with respect to $x$. This also requires the chain rule. We differentiate $e^u$ with respect to $u$, where $u = x+y$, and then multiply by $\frac{d}{dx}(x+y)$. The derivative of $e^u$ with respect to $u$ is $e^u$. So, $\frac{d}{dx}(e^{x+y}) = e^{x+y} \cdot \frac{d}{dx}(x+y)$.

Now, differentiate the term $(x+y)$ with respect to $x$:

$\frac{d}{dx}(x+y) = \frac{d}{dx}(x) + \frac{d}{dx}(y) = 1 + \frac{dy}{dx}$

So, the right side becomes:

$e^{x+y} \cdot (1 + \frac{dy}{dx}) = e^{x+y} + e^{x+y} \frac{dy}{dx}$

Equate the derivatives of the left and right sides:

$e^x + e^y \frac{dy}{dx} = e^{x+y} + e^{x+y} \frac{dy}{dx}$

Now, we need to rearrange this equation to isolate $\frac{dy}{dx}$. Gather all terms containing $\frac{dy}{dx}$ on one side and other terms on the other side.

$e^y \frac{dy}{dx} - e^{x+y} \frac{dy}{dx} = e^{x+y} - e^x$

Factor out $\frac{dy}{dx}$ on the left side:

$\frac{dy}{dx} (e^y - e^{x+y}) = e^{x+y} - e^x$

Now, solve for $\frac{dy}{dx}$ by dividing both sides by $(e^y - e^{x+y})$:

$\frac{dy}{dx} = \frac{e^{x+y} - e^x}{e^y - e^{x+y}}$

We are asked to prove that $\frac{dy}{dx} = -e^{y-x}$. Let's try to simplify the expression we found using the original given equation $e^x + e^y = e^{x+y}$.

From the given equation, we can write $e^{x+y} - e^x = e^y$. This is the numerator of our $\frac{dy}{dx}$ expression.

So, the numerator is $e^y$.

Also from the given equation, we can write $e^y - e^{x+y} = -e^x$. This is the denominator of our $\frac{dy}{dx}$ expression.

So, the denominator is $-e^x$.

Substitute these simplified terms back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{e^y}{-e^x}$

Using the property of exponents $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{dy}{dx} = -e^{y-x}$

This matches the expression we were asked to prove.

Therefore, we have shown that if $e^x + e^y = e^{x+y}$, then $\frac{dy}{dx} = -e^{y-x}$.

Alternate Method (Simplifying the equation first)

Start with the given equation:

$e^x + e^y = e^{x+y}$

We can use the property $e^{x+y} = e^x e^y$.

$e^x + e^y = e^x e^y$

Divide the entire equation by $e^x e^y$ (assuming $e^x e^y \neq 0$, which is always true for real $x, y$).

$\frac{e^x}{e^x e^y} + \frac{e^y}{e^x e^y} = \frac{e^x e^y}{e^x e^y}$

$\frac{1}{e^y} + \frac{1}{e^x} = 1$

Using the property $\frac{1}{a^m} = a^{-m}$:

$e^{-y} + e^{-x} = 1$

Now, differentiate this simplified equation implicitly with respect to $x$:

$\frac{d}{dx}(e^{-y} + e^{-x}) = \frac{d}{dx}(1)$

$\frac{d}{dx}(e^{-y}) + \frac{d}{dx}(e^{-x}) = 0$

Differentiate $e^{-y}$ with respect to $x$ using the chain rule (derivative of $e^u$ is $e^u \frac{du}{dx}$, with $u=-y$):

$e^{-y} \cdot \frac{d}{dx}(-y) = e^{-y} \cdot (-1) \frac{dy}{dx} = -e^{-y} \frac{dy}{dx}$

Differentiate $e^{-x}$ with respect to $x$ (derivative of $e^u$ is $e^u \frac{du}{dx}$, with $u=-x$):

$e^{-x} \cdot \frac{d}{dx}(-x) = e^{-x} \cdot (-1) = -e^{-x}$

Substitute these derivatives back into the differentiated equation:

$-e^{-y} \frac{dy}{dx} - e^{-x} = 0$

Move the term without $\frac{dy}{dx}$ to the right side:

$-e^{-y} \frac{dy}{dx} = e^{-x}$

Solve for $\frac{dy}{dx}$ by dividing by $-e^{-y}$:

$\frac{dy}{dx} = \frac{e^{-x}}{-e^{-y}}$

$\frac{dy}{dx} = -\frac{e^{-x}}{e^{-y}}$

Using the property $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{dy}{dx} = -e^{-x - (-y)} = -e^{-x + y} = -e^{y-x}$

This also proves the required result.

Given:

The function $y = \tan^{-1} \left( \frac{3x − x^3}{1 − 3x^2} \right)$.

The domain for $x$ is $- \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

To simplify the expression inside the inverse tangent function, we use a trigonometric substitution.

Let $x = \tan \theta$.

The given domain for $x$ is $- \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$. Substituting $x = \tan \theta$, we get:

$- \frac{1}{\sqrt{3}} < \tan \theta < \frac{1}{\sqrt{3}}$

Taking the inverse tangent of this inequality, and recalling that the principal value branch of $\tan^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, we find the range for $\theta$:

$\tan^{-1}\left(- \frac{1}{\sqrt{3}}\right) < \tan^{-1}(\tan \theta) < \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$-\frac{\pi}{6} < \theta < \frac{\pi}{6}$

Now, substitute $x = \tan \theta$ into the expression inside the $\tan^{-1}$ function:

$\frac{3x − x^3}{1 − 3x^2} = \frac{3\tan \theta − \tan^3 \theta}{1 − 3\tan^2 \theta}$

This expression is a standard trigonometric identity for $\tan 3\theta$:

$\frac{3\tan \theta − \tan^3 \theta}{1 − 3\tan^2 \theta} = \tan 3\theta$

Substitute this back into the expression for $y$:

$y = \tan^{-1}(\tan 3\theta)$

Now, we need to evaluate $\tan^{-1}(\tan A)$ where $A = 3\theta$. The property $\tan^{-1}(\tan A) = A$ is valid if and only if $A$ lies in the principal value range of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$, multiply the inequality by 3:

$3 \times (-\frac{\pi}{6}) < 3\theta < 3 \times (\frac{\pi}{6})$

$-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$

The range of $3\theta$ is indeed $(-\frac{\pi}{2}, \frac{\pi}{2})$, which is the principal value range of $\tan^{-1}$. Therefore, we can simplify $y$ as:

$y = 3\theta$

Now, we need to express $y$ in terms of $x$. From our substitution $x = \tan \theta$, we have $\theta = \tan^{-1} x$.

Substitute this back into the simplified expression for $y$:

$y = 3 \tan^{-1} x$

Now, we can easily find the derivative of $y$ with respect to $x$. We differentiate $y$ using the standard derivative formula for $\tan^{-1} x$, which is $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$.

$\frac{dy}{dx} = \frac{d}{dx}(3 \tan^{-1} x)$

$\frac{dy}{dx} = 3 \frac{d}{dx}(\tan^{-1} x)$

$\frac{dy}{dx} = 3 \cdot \frac{1}{1+x^2}$

So, the derivative is:

$\frac{dy}{dx} = \frac{3}{1+x^2}$

Given:

The function $y = \sin^{-1} \left\{ x\sqrt{1−x}−\sqrt{x} \sqrt{1−x^2} \right\}$.

The domain for $x$ is $0 < x < 1$.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We aim to simplify the expression inside the inverse sine function using trigonometric substitution and relevant identities. The expression inside the $\sin^{-1}$ is $x\sqrt{1−x}−\sqrt{x} \sqrt{1−x^2}$. This expression resembles the form $\sin A \cos B - \cos A \sin B = \sin(A-B)$ or $\cos A \sin B - \sin A \cos B = \sin(B-A)$.

Consider the terms $x$ and $\sqrt{x}$. Also consider the terms $\sqrt{1-x^2}$ and $\sqrt{1-x}$.

Let's try the substitution $x = \sin A$ and $\sqrt{x} = \sin B$.

If $x = \sin A$, then for $0 < x < 1$, we can choose the principal value $A = \sin^{-1} x$, where $A \in (0, \pi/2)$.

If $\sqrt{x} = \sin B$, then for $0 < \sqrt{x} < 1$, we can choose the principal value $B = \sin^{-1} \sqrt{x}$, where $B \in (0, \pi/2)$.

From $x = \sin A$, we have $\sqrt{1-x^2} = \sqrt{1-\sin^2 A} = \sqrt{\cos^2 A} = |\cos A|$. Since $A \in (0, \pi/2)$, $\cos A > 0$, so $\sqrt{1-x^2} = \cos A$.

From $\sqrt{x} = \sin B$, we have $x = \sin^2 B$. Then $\sqrt{1-x} = \sqrt{1-\sin^2 B} = \sqrt{\cos^2 B} = |\cos B|$. Since $B \in (0, \pi/2)$, $\cos B > 0$, so $\sqrt{1-x} = \cos B$.

Now, substitute these into the expression inside the $\sin^{-1}$ function:

$x\sqrt{1−x}−\sqrt{x} \sqrt{1−x^2} = (\sin A)(\cos B) - (\sin B)(\cos A)$

This is the expansion of $\sin(A-B)$:

$\sin(A-B) = \sin A \cos B - \cos A \sin B$

So, the expression inside the inverse sine is $\sin(A-B)$.

$y = \sin^{-1}(\sin(A-B))$

For $\sin^{-1}(\sin \theta) = \theta$ to be valid, $\theta$ must be in the range $[-\pi/2, \pi/2]$. We need to check if $A-B$ is in this range.

We have $A = \sin^{-1} x$ and $B = \sin^{-1} \sqrt{x}$. For $0 < x < 1$, $A \in (0, \pi/2)$ and $B \in (0, \pi/2)$.

Also, from our substitutions, $x = \sin A$ and $x = \sin^2 B$. This implies $\sin A = \sin^2 B$.

Since $B \in (0, \pi/2)$, $0 < \sin B < 1$. Thus, $\sin^2 B < \sin B$.

So, $\sin A < \sin B$.

Since the sine function is strictly increasing on $(0, \pi/2)$, the inequality $\sin A < \sin B$ for $A, B \in (0, \pi/2)$ implies $A < B$.

Since $A \in (0, \pi/2)$, $B \in (0, \pi/2)$ and $A < B$, the difference $A-B$ must be in the interval $(-\pi/2, 0)$.

The interval $(-\pi/2, 0)$ is contained within the principal value range $[-\pi/2, \pi/2]$ of $\sin^{-1}$.

Therefore, we can simplify $y$ as:

$y = A - B$

Substitute back the expressions for $A$ and $B$ in terms of $x$:

$y = \sin^{-1} x - \sin^{-1} \sqrt{x}$

Now, we differentiate this simplified expression with respect to $x$. We use the standard derivative formula for $\sin^{-1} u$, which is $\frac{d}{dx}(\sin^{-1} u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$.

$\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x - \sin^{-1} \sqrt{x})$

$\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x) - \frac{d}{dx}(\sin^{-1} \sqrt{x})$

For the first term, $u=x$:

$\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \cdot \frac{d}{dx}(x) = \frac{1}{\sqrt{1-x^2}} \cdot 1 = \frac{1}{\sqrt{1-x^2}}$

For the second term, $u=\sqrt{x}$. We apply the chain rule:

$\frac{d}{dx}(\sin^{-1} \sqrt{x}) = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$

$= \frac{1}{\sqrt{1-x}} \cdot \frac{d}{dx}(x^{1/2})$

$= \frac{1}{\sqrt{1-x}} \cdot \left(\frac{1}{2} x^{1/2 - 1}\right)$

$= \frac{1}{\sqrt{1-x}} \cdot \left(\frac{1}{2} x^{-1/2}\right)$

$= \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}}$

$= \frac{1}{2\sqrt{x}\sqrt{1-x}} = \frac{1}{2\sqrt{x(1-x)}}$

Combine the derivatives of the two terms:

$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{2\sqrt{x(1-x)}}$

This is the derivative of the given function with respect to $x$.

Given:

The parametric equations are $x = a \sec^3 \theta$ and $y = a \tan^3 \theta$.

To Find:

The derivative $\frac{dy}{dx}$ at $\theta = \frac{\pi}{3}$.

Solution:

We are given $x$ and $y$ in terms of a parameter $\theta$. To find $\frac{dy}{dx}$, we use the formula for the derivative of parametric functions:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$

provided that $\frac{dx}{d\theta} \neq 0$.

First, find the derivative of $x$ with respect to $\theta$:

$x = a \sec^3 \theta = a (\sec \theta)^3$

Using the chain rule, $\frac{d}{d\theta}(u^n) = nu^{n-1} \frac{du}{d\theta}$ and $\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta} (a (\sec \theta)^3)$

$= a \cdot 3 (\sec \theta)^{3-1} \cdot \frac{d}{d\theta}(\sec \theta)$

$= 3a \sec^2 \theta \cdot (\sec \theta \tan \theta)$

$= 3a \sec^3 \theta \tan \theta$

Next, find the derivative of $y$ with respect to $\theta$:

$y = a \tan^3 \theta = a (\tan \theta)^3$

Using the chain rule, $\frac{d}{d\theta}(v^n) = nv^{n-1} \frac{dv}{d\theta}$ and $\frac{d}{d\theta}(\tan \theta) = \sec^2 \theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta} (a (\tan \theta)^3)$

$= a \cdot 3 (\tan \theta)^{3-1} \cdot \frac{d}{d\theta}(\tan \theta)$

$= 3a \tan^2 \theta \cdot (\sec^2 \theta)$

$= 3a \tan^2 \theta \sec^2 \theta$

Now, calculate $\frac{dy}{dx}$ using the formula $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$:

$\frac{dy}{dx} = \frac{3a \tan^2 \theta \sec^2 \theta}{3a \sec^3 \theta \tan \theta}$

Assuming $\sec \theta \neq 0$ and $\tan \theta \neq 0$ (which is true for $\theta = \pi/3$), we can cancel common terms:

$\frac{dy}{dx} = \frac{\cancel{3a} \tan^2 \theta \sec^2 \theta}{\cancel{3a} \sec^3 \theta \tan \theta}$

$= \frac{\tan^2 \theta}{\tan \theta} \cdot \frac{\sec^2 \theta}{\sec^3 \theta}$

$= (\tan \theta) \cdot \left(\frac{1}{\sec \theta}\right)$

$= \tan \theta \cdot \cos \theta$

Substitute $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$= \frac{\sin \theta}{\cos \theta} \cdot \cos \theta$

$= \sin \theta$

So, $\frac{dy}{dx} = \sin \theta$.

Finally, evaluate $\frac{dy}{dx}$ at the given value of $\theta = \frac{\pi}{3}$:

$\left.\frac{dy}{dx}\right|_{\theta=\frac{\pi}{3}} = \sin \left(\frac{\pi}{3}\right)$

We know that $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

Therefore, $\frac{dy}{dx}$ at $\theta = \frac{\pi}{3}$ is $\frac{\sqrt{3}}{2}$.

Given:

The equation $x^y = e^{x–y}$.

To Prove:

$\frac{dy}{dx} = \frac{\log x}{(1+ \log x)^2}$.

Proof:

We are given the equation $x^y = e^{x–y}$. To differentiate this implicitly, it is easier to first take the natural logarithm of both sides of the equation. We use the property $\log(a^b) = b \log a$ and $\log(e^u) = u$.

Taking the natural logarithm of both sides:

$\log(x^y) = \log(e^{x–y})$

$y \log x = x - y$

Now, we differentiate both sides of this equation implicitly with respect to $x$. Remember that $y$ is a function of $x$, so we use the product rule for the term $y \log x$ and the chain rule for terms involving $y$.

Differentiate the left side, $y \log x$, using the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$ (here $u=y$, $v=\log x$):

$\frac{d}{dx}(y \log x) = y \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(y)$

$= y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx}$

$= \frac{y}{x} + (\log x) \frac{dy}{dx}$

Differentiate the right side, $x - y$, with respect to $x$:

$\frac{d}{dx}(x - y) = \frac{d}{dx}(x) - \frac{d}{dx}(y)$

$= 1 - \frac{dy}{dx}$

Equate the derivatives of the left and right sides:

$\frac{y}{x} + (\log x) \frac{dy}{dx} = 1 - \frac{dy}{dx}$

Now, we need to rearrange this equation to isolate $\frac{dy}{dx}$. Gather all terms containing $\frac{dy}{dx}$ on one side and other terms on the other side.

$(\log x) \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{y}{x}$

Factor out $\frac{dy}{dx}$ on the left side:

$\frac{dy}{dx} (\log x + 1) = 1 - \frac{y}{x}$

Express the right side with a common denominator:

$\frac{dy}{dx} (1 + \log x) = \frac{x - y}{x}$

Now, solve for $\frac{dy}{dx}$ by dividing both sides by $(1 + \log x)$:

$\frac{dy}{dx} = \frac{\frac{x - y}{x}}{1 + \log x}$

$\frac{dy}{dx} = \frac{x - y}{x(1 + \log x)}$

The expression for $\frac{dy}{dx}$ is currently in terms of both $x$ and $y$. We need to express it solely in terms of $x$. We can use the equation (i), $y \log x = x - y$, to substitute for the term $(x-y)$ in the numerator.

Substitute this into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{y \log x}{x(1 + \log x)}$

This expression still contains $y$. We can solve equation (i) for $y$ to substitute it into the expression. From $y \log x = x - y$, add $y$ to both sides:

$y \log x + y = x$

Factor out $y$ on the left side:

$y (\log x + 1) = x$

$y = \frac{x}{1 + \log x}$

Now, substitute this expression for $y$ into the current expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{\left(\frac{x}{1 + \log x}\right) \log x}{x(1 + \log x)}$

Multiply the terms in the numerator:

$\frac{dy}{dx} = \frac{\frac{x \log x}{1 + \log x}}{x(1 + \log x)}$

Divide the numerator by the denominator (multiply by the reciprocal of the denominator):

$\frac{dy}{dx} = \frac{x \log x}{1 + \log x} \cdot \frac{1}{x(1 + \log x)}$

Cancel the common factor $x$ (assuming $x \neq 0$; from the original equation $x^y = e^{x-y}$, $x$ must be positive for $\log x$ to be real, so $x \neq 0$):

$\frac{dy}{dx} = \frac{\cancel{x} \log x}{(1 + \log x) \cancel{x} (1 + \log x)}$

$\frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}$

This matches the expression we were asked to prove.

Therefore, we have proved that if $x^y = e^{x-y}$, then $\frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}$.

Given:

The function $y = \tan x + \sec x$.

To Prove:

$\frac{d^2y}{dx^2} = \frac{\cos x}{(1 − \sin x)^2}$.

Proof:

First, we find the first derivative of $y$ with respect to $x$, $\frac{dy}{dx}$.

$y = \tan x + \sec x$

Differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(\tan x + \sec x)$

Using the sum rule, this is the sum of the derivatives of $\tan x$ and $\sec x$:

$\frac{dy}{dx} = \frac{d}{dx}(\tan x) + \frac{d}{dx}(\sec x)$

We know that $\frac{d}{dx}(\tan x) = \sec^2 x$ and $\frac{d}{dx}(\sec x) = \sec x \tan x$.

So, the first derivative is:

$\frac{dy}{dx} = \sec^2 x + \sec x \tan x$

Next, we find the second derivative, $\frac{d^2y}{dx^2}$, by differentiating the first derivative with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(\sec^2 x + \sec x \tan x)$

Using the sum rule again:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\sec^2 x) + \frac{d}{dx}(\sec x \tan x)$

Differentiate $\sec^2 x = (\sec x)^2$ using the chain rule:

$\frac{d}{dx}((\sec x)^2) = 2(\sec x)^{2-1} \cdot \frac{d}{dx}(\sec x) = 2 \sec x \cdot (\sec x \tan x) = 2 \sec^2 x \tan x$

Differentiate $\sec x \tan x$ using the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$, where $u = \sec x$ and $v = \tan x$.

$\frac{du}{dx} = \frac{d}{dx}(\sec x) = \sec x \tan x$

$\frac{dv}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x$

So, $\frac{d}{dx}(\sec x \tan x) = \sec x (\sec^2 x) + \tan x (\sec x \tan x) = \sec^3 x + \sec x \tan^2 x$

Combine the results for the second derivative:

$\frac{d^2y}{dx^2} = (2 \sec^2 x \tan x) + (\sec^3 x + \sec x \tan^2 x)$

$\frac{d^2y}{dx^2} = 2 \sec^2 x \tan x + \sec x \tan^2 x + \sec^3 x$

Now, we need to simplify this expression to match the target $\frac{\cos x}{(1 − \sin x)^2}$. Convert the expression in terms of $\sin x$ and $\cos x$ using $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$.

$\frac{d^2y}{dx^2} = 2 \left(\frac{1}{\cos x}\right)^2 \left(\frac{\sin x}{\cos x}\right) + \left(\frac{1}{\cos x}\right) \left(\frac{\sin x}{\cos x}\right)^2 + \left(\frac{1}{\cos x}\right)^3$

$= 2 \frac{1}{\cos^2 x} \frac{\sin x}{\cos x} + \frac{1}{\cos x} \frac{\sin^2 x}{\cos^2 x} + \frac{1}{\cos^3 x}$

$= \frac{2 \sin x}{\cos^3 x} + \frac{\sin^2 x}{\cos^3 x} + \frac{1}{\cos^3 x}$

Combine the terms over the common denominator $\cos^3 x$:

$= \frac{2 \sin x + \sin^2 x + 1}{\cos^3 x}$

Rearrange the terms in the numerator:

$= \frac{1 + 2 \sin x + \sin^2 x}{\cos^3 x}$

Recognize the numerator as a perfect square, $(1 + \sin x)^2$:

$= \frac{(1 + \sin x)^2}{\cos^3 x}$

Now, we need to manipulate this expression to get $\frac{\cos x}{(1 − \sin x)^2}$. We know that $\cos^2 x = 1 - \sin^2 x$. The denominator $\cos^3 x$ can be written as $\cos x \cdot \cos^2 x$.

$= \frac{(1 + \sin x)^2}{\cos x \cdot \cos^2 x}$

Substitute $\cos^2 x = 1 - \sin^2 x$ in the denominator:

$= \frac{(1 + \sin x)^2}{\cos x (1 - \sin^2 x)}$

Factor the difference of squares in the denominator: $1 - \sin^2 x = (1 - \sin x)(1 + \sin x)$.

$= \frac{(1 + \sin x)^2}{\cos x (1 - \sin x)(1 + \sin x)}$

Assuming $1 + \sin x \neq 0$ (which is true unless $\sin x = -1$, where the original function is undefined), we can cancel one factor of $(1 + \sin x)$ from the numerator and denominator:

$= \frac{1 + \sin x}{\cos x (1 - \sin x)}$

We are close to the target expression. We need $\cos x$ in the numerator and $(1 - \sin x)^2$ in the denominator. We can achieve this by multiplying the numerator and the denominator by $\cos x$.

$= \frac{(1 + \sin x) \cdot \cos x}{\cos x (1 - \sin x) \cdot \cos x}$

$= \frac{(1 + \sin x) \cos x}{\cos^2 x (1 - \sin x)}$

Substitute $\cos^2 x = 1 - \sin^2 x$ in the denominator again:

$= \frac{(1 + \sin x) \cos x}{(1 - \sin^2 x) (1 - \sin x)}$

Factor the difference of squares in the denominator: $1 - \sin^2 x = (1 - \sin x)(1 + \sin x)$.

$= \frac{(1 + \sin x) \cos x}{(1 - \sin x)(1 + \sin x) (1 - \sin x)}$

Assuming $1 + \sin x \neq 0$, cancel the factor $(1 + \sin x)$:

$= \frac{\cos x}{(1 - \sin x) (1 - \sin x)}$

$= \frac{\cos x}{(1 - \sin x)^2}$

This matches the expression we were asked to prove.

Therefore, we have shown that $\frac{d^2y}{dx^2} = \frac{\cos x}{(1 − \sin x)^2}$.

Given:

The function $f(x) = |\cos x|$.

To Find:

The value of the derivative $f'(x)$ at $x = \frac{3\pi}{4}$, i.e., $f'\left(\frac{3\pi}{4}\right)$.

Solution:

The function $f(x) = |\cos x|$ is defined as:

$f(x) = \begin{cases} \cos x, & \text{if } \cos x \geq 0 \\ -\cos x, & \text{if } \cos x < 0 \end{cases}$

We need to find the derivative at $x = \frac{3\pi}{4}$. First, let's determine the value and the sign of $\cos x$ at $x = \frac{3\pi}{4}$.

$\cos \left(\frac{3\pi}{4}\right) = \cos \left(\pi - \frac{\pi}{4}\right) = -\cos \left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$.

Since $\cos \left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}$, which is a negative value, the point $x = \frac{3\pi}{4}$ lies in an interval where $\cos x < 0$. For example, in the interval $(\frac{\pi}{2}, \pi)$, $\cos x$ is negative.

Since $x = \frac{3\pi}{4}$ is in the interval $(\frac{\pi}{2}, \pi)$, for $x$ values near $\frac{3\pi}{4}$, $\cos x < 0$.

Therefore, in a neighborhood around $x = \frac{3\pi}{4}$, the function $f(x) = |\cos x|$ can be written as $f(x) = -\cos x$.

We need to find the derivative $f'(x)$. For $x$ in the interval where $\cos x < 0$, we differentiate $f(x) = -\cos x$:

$f'(x) = \frac{d}{dx}(-\cos x)$

$f'(x) = - \frac{d}{dx}(\cos x)$

We know that $\frac{d}{dx}(\cos x) = -\sin x$.

$f'(x) = - (-\sin x) = \sin x$

Note that this derivative is valid at any point where $\cos x \neq 0$. Since $\cos \left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}} \neq 0$, the function is differentiable at $x = \frac{3\pi}{4}$, and the derivative $f'(x)$ at this point is given by the formula $f'(x) = \sin x$ (derived from the local form of the function).

Now, evaluate $f'\left(\frac{3\pi}{4}\right)$ by substituting $x = \frac{3\pi}{4}$ into the derivative formula $f'(x) = \sin x$:

$f'\left(\frac{3\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right)$

We know that $\sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

Thus, $f'\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

The value of the derivative of $f(x) = |\cos x|$ at $x = \frac{3\pi}{4}$ is $\frac{1}{\sqrt{2}}$.

Given:

The function $f(x) = |\cos x - \sin x|$.

To Find:

The value of the derivative $f'(x)$ at $x = \frac{\pi}{6}$, i.e., $f'\left(\frac{\pi}{6}\right)$.

Solution:

The function $f(x) = |\cos x - \sin x|$ is defined as:

$f(x) = \begin{cases} \cos x - \sin x, & \text{if } \cos x - \sin x \geq 0 \\ -(\cos x - \sin x), & \text{if } \cos x - \sin x < 0 \end{cases}$

$f(x) = \begin{cases} \cos x - \sin x, & \text{if } \cos x \geq \sin x \\ \sin x - \cos x, & \text{if } \cos x < \sin x \end{cases}$

We need to find the derivative at $x = \frac{\pi}{6}$. First, let's determine the sign of the expression inside the absolute value, $\cos x - \sin x$, at $x = \frac{\pi}{6}$.

Evaluate $\cos x - \sin x$ at $x = \frac{\pi}{6}$:

$\cos \left(\frac{\pi}{6}\right) - \sin \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2}$

Since $\sqrt{3} \approx 1.732$, $\sqrt{3} - 1 \approx 0.732$, which is a positive value. So, $\cos \left(\frac{\pi}{6}\right) - \sin \left(\frac{\pi}{6}\right) > 0$.

This means that the point $x = \frac{\pi}{6}$ lies in an interval where $\cos x - \sin x > 0$. For instance, for $x \in (0, \frac{\pi}{4})$, $\cos x > \sin x$, so $\cos x - \sin x > 0$. Since $\frac{\pi}{6} \in (0, \frac{\pi}{4})$, for $x$ values near $\frac{\pi}{6}$, $\cos x - \sin x$ is positive.

Therefore, in a neighborhood around $x = \frac{\pi}{6}$, the function $f(x) = |\cos x - \sin x|$ can be written as $f(x) = \cos x - \sin x$.

We need to find the derivative $f'(x)$. For $x$ in this neighborhood, we differentiate $f(x) = \cos x - \sin x$:

$f'(x) = \frac{d}{dx}(\cos x - \sin x)$

$f'(x) = \frac{d}{dx}(\cos x) - \frac{d}{dx}(\sin x)$

$f'(x) = (-\sin x) - (\cos x)$

$f'(x) = -\sin x - \cos x$

This derivative is valid at any point where $\cos x - \sin x \neq 0$. Since $\cos \left(\frac{\pi}{6}\right) - \sin \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}-1}{2} \neq 0$, the function is differentiable at $x = \frac{\pi}{6}$, and the derivative $f'(x)$ at this point is given by the formula $f'(x) = -\sin x - \cos x$.

Now, evaluate $f'\left(\frac{\pi}{6}\right)$ by substituting $x = \frac{\pi}{6}$ into the derivative formula:

$f'\left(\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{6}\right)$

$f'\left(\frac{\pi}{6}\right) = -\left(\frac{1}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)$

$f'\left(\frac{\pi}{6}\right) = -\frac{1}{2} - \frac{\sqrt{3}}{2}$

$f'\left(\frac{\pi}{6}\right) = -\frac{1 + \sqrt{3}}{2}$

The value of the derivative of $f(x) = |\cos x - \sin x|$ at $x = \frac{\pi}{6}$ is $-\frac{1 + \sqrt{3}}{2}$.

Given:

The function $f(x) = \sin 2x$ on the interval $\left[ 0, \frac{\pi}{2} \right]$.

To Verify:

Rolle’s theorem for the function $f(x)$ on the given interval.

Solution:

Rolle's Theorem states that if a function $f(x)$ satisfies the following three conditions on a closed interval $[a, b]$:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

3. $f(a) = f(b)$.

Then there exists at least one point $c \in (a, b)$ such that $f'(c) = 0$.

We will check these three conditions for $f(x) = \sin 2x$ on the interval $\left[ 0, \frac{\pi}{2} \right]$. Here, $a = 0$ and $b = \frac{\pi}{2}$.

Condition 1: Continuity on $[0, \frac{\pi}{2}]$

The sine function is continuous for all real numbers. The function $2x$ is a polynomial function and is continuous for all real numbers. The composition of two continuous functions is continuous.

Therefore, $f(x) = \sin 2x$ is continuous on the closed interval $\left[ 0, \frac{\pi}{2} \right]$.

Condition 1 is satisfied.

Condition 2: Differentiability on $(0, \frac{\pi}{2})$

The sine function is differentiable for all real numbers, and its derivative is $\cos x$. The function $2x$ is differentiable for all real numbers, and its derivative is $2$. The composition of two differentiable functions is differentiable.

The derivative of $f(x) = \sin 2x$ is $f'(x) = \frac{d}{dx}(\sin 2x) = \cos(2x) \cdot \frac{d}{dx}(2x) = 2 \cos(2x)$.

Since $f'(x) = 2 \cos(2x)$ exists for all $x \in \mathbb{R}$, it exists for all $x$ in the open interval $\left( 0, \frac{\pi}{2} \right)$.

Therefore, $f(x) = \sin 2x$ is differentiable on the open interval $\left( 0, \frac{\pi}{2} \right)$.

Condition 2 is satisfied.

Condition 3: $f(a) = f(b)$

We need to check if $f(0) = f\left(\frac{\pi}{2}\right)$.

Evaluate $f(0)$:

$f(0) = \sin(2 \cdot 0) = \sin(0) = 0$

Evaluate $f\left(\frac{\pi}{2}\right)$:

$f\left(\frac{\pi}{2}\right) = \sin\left(2 \cdot \frac{\pi}{2}\right) = \sin(\pi) = 0$

Since $f(0) = 0$ and $f\left(\frac{\pi}{2}\right) = 0$, we have $f(0) = f\left(\frac{\pi}{2}\right)$.

Condition 3 is satisfied.

Since all three conditions of Rolle's Theorem are satisfied, there must exist at least one value $c$ in the open interval $\left( 0, \frac{\pi}{2} \right)$ such that $f'(c) = 0$.

We found the derivative $f'(x) = 2 \cos(2x)$. Set $f'(c) = 0$:

$2 \cos(2c) = 0$

$\cos(2c) = 0$

The general solutions for $\cos \theta = 0$ are $\theta = (2n + 1) \frac{\pi}{2}$, where $n$ is an integer.

So, $2c = (2n + 1) \frac{\pi}{2}$

Solving for $c$: $c = (2n + 1) \frac{\pi}{4}$

We need to find the value(s) of $c$ that lie in the open interval $\left( 0, \frac{\pi}{2} \right)$.

For $n=0$, $c = (2(0) + 1) \frac{\pi}{4} = \frac{\pi}{4}$.

Check if $c = \frac{\pi}{4}$ is in $\left( 0, \frac{\pi}{2} \right)$: $0 < \frac{\pi}{4} < \frac{\pi}{2}$. This is true.

For $n=1$, $c = (2(1) + 1) \frac{\pi}{4} = \frac{3\pi}{4}$.

Check if $c = \frac{3\pi}{4}$ is in $\left( 0, \frac{\pi}{2} \right)$: $\frac{3\pi}{4} > \frac{\pi}{2}$. This is false.

For $n=-1$, $c = (2(-1) + 1) \frac{\pi}{4} = -\frac{\pi}{4}$.

Check if $c = -\frac{\pi}{4}$ is in $\left( 0, \frac{\pi}{2} \right)$: $-\frac{\pi}{4} < 0$. This is false.

The only value of $c$ in the interval $\left( 0, \frac{\pi}{2} \right)$ is $c = \frac{\pi}{4}$.

Since we found a value $c = \frac{\pi}{4} \in \left( 0, \frac{\pi}{2} \right)$ such that $f'(\frac{\pi}{4}) = 0$, Rolle's theorem is verified for the function $f(x) = \sin 2x$ on the interval $\left[ 0, \frac{\pi}{2} \right]$.

Given:

The function $f(x) = (x – 3) (x – 6) (x – 9)$.

The interval is $[3, 5]$.

To Verify:

Mean Value Theorem for the function $f(x)$ on the given interval $[3, 5]$.

Solution:

The Mean Value Theorem (MVT) states that if a function $f(x)$ satisfies the following two conditions on a closed interval $[a, b]$:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

Then there exists at least one point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

In this problem, $a = 3$ and $b = 5$. The function is $f(x) = (x-3)(x-6)(x-9)$.

Condition 1: Continuity on $[3, 5]$

The function $f(x)$ is a product of three linear polynomials. Expanding this product gives a polynomial function:

$f(x) = (x^2 - 6x - 3x + 18)(x-9)$

$f(x) = (x^2 - 9x + 18)(x-9)$

$f(x) = x(x^2 - 9x + 18) - 9(x^2 - 9x + 18)$

$f(x) = x^3 - 9x^2 + 18x - 9x^2 + 81x - 162$

$f(x) = x^3 - 18x^2 + 99x - 162$

Since $f(x)$ is a polynomial function, it is continuous for all real numbers. Therefore, $f(x)$ is continuous on the closed interval $[3, 5]$.

Condition 1 is satisfied.

Condition 2: Differentiability on $(3, 5)$

Since $f(x)$ is a polynomial function, it is differentiable for all real numbers. Its derivative $f'(x)$ exists for all $x \in \mathbb{R}$.

Differentiating $f(x) = x^3 - 18x^2 + 99x - 162$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^3) - 18\frac{d}{dx}(x^2) + 99\frac{d}{dx}(x) - \frac{d}{dx}(162)$

$f'(x) = 3x^2 - 18(2x) + 99(1) - 0$

$f'(x) = 3x^2 - 36x + 99$

Since $f'(x)$ exists for all $x \in \mathbb{R}$, it exists for all $x$ in the open interval $(3, 5)$.

Therefore, $f(x)$ is differentiable on the open interval $(3, 5)$.

Condition 2 is satisfied.

Since both conditions of the Mean Value Theorem are satisfied, there must exist at least one value $c$ in the open interval $(3, 5)$ such that $f'(c) = \frac{f(5) - f(3)}{5 - 3}$.

First, evaluate $f(a)$ and $f(b)$:

$f(3) = (3-3)(3-6)(3-9) = 0 \cdot (-3) \cdot (-6) = 0$

$f(5) = (5-3)(5-6)(5-9) = (2)(-1)(-4) = 8$

Now, calculate the slope of the secant line connecting the points $(3, f(3))$ and $(5, f(5))$:

$\frac{f(b) - f(a)}{b - a} = \frac{f(5) - f(3)}{5 - 3} = \frac{8 - 0}{2} = \frac{8}{2} = 4$

Set $f'(c)$ equal to this value:

$f'(c) = 3c^2 - 36c + 99 = 4$

Rearrange into a quadratic equation:

$3c^2 - 36c + 99 - 4 = 0$

$3c^2 - 36c + 95 = 0$

Solve this quadratic equation for $c$ using the quadratic formula $c = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$, where $A=3$, $B=-36$, $C=95$.

$c = \frac{-(-36) \pm \sqrt{(-36)^2 - 4(3)(95)}}{2(3)}$

$c = \frac{36 \pm \sqrt{1296 - 1140}}{6}$

$c = \frac{36 \pm \sqrt{156}}{6}$

$c = \frac{36 \pm \sqrt{4 \cdot 39}}{6}$

$c = \frac{36 \pm 2\sqrt{39}}{6}$

$c = \frac{2(18 \pm \sqrt{39})}{6}$

$c = \frac{18 \pm \sqrt{39}}{3}$

We have two possible values for $c$:

$c_1 = \frac{18 + \sqrt{39}}{3}$ and $c_2 = \frac{18 - \sqrt{39}}{3}$.

We need to check if either of these values lies in the open interval $(3, 5)$.

We know that $6^2 = 36$ and $7^2 = 49$, so $6 < \sqrt{39} < 7$.

For $c_1 = \frac{18 + \sqrt{39}}{3}$:

Since $\sqrt{39} > 6$, $18 + \sqrt{39} > 18 + 6 = 24$.

So, $c_1 = \frac{18 + \sqrt{39}}{3} > \frac{24}{3} = 8$.

Since $c_1 > 8$, it is not in the interval $(3, 5)$.

For $c_2 = \frac{18 - \sqrt{39}}{3}$:

Check if $c_2 > 3$:

$\frac{18 - \sqrt{39}}{3} > 3 \implies 18 - \sqrt{39} > 9 \implies 9 > \sqrt{39}$. Since $9^2 = 81$ and $81 > 39$, this inequality is true. So $c_2 > 3$.

Check if $c_2 < 5$:

$\frac{18 - \sqrt{39}}{3} < 5 \implies 18 - \sqrt{39} < 15 \implies 3 < \sqrt{39}$. Since $3^2 = 9$ and $9 < 39$, this inequality is true. So $c_2 < 5$.

Thus, $3 < c_2 < 5$. The value $c_2 = \frac{18 - \sqrt{39}}{3}$ lies in the open interval $(3, 5)$.

Since we found a value $c = \frac{18 - \sqrt{39}}{3} \in (3, 5)$ such that $f'(c) = 4 = \frac{f(5) - f(3)}{5 - 3}$, the Mean Value Theorem is verified for the function $f(x) = (x-3)(x-6)(x-9)$ on the interval $[3, 5]$.

Given:

The function $f(x) = \frac{\sqrt{2} \cos x−1}{\cot x−1}$ for $x ≠ \frac{\pi}{4}$.

To Find:

The value of $f \left( \frac{\pi}{4} \right)$ such that $f(x)$ is continuous at $x = \frac{\pi}{4}$.

Solution:

For the function $f(x)$ to be continuous at $x = \frac{\pi}{4}$, the limit of the function as $x$ approaches $\frac{\pi}{4}$ must exist and be equal to the value of the function at $x = \frac{\pi}{4}$. That is:

The value $f \left( \frac{\pi}{4} \right)$ is the value we need to find to make the function continuous at this point. Let's denote this value as $k$ for a moment, so $f(\frac{\pi}{4}) = k$. According to the condition (i), this value $k$ must be equal to the limit of $f(x)$ as $x \to \frac{\pi}{4}$.

We need to evaluate the limit of $f(x)$ as $x$ approaches $\frac{\pi}{4}$. For $x \neq \frac{\pi}{4}$, the function is defined as $\frac{\sqrt{2} \cos x−1}{\cot x−1}$. So, we consider the limit:

$\lim\limits_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \cos x−1}{\cot x−1}$

Let's evaluate the numerator and denominator at $x = \frac{\pi}{4}$:

Numerator at $x=\frac{\pi}{4}$: $\sqrt{2} \cos \left(\frac{\pi}{4}\right) - 1 = \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) - 1 = 1 - 1 = 0$.

Denominator at $x=\frac{\pi}{4}$: $\cot \left(\frac{\pi}{4}\right) - 1 = 1 - 1 = 0$.

The limit is in the indeterminate form $\frac{0}{0}$. We can use L'Hopital's Rule.

Apply L'Hopital's Rule by differentiating the numerator and the denominator with respect to $x$:

$\lim\limits_{x \to \frac{\pi}{4}} \frac{\frac{d}{dx}(\sqrt{2} \cos x−1)}{\frac{d}{dx}(\cot x−1)}$

$= \lim\limits_{x \to \frac{\pi}{4}} \frac{\sqrt{2}(-\sin x)}{(-\text{cosec}^2 x)}$

$= \lim\limits_{x \to \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\text{cosec}^2 x}$

$= \lim\limits_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \sin x}{\text{cosec}^2 x}$

Recall that $\text{cosec } x = \frac{1}{\sin x}$, so $\text{cosec}^2 x = \frac{1}{\sin^2 x}$.

$= \lim\limits_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \sin x}{1/\sin^2 x}$

$= \lim\limits_{x \to \frac{\pi}{4}} \sqrt{2} \sin x \cdot \sin^2 x$

$= \lim\limits_{x \to \frac{\pi}{4}} \sqrt{2} \sin^3 x$

Now, substitute $x = \frac{\pi}{4}$ into the simplified expression:

$= \sqrt{2} \left(\sin \frac{\pi}{4}\right)^3$

$= \sqrt{2} \left(\frac{1}{\sqrt{2}}\right)^3$

$= \sqrt{2} \left(\frac{1}{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2}}\right)$

$= \sqrt{2} \left(\frac{1}{2\sqrt{2}}\right)$

Cancel the $\sqrt{2}$ term:

$= \frac{1}{2}$

So, the limit of the function as $x$ approaches $\frac{\pi}{4}$ is:

$\lim\limits_{x \to \frac{\pi}{4}} f(x) = \frac{1}{2}$

For the function to be continuous at $x = \frac{\pi}{4}$, the value of $f \left( \frac{\pi}{4} \right)$ must be equal to this limit.

$f\left(\frac{\pi}{4}\right) = \frac{1}{2}$

Therefore, the value of $f \left( \frac{\pi}{4} \right)$ that makes the function continuous at $x = \frac{\pi}{4}$ is $\frac{1}{2}$.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases} \frac{e^{\frac{1}{x}} − 1}{e^{\frac{1}{x}} + 1},& if \;x≠0\\0,& if\;x=0\end{cases}$

To Show:

The function $f(x)$ is discontinuous at $x = 0$.

Solution:

For a function $f(x)$ to be continuous at a point $x=c$, the following condition must be met:

If this condition is not met, the function is discontinuous at $x=c$. We need to check this condition at $x=0$.

First, let's find the value of the function at $x=0$. From the definition of $f(x)$ for $x=0$:

$f(0) = 0$

Next, we need to evaluate the limit of $f(x)$ as $x$ approaches $0$. Since the function is defined differently for $x \neq 0$, we need to check the left-hand limit (LHL) and the right-hand limit (RHL) at $x=0$. For $x \neq 0$, $f(x) = \frac{e^{\frac{1}{x}} − 1}{e^{\frac{1}{x}} + 1}$.

Left-Hand Limit (LHL) at $x=0$:

We evaluate the limit as $x \to 0^-$ (as $x$ approaches 0 from the left side, i.e., $x < 0$).

$\text{LHL} = \lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \frac{e^{\frac{1}{x}} − 1}{e^{\frac{1}{x}} + 1}$

As $x \to 0^-$ through negative values, the term $\frac{1}{x} \to -\infty$.

We know that $\lim\limits_{z \to -\infty} e^z = 0$. Let $z = \frac{1}{x}$. As $x \to 0^-$, $z \to -\infty$.

So, $\lim\limits_{x \to 0^-} e^{\frac{1}{x}} = \lim\limits_{z \to -\infty} e^z = 0$.

Substitute this limit back into the LHL expression:

$\text{LHL} = \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1$

So, the left-hand limit is $\text{LHL} = -1$.

Right-Hand Limit (RHL) at $x=0$:

We evaluate the limit as $x \to 0^+$ (as $x$ approaches 0 from the right side, i.e., $x > 0$).

$\text{RHL} = \lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} \frac{e^{\frac{1}{x}} − 1}{e^{\frac{1}{x}} + 1}$

As $x \to 0^+$ through positive values, the term $\frac{1}{x} \to +\infty$.

We know that $\lim\limits_{z \to +\infty} e^z = +\infty$. Let $z = \frac{1}{x}$. As $x \to 0^+$, $z \to +\infty$.

So, $\lim\limits_{x \to 0^+} e^{\frac{1}{x}} = \lim\limits_{z \to +\infty} e^z = +\infty$.

The limit expression is of the form $\frac{\infty}{\infty}$, which is an indeterminate form. To handle this, we can divide the numerator and the denominator by the dominant term, which is $e^{\frac{1}{x}}$.

$\text{RHL} = \lim\limits_{x \to 0^+} \frac{e^{\frac{1}{x}}(1 − e^{-\frac{1}{x}})}{e^{\frac{1}{x}}(1 + e^{-\frac{1}{x}})}$

$= \lim\limits_{x \to 0^+} \frac{1 − e^{-\frac{1}{x}}}{1 + e^{-\frac{1}{x}}}$

As $x \to 0^+$, $\frac{1}{x} \to +\infty$, so $-\frac{1}{x} \to -\infty$.

We know that $\lim\limits_{w \to -\infty} e^w = 0$. Let $w = -\frac{1}{x}$. As $x \to 0^+$, $w \to -\infty$.

So, $\lim\limits_{x \to 0^+} e^{-\frac{1}{x}} = \lim\limits_{w \to -\infty} e^w = 0$.

Substitute this limit back into the RHL expression:

$\text{RHL} = \frac{1 - 0}{1 + 0} = \frac{1}{1} = 1$

So, the right-hand limit is $\text{RHL} = 1$.

We compare the LHL and RHL:

$\text{LHL} = -1$

$\text{RHL} = 1$

Since $\text{LHL} \neq \text{RHL}$, the limit of the function as $x$ approaches 0 does not exist.

For continuity at $x=0$, the condition is $\lim\limits_{x \to 0} f(x) = f(0)$. From (ii), the limit on the left side does not exist, while $f(0) = 0$.

Since the limit of the function as $x$ approaches $0$ does not exist, the condition for continuity at $x=0$ is not met.

Therefore, the function $f(x)$ is discontinuous at $x = 0$.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}\frac{1 − \cos 4x}{x^2},& x<0\\a,& x=0 \\ \frac{\sqrt{x}}{\sqrt{16} + \sqrt{x − 4}},& x>0\end{cases}$

To Find:

The value of the constant $a$ such that the function $f(x)$ is continuous at $x=0$.

Solution:

For the function $f(x)$ to be continuous at $x=0$, the limit of the function as $x$ approaches $0$ must exist and be equal to the value of the function at $x=0$. This requires that the left-hand limit (LHL) and the right-hand limit (RHL) at $x=0$ both exist and are equal to $f(0)$.

First, let's find the value of the function at $x=0$. From the definition of $f(x)$ for $x=0$:

$f(0) = a$

Next, we evaluate the Left-Hand Limit (LHL) at $x=0$. For $x < 0$, $f(x) = \frac{1 − \cos 4x}{x^2}$.

$\text{LHL} = \lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \frac{1 − \cos 4x}{x^2}$

This limit is in the indeterminate form $\frac{0}{0}$ as $x \to 0$. We use the trigonometric identity $1 - \cos 2\theta = 2\sin^2\theta$. Let $2\theta = 4x$, so $\theta = 2x$.

$1 - \cos 4x = 2\sin^2(2x)$

Substitute this into the limit expression:

$\text{LHL} = \lim\limits_{x \to 0^-} \frac{2\sin^2(2x)}{x^2}$

We can rewrite this to use the standard limit $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$.

$\text{LHL} = \lim\limits_{x \to 0^-} 2 \left(\frac{\sin(2x)}{x}\right)^2$

Multiply the numerator and denominator inside the parenthesis by 2:

$= \lim\limits_{x \to 0^-} 2 \left(\frac{\sin(2x)}{2x} \cdot 2\right)^2$

$= \lim\limits_{x \to 0^-} 2 \cdot 4 \left(\frac{\sin(2x)}{2x}\right)^2$

$= 8 \lim\limits_{x \to 0^-} \left(\frac{\sin(2x)}{2x}\right)^2$

Let $y = 2x$. As $x \to 0^-$, $y \to 0^-$. Using $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$ (which holds for one-sided limits as well):

$= 8 (1)^2 = 8$

So, the left-hand limit is $\text{LHL} = 8$.

Next, we evaluate the Right-Hand Limit (RHL) at $x=0$. For $x > 0$, $f(x) = \frac{\sqrt{x}}{\sqrt{16} + \sqrt{x − 4}}$.

$\text{RHL} = \lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16} + \sqrt{x − 4}}$

For this limit to exist, the function $f(x)$ must be defined for values of $x$ in some interval $(0, \delta)$ where $\delta > 0$. Let's consider the domain of the expression $\frac{\sqrt{x}}{\sqrt{16} + \sqrt{x − 4}}$.

The term $\sqrt{x}$ requires $x \geq 0$.

The term $\sqrt{x-4}$ requires $x-4 \geq 0$, which means $x \geq 4$.

The denominator $\sqrt{16} + \sqrt{x-4} = 4 + \sqrt{x-4}$ must be non-zero. This is always true for real values of $x$ where $\sqrt{x-4}$ is defined.

The third case of the function is defined for $x > 0$ using this expression. Combining the condition $x > 0$ with the domain of the expression ($x \geq 4$), this part of the function is only defined for $x \ge 4$.

This means that for any small $\delta > 0$, there are no values of $x$ in the interval $(0, \delta)$ for which $f(x)$ is defined according to the rule for $x>0$.

Therefore, the right-hand limit $\lim\limits_{x \to 0^+} f(x)$ does not exist.

For the function to be continuous at $x=0$, the condition $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$ must be satisfied.

We found $\text{LHL} = 8$.

We found that $\text{RHL}$ does not exist.

Since the right-hand limit does not exist, the overall limit $\lim\limits_{x \to 0} f(x)$ does not exist.

As the limit of the function at $x=0$ does not exist, it cannot be equal to $f(0)$, regardless of the value of $a$.

Therefore, the function $f(x)$ is discontinuous at $x=0$ for any value of $a$. There is no value of $a$ that makes the function continuous at $x=0$, as the question is stated.

Note on Potential Typo:

The phrasing of the question ("For what value of a...") suggests that a value of $a$ exists that makes the function continuous. This implies that $\lim\limits_{x \to 0} f(x)$ is expected to exist and be finite. Given that the LHL is 8, it is likely that the definition of the function for $x>0$ was intended to result in a RHL equal to 8, although the expression provided $\frac{\sqrt{x}}{\sqrt{16} + \sqrt{x − 4}}$ does not achieve this due to its domain restriction $[4, \infty)$ for $x>0$. If we assume the question implicitly requires LHL = RHL = f(0) for continuity, then the value of $a$ must be equal to the common limit. Since LHL = 8, it would imply RHL must be 8 and $f(0)$ must be 8. In that hypothetical scenario, the required value of $a$ would be 8.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}2x+3,& -3 \leq x < -2\\x+1,& -2 \leq x < 0 \\x+2,& 0 \leq x \leq 1 \end{cases}$

To Examine:

The differentiability of the function $f(x)$.

Solution:

A function is differentiable at a point if and only if it is continuous at that point, and its left-hand derivative (LHD) is equal to its right-hand derivative (RHD) at that point. The function is piecewise defined, changing its definition at $x = -2$ and $x = 0$. The individual pieces ($2x+3$, $x+1$, $x+2$) are polynomial functions, which are differentiable everywhere within their respective open intervals $(-3, -2)$, $(-2, 0)$, and $(0, 1)$. Therefore, we only need to examine the differentiability at the transition points $x = -2$ and $x = 0$.

Examination of Differentiability at $x = -2$:

Step 1: Check for Continuity at $x = -2$.

For continuity at $x = -2$, we need $\lim\limits_{x \to -2^-} f(x) = \lim\limits_{x \to -2^+} f(x) = f(-2)$.

• Evaluate $f(-2)$: Using the definition for $-2 \leq x < 0$, $f(-2) = (-2) + 1 = -1$.
• Evaluate the Left-Hand Limit (LHL) at $x = -2$: $\lim\limits_{x \to -2^-} f(x) = \lim\limits_{x \to -2^-} (2x+3) = 2(-2) + 3 = -4 + 3 = -1$.
• Evaluate the Right-Hand Limit (RHL) at $x = -2$: $\lim\limits_{x \to -2^+} f(x) = \lim\limits_{x \to -2^+} (x+1) = -2 + 1 = -1$.

Since $\lim\limits_{x \to -2^-} f(x) = \lim\limits_{x \to -2^+} f(x) = f(-2) = -1$, the function $f(x)$ is continuous at $x = -2$.

Step 2: Check for Differentiability at $x = -2$.

We calculate the LHD and RHD at $x = -2$.

• Left-Hand Derivative (LHD) at $x = -2$:

  $f'(-2^-) = \lim\limits_{h \to 0^-} \frac{f(-2+h) - f(-2)}{h}$

  For small $h < 0$, $-2+h < -2$. We use $f(x) = 2x+3$.

  $f'(-2^-) = \lim\limits_{h \to 0^-} \frac{(2(-2+h)+3) - (-1)}{h} = \lim\limits_{h \to 0^-} \frac{(-4+2h+3) + 1}{h} = \lim\limits_{h \to 0^-} \frac{2h-1+1}{h} = \lim\limits_{h \to 0^-} \frac{2h}{h} = \lim\limits_{h \to 0^-} 2 = 2$

  So, LHD at $x=-2$ is 2.

• Right-Hand Derivative (RHD) at $x = -2$:

  $f'(-2^+) = \lim\limits_{h \to 0^+} \frac{f(-2+h) - f(-2)}{h}$

  For small $h > 0$, $-2 < -2+h < 0$. We use $f(x) = x+1$.

  $f'(-2^+) = \lim\limits_{h \to 0^+} \frac{((-2+h)+1) - (-1)}{h} = \lim\limits_{h \to 0^+} \frac{(-1+h) + 1}{h} = \lim\limits_{h \to 0^+} \frac{h}{h} = \lim\limits_{h \to 0^+} 1 = 1$

  So, RHD at $x=-2$ is 1.

Since LHD at $x = -2$ (which is 2) $\neq$ RHD at $x = -2$ (which is 1), the function $f(x)$ is not differentiable at $x = -2$.

Examination of Differentiability at $x = 0$:

Step 1: Check for Continuity at $x = 0$.

For continuity at $x = 0$, we need $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$.

• Evaluate $f(0)$: Using the definition for $0 \leq x \leq 1$, $f(0) = 0 + 2 = 2$.
• Evaluate the Left-Hand Limit (LHL) at $x = 0$: $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (x+1) = 0 + 1 = 1$.
• Evaluate the Right-Hand Limit (RHL) at $x = 0$: $\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (x+2) = 0 + 2 = 2$.

Since $\lim\limits_{x \to 0^-} f(x) = 1$ and $\lim\limits_{x \to 0^+} f(x) = 2$, the left-hand limit is not equal to the right-hand limit ($\text{LHL} \neq \text{RHL}$). Therefore, the limit $\lim\limits_{x \to 0} f(x)$ does not exist.

Since the function is not continuous at $x = 0$, it is not differentiable at $x = 0$. (Continuity is a necessary condition for differentiability).

Conclusion:

The function $f(x)$ is not differentiable at $x = -2$ and not differentiable at $x = 0$. The function is differentiable at all other points in its domain $[-3, 1]$, specifically on the open intervals $(-3, -2)$, $(-2, 0)$, and $(0, 1)$, because it is defined by polynomials in these intervals.

Given:

Let $u = \tan^{-1} \left( \frac{\sqrt{1 − x^2}}{x} \right)$

Let $v = \cos^{-1} \left( 2x\sqrt{1−x^2} \right)$

The domain is $x \in \left( \frac{1}{\sqrt{2}}\;,\; 1 \right)$.

To Find:

The derivative of $u$ with respect to $v$, i.e., $\frac{du}{dv}$.

Solution:

To find $\frac{du}{dv}$ where both $u$ and $v$ are functions of $x$, we use the formula:

$\frac{du}{dv} = \frac{du/dx}{dv/dx}$

We will first simplify $u$ and $v$ using trigonometric substitutions and then find their derivatives with respect to $x$.

Step 1: Simplify $u$ and find $\frac{du}{dx}$

Let $u = \tan^{-1} \left( \frac{\sqrt{1 − x^2}}{x} \right)$.

Let $x = \sin \theta$. Since $x \in \left( \frac{1}{\sqrt{2}}\;,\; 1 \right)$, we have $\sin \theta \in \left( \frac{1}{\sqrt{2}}\;,\; 1 \right)$. This implies $\theta \in \left( \frac{\pi}{4}\;,\; \frac{\pi}{2} \right)$.

Substitute $x = \sin \theta$ into the expression for $u$:

$u = \tan^{-1} \left( \frac{\sqrt{1 − \sin^2 \theta}}{\sin \theta} \right) = \tan^{-1} \left( \frac{\sqrt{\cos^2 \theta}}{\sin \theta} \right) = \tan^{-1} \left( \frac{|\cos \theta|}{\sin \theta} \right)$

Since $\theta \in \left( \frac{\pi}{4}\;,\; \frac{\pi}{2} \right)$, $\cos \theta > 0$ and $\sin \theta > 0$. So, $|\cos \theta| = \cos \theta$.

$u = \tan^{-1} \left( \frac{\cos \theta}{\sin \theta} \right) = \tan^{-1}(\cot \theta)$

We use the identity $\cot \theta = \tan \left( \frac{\pi}{2} - \theta \right)$:

$u = \tan^{-1} \left( \tan \left( \frac{\pi}{2} - \theta \right) \right)$

For $\tan^{-1}(\tan A) = A$ to be valid, $A$ must be in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $\theta \in \left( \frac{\pi}{4}\;,\; \frac{\pi}{2} \right)$, we have:

$-\frac{\pi}{2} < -\theta < -\frac{\pi}{4}$

Adding $\frac{\pi}{2}$ to all parts:

$\frac{\pi}{2} - \frac{\pi}{2} < \frac{\pi}{2} - \theta < \frac{\pi}{2} - \frac{\pi}{4}$

$0 < \frac{\pi}{2} - \theta < \frac{\pi}{4}$

The range of $\frac{\pi}{2} - \theta$ is $(0, \frac{\pi}{4})$, which is contained in $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Therefore, $u = \frac{\pi}{2} - \theta$.

Since $x = \sin \theta$ and $\theta \in \left( \frac{\pi}{4}\;,\; \frac{\pi}{2} \right)$, $\theta = \sin^{-1} x$.

$u = \frac{\pi}{2} - \sin^{-1} x$

Now, differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx} \left( \frac{\pi}{2} - \sin^{-1} x \right) = \frac{d}{dx} \left( \frac{\pi}{2} \right) - \frac{d}{dx} \left( \sin^{-1} x \right)$

$\frac{du}{dx} = 0 - \frac{1}{\sqrt{1 - x^2}} = -\frac{1}{\sqrt{1 - x^2}}$

Step 2: Simplify $v$ and find $\frac{dv}{dx}$

Let $v = \cos^{-1} \left( 2x\sqrt{1−x^2} \right)$.

Let $x = \sin \phi$. Since $x \in \left( \frac{1}{\sqrt{2}}\;,\; 1 \right)$, we have $\sin \phi \in \left( \frac{1}{\sqrt{2}}\;,\; 1 \right)$. This implies $\phi \in \left( \frac{\pi}{4}\;,\; \frac{\pi}{2} \right)$.

Substitute $x = \sin \phi$ into the expression for $v$:

$v = \cos^{-1} \left( 2 \sin \phi \sqrt{1−\sin^2 \phi} \right) = \cos^{-1} \left( 2 \sin \phi \sqrt{\cos^2 \phi} \right) = \cos^{-1} \left( 2 \sin \phi |\cos \phi| \right)$

Since $\phi \in \left( \frac{\pi}{4}\;,\; \frac{\pi}{2} \right)$, $\cos \phi > 0$. So, $|\cos \phi| = \cos \phi$.

$v = \cos^{-1} (2 \sin \phi \cos \phi)$

We use the identity $2 \sin \phi \cos \phi = \sin(2\phi)$:

$v = \cos^{-1}(\sin(2\phi))$

We use the identity $\sin A = \cos \left( \frac{\pi}{2} - A \right)$:

$v = \cos^{-1} \left( \cos \left( \frac{\pi}{2} - 2\phi \right) \right)$

For $\cos^{-1}(\cos A) = A$ to be valid, $A$ must be in the range $[0, \pi]$.

Since $\phi \in \left( \frac{\pi}{4}\;,\; \frac{\pi}{2} \right)$, we have:

$\frac{\pi}{2} < 2\phi < \pi$

Subtracting $\frac{\pi}{2}$ from all parts:

$\frac{\pi}{2} - \pi < \frac{\pi}{2} - 2\phi < \frac{\pi}{2} - \frac{\pi}{2}$

$-\frac{\pi}{2} < \frac{\pi}{2} - 2\phi < 0$

The range of $\frac{\pi}{2} - 2\phi$ is $(-\frac{\pi}{2}, 0)$, which is not in the principal value range $[0, \pi]$ of $\cos^{-1}$.

However, for $A \in (-\pi, 0)$, $\cos^{-1}(\cos A) = -A$. Here $A = \frac{\pi}{2} - 2\phi$, which is in $(-\frac{\pi}{2}, 0)$.

So, $v = - \left( \frac{\pi}{2} - 2\phi \right) = 2\phi - \frac{\pi}{2}$.

Since $x = \sin \phi$ and $\phi \in \left( \frac{\pi}{4}\;,\; \frac{\pi}{2} \right)$, $\phi = \sin^{-1} x$.

$v = 2 \sin^{-1} x - \frac{\pi}{2}$

Now, differentiate $v$ with respect to $x$:

$\frac{dv}{dx} = \frac{d}{dx} \left( 2 \sin^{-1} x - \frac{\pi}{2} \right) = 2 \frac{d}{dx} \left( \sin^{-1} x \right) - \frac{d}{dx} \left( \frac{\pi}{2} \right)$

$\frac{dv}{dx} = 2 \cdot \frac{1}{\sqrt{1 - x^2}} - 0 = \frac{2}{\sqrt{1 - x^2}}$

Step 3: Calculate $\frac{du}{dv}$

Now, divide $\frac{du}{dx}$ by $\frac{dv}{dx}$:

$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-\frac{1}{\sqrt{1 - x^2}}}{\frac{2}{\sqrt{1 - x^2}}}$

$\frac{du}{dv} = -\frac{1}{\sqrt{1 - x^2}} \cdot \frac{\sqrt{1 - x^2}}{2}$

Cancel the common term $\sqrt{1 - x^2}$ (since $x \in (\frac{1}{\sqrt{2}}, 1)$, $\sqrt{1-x^2} \neq 0$):

$\frac{du}{dv} = -\frac{1}{2}$

The derivative of $\tan^{-1} \left( \frac{\sqrt{1 − x^2}}{x} \right)$ with respect to $\cos^{-1} \left( 2x\sqrt{1−x^2} \right)$ is $-\frac{1}{2}$.

Given:

The function $f (x)$ is defined as:

$f (x) = \begin{cases}\frac{\sin x}{x} + \cos x,& if\; x≠0\\k ,&if\; x=0\end{cases}$

The function $f(x)$ is continuous at $x=0$.

To Find:

The value of $k$.

Solution:

For a function $f(x)$ to be continuous at a point $x=a$, the limit of the function as $x$ approaches $a$ must be equal to the value of the function at $a$.

That is, $\lim\limits_{x \to a} f(x) = f(a)$.

In this problem, the point is $a=0$. So, for $f(x)$ to be continuous at $x=0$, we must have:

From the definition of the function, we are given that $f(0) = k$.

Now, we need to evaluate the limit $\lim\limits_{x \to 0} f(x)$. Since the function definition for $x \neq 0$ is $\frac{\sin x}{x} + \cos x$, we consider the limit as $x$ approaches $0$ from both sides (or simply as $x \to 0$).

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \left(\frac{\sin x}{x} + \cos x\right)$

Using the property of limits that the limit of a sum is the sum of the limits (if they exist), we can write:

$\lim\limits_{x \to 0} \left(\frac{\sin x}{x} + \cos x\right) = \lim\limits_{x \to 0} \frac{\sin x}{x} + \lim\limits_{x \to 0} \cos x$

We know the standard limits:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

$\lim\limits_{x \to 0} \cos x = \cos(0) = 1$

Substituting these values into the limit expression:

$\lim\limits_{x \to 0} f(x) = 1 + 1 = 2$

Now, using the continuity condition from (i):

$\lim\limits_{x \to 0} f(x) = f(0)$

$2 = k$

Thus, the value of $k$ is 2.

The correct option is (B) 2.

Given:

The function $f(x) = [x]$, where $[x]$ denotes the greatest integer function.

To Find:

The point among the given options where the function is continuous.

Solution:

The greatest integer function $f(x) = [x]$ is defined as the greatest integer less than or equal to $x$.

For a function to be continuous at a point $x=a$, the limit of the function as $x$ approaches $a$ must be equal to the value of the function at $a$. That is, $\lim\limits_{x \to a} f(x) = f(a)$.

Let's examine the continuity of $f(x) = [x]$ at integer and non-integer points.

Case 1: At an integer point $a = n$ (where $n$ is an integer)

The value of the function at $x=n$ is $f(n) = [n] = n$.

Consider the left-hand limit (LHL) as $x$ approaches $n$ from the left ($x < n$).

For $x$ slightly less than $n$ (e.g., $n-0.1$), $[x] = n-1$.

So, $\lim\limits_{x \to n^-} [x] = n-1$.

Consider the right-hand limit (RHL) as $x$ approaches $n$ from the right ($x > n$).

For $x$ slightly greater than $n$ (e.g., $n+0.1$), $[x] = n$.

So, $\lim\limits_{x \to n^+} [x] = n$.

Since the LHL ($\lim\limits_{x \to n^-} f(x) = n-1$) is not equal to the RHL ($\lim\limits_{x \to n^+} f(x) = n$), the limit $\lim\limits_{x \to n} f(x)$ does not exist at integer points. Therefore, the function $f(x) = [x]$ is discontinuous at all integer points.

Case 2: At a non-integer point $a = c$ (where $c$ is not an integer)

Let $n$ be the integer such that $n < c < n+1$.

The value of the function at $x=c$ is $f(c) = [c] = n$.

Consider the limit as $x$ approaches $c$. For $x$ values in a small neighborhood around $c$ (such that $n < x < n+1$), the value of $[x]$ is $n$.

So, $\lim\limits_{x \to c} [x] = n$.

Since $\lim\limits_{x \to c} f(x) = n$ and $f(c) = n$, we have $\lim\limits_{x \to c} f(x) = f(c)$. Therefore, the function $f(x) = [x]$ is continuous at all non-integer points.

Now let's check the given options:

(A) 4 is an integer.

(B) – 2 is an integer.

(C) 1 is an integer.

(D) 1.5 is a non-integer.

Based on our analysis, the function $f(x) = [x]$ is continuous at non-integer points. Only option (D) is a non-integer.

The function is continuous at 1.5.

The correct option is (D) 1.5.

Given:

The function $f (x) = \frac{1}{x − [x]}$, where $[x]$ denotes the greatest integer function.

To Find:

The number of points at which the function $f(x)$ is not continuous.

Solution:

A function is not continuous at points where it is undefined or where the limit does not exist or is not equal to the function value.

The given function is $f(x) = \frac{1}{x - [x]}$.

This function is undefined when the denominator is equal to zero.

The denominator is zero when $x - [x] = 0$, which means $x = [x]$.

The equation $x = [x]$ holds true exactly when $x$ is an integer.

For example, if $x=3$, $[x]=3$, and $x=[x]$. If $x=3.5$, $[x]=3$, and $x \neq [x]$.

Therefore, the denominator $x - [x]$ is zero for all integer values of $x$.

This means the function $f(x) = \frac{1}{x - [x]}$ is undefined at all integer points.

A function is discontinuous at any point where it is undefined.

The set of integer points is $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$. This set is infinite.

So, the function $f(x)$ is not continuous at all integer points, and there are infinitely many integer points.

Let's check the options:

(A) 1 - There are more than 1 points of discontinuity.

(B) 2 - There are more than 2 points of discontinuity.

(C) 3 - There are more than 3 points of discontinuity.

(D) none of these - This option is correct because the number of points of discontinuity (all integers) is infinite, which is not covered by options (A), (B), or (C).

The function $f(x)$ is discontinuous at all integer values of $x$. The number of such points is infinite.

The correct option is (D) none of these.

Given:

The function $f(x) = \tan x$.

To Find:

The set of points where the function $f(x) = \tan x$ is discontinuous.

Solution:

The function $\tan x$ is defined as the ratio of $\sin x$ to $\cos x$.

A rational function (a ratio of two functions) is discontinuous at points where the denominator is zero, because the function is undefined at such points.

In the case of $f(x) = \tan x$, the denominator is $\cos x$. The function is undefined when $\cos x = 0$.

The cosine function, $\cos x$, is zero at points where the angle $x$ is an odd multiple of $\frac{\pi}{2}$.

These points are $x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{5\pi}{2}, ...$

The general form for an odd multiple of $\frac{\pi}{2}$ is $(2n+1)\frac{\pi}{2}$, where $n$ is any integer ($n \in Z$).

So, $\cos x = 0$ when $x = (2n+1)\frac{\pi}{2}$ for $n \in Z$.

Therefore, the function $f(x) = \tan x$ is discontinuous at the set of points $\left\{ (2n+1)\frac{π}{2} : n∈Z \right\}$.

Let's examine the given options:

(A) $\{nπ : n ∈ Z\}$: At these points ($\dots, -\pi, 0, \pi, 2\pi, \dots$), $\cos x = \pm 1$, so $\tan x = 0$ or undefined (not the case here). The function is continuous here.

(B) $\{2nπ : n ∈ Z\}$: At these points ($\dots, -2\pi, 0, 2\pi, 4\pi, \dots$), $\cos x = 1$, so $\tan x = 0$. The function is continuous here.

(C) $\left\{ (2n+1)\frac{π}{2} : n∈Z \right\}$: At these points ($\dots, -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \dots$), $\cos x = 0$, so $\tan x$ is undefined. The function is discontinuous here.

(D) $\left\{\frac{nπ}{2} : n∈Z \right\}$: This set includes points where $n$ is even (giving multiples of $\pi$) and points where $n$ is odd (giving odd multiples of $\frac{\pi}{2}$). The function is only discontinuous at the points where $\cos x = 0$, which are the odd multiples of $\frac{\pi}{2}$. This set contains points where the function is continuous (multiples of $\pi$). The question asks for the set where it is discontinuous.

The set of points where $\tan x$ is discontinuous is $\left\{ (2n+1)\frac{π}{2} : n∈Z \right\}$.

The correct option is (C) $\left\{ (2n+1)\frac{π}{2} : n∈Z \right\}$.

Given:

The function $f(x) = |\cos x|$.

To Analyze:

The continuity and differentiability of the function $f(x) = |\cos x|$ based on the given options.

Solution:

Let's analyze the continuity and differentiability of $f(x) = |\cos x|$.

Continuity:

The function $\cos x$ is continuous for all real numbers $x$.

The absolute value function, $g(u) = |u|$, is continuous for all real numbers $u$.

The function $f(x) = |\cos x|$ is a composite function, $f(x) = g(h(x))$, where $h(x) = \cos x$ and $g(u) = |u|$.

Since the composition of two continuous functions is continuous, and both $\cos x$ and $|u|$ are continuous everywhere, the function $f(x) = |\cos x|$ is continuous everywhere on $\mathbb{R}$.

Differentiability:

The function $h(x) = \cos x$ is differentiable for all real numbers $x$, with $h'(x) = -\sin x$.

The function $g(u) = |u|$ is differentiable for all real numbers $u$ except at $u=0$. The derivative is $g'(u) = \begin{cases} 1 & , u > 0 \\ -1 & , u < 0 \end{cases}$, or $\text{sgn}(u)$.

The composite function $f(x) = g(h(x))$ is differentiable at $x$ if $h(x)$ is differentiable at $x$ and $g(u)$ is differentiable at $u = h(x)$.

So, $f(x) = |\cos x|$ is differentiable for all $x$ where $\cos x \neq 0$.

The points where $\cos x = 0$ are $x = (2n+1)\frac{\pi}{2}$, where $n$ is an integer ($n \in Z$).

These are the points where the function $|u|$ is not differentiable, with $u = \cos x = 0$. We need to examine the differentiability of $f(x) = |\cos x|$ specifically at these points using the definition of the derivative.

Let's consider a generic point $x_0 = (2n+1)\frac{\pi}{2}$ where $\cos x_0 = 0$.

We need to evaluate the limit of the difference quotient:

$\lim\limits_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} = \lim\limits_{h \to 0} \frac{|\cos(x_0 + h)| - |\cos(x_0)|}{h}$

Since $\cos x_0 = 0$, this becomes:

$\lim\limits_{h \to 0} \frac{|\cos(x_0 + h)| - 0}{h} = \lim\limits_{h \to 0} \frac{|\cos(x_0 + h)|}{h}$

Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$, with $A = x_0 = (2n+1)\frac{\pi}{2}$ and $B=h$:

$\cos((2n+1)\frac{\pi}{2} + h) = \cos((2n+1)\frac{\pi}{2})\cos h - \sin((2n+1)\frac{\pi}{2})\sin h$

Since $\cos((2n+1)\frac{\pi}{2}) = 0$ for any integer $n$, and $\sin((2n+1)\frac{\pi}{2}) = (-1)^n$, we get:

$\cos((2n+1)\frac{\pi}{2} + h) = 0 \cdot \cos h - (-1)^n \cdot \sin h = -(-1)^n \sin h = (-1)^{n+1} \sin h$

So, $|\cos(x_0 + h)| = |(-1)^{n+1} \sin h| = |\sin h|$.

The limit becomes $\lim\limits_{h \to 0} \frac{|\sin h|}{h}$.

We evaluate the left-hand limit (LHL) and the right-hand limit (RHL) at $h=0$ for $\frac{|\sin h|}{h}$.

LHL: $\lim\limits_{h \to 0^-} \frac{|\sin h|}{h}$. For $h < 0$ and close to 0, $\sin h < 0$. So $|\sin h| = -\sin h$.

LHL $= \lim\limits_{h \to 0^-} \frac{-\sin h}{h} = -1 \cdot \lim\limits_{h \to 0^-} \frac{\sin h}{h} = -1 \cdot 1 = -1$.

RHL: $\lim\limits_{h \to 0^+} \frac{|\sin h|}{h}$. For $h > 0$ and close to 0, $\sin h > 0$. So $|\sin h| = \sin h$.

RHL $= \lim\limits_{h \to 0^+} \frac{\sin h}{h} = 1$.

Since the LHL (-1) is not equal to the RHL (1), the limit $\lim\limits_{h \to 0} \frac{|\cos(x_0 + h)|}{h}$ does not exist at $x_0 = (2n+1)\frac{\pi}{2}$.

Thus, the function $f(x) = |\cos x|$ is not differentiable at the points $x = (2n+1)\frac{\pi}{2}$ for $n \in Z$.

These are the points where the graph of $\cos x$ crosses the x-axis, resulting in sharp points (cusps) in the graph of $|\cos x|$.

Comparing our findings with the options:

(A) f is everywhere differentiable. False, as it's not differentiable at $(2n+1)\frac{\pi}{2}$.

(B) f is everywhere continuous but not differentiable at x = nπ, n ∈ Z. False. It is continuous everywhere. The points of non-differentiability are not $n\pi$ (where $\cos(n\pi) = \pm 1$), but $(2n+1)\frac{\pi}{2}$ (where $\cos((2n+1)\frac{\pi}{2}) = 0$).

(C) f is everywhere continuous but not differentiable at x = (2n + 1) $\frac{π}{2}$ , n ∈ Z . True. It is continuous everywhere, and not differentiable at the points $(2n+1)\frac{\pi}{2}$.

(D) none of these. False, as (C) is correct.

The function $f(x) = |\cos x|$ is everywhere continuous but not differentiable at $x = (2n+1)\frac{\pi}{2}$ for $n \in Z$.

The correct option is (C) f is everywhere continuous but not differentiable at x = (2n + 1) $\frac{π}{2}$ , n ∈ Z .

Given:

The function $f(x) = |x| + |x – 1|$.

To Analyze:

The continuity of the function $f(x)$ at $x=0$ and $x=1$.

Solution:

The function $f(x)$ involves absolute value terms. We can rewrite the function by removing the absolute values based on the sign of the expressions inside them.

The critical points where the expressions inside the absolute values change sign are $x=0$ (for $|x|$) and $x=1$ (for $|x-1|$).

We define the function piecewise based on these critical points:

• If $x < 0$: $|x| = -x$ and $|x-1| = -(x-1) = 1-x$.
• If $0 \leq x < 1$: $|x| = x$ and $|x-1| = -(x-1) = 1-x$.
• If $x \geq 1$: $|x| = x$ and $|x-1| = x-1$.

So, the piecewise definition of $f(x)$ is:

If $x < 0$: $f(x) = -x + (1-x) = 1 - 2x$

If $0 \leq x < 1$: $f(x) = x + (1-x) = 1$

If $x \geq 1$: $f(x) = x + (x-1) = 2x - 1$

Thus, $f(x)$ can be written as:

$f(x) = \begin{cases} 1 - 2x & , x < 0 \\ 1 & , 0 \leq x < 1 \\ 2x - 1 & , x \geq 1 \end{cases}$

Now, let's check the continuity at $x=0$ and $x=1$. A function is continuous at a point $a$ if $\lim\limits_{x \to a} f(x) = f(a)$. This requires the left-hand limit, the right-hand limit, and the function value at $a$ to be equal.

Continuity at $x=0$:

Function value at $x=0$:

$f(0) = 1$ (from the definition for $0 \leq x < 1$)

Left-hand limit at $x=0$:

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (1 - 2x)$ (using the definition for $x < 0$)

$= 1 - 2(0) = 1$

Right-hand limit at $x=0$:

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} 1$ (using the definition for $0 \leq x < 1$)

$= 1$

Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0) = 1$, the function $f(x)$ is continuous at $x=0$.

Continuity at $x=1$:

Function value at $x=1$:

$f(1) = 2(1) - 1 = 2 - 1 = 1$ (from the definition for $x \geq 1$)

Left-hand limit at $x=1$:

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} 1$ (using the definition for $0 \leq x < 1$)

$= 1$

Right-hand limit at $x=1$:

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (2x - 1)$ (using the definition for $x \geq 1$)

$= 2(1) - 1 = 2 - 1 = 1$

Since $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x) = f(1) = 1$, the function $f(x)$ is continuous at $x=1$.

The function is continuous at both $x=0$ and $x=1$.

Let's check the options:

(A) continuous at x = 0 as well as at x = 1. This matches our conclusion.

(B) continuous at x = 1 but not at x = 0. Incorrect.

(C) discontinuous at x = 0 as well as at x = 1. Incorrect.

(D) continuous at x = 0 but not at x = 1. Incorrect.

The correct option is (A) continuous at x = 0 as well as at x = 1.

Given:

The function is defined by:

$f(x) = \begin{cases}\sin \frac{1}{x},& if\; x≠0\\k,& if\; x=0\end{cases}$

To Find:

The value of $k$ that makes the function continuous at $x=0$.

Solution:

For a function $f(x)$ to be continuous at $x=0$, the limit of the function as $x$ approaches $0$ must be equal to the value of the function at $0$.

That is, $\lim\limits_{x \to 0} f(x) = f(0)$.

From the definition of the function, we are given that $f(0) = k$.

Now, we need to evaluate the limit $\lim\limits_{x \to 0} f(x)$. Since the function definition for $x \neq 0$ is $\sin \frac{1}{x}$, we consider the limit:

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \sin \frac{1}{x}$.

Let's analyze the behavior of $\sin \frac{1}{x}$ as $x$ approaches 0.

As $x \to 0$, the argument $\frac{1}{x}$ approaches $\pm \infty$.

Consider the values of $\frac{1}{x}$ as $x$ gets closer to 0:

If $x = \frac{1}{2n\pi}$ for large integer $n$, then $x \to 0$ and $\frac{1}{x} = 2n\pi$. In this case, $\sin \frac{1}{x} = \sin(2n\pi) = 0$.

If $x = \frac{1}{(2n + \frac{1}{2})\pi}$ for large integer $n$, then $x \to 0$ and $\frac{1}{x} = (2n + \frac{1}{2})\pi$. In this case, $\sin \frac{1}{x} = \sin((2n + \frac{1}{2})\pi) = 1$.

If $x = \frac{1}{(2n + \frac{3}{2})\pi}$ for large integer $n$, then $x \to 0$ and $\frac{1}{x} = (2n + \frac{3}{2})\pi$. In this case, $\sin \frac{1}{x} = \sin((2n + \frac{3}{2})\pi) = -1$.

As $x$ approaches 0, the value of $\frac{1}{x}$ covers a wide range of values extending to infinity (both positive and negative). The sine function oscillates between -1 and 1 as its argument tends to infinity.

Because $\sin \frac{1}{x}$ does not approach a single value as $x \to 0$, the limit $\lim\limits_{x \to 0} \sin \frac{1}{x}$ does not exist.

For the function to be continuous at $x=0$, the limit $\lim\limits_{x \to 0} f(x)$ must exist and be equal to $f(0) = k$.

Since the limit $\lim\limits_{x \to 0} f(x)$ does not exist, there is no value of $k$ that can make the function continuous at $x=0$.

Let's check the options:

(A) 8 - The limit does not exist, so $k$ cannot be 8.

(B) 1 - The limit does not exist, so $k$ cannot be 1.

(C) –1 - The limit does not exist, so $k$ cannot be -1.

(D) none of these - This option is correct because no value of $k$ can make the function continuous at $x=0$ as the required limit does not exist.

The correct option is (D) none of these.

Given:

The function $f(x) = |x – 3| \cos x$.

To Find:

The set of points where the function $f(x)$ is differentiable.

Solution:

Let $g(x) = |x – 3|$ and $h(x) = \cos x$. Then $f(x) = g(x) h(x)$.

We know that the function $h(x) = \cos x$ is differentiable for all real numbers $x$.

The function $g(x) = |x – 3|$ can be written piecewise as:

$g(x) = \begin{cases} -(x - 3) = 3 - x & , x < 3 \\ x - 3 & , x \geq 3 \end{cases}$

The derivative of $g(x)$ is:

$g'(x) = \begin{cases} -1 & , x < 3 \\ 1 & , x > 3 \end{cases}$

The function $g(x) = |x-3|$ is differentiable for all $x$ except possibly at $x=3$. At $x=3$, the left-hand derivative is $\lim\limits_{x \to 3^-} \frac{|x-3|-|3-3|}{x-3} = \lim\limits_{x \to 3^-} \frac{-(x-3)}{x-3} = -1$. The right-hand derivative is $\lim\limits_{x \to 3^+} \frac{|x-3|-|3-3|}{x-3} = \lim\limits_{x \to 3^+} \frac{x-3}{x-3} = 1$. Since the LHD $\neq$ RHD, $g(x)$ is not differentiable at $x=3$.

The function $f(x) = g(x)h(x)$ is the product of two functions. The product rule for differentiation states that if $g(x)$ and $h(x)$ are differentiable at a point $x$, then $f(x) = g(x)h(x)$ is also differentiable at $x$, and $f'(x) = g'(x)h(x) + g(x)h'(x)$.

For $x \neq 3$, both $g(x) = |x-3|$ and $h(x) = \cos x$ are differentiable. Thus, their product $f(x) = |x-3| \cos x$ is differentiable for all $x \in \mathbb{R} \setminus \{3\}$.

Now, we need to check the differentiability of $f(x)$ at the point $x=3$ where $g(x)$ is not differentiable. We use the definition of the derivative:

$f'(3) = \lim\limits_{h \to 0} \frac{f(3+h) - f(3)}{h}$

$f(3) = |3 - 3| \cos(3) = 0 \cdot \cos 3 = 0$

$f(3+h) = |(3+h) - 3| \cos(3+h) = |h| \cos(3+h)$

So, $f'(3) = \lim\limits_{h \to 0} \frac{|h| \cos(3+h) - 0}{h} = \lim\limits_{h \to 0} \frac{|h| \cos(3+h)}{h}$.

Let's evaluate the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x=3$ (or $h=0$):

LHD at $x=3$: $\lim\limits_{h \to 0^-} \frac{|h| \cos(3+h)}{h}$. For $h < 0$, $|h| = -h$.

LHD $= \lim\limits_{h \to 0^-} \frac{-h \cos(3+h)}{h} = \lim\limits_{h \to 0^-} (-\cos(3+h))$

As $h \to 0^-$, $3+h \to 3$. Since $\cos x$ is continuous, $\lim\limits_{h \to 0^-} (-\cos(3+h)) = -\cos 3$.

RHD at $x=3$: $\lim\limits_{h \to 0^+} \frac{|h| \cos(3+h)}{h}$. For $h > 0$, $|h| = h$.

RHD $= \lim\limits_{h \to 0^+} \frac{h \cos(3+h)}{h} = \lim\limits_{h \to 0^+} \cos(3+h)$

As $h \to 0^+$, $3+h \to 3$. Since $\cos x$ is continuous, $\lim\limits_{h \to 0^+} \cos(3+h) = \cos 3$.

For the derivative $f'(3)$ to exist, LHD must equal RHD:

This implies $2 \cos 3 = 0$, which means $\cos 3 = 0$.

However, the value $x=3$ radians is approximately $3 \times \frac{180^\circ}{\pi} \approx 3 \times 57.3^\circ \approx 171.9^\circ$. The cosine of $171.9^\circ$ is not zero. The values of $x$ where $\cos x = 0$ are $x = (2n+1)\frac{\pi}{2}$ for $n \in Z$. Since $\frac{\pi}{2} \approx 1.57$ and $\frac{3\pi}{2} \approx 4.71$, $3$ is not an odd multiple of $\frac{\pi}{2}$. Therefore, $\cos 3 \neq 0$.

Since $-\cos 3 \neq \cos 3$, the derivative $f'(3)$ does not exist at $x=3$.

Thus, the function $f(x) = |x-3| \cos x$ is differentiable at all points in $\mathbb{R}$ except at $x=3$.

The set of points where $f(x)$ is differentiable is $\mathbb{R} - \{3\}$.

Let's check the options:

(A) R - Incorrect.

(B) R – {3} - Correct.

(C) (0, ∞) - Incorrect.

(D) none of these - Incorrect.

The correct option is (B) R – {3}.

Given:

The function $f(x) = \sec (\tan^{–1} x)$.

To Find:

The differential coefficient (derivative) of $f(x)$ with respect to $x$, i.e., $\frac{d}{dx} (\sec (\tan^{–1} x))$.

Solution:

Let $y = \sec (\tan^{–1} x)$.

To simplify the expression, let $\theta = \tan^{–1} x$. This implies $\tan \theta = x$.

We can construct a right-angled triangle where $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{x}{1}$.

Using the Pythagorean theorem, the Hypotenuse $= \sqrt{\text{Opposite}^2 + \text{Adjacent}^2} = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1}$.

Now, we need to find $\sec \theta$. By definition, $\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}}$.

$\sec \theta = \frac{\sqrt{x^2 + 1}}{1} = \sqrt{x^2 + 1}$.

Substituting $\theta = \tan^{-1} x$ back, we get $\sec (\tan^{–1} x) = \sqrt{x^2 + 1}$.

So, the function can be rewritten as $y = \sqrt{x^2 + 1}$.

Now we need to find the derivative of $y$ with respect to $x$.

$y = (x^2 + 1)^{\frac{1}{2}}$

Using the chain rule, $\frac{dy}{dx} = \frac{d}{dx} [(x^2 + 1)^{\frac{1}{2}}]$.

Let $u = x^2 + 1$. Then $y = u^{\frac{1}{2}}$.

$\frac{dy}{du} = \frac{1}{2} u^{\frac{1}{2} - 1} = \frac{1}{2} u^{-\frac{1}{2}} = \frac{1}{2\sqrt{u}}$

$\frac{du}{dx} = \frac{d}{dx} (x^2 + 1) = 2x + 0 = 2x$

By the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

$\frac{dy}{dx} = \left(\frac{1}{2\sqrt{u}}\right) \cdot (2x)$

Substitute $u = x^2 + 1$ back:

$\frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x$

$\frac{dy}{dx} = \frac{2x}{2\sqrt{x^2 + 1}}$

$\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}}$

Comparing with the given options:

(A) $\frac{x}{\sqrt{1+x^2}}$ - Matches our result.

(B) $\frac{x}{1+x^2}$ - Does not match.

(C) $x\sqrt{1+x^2}$ - Does not match.

(D) $\frac{1}{\sqrt{1+x^2}}$ - Does not match.

The correct option is (A) $\frac{x}{\sqrt{1+x^2}}$.

Alternate Solution (Using Chain Rule Directly):

Let $y = \sec (\tan^{–1} x)$.

Using the chain rule, $\frac{dy}{dx} = \frac{d}{d(\tan^{–1} x)} (\sec (\tan^{–1} x)) \cdot \frac{d}{dx} (\tan^{–1} x)$.

The derivative of $\sec u$ with respect to $u$ is $\sec u \tan u$. So, $\frac{d}{d(\tan^{–1} x)} (\sec (\tan^{–1} x)) = \sec (\tan^{–1} x) \tan (\tan^{–1} x)$.

The derivative of $\tan^{–1} x$ with respect to $x$ is $\frac{1}{1+x^2}$.

So, $\frac{dy}{dx} = [\sec (\tan^{–1} x) \tan (\tan^{–1} x)] \cdot \left(\frac{1}{1+x^2}\right)$.

We know that $\tan (\tan^{–1} x) = x$ (for all real $x$).

And, as shown in the previous method, $\sec (\tan^{–1} x) = \sqrt{x^2 + 1}$.

Substitute these into the derivative expression:

$\frac{dy}{dx} = (\sqrt{x^2 + 1}) \cdot (x) \cdot \left(\frac{1}{1+x^2}\right)$

$\frac{dy}{dx} = \frac{x\sqrt{x^2 + 1}}{1+x^2}$

Since $1+x^2 = (\sqrt{1+x^2})^2$, we can simplify:

$\frac{dy}{dx} = \frac{x\sqrt{x^2 + 1}}{(\sqrt{x^2 + 1})^2} = \frac{x}{\sqrt{x^2 + 1}}$.

This confirms the result obtained by the first method.

The correct option is (A) $\frac{x}{\sqrt{1+x^2}}$.

Given:

$u = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$

$v = \tan^{-1} \left( \frac{2x}{1−x^2} \right)$

To Find:

The value of $\frac{du}{dv}$.

Solution:

We need to find the derivative of $u$ with respect to $v$. We can use parametric differentiation by finding the derivatives of $u$ and $v$ with respect to $x$, and then computing $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.

Let's use the substitution $x = \tan \theta$. Then $\theta = \tan^{-1} x$.

Consider the expression for $u$:

$u = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$

Substitute $x = \tan \theta$:

$u = \sin^{-1} \left( \frac{2 \tan \theta}{1+\tan^2 \theta} \right)$

Using the trigonometric identity $\sin(2\theta) = \frac{2 \tan \theta}{1+\tan^2 \theta}$, we get:

$u = \sin^{-1} (\sin(2\theta))$

For the standard principal value branch of $\sin^{-1}$ (which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$), $\sin^{-1}(\sin y) = y$ if $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

If we consider the domain $|x| \leq 1$, then $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$, which implies $-\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2}$. In this range, $u = 2\theta$.

So, $u = 2 \tan^{-1} x$ for $|x| \leq 1$.

Consider the expression for $v$:

$v = \tan^{-1} \left( \frac{2x}{1−x^2} \right)$

Substitute $x = \tan \theta$:

$v = \tan^{-1} \left( \frac{2 \tan \theta}{1-\tan^2 \theta} \right)$

Using the trigonometric identity $\tan(2\theta) = \frac{2 \tan \theta}{1-\tan^2 \theta}$, we get:

$v = \tan^{-1} (\tan(2\theta))$

For the standard principal value branch of $\tan^{-1}$ (which is $(-\frac{\pi}{2}, \frac{\pi}{2})$), $\tan^{-1}(\tan y) = y$ if $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

If we consider the domain $|x| < 1$, then $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, which implies $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$. In this range, $v = 2\theta$.

So, $v = 2 \tan^{-1} x$ for $|x| < 1$.

Assuming the domain where both simplifications hold simultaneously, i.e., for $|x|<1$, we have:

$u = 2 \tan^{-1} x$

$v = 2 \tan^{-1} x$

Now, let's find the derivatives with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx} (2 \tan^{-1} x) = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$

$\frac{dv}{dx} = \frac{d}{dx} (2 \tan^{-1} x) = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$

Now, we can find $\frac{du}{dv}$:

$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}}$

$\frac{du}{dv} = 1$

This result is obtained under the assumption that the domain of $x$ is restricted such that the principal value simplifications $u = 2\theta$ and $v = 2\theta$ are valid. Given the options, a constant answer is expected, which supports this assumption for the intended scope of the problem.

Comparing with the given options:

(A) $\frac{1}{2}$ - Incorrect.

(B) x - Incorrect.

(C) $\frac{1−x^2}{1+x^2}$ - Incorrect.

(D) 1 - Matches our result.

The correct option is (D) 1.

Given:

The function $f(x) = e^x \sin x$ on the interval $[0, \pi]$.

To Find:

The value of $c$ in the interval $(0, \pi)$ that satisfies the conclusion of Rolle's Theorem.

Solution:

Rolle's Theorem states that if a function $f(x)$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.

Let's check the conditions for $f(x) = e^x \sin x$ on $[0, \pi]$:

1. Continuity: The function $e^x$ is continuous for all real $x$. The function $\sin x$ is continuous for all real $x$. The product of two continuous functions is continuous. Thus, $f(x) = e^x \sin x$ is continuous on $[0, \pi]$.

2. Differentiability: The function $e^x$ is differentiable for all real $x$. The function $\sin x$ is differentiable for all real $x$. The product of two differentiable functions is differentiable. Thus, $f(x) = e^x \sin x$ is differentiable on $(0, \pi)$.

3. Check $f(a) = f(b)$:

$f(0) = e^0 \sin 0 = 1 \cdot 0 = 0$

$f(\pi) = e^\pi \sin \pi = e^\pi \cdot 0 = 0$

Since $f(0) = f(\pi) = 0$, the third condition is satisfied.

All conditions of Rolle's Theorem are met, so there exists at least one $c \in (0, \pi)$ such that $f'(c) = 0$.

Now, we find the derivative of $f(x)$ using the product rule:

$f'(x) = \frac{d}{dx} (e^x \sin x)$

$f'(x) = (\frac{d}{dx} e^x) \sin x + e^x (\frac{d}{dx} \sin x)$

$f'(x) = e^x \sin x + e^x \cos x$

$f'(x) = e^x (\sin x + \cos x)$

Set $f'(c) = 0$ to find the value(s) of $c$:

Since $e^c$ is always positive for any real $c$, we must have:

This implies $\sin c = -\cos c$.

If $\cos c \neq 0$, we can divide both sides by $\cos c$:

We need to find the value(s) of $c$ in the interval $(0, \pi)$ for which $\tan c = -1$.

In the interval $(0, \pi)$, the tangent function is negative in the second quadrant.

The general solution for $\tan x = -1$ is $x = n\pi - \frac{\pi}{4}$, where $n \in Z$.

For $n=1$, $x = 1\pi - \frac{\pi}{4} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Let's check if $c = \frac{3\pi}{4}$ is in the interval $(0, \pi)$: $0 < \frac{3\pi}{4} < \pi$. This is true.

For other integer values of $n$, the values of $c$ will fall outside the interval $(0, \pi)$. For example, if $n=0$, $c = -\frac{\pi}{4} \notin (0, \pi)$. If $n=2$, $c = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \notin (0, \pi)$.

The only value of $c$ in $(0, \pi)$ that satisfies $f'(c) = 0$ is $c = \frac{3\pi}{4}$.

Comparing with the given options:

(A) $\frac{π}{6}$ - Incorrect.

(B) $\frac{π}{4}$ - Incorrect ($\tan \frac{\pi}{4} = 1$).

(C) $\frac{π}{2}$ - Incorrect ($\tan \frac{\pi}{2}$ is undefined, $\cos \frac{\pi}{2} = 0$). If $\cos c = 0$, $\sin c = \pm 1$, so $\sin c + \cos c \neq 0$.

(D) $\frac{3π}{4}$ - Correct ($\tan \frac{3\pi}{4} = -1$).

The correct option is (D) $\frac{3π}{4}$.

Given:

The function $f(x) = x(x – 2) = x^2 - 2x$.

The interval is $[1, 2]$.

To Find:

The value of $c$ in the interval $(1, 2)$ that satisfies the conclusion of the Mean Value Theorem.

Solution:

The Mean Value Theorem (MVT) states that if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one value $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

Let's check the conditions for the given function $f(x) = x^2 - 2x$ on the interval $[1, 2]$:

1. Continuity: The function $f(x) = x^2 - 2x$ is a polynomial function. Polynomial functions are continuous for all real numbers. Therefore, $f(x)$ is continuous on the closed interval $[1, 2]$.

2. Differentiability: The function $f(x) = x^2 - 2x$ is a polynomial function. Polynomial functions are differentiable for all real numbers. Therefore, $f(x)$ is differentiable on the open interval $(1, 2)$.

Both conditions of the MVT are satisfied. Thus, there exists a value $c \in (1, 2)$ such that $f'(c) = \frac{f(2) - f(1)}{2 - 1}$.

First, let's calculate the values of the function at the endpoints of the interval:

$f(1) = 1(1 - 2) = 1(-1) = -1$

$f(2) = 2(2 - 2) = 2(0) = 0$

Next, let's calculate the slope of the secant line connecting the endpoints:

$\frac{f(b) - f(a)}{b - a} = \frac{f(2) - f(1)}{2 - 1} = \frac{0 - (-1)}{1} = \frac{1}{1} = 1$

Now, let's find the derivative of the function $f(x)$:

$f'(x) = \frac{d}{dx} (x^2 - 2x) = 2x - 2$

According to the MVT, there exists a value $c \in (1, 2)$ such that $f'(c)$ is equal to the slope of the secant line.

Substitute $f'(c) = 2c - 2$ and the calculated slope (1) into equation (i):

Solve for $c$:

$2c = 1 + 2$

$2c = 3$

$c = \frac{3}{2}$

Finally, we must check if this value of $c$ lies within the open interval $(1, 2)$.

$1 < \frac{3}{2} < 2$ is true, since $\frac{3}{2} = 1.5$.

The value $c = \frac{3}{2}$ is indeed in the interval $(1, 2)$.

Comparing with the given options:

(A) $\frac{3}{2}$ - Matches our result.

(B) $\frac{2}{3}$ - Incorrect ($\frac{2}{3} \approx 0.67$, which is not in $(1, 2)$).

(C) $\frac{1}{2}$ - Incorrect ($\frac{1}{2} = 0.5$, which is not in $(1, 2)$).

(D) $\frac{3}{2}$ - Matches our result (Option (A) and (D) are the same).

The correct option is (A) $\frac{3}{2}$ (or (D) $\frac{3}{2}$).

Let's analyze each statement in COLUMN-I and match it with the appropriate item in COLUMN-II.

Analysis of (A):

The function is given by $f(x) = \begin{cases}\frac{\sin 3x}{x},& if\; x≠0\\ \frac{k}{2},& if x=0\end{cases}$.

For the function to be continuous at $x=0$, the limit as $x \to 0$ must equal the function value at $x=0$.

$\lim\limits_{x \to 0} f(x) = f(0)$

We are given $f(0) = \frac{k}{2}$.

We evaluate the limit $\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{\sin 3x}{x}$.

Multiply and divide by 3:

$\lim\limits_{x \to 0} \frac{\sin 3x}{x} = \lim\limits_{x \to 0} 3 \cdot \frac{\sin 3x}{3x}$

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, let $\theta = 3x$. As $x \to 0$, $\theta \to 0$.

$\lim\limits_{x \to 0} 3 \cdot \frac{\sin 3x}{3x} = 3 \cdot \lim\limits_{3x \to 0} \frac{\sin 3x}{3x} = 3 \cdot 1 = 3$.

So, $\lim\limits_{x \to 0} f(x) = 3$.

For continuity, we set $\frac{k}{2} = 3$.

$k = 3 \times 2 = 6$.

This matches option (c) 6 in COLUMN-II.

(A) matches (c).

Analysis of (B):

The statement is "Every continuous function is differentiable".

This statement is false. Differentiability implies continuity, but the converse is not true. A classic counterexample is the function $f(x) = |x|$ at $x=0$, which is continuous at $x=0$ but not differentiable at $x=0$.

This matches option (d) False in COLUMN-II.

(B) matches (d).

Analysis of (C):

We need an example of a function which is continuous everywhere but not differentiable at exactly one point.

Consider the function $f(x) = |x|$.

We know that $f(x) = |x|$ is continuous for all $x \in \mathbb{R}$.

The derivative of $f(x) = |x|$ is $f'(x) = \begin{cases} -1 & , x < 0 \\ 1 & , x > 0 \end{cases}$.

The left-hand derivative at $x=0$ is $\lim\limits_{h \to 0^-} \frac{|0+h| - |0|}{h} = \lim\limits_{h \to 0^-} \frac{-h}{h} = -1$.

The right-hand derivative at $x=0$ is $\lim\limits_{h \to 0^+} \frac{|0+h| - |0|}{h} = \lim\limits_{h \to 0^+} \frac{h}{h} = 1$.

Since the LHD $\neq$ RHD at $x=0$, the function $f(x) = |x|$ is not differentiable at $x=0$. It is differentiable everywhere else.

Thus, $f(x) = |x|$ is continuous everywhere but not differentiable at exactly one point ($x=0$).

This matches option (a) |x| in COLUMN-II.

(C) matches (a).

Analysis of (D):

The statement is "The identity function i.e. f (x) = x ∀ x ∈ R is a continuous function".

The identity function $f(x) = x$ is a polynomial function. Polynomial functions are known to be continuous for all real numbers.

Alternatively, for any point $a \in \mathbb{R}$, $\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} x = a$. Also, $f(a) = a$. Since $\lim\limits_{x \to a} f(x) = f(a)$ for all $a \in \mathbb{R}$, the function is continuous everywhere.

Thus, the statement is true.

This matches option (b) True in COLUMN-II.

(D) matches (b).

Summary of matches:

(A) matches (c)

(B) matches (d)

(C) matches (a)

(D) matches (b)

Given:

The function $f (x) = \frac{1}{\log |x|}$.

To Find:

The number of points at which the function is discontinuous.

Solution:

A function is discontinuous at points where it is not defined.

For the function $f(x) = \frac{1}{\log |x|}$ to be defined, two conditions must be met:

1. The argument of the logarithm must be strictly positive: $|x| > 0$. This means $x \neq 0$.

2. The denominator must not be zero: $\log |x| \neq 0$.

We need to find the values of $x$ for which $\log |x| = 0$.

This is equivalent to $|x| = e^0$.

This holds when $x = 1$ or $x = -1$.

So, the function $f(x)$ is undefined (and thus discontinuous) at the points where $|x| \leq 0$ (which is only $x=0$) or where $\log |x| = 0$ (which are $x=1$ and $x=-1$).

The points of discontinuity are $x = 0$, $x = 1$, and $x = -1$.

These are three distinct points.

The number of points at which the function is discontinuous is 3.

The number of points at which the function $f (x) = \frac{1}{\log |x|}$ is discontinuous is 3.

Given:

The function $f(x) = \begin{cases}ax+1& if\; x≥1\\x+2 & if x<1\end{cases}$ is continuous at $x=1$ (since the definition changes at $x=1$, this is the point where continuity needs to be explicitly ensured for the function to be continuous everywhere).

To Find:

The value of $a$ that makes the function continuous.

Solution:

For the function $f(x)$ to be continuous at $x=1$, the following condition must be satisfied:

Let's calculate each part:

1. Function value at $x=1$:

When $x \geq 1$, $f(x) = ax+1$. So, $f(1) = a(1) + 1 = a+1$.

2. Left-hand limit at $x=1$:

When $x < 1$, $f(x) = x+2$. So, $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x+2)$.

Since $x+2$ is a polynomial, the limit can be evaluated by direct substitution:

3. Right-hand limit at $x=1$:

When $x \geq 1$, $f(x) = ax+1$. So, $\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (ax+1)$.

Since $ax+1$ is a polynomial, the limit can be evaluated by direct substitution:

Using the continuity condition (i), we set the left-hand limit, right-hand limit, and function value equal:

From the equality $3 = a+1$, we can solve for $a$:

$a = 3 - 1$

Thus, the value of $a$ that makes the function continuous is 2.

If $f(x) = \begin{cases}ax+1& if\; x≥1\\x+2 & if x<1\end{cases}$ is continuous, then a should be equal to 2.

Given:

The function $f(x) = \log_{10} x$.

To Find:

The derivative of $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}(\log_{10} x)$.

Solution:

We use the change of base formula for logarithms to express $\log_{10} x$ in terms of the natural logarithm (logarithm with base $e$, denoted as $\ln$).

Using base $e$, we have:

Now we need to find the derivative of $f(x) = \frac{\ln x}{\ln 10}$ with respect to $x$.

The term $\frac{1}{\ln 10}$ is a constant.

Using the constant multiple rule for differentiation:

The derivative of $\ln x$ with respect to $x$ is $\frac{1}{x}$.

Substitute (iii) into the derivative expression:

The derivative of $\log_{10} x$ w.r.t. x is $\frac{1}{x \ln 10}$.

Given:

$y = \sec^{-1} \left( \frac{\sqrt{x} + 1}{\sqrt{x} − 1} \right) + \sin^{-1} \left( \frac{\sqrt{x} − 1}{\sqrt{x} + 1} \right)$

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We can simplify the expression for $y$ before differentiating.

Recall the inverse trigonometric identity: $\sec^{-1} u = \cos^{-1} \left(\frac{1}{u}\right)$ for $|u| \geq 1$.

Let $u = \frac{\sqrt{x} + 1}{\sqrt{x} − 1}$. Then $\frac{1}{u} = \frac{\sqrt{x} − 1}{\sqrt{x} + 1}$.

Assuming $\sqrt{x} - 1 \neq 0$, which means $x \neq 1$, and also assuming the range where the identity holds.

The first term is $\sec^{-1} \left( \frac{\sqrt{x} + 1}{\sqrt{x} − 1} \right)$.

Using the identity, this term can be written as $\cos^{-1} \left( \frac{\sqrt{x} − 1}{\sqrt{x} + 1} \right)$.

So, the expression for $y$ becomes:

$y = \cos^{-1} \left( \frac{\sqrt{x} − 1}{\sqrt{x} + 1} \right) + \sin^{-1} \left( \frac{\sqrt{x} − 1}{\sqrt{x} + 1} \right)$

Now, recall another inverse trigonometric identity: $\sin^{-1} z + \cos^{-1} z = \frac{\pi}{2}$ for $-1 \leq z \leq 1$.

Let $z = \frac{\sqrt{x} − 1}{\sqrt{x} + 1}$.

The domain of the original function requires $\sqrt{x} \geq 0$, so $x \geq 0$. Also, the arguments of $\sec^{-1}$ and $\sin^{-1}$ must be defined. The argument of $\sec^{-1}$ requires $\sqrt{x}-1 \neq 0$, so $x \neq 1$. The argument of $\log |x|$ in previous problem needs $|x|>0$. $\log_{10}x$ needs $x>0$. $\sqrt{x}$ needs $x \ge 0$. For $\sec^{-1} u$, $|u| \ge 1$ is required, $\frac{\sqrt{x} + 1}{|\sqrt{x} − 1|} \ge 1$. The argument of $\sin^{-1} z$ requires $-1 \leq z \leq 1$. $\frac{\sqrt{x} − 1}{\sqrt{x} + 1}$ should be in $[-1, 1]$.

For $x > 0$, $\sqrt{x} > 0$. Then $\sqrt{x} + 1 > 0$.

The condition $\frac{\sqrt{x} − 1}{\sqrt{x} + 1} \geq -1$ is $\sqrt{x} - 1 \geq -(\sqrt{x} + 1) = -\sqrt{x} - 1$, which simplifies to $2\sqrt{x} \geq 0$, or $\sqrt{x} \geq 0$, which is true for $x \geq 0$.

The condition $\frac{\sqrt{x} − 1}{\sqrt{x} + 1} \leq 1$ is $\sqrt{x} - 1 \leq \sqrt{x} + 1$, which simplifies to $-1 \leq 1$, which is always true.

So, for $x > 0$, the argument $\frac{\sqrt{x} − 1}{\sqrt{x} + 1}$ is in the interval $[-1, 1)$.

The condition for $\sec^{-1} u = \cos^{-1} (1/u)$ is $|u| \geq 1$. $|\frac{\sqrt{x} + 1}{\sqrt{x} − 1}| \geq 1$. This means $\sqrt{x}+1 \geq |\sqrt{x}-1|$. If $\sqrt{x} \ge 1$, this is $\sqrt{x}+1 \ge \sqrt{x}-1$ which is $1 \ge -1$, true. If $0 \le \sqrt{x} < 1$, this is $\sqrt{x}+1 \ge -(\sqrt{x}-1) = 1-\sqrt{x}$ which is $2\sqrt{x} \ge 0$, true. So the identity $\sec^{-1} u = \cos^{-1} (1/u)$ holds for $x \ge 0, x \ne 1$.

Thus, for $x > 0$ and $x \neq 1$, we can apply the identity $\sin^{-1} z + \cos^{-1} z = \frac{\pi}{2}$.

$y = \cos^{-1} \left( \frac{\sqrt{x} − 1}{\sqrt{x} + 1} \right) + \sin^{-1} \left( \frac{\sqrt{x} − 1}{\sqrt{x} + 1} \right) = \frac{\pi}{2}$

So, for $x > 0$ and $x \neq 1$, the function $y$ is a constant, $y = \frac{\pi}{2}$.

The derivative of a constant is 0.

The derivative is 0 for $x > 0$ and $x \neq 1$. Since this is a fill-in-the-blank question, the expected answer is the value of the derivative where it exists.

If $y = \sec^{-1} \left( \frac{\sqrt{x} + 1}{\sqrt{x} − 1} \right) + \sin^{-1} \left( \frac{\sqrt{x} − 1}{\sqrt{x} + 1} \right)$, then $\frac{dy}{dx}$ is equal to 0.

Given:

We are asked to find the derivative of $\sin x$ with respect to $\cos x$.

To Find:

$\frac{d(\sin x)}{d(\cos x)}$

Solution:

To find the derivative of one function with respect to another, we can use the formula:

Here, we let $u = \sin x$ and $v = \cos x$.

First, find the derivative of $u$ with respect to $x$:

Next, find the derivative of $v$ with respect to $x$:

Now, substitute the results from (ii) and (iii) into the formula (i):

Simplify the expression:

The derivative is defined when $\sin x \neq 0$, i.e., $x \neq n\pi$ for $n \in Z$.

The derivative of sin x w.r.t. cos x is $-\cot x$.

Given:

The statement: For continuity, at $x = a$, each of $\lim\limits_{x \to a^+} f(x)$ and $\lim\limits_{x \to a^-} f(x)$ is equal to $f(a)$.

To Evaluate:

Whether the given statement is True or False.

Solution:

The definition of continuity of a function $f(x)$ at a point $x=a$ requires three conditions to be met:

1. The function $f(a)$ is defined.

2. The limit of the function as $x$ approaches $a$, $\lim\limits_{x \to a} f(x)$, exists.

3. The limit of the function is equal to the function value at $a$, i.e., $\lim\limits_{x \to a} f(x) = f(a)$.

The second condition, that the limit $\lim\limits_{x \to a} f(x)$ exists, is equivalent to stating that both the left-hand limit and the right-hand limit exist and are equal to each other.

Let this common limit value be $L$. So, $\lim\limits_{x \to a} f(x) = L$.

The third condition for continuity states that this limit $L$ must be equal to the function value $f(a)$.

Combining conditions (i) and (ii), for $f(x)$ to be continuous at $x=a$, it is necessary and sufficient that:

This means that the left-hand limit must be equal to $f(a)$, and the right-hand limit must also be equal to $f(a)$. This is exactly what the given statement says.

Therefore, the statement is true.

The statement is True.

Given:

The statement: $y = |x – 1|$ is a continuous function.

To Evaluate:

Whether the given statement is True or False.

Solution:

The function is given by $f(x) = |x – 1|$.

The absolute value function, $g(u) = |u|$, is known to be continuous for all real numbers $u$.

The function $f(x) = |x-1|$ is a composite function where the outer function is the absolute value function and the inner function is $h(x) = x-1$.

$f(x) = g(h(x))$, where $g(u) = |u|$ and $h(x) = x-1$.

The function $h(x) = x-1$ is a polynomial function, which is continuous for all real numbers $x$.

Since $g(u) = |u|$ is continuous for all real numbers $u$ and $h(x) = x-1$ is continuous for all real numbers $x$, the composite function $f(x) = g(h(x)) = |x-1|$ is continuous for all real numbers $x$ where $h(x)$ is defined (which is everywhere).

Alternatively, we can examine the function piecewise:

$f(x) = |x-1| = \begin{cases} -(x-1) = 1-x & , x < 1 \\ x-1 & , x \geq 1 \end{cases}$

For $x < 1$, $f(x) = 1-x$, which is a polynomial and continuous.

For $x > 1$, $f(x) = x-1$, which is a polynomial and continuous.

We check continuity at the point where the definition changes, $x=1$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (1-x) = 1 - 1 = 0$

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (x-1) = 1 - 1 = 0$

$f(1) = 1 - 1 = 0$

Since $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x) = f(1) = 0$, the function is continuous at $x=1$.

Since the function is continuous for $x<1$, for $x>1$, and at $x=1$, it is continuous for all real numbers.

Therefore, the statement is true.

The statement is True.

Given:

The statement: A continuous function can have some points where limit does not exist.

To Evaluate:

Whether the given statement is True or False.

Solution:

Let's consider the definition of continuity of a function $f(x)$ at a point $x=a$.

A function $f(x)$ is said to be continuous at $x=a$ if and only if the following conditions are satisfied:

1. $f(a)$ is defined.

2. The limit of the function as $x$ approaches $a$, $\lim\limits_{x \to a} f(x)$, exists.

3. The limit is equal to the function value at $a$, i.e., $\lim\limits_{x \to a} f(x) = f(a)$.

The second condition for continuity at $x=a$ explicitly states that the limit of the function at $x=a$ *must* exist.

If the limit of the function does not exist at a point $a$, then the function cannot satisfy the definition of continuity at that point $a$.

Therefore, a function cannot be continuous at a point where its limit does not exist.

The statement claims that a *continuous function* can have points where the limit does not exist. If a function is continuous, it means it is continuous at every point in its domain (or on a specific interval). By the definition of continuity, the limit must exist at every point where the function is continuous.

The statement is a contradiction of the definition of continuity.

Therefore, the statement is false.

The statement is False.

Given:

The statement: $|\sin x|$ is a differentiable function for every value of $x$.

To Evaluate:

Whether the given statement is True or False.

Solution:

The function is $f(x) = |\sin x|$.

The absolute value function $|u|$ is not differentiable at $u=0$.

The function $f(x) = |\sin x|$ is a composite function, $f(x) = g(h(x))$, where $g(u) = |u|$ and $h(x) = \sin x$.

The function $f(x)$ is differentiable at points where $\sin x \neq 0$ and also where the derivative exists even if $\sin x = 0$.

The points where $\sin x = 0$ are $x = n\pi$, where $n$ is an integer ($n \in Z$).

Let's check the differentiability at these points, $x_0 = n\pi$, using the definition of the derivative:

$f'(x_0) = \lim\limits_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$

$f(n\pi) = |\sin(n\pi)| = |0| = 0$.

$f(n\pi + h) = |\sin(n\pi + h)|$.

Using the identity $\sin(n\pi + h) = (-1)^n \sin h$, we get:

$|\sin(n\pi + h)| = |(-1)^n \sin h| = |\sin h|$.

So, the limit of the difference quotient at $x=n\pi$ is:

$\lim\limits_{h \to 0} \frac{|\sin h| - 0}{h} = \lim\limits_{h \to 0} \frac{|\sin h|}{h}$.

We evaluate the left-hand limit (LHL) and the right-hand limit (RHL) at $h=0$ for $\frac{|\sin h|}{h}$.

LHL: $\lim\limits_{h \to 0^-} \frac{|\sin h|}{h}$. For $h < 0$ and close to 0, $\sin h < 0$. So $|\sin h| = -\sin h$.

LHL $= \lim\limits_{h \to 0^-} \frac{-\sin h}{h} = -1 \cdot \lim\limits_{h \to 0^-} \frac{\sin h}{h} = -1 \cdot 1 = -1$.

RHL: $\lim\limits_{h \to 0^+} \frac{|\sin h|}{h}$. For $h > 0$ and close to 0, $\sin h > 0$. So $|\sin h| = \sin h$.

RHL $= \lim\limits_{h \to 0^+} \frac{\sin h}{h} = 1$.

Since the LHL (-1) is not equal to the RHL (1), the limit $\lim\limits_{h \to 0} \frac{|\sin h|}{h}$ does not exist at $h=0$.

Thus, the function $f(x) = |\sin x|$ is not differentiable at the points $x = n\pi$ for $n \in Z$.

These are the points where the graph of $\sin x$ crosses the x-axis, resulting in sharp points (cusps) in the graph of $|\sin x|$.

The statement claims that $|\sin x|$ is differentiable for *every* value of $x$. This is false because it is not differentiable at $x = n\pi$ for integer values of $n$.

The statement is False.

Given:

The statement: $\cos |x|$ is differentiable everywhere.

To Evaluate:

Whether the given statement is True or False.

Solution:

Let the function be $f(x) = \cos |x|$.

We can write the function piecewise based on the definition of the absolute value:

$f(x) = \begin{cases} \cos(-x) & , x < 0 \\ \cos(x) & , x \geq 0 \end{cases}$

Since $\cos(-x) = \cos x$, the function simplifies to:

$f(x) = \begin{cases} \cos x & , x < 0 \\ \cos x & , x \geq 0 \end{cases}$

This means $f(x) = \cos x$ for all real numbers $x$.

The function $\cos x$ is known to be differentiable for all real numbers $x$.

The derivative of $\cos x$ is $-\sin x$.

Thus, the function $f(x) = \cos |x| = \cos x$ is differentiable everywhere, and its derivative is $f'(x) = -\sin x$.

Alternatively, we can use the definition of the derivative at $x=0$, which is the point where the absolute value function $|x|$ is not differentiable.

We need to evaluate $f'(0) = \lim\limits_{h \to 0} \frac{f(0+h) - f(0)}{h}$.

$f(0) = \cos |0| = \cos 0 = 1$.

$f(0+h) = \cos |h|$.

So, $f'(0) = \lim\limits_{h \to 0} \frac{\cos |h| - 1}{h}$.

Let's consider the left-hand limit (LHL) and the right-hand limit (RHL):

LHL = $\lim\limits_{h \to 0^-} \frac{\cos |h| - 1}{h} = \lim\limits_{h \to 0^-} \frac{\cos (-h) - 1}{h}$ (since $|h| = -h$ for $h < 0$)

= $\lim\limits_{h \to 0^-} \frac{\cos h - 1}{h}$ (since $\cos(-h) = \cos h$)

This is a standard limit, $\lim\limits_{\theta \to 0} \frac{\cos \theta - 1}{\theta} = 0$. So, LHL = 0.

RHL = $\lim\limits_{h \to 0^+} \frac{\cos |h| - 1}{h} = \lim\limits_{h \to 0^+} \frac{\cos h - 1}{h}$ (since $|h| = h$ for $h > 0$)

This is also the standard limit, which is 0. So, RHL = 0.

Since LHL = RHL = 0, the limit exists and is equal to 0. Thus, $f'(0) = 0$.

For $x \neq 0$, $f(x) = \cos x$, and its derivative is $-\sin x$.

At $x=0$, $f'(0) = -\sin 0 = 0$. The derivative matches the form $-\sin x$ even at $x=0$.

Therefore, the function $f(x) = \cos |x|$ is differentiable everywhere.

The statement is true.

The statement is True.

To examine the continuity of the function $f(x) = x^3 + 2x^2 - 1$ at $x = 1$, we need to check the following conditions:

1. $f(1)$ is defined.

2. $\lim\limits_{x \to 1} f(x)$ exists.

3. $\lim\limits_{x \to 1} f(x) = f(1)$.

Solution:

Let the given function be $f(x) = x^3 + 2x^2 - 1$.

We need to examine its continuity at $x=1$.

Step 1: Evaluate $f(1)$

$f(1) = (1)^3 + 2(1)^2 - 1$

$f(1) = 1 + 2(1) - 1$

$f(1) = 1 + 2 - 1$

$f(1) = 2$

So, $f(1)$ is defined and $f(1) = 2$.

Step 2: Evaluate $\lim\limits_{x \to 1} f(x)$

The function $f(x) = x^3 + 2x^2 - 1$ is a polynomial function.

The limit of a polynomial function at any point is equal to the value of the function at that point.

So, $\lim\limits_{x \to 1} f(x) = \lim\limits_{x \to 1} (x^3 + 2x^2 - 1)$

$\lim\limits_{x \to 1} f(x) = (1)^3 + 2(1)^2 - 1$

$\lim\limits_{x \to 1} f(x) = 1 + 2 - 1$

$\lim\limits_{x \to 1} f(x) = 2$

So, $\lim\limits_{x \to 1} f(x)$ exists and $\lim\limits_{x \to 1} f(x) = 2$.

Step 3: Compare $f(1)$ and $\lim\limits_{x \to 1} f(x)$

From Step 1, we have $f(1) = 2$.

From Step 2, we have $\lim\limits_{x \to 1} f(x) = 2$.

Since $\lim\limits_{x \to 1} f(x) = f(1)$, the third condition for continuity is satisfied.

Conclusion:

Since all three conditions for continuity are satisfied at $x=1$, the function $f(x) = x^3 + 2x^2 - 1$ is continuous at $x=1$.

To examine the continuity of the function $f(x)$ at $x=2$, we need to check the following conditions:

1. $f(2)$ is defined.

2. $\lim\limits_{x \to 2} f(x)$ exists (i.e., the left-hand limit and the right-hand limit are equal).

3. $\lim\limits_{x \to 2} f(x) = f(2)$.

Solution:

The given function is $f(x) = \begin{cases}3x+5,& if\; x \geq 2\\x^2,& if\; x < 2\end{cases}$.

We need to examine its continuity at $x=2$.

Step 1: Evaluate $f(2)$

Since $x=2$ falls in the condition $x \geq 2$, we use the first part of the function:

$f(x) = 3x+5$ for $x \geq 2$.

$f(2) = 3(2) + 5$

$f(2) = 6 + 5$

$f(2) = 11$

$f(2)$ is defined and its value is $11$.

Step 2: Evaluate the left-hand limit (LHL) at $x=2$

For the limit as $x$ approaches $2$ from the left ($x < 2$), we use the second part of the function:

$f(x) = x^2$ for $x < 2$.

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} x^2$

$\lim\limits_{x \to 2^-} f(x) = (2)^2$

$\lim\limits_{x \to 2^-} f(x) = 4$

Step 3: Evaluate the right-hand limit (RHL) at $x=2$

For the limit as $x$ approaches $2$ from the right ($x > 2$), we use the first part of the function (since $x>2$ is covered by $x \geq 2$):

$f(x) = 3x+5$ for $x \geq 2$.

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (3x+5)$

$\lim\limits_{x \to 2^+} f(x) = 3(2) + 5$

$\lim\limits_{x \to 2^+} f(x) = 6 + 5$

$\lim\limits_{x \to 2^+} f(x) = 11$

Step 4: Check if the limit $\lim\limits_{x \to 2} f(x)$ exists

For the limit $\lim\limits_{x \to 2} f(x)$ to exist, the LHL must be equal to the RHL.

LHL = $\lim\limits_{x \to 2^-} f(x) = 4$

RHL = $\lim\limits_{x \to 2^+} f(x) = 11$

Since LHL $\neq$ RHL ($4 \neq 11$), the limit $\lim\limits_{x \to 2} f(x)$ does not exist.

Conclusion:

The function $f(x)$ is discontinuous at $x=2$ because the limit of the function as $x$ approaches $2$ does not exist (the left-hand limit and the right-hand limit are not equal).

To examine the continuity of the function $f(x)$ at $x=0$, we need to check the following conditions:

1. $f(0)$ is defined.

2. $\lim\limits_{x \to 0} f(x)$ exists.

3. $\lim\limits_{x \to 0} f(x) = f(0)$.

Solution:

The given function is $f(x) = \begin{cases}\frac{1 − \cos 2x}{x^2},& if\; x \neq 0\\5,& if\; x = 0\end{cases}$.

We need to examine its continuity at $x=0$.

Step 1: Evaluate $f(0)$

According to the definition of the function, when $x=0$, $f(x)=5$.

$f(0) = 5$

$f(0)$ is defined and its value is $5$.

Step 2: Evaluate $\lim\limits_{x \to 0} f(x)$

For $x \neq 0$, $f(x) = \frac{1 - \cos 2x}{x^2}$. We need to evaluate the limit of this expression as $x \to 0$.

We know the trigonometric identity $1 - \cos 2x = 2 \sin^2 x$.

So, $\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{1 - \cos 2x}{x^2}$

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{2 \sin^2 x}{x^2}$

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} 2 \left(\frac{\sin x}{x}\right)^2$

We use the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

$\lim\limits_{x \to 0} f(x) = 2 \times (1)^2$

$\lim\limits_{x \to 0} f(x) = 2 \times 1$

$\lim\limits_{x \to 0} f(x) = 2$

So, $\lim\limits_{x \to 0} f(x)$ exists and its value is $2$.

Step 3: Compare $f(0)$ and $\lim\limits_{x \to 0} f(x)$

From Step 1, we have $f(0) = 5$.

From Step 2, we have $\lim\limits_{x \to 0} f(x) = 2$.

Since $\lim\limits_{x \to 0} f(x) \neq f(0)$, the third condition for continuity is not satisfied.

Conclusion:

Since the limit of the function as $x$ approaches $0$ is not equal to the value of the function at $x=0$, the function $f(x)$ is discontinuous at $x=0$.

To examine the continuity of the function $f(x)$ at $x=2$, we need to check the following conditions:

1. $f(2)$ is defined.

2. $\lim\limits_{x \to 2} f(x)$ exists.

3. $\lim\limits_{x \to 2} f(x) = f(2)$.

Solution:

The given function is $f(x) = \begin{cases}\frac{2x^2 − 3x − 2}{x − 2}, & if\; x \neq 2\\5, & if\; x = 2\end{cases}$.

We need to examine its continuity at $x=2$.

Step 1: Evaluate $f(2)$

According to the definition of the function, when $x=2$, $f(x)=5$.

$f(2) = 5$

$f(2)$ is defined and its value is $5$.

Step 2: Evaluate $\lim\limits_{x \to 2} f(x)$

For $x \neq 2$, $f(x) = \frac{2x^2 - 3x - 2}{x - 2}$. We need to evaluate the limit of this expression as $x \to 2$.

First, let's factor the numerator $2x^2 - 3x - 2$.

$2x^2 - 3x - 2 = 2x^2 - 4x + x - 2$

$= 2x(x - 2) + 1(x - 2)$

$= (x - 2)(2x + 1)$

So, for $x \neq 2$, $f(x) = \frac{(x - 2)(2x + 1)}{x - 2}$.

Since $x \to 2$, we are considering values of $x$ close to $2$ but not equal to $2$. Thus, $x-2 \neq 0$, and we can cancel the $(x-2)$ term from the numerator and the denominator.

$f(x) = 2x + 1$ for $x \neq 2$.

Now, we evaluate the limit:

$\lim\limits_{x \to 2} f(x) = \lim\limits_{x \to 2} (2x + 1)$

Since $2x+1$ is a polynomial, we can substitute the value of $x$ directly:

$\lim\limits_{x \to 2} f(x) = 2(2) + 1$

$\lim\limits_{x \to 2} f(x) = 4 + 1$

$\lim\limits_{x \to 2} f(x) = 5$

So, $\lim\limits_{x \to 2} f(x)$ exists and its value is $5$.

Step 3: Compare $f(2)$ and $\lim\limits_{x \to 2} f(x)$

From Step 1, we have $f(2) = 5$.

From Step 2, we have $\lim\limits_{x \to 2} f(x) = 5$.

Since $\lim\limits_{x \to 2} f(x) = f(2)$, the third condition for continuity is satisfied.

Conclusion:

Since all three conditions for continuity are satisfied at $x=2$, the function $f(x)$ is continuous at $x=2$.

To examine the continuity of the function $f(x)$ at $x=4$, we need to check the following conditions:

1. $f(4)$ is defined.

2. $\lim\limits_{x \to 4} f(x)$ exists (i.e., the left-hand limit and the right-hand limit are equal).

3. $\lim\limits_{x \to 4} f(x) = f(4)$.

Solution:

The given function is $f(x) = \begin{cases}\frac{|x−4|}{2(x−4)},& if\; x \neq 4\\0,& if\; x = 4\end{cases}$.

We need to examine its continuity at $x=4$.

Step 1: Evaluate $f(4)$

According to the definition of the function, when $x=4$, $f(x)=0$.

$f(4) = 0$

$f(4)$ is defined and its value is $0$.

Step 2: Evaluate the left-hand limit (LHL) at $x=4$

For the limit as $x$ approaches $4$ from the left ($x < 4$), we use the first part of the function $f(x) = \frac{|x-4|}{2(x-4)}$.

When $x < 4$, $x-4$ is negative. Therefore, $|x-4| = -(x-4)$.

$f(x) = \frac{-(x-4)}{2(x-4)} = -\frac{1}{2}$ for $x < 4$ and $x \neq 4$.

$\lim\limits_{x \to 4^-} f(x) = \lim\limits_{x \to 4^-} \left(-\frac{1}{2}\right)$

$\lim\limits_{x \to 4^-} f(x) = -\frac{1}{2}$

Step 3: Evaluate the right-hand limit (RHL) at $x=4$

For the limit as $x$ approaches $4$ from the right ($x > 4$), we use the first part of the function $f(x) = \frac{|x-4|}{2(x-4)}$.

When $x > 4$, $x-4$ is positive. Therefore, $|x-4| = x-4$.

$f(x) = \frac{x-4}{2(x-4)} = \frac{1}{2}$ for $x > 4$ and $x \neq 4$.

$\lim\limits_{x \to 4^+} f(x) = \lim\limits_{x \to 4^+} \left(\frac{1}{2}\right)$

$\lim\limits_{x \to 4^+} f(x) = \frac{1}{2}$

Step 4: Check if the limit $\lim\limits_{x \to 4} f(x)$ exists

For the limit $\lim\limits_{x \to 4} f(x)$ to exist, the LHL must be equal to the RHL.

LHL = $\lim\limits_{x \to 4^-} f(x) = -\frac{1}{2}$

RHL = $\lim\limits_{x \to 4^+} f(x) = \frac{1}{2}$

Since LHL $\neq$ RHL ($-\frac{1}{2} \neq \frac{1}{2}$), the limit $\lim\limits_{x \to 4} f(x)$ does not exist.

Conclusion:

Since the limit of the function as $x$ approaches $4$ does not exist, the function $f(x)$ is discontinuous at $x=4$.

To examine the continuity of the function $f(x)$ at $x=0$, we need to check the following conditions:

1. $f(0)$ is defined.

2. $\lim\limits_{x \to 0} f(x)$ exists.

3. $\lim\limits_{x \to 0} f(x) = f(0)$.

Solution:

The given function is $f(x) = \begin{cases}|x| \cos \frac{1}{x},& if\; x \neq 0\\0, & if\; x = 0\end{cases}$.

We need to examine its continuity at $x=0$.

Step 1: Evaluate $f(0)$

According to the definition of the function, when $x=0$, $f(x)=0$.

$f(0) = 0$

$f(0)$ is defined and its value is $0$.

Step 2: Evaluate $\lim\limits_{x \to 0} f(x)$

For $x \neq 0$, $f(x) = |x| \cos \frac{1}{x}$. We need to evaluate the limit of this expression as $x \to 0$.

We know that for any real number $\theta$, $-1 \leq \cos \theta \leq 1$.

So, for $x \neq 0$, we have $-1 \leq \cos \frac{1}{x} \leq 1$.

Multiplying the inequality by $|x|$ (which is non-negative), we get:

$-|x| \leq |x| \cos \frac{1}{x} \leq |x|$

Let $g(x) = -|x|$ and $h(x) = |x|$. We know that:

$\lim\limits_{x \to 0} g(x) = \lim\limits_{x \to 0} (-|x|) = 0$

$\lim\limits_{x \to 0} h(x) = \lim\limits_{x \to 0} |x| = 0$

Since $-|x| \leq |x| \cos \frac{1}{x} \leq |x|$ for $x \neq 0$, and $\lim\limits_{x \to 0} (-|x|) = 0$ and $\lim\limits_{x \to 0} |x| = 0$, by the Squeeze Theorem, the limit of the function $f(x)$ as $x$ approaches $0$ must also be $0$.

$\lim\limits_{x \to 0} |x| \cos \frac{1}{x} = 0$

So, $\lim\limits_{x \to 0} f(x)$ exists and its value is $0$.

Step 3: Compare $f(0)$ and $\lim\limits_{x \to 0} f(x)$

From Step 1, we have $f(0) = 0$.

From Step 2, we have $\lim\limits_{x \to 0} f(x) = 0$.

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the third condition for continuity is satisfied.

Conclusion:

Since all three conditions for continuity are satisfied at $x=0$, the function $f(x)$ is continuous at $x=0$.

To examine the continuity of the function $f(x)$ at $x=a$, we need to check the following conditions:

1. $f(a)$ is defined.

2. $\lim\limits_{x \to a} f(x)$ exists.

3. $\lim\limits_{x \to a} f(x) = f(a)$.

Solution:

The given function is $f(x) = \begin{cases}|x−a| \sin \frac{1}{x−a},& if\; x \neq a\\0,& if\; x = a\end{cases}$.

We need to examine its continuity at $x=a$.

Step 1: Evaluate $f(a)$

According to the definition of the function, when $x=a$, $f(x)=0$.

$f(a) = 0$

$f(a)$ is defined and its value is $0$.

Step 2: Evaluate $\lim\limits_{x \to a} f(x)$

For $x \neq a$, $f(x) = |x-a| \sin \frac{1}{x-a}$. We need to evaluate the limit of this expression as $x \to a$.

We know that the range of the sine function is $[-1, 1]$. Thus, for any real number $\theta$, $-1 \leq \sin \theta \leq 1$.

So, for $x \neq a$, we have $-1 \leq \sin \frac{1}{x-a} \leq 1$.

Multiplying the inequality by $|x-a|$ (which is non-negative since $|x-a| \geq 0$), we get:

$-|x-a| \leq |x-a| \sin \frac{1}{x-a} \leq |x-a|$ for $x \neq a$.

As $x \to a$, $|x-a| \to |a-a| = |0| = 0$.

So, $\lim\limits_{x \to a} (-|x-a|) = 0$ and $\lim\limits_{x \to a} |x-a| = 0$.

Since $-|x-a| \leq f(x) \leq |x-a|$ for $x \neq a$, and both bounding functions tend to $0$ as $x \to a$, by the Squeeze Theorem, the limit of $f(x)$ as $x \to a$ must also be $0$.

$\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} |x-a| \sin \frac{1}{x-a} = 0$

So, $\lim\limits_{x \to a} f(x)$ exists and its value is $0$.

Step 3: Compare $f(a)$ and $\lim\limits_{x \to a} f(x)$

From Step 1, we have $f(a) = 0$.

From Step 2, we have $\lim\limits_{x \to a} f(x) = 0$.

Since $\lim\limits_{x \to a} f(x) = f(a)$, the third condition for continuity is satisfied.

Conclusion:

Since all three conditions for continuity are satisfied at $x=a$, the function $f(x)$ is continuous at $x=a$.

To examine the continuity of the function $f(x)$ at $x=0$, we need to check the following conditions:

1. $f(0)$ is defined.

2. $\lim\limits_{x \to 0} f(x)$ exists (i.e., the left-hand limit and the right-hand limit are equal).

3. $\lim\limits_{x \to 0} f(x) = f(0)$.

Solution:

The given function is $f(x) = \begin{cases}\frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}},& if\; x \neq 0\\0,& if\; x = 0\end{cases}$.

We need to examine its continuity at $x=0$.

Step 1: Evaluate $f(0)$

According to the definition of the function, when $x=0$, $f(x)=0$.

$f(0) = 0$

$f(0)$ is defined and its value is $0$.

Step 2: Evaluate the left-hand limit (LHL) at $x=0$

For the limit as $x$ approaches $0$ from the left ($x < 0$), we use the first part of the function $f(x) = \frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}$.

As $x \to 0^-$, the value of $\frac{1}{x}$ approaches $-\infty$.

$\lim\limits_{x \to 0^-} \frac{1}{x} = -\infty$

Therefore, $\lim\limits_{x \to 0^-} e^{\frac{1}{x}} = e^{-\infty} = 0$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}} = \frac{\lim\limits_{x \to 0^-} e^{\frac{1}{x}}}{1+\lim\limits_{x \to 0^-} e^{\frac{1}{x}}}$

$\lim\limits_{x \to 0^-} f(x) = \frac{0}{1+0} = 0$

So, LHL = $0$.

Step 3: Evaluate the right-hand limit (RHL) at $x=0$

For the limit as $x$ approaches $0$ from the right ($x > 0$), we use the first part of the function $f(x) = \frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}$.

As $x \to 0^+$, the value of $\frac{1}{x}$ approaches $+\infty$.

$\lim\limits_{x \to 0^+} \frac{1}{x} = +\infty$

Therefore, $\lim\limits_{x \to 0^+} e^{\frac{1}{x}} = e^{+\infty} = +\infty$.

The limit is of the form $\frac{\infty}{1+\infty}$, which is indeterminate. We can divide the numerator and the denominator by $e^{\frac{1}{x}}$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} \frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}$

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} \frac{\frac{e^{\frac{1}{x}}}{e^{\frac{1}{x}}}}{\frac{1}{e^{\frac{1}{x}}}+\frac{e^{\frac{1}{x}}}{e^{\frac{1}{x}}}}$

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} \frac{1}{e^{-\frac{1}{x}} + 1}$

As $x \to 0^+$, $\frac{1}{x} \to +\infty$, so $-\frac{1}{x} \to -\infty$.

Therefore, $\lim\limits_{x \to 0^+} e^{-\frac{1}{x}} = e^{-\infty} = 0$.

$\lim\limits_{x \to 0^+} f(x) = \frac{1}{\lim\limits_{x \to 0^+} e^{-\frac{1}{x}} + 1} = \frac{1}{0+1} = 1$

So, RHL = $1$.

Step 4: Check if the limit $\lim\limits_{x \to 0} f(x)$ exists

For the limit $\lim\limits_{x \to 0} f(x)$ to exist, the LHL must be equal to the RHL.

LHL = $0$

RHL = $1$

Since LHL $\neq$ RHL ($0 \neq 1$), the limit $\lim\limits_{x \to 0} f(x)$ does not exist.

Conclusion:

Since the limit of the function as $x$ approaches $0$ does not exist, the function $f(x)$ is discontinuous at $x=0$.

To examine the continuity of the function $f(x)$ at $x=1$, we need to check the following conditions:

1. $f(1)$ is defined.

2. $\lim\limits_{x \to 1} f(x)$ exists (i.e., the left-hand limit and the right-hand limit are equal).

3. $\lim\limits_{x \to 1} f(x) = f(1)$.

Solution:

The given function is $f(x) = \begin{cases}\frac{x^2}{2},& if\; 0 \leq x \leq 1 \\ 2x^2 − 3x + \frac{3}{2},& if\; 1 < x \leq 2 \end{cases}$.

We need to examine its continuity at $x=1$.

Step 1: Evaluate $f(1)$

Since $x=1$ falls in the condition $0 \leq x \leq 1$, we use the first part of the function:

$f(x) = \frac{x^2}{2}$ for $0 \leq x \leq 1$.

$f(1) = \frac{(1)^2}{2}$

$f(1) = \frac{1}{2}$

$f(1)$ is defined and its value is $\frac{1}{2}$.

Step 2: Evaluate the left-hand limit (LHL) at $x=1$

For the limit as $x$ approaches $1$ from the left ($x < 1$), we use the first part of the function:

$f(x) = \frac{x^2}{2}$ for $0 \leq x \leq 1$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} \frac{x^2}{2}$

$\lim\limits_{x \to 1^-} f(x) = \frac{(1)^2}{2}$

$\lim\limits_{x \to 1^-} f(x) = \frac{1}{2}$

Step 3: Evaluate the right-hand limit (RHL) at $x=1$

For the limit as $x$ approaches $1$ from the right ($x > 1$), we use the second part of the function:

$f(x) = 2x^2 - 3x + \frac{3}{2}$ for $1 < x \leq 2$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (2x^2 - 3x + \frac{3}{2})$

Since $2x^2 - 3x + \frac{3}{2}$ is a polynomial, we can substitute the value of $x$ directly:

$\lim\limits_{x \to 1^+} f(x) = 2(1)^2 - 3(1) + \frac{3}{2}$

$\lim\limits_{x \to 1^+} f(x) = 2 - 3 + \frac{3}{2}$

$\lim\limits_{x \to 1^+} f(x) = -1 + \frac{3}{2}$

$\lim\limits_{x \to 1^+} f(x) = \frac{-2 + 3}{2}$

$\lim\limits_{x \to 1^+} f(x) = \frac{1}{2}$

Step 4: Check if the limit $\lim\limits_{x \to 1} f(x)$ exists

For the limit $\lim\limits_{x \to 1} f(x)$ to exist, the LHL must be equal to the RHL.

LHL = $\lim\limits_{x \to 1^-} f(x) = \frac{1}{2}$

RHL = $\lim\limits_{x \to 1^+} f(x) = \frac{1}{2}$

Since LHL = RHL, the limit $\lim\limits_{x \to 1} f(x)$ exists and $\lim\limits_{x \to 1} f(x) = \frac{1}{2}$.

Step 5: Compare $f(1)$ and $\lim\limits_{x \to 1} f(x)$

From Step 1, we have $f(1) = \frac{1}{2}$.

From Step 4, we have $\lim\limits_{x \to 1} f(x) = \frac{1}{2}$.

Since $\lim\limits_{x \to 1} f(x) = f(1)$, the third condition for continuity is satisfied.

Conclusion:

Since all three conditions for continuity are satisfied at $x=1$, the function $f(x)$ is continuous at $x=1$.

To examine the continuity of the function $f(x) = |x| + |x-1|$ at $x=1$, we need to check the following conditions:

1. $f(1)$ is defined.

2. $\lim\limits_{x \to 1} f(x)$ exists (i.e., the left-hand limit and the right-hand limit are equal).

3. $\lim\limits_{x \to 1} f(x) = f(1)$.

Solution:

The given function is $f(x) = |x| + |x-1|$.

We need to examine its continuity at $x=1$.

First, let's rewrite the function $f(x)$ as a piecewise function around $x=1$. The critical points for the absolute values are $x=0$ and $x=1$. We are interested in the behavior around $x=1$, so we consider the intervals $x < 1$ and $x \geq 1$.

For $x < 1$: $|x-1| = -(x-1)$. $|x|$ depends on whether $x$ is less than 0 or greater than or equal to 0. However, when we consider the limit as $x \to 1^-$, we are looking at values of $x$ very close to 1 but less than 1. In this neighbourhood (e.g., $0 < x < 1$), $|x| = x$. So, for $x$ slightly less than 1 (and $x>0$), $f(x) = x + (-(x-1)) = x - x + 1 = 1$.

For $x \geq 1$: $|x| = x$ and $|x-1| = x-1$. So, for $x \geq 1$, $f(x) = x + (x-1) = 2x - 1$.

Thus, the piecewise definition of the function around $x=1$ can be written as:

$f(x) = \begin{cases} 1 & , & x < 1 \\ 2x - 1 & , & x \geq 1 \end{cases}$

Step 1: Evaluate $f(1)$

Since $x=1$ falls in the condition $x \geq 1$, we use the second part of the function:

$f(x) = 2x-1$ for $x \geq 1$.

$f(1) = 2(1) - 1$

$f(1) = 2 - 1$

$f(1) = 1$

$f(1)$ is defined and its value is $1$.

Step 2: Evaluate the left-hand limit (LHL) at $x=1$

For the limit as $x$ approaches $1$ from the left ($x < 1$), we use the first part of the function:

$f(x) = 1$ for $x < 1$ (in the vicinity of $x=1$).

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} 1$

$\lim\limits_{x \to 1^-} f(x) = 1$

Step 3: Evaluate the right-hand limit (RHL) at $x=1$

For the limit as $x$ approaches $1$ from the right ($x > 1$), we use the second part of the function:

$f(x) = 2x-1$ for $x \geq 1$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (2x-1)$

Since $2x-1$ is a polynomial, we can substitute the value of $x$ directly:

$\lim\limits_{x \to 1^+} f(x) = 2(1) - 1$

$\lim\limits_{x \to 1^+} f(x) = 2 - 1$

$\lim\limits_{x \to 1^+} f(x) = 1$

Step 4: Check if the limit $\lim\limits_{x \to 1} f(x)$ exists

For the limit $\lim\limits_{x \to 1} f(x)$ to exist, the LHL must be equal to the RHL.

LHL = $\lim\limits_{x \to 1^-} f(x) = 1$

RHL = $\lim\limits_{x \to 1^+} f(x) = 1$

Since LHL = RHL, the limit $\lim\limits_{x \to 1} f(x)$ exists and $\lim\limits_{x \to 1} f(x) = 1$.

Step 5: Compare $f(1)$ and $\lim\limits_{x \to 1} f(x)$

From Step 1, we have $f(1) = 1$.

From Step 4, we have $\lim\limits_{x \to 1} f(x) = 1$.

Since $\lim\limits_{x \to 1} f(x) = f(1)$, the third condition for continuity is satisfied.

Conclusion:

Since all three conditions for continuity are satisfied at $x=1$, the function $f(x) = |x| + |x-1|$ is continuous at $x=1$.

For the function $f(x)$ to be continuous at $x=5$, the following condition must be satisfied:

$\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^+} f(x) = f(5)$

Solution:

The given function is $f(x) = \begin{cases}3x−8,& if\; x \leq 5 \\ 2k,& if\; x > 5\end{cases}$.

We need to find the value of $k$ for which $f(x)$ is continuous at $x=5$.

Step 1: Evaluate $f(5)$

Since $x=5$ falls in the condition $x \leq 5$, we use the first part of the function:

$f(x) = 3x-8$ for $x \leq 5$.

$f(5) = 3(5) - 8$

$f(5) = 15 - 8$

$f(5) = 7$

Step 2: Evaluate the left-hand limit (LHL) at $x=5$

For the limit as $x$ approaches $5$ from the left ($x < 5$), we use the first part of the function:

$f(x) = 3x-8$ for $x \leq 5$.

$\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^-} (3x-8)$

Since $3x-8$ is a polynomial, we can substitute the value of $x$ directly:

$\lim\limits_{x \to 5^-} f(x) = 3(5) - 8$

$\lim\limits_{x \to 5^-} f(x) = 15 - 8$

$\lim\limits_{x \to 5^-} f(x) = 7$

Step 3: Evaluate the right-hand limit (RHL) at $x=5$

For the limit as $x$ approaches $5$ from the right ($x > 5$), we use the second part of the function:

$f(x) = 2k$ for $x > 5$.

$\lim\limits_{x \to 5^+} f(x) = \lim\limits_{x \to 5^+} (2k)$

Since $2k$ is a constant with respect to $x$, the limit is the constant value:

$\lim\limits_{x \to 5^+} f(x) = 2k$

Step 4: Set the limits and function value equal for continuity

For $f(x)$ to be continuous at $x=5$, we must have LHL = RHL = $f(5)$.

From Step 1, $f(5) = 7$.

From Step 2, LHL = $7$.

From Step 3, RHL = $2k$.

Equating LHL and RHL:

$7 = 2k$

Dividing both sides by 2:

$k = \frac{7}{2}$

Conclusion:

For the function $f(x)$ to be continuous at $x=5$, the value of $k$ must be $\frac{7}{2}$.

For the function $f(x)$ to be continuous at $x=2$, the following condition must be satisfied:

$\lim\limits_{x \to 2} f(x) = f(2)$

Solution:

The given function is $f(x) = \begin{cases}\frac{2^{x+2} − 16}{4^x − 16},& if\; x \neq 2 \\ k,& if\; x = 2\end{cases}$.

We need to find the value of $k$ for which $f(x)$ is continuous at $x=2$.

Step 1: Evaluate $f(2)$

According to the definition of the function, when $x=2$, $f(x)=k$.

$f(2) = k$

Step 2: Evaluate $\lim\limits_{x \to 2} f(x)$

For $x \neq 2$, $f(x) = \frac{2^{x+2} - 16}{4^x - 16}$. We need to evaluate the limit of this expression as $x \to 2$.

Let's simplify the expression for $f(x)$ when $x \neq 2$:

$f(x) = \frac{2^x \cdot 2^2 - 16}{(2^2)^x - 16}$

$f(x) = \frac{4 \cdot 2^x - 16}{(2^x)^2 - 4^2}$

Factor out 4 from the numerator and use the difference of squares formula in the denominator:

$f(x) = \frac{4(2^x - 4)}{(2^x - 4)(2^x + 4)}$

Since $x \to 2$, $x \neq 2$, which means $2^x \neq 2^2 = 4$. Therefore, $2^x - 4 \neq 0$. We can cancel the term $(2^x - 4)$ from the numerator and the denominator.

$f(x) = \frac{4}{2^x + 4}$ for $x \neq 2$.

Now, we evaluate the limit:

$\lim\limits_{x \to 2} f(x) = \lim\limits_{x \to 2} \frac{4}{2^x + 4}$

Since the denominator $2^x + 4$ is a continuous function and is non-zero at $x=2$, we can substitute the value of $x$ directly:

$\lim\limits_{x \to 2} f(x) = \frac{4}{2^2 + 4}$

$\lim\limits_{x \to 2} f(x) = \frac{4}{4 + 4}$

$\lim\limits_{x \to 2} f(x) = \frac{4}{8}$

$\lim\limits_{x \to 2} f(x) = \frac{1}{2}$

Step 3: Set the limit equal to the function value for continuity

For $f(x)$ to be continuous at $x=2$, we must have $\lim\limits_{x \to 2} f(x) = f(2)$.

From Step 1, $f(2) = k$.

From Step 2, $\lim\limits_{x \to 2} f(x) = \frac{1}{2}$.

Equating these values:

$k = \frac{1}{2}$

Conclusion:

For the function $f(x)$ to be continuous at $x=2$, the value of $k$ must be $\frac{1}{2}$.

Given:

The function $f(x) = \begin{cases}\frac{\sqrt{1+kx} − \sqrt{1 − kx}}{x},& if\; −1 \leq x < 0 \\ \frac{2x+1}{x−1},& if\;\; 0 \leq x \leq 1 \end{cases}$.

The point at which continuity is to be examined is $x=0$.

To Find:

The value of $k$ for which the function $f$ is continuous at $x=0$.

Solution:

For the function $f(x)$ to be continuous at $x=0$, the following condition must be satisfied:

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$

Step 1: Evaluate $f(0)$

When $x=0$, the function is defined by the second case: $f(x) = \frac{2x+1}{x-1}$.

$f(0) = \frac{2(0)+1}{0-1}$

$f(0) = \frac{0+1}{-1}$

$f(0) = \frac{1}{-1}$

$f(0) = -1$

Step 2: Evaluate the left-hand limit (LHL) at $x=0$

For the limit as $x$ approaches $0$ from the left ($x < 0$), we use the first part of the function: $f(x) = \frac{\sqrt{1+kx} − \sqrt{1 − kx}}{x}$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \frac{\sqrt{1+kx} − \sqrt{1 − kx}}{x}$

This limit is of the indeterminate form $\frac{0}{0}$. We can rationalize the numerator:

$\lim\limits_{x \to 0^-} \frac{\sqrt{1+kx} − \sqrt{1 − kx}}{x} \times \frac{\sqrt{1+kx} + \sqrt{1 − kx}}{\sqrt{1+kx} + \sqrt{1 − kx}}$

$= \lim\limits_{x \to 0^-} \frac{(1+kx) - (1-kx)}{x(\sqrt{1+kx} + \sqrt{1 − kx})}$

$= \lim\limits_{x \to 0^-} \frac{1+kx - 1+kx}{x(\sqrt{1+kx} + \sqrt{1 − kx})}$

$= \lim\limits_{x \to 0^-} \frac{2kx}{x(\sqrt{1+kx} + \sqrt{1 − kx})}$

For $x \neq 0$, we can cancel $x$ from the numerator and denominator:

$= \lim\limits_{x \to 0^-} \frac{2k}{\sqrt{1+kx} + \sqrt{1 − kx}}$

Now, we can substitute $x=0$ into the simplified expression:

$= \frac{2k}{\sqrt{1+k(0)} + \sqrt{1 − k(0)}}$

$= \frac{2k}{\sqrt{1} + \sqrt{1}}$

$= \frac{2k}{1 + 1}$

$= \frac{2k}{2}$

$= k$

So, LHL = $k$.

Step 3: Evaluate the right-hand limit (RHL) at $x=0$

For the limit as $x$ approaches $0$ from the right ($x > 0$), we use the second part of the function: $f(x) = \frac{2x+1}{x-1}$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} \frac{2x+1}{x-1}$

Since $\frac{2x+1}{x-1}$ is a rational function and the denominator is non-zero at $x=0$, we can substitute $x=0$ directly:

$= \frac{2(0)+1}{0-1}$

$= \frac{1}{-1}$

$= -1$

So, RHL = $-1$.

Step 4: Set the limits and function value equal for continuity

For $f(x)$ to be continuous at $x=0$, we must have LHL = RHL = $f(0)$.

From Step 1, $f(0) = -1$.

From Step 2, LHL = $k$.

From Step 3, RHL = $-1$.

Equating LHL and RHL for the limit to exist:

$k = -1$

Equating the limit to the function value:

$\lim\limits_{x \to 0} f(x) = f(0)$

Since LHL = RHL = $-1$ when $k=-1$, the limit $\lim\limits_{x \to 0} f(x)$ exists and is equal to $-1$.

We have $\lim\limits_{x \to 0} f(x) = -1$ and $f(0) = -1$.

Thus, $\lim\limits_{x \to 0} f(x) = f(0)$ is satisfied if $k = -1$.

Conclusion:

For the function $f(x)$ to be continuous at $x=0$, the value of $k$ must be $-1$.

Given:

The function $f(x) = \begin{cases}\frac{1− \cos kx}{x \sin x},& if\; x \neq 0 \\ \frac{1}{2},& if\; x = 0\end{cases}$.

The point at which continuity is to be examined is $x=0$.

To Find:

The value of $k$ for which the function $f$ is continuous at $x=0$.

Solution:

For the function $f(x)$ to be continuous at $x=0$, the following condition must be satisfied:

$\lim\limits_{x \to 0} f(x) = f(0)$

Step 1: Evaluate $f(0)$

According to the definition of the function, when $x=0$, $f(x)=\frac{1}{2}$.

$f(0) = \frac{1}{2}$

Step 2: Evaluate $\lim\limits_{x \to 0} f(x)$

For $x \neq 0$, $f(x) = \frac{1 - \cos kx}{x \sin x}$. We need to evaluate the limit of this expression as $x \to 0$.

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{1 - \cos kx}{x \sin x}$

We use the trigonometric identity $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$.

So, $1 - \cos kx = 2 \sin^2 \left(\frac{kx}{2}\right)$.

Also, we can rewrite the denominator $x \sin x$ using the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{2 \sin^2 \left(\frac{kx}{2}\right)}{x \sin x}$

We can rewrite the expression to make use of the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{2 \sin^2 \left(\frac{kx}{2}\right)}{x \left(\frac{\sin x}{x}\right) x}$

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{2 \sin^2 \left(\frac{kx}{2}\right)}{x^2 \left(\frac{\sin x}{x}\right)}$

Now, focus on the numerator. We need a $(\frac{kx}{2})^2$ term in the denominator for $\sin^2 (\frac{kx}{2})$.

$\sin^2 \left(\frac{kx}{2}\right) = \left(\sin \left(\frac{kx}{2}\right)\right)^2$. To get the standard limit form, we need $(\frac{kx}{2})^2$ in the denominator of the sine term.

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{2 \cdot \left(\frac{\sin (\frac{kx}{2})}{\frac{kx}{2}}\right)^2 \cdot \left(\frac{kx}{2}\right)^2}{x^2 \left(\frac{\sin x}{x}\right)}$

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{2 \cdot \left(\frac{\sin (\frac{kx}{2})}{\frac{kx}{2}}\right)^2 \cdot \frac{k^2 x^2}{4}}{x^2 \left(\frac{\sin x}{x}\right)}$

Cancel the $x^2$ term (since $x \neq 0$ as $x \to 0$):

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{2 \cdot \left(\frac{\sin (\frac{kx}{2})}{\frac{kx}{2}}\right)^2 \cdot \frac{k^2}{4}}{\left(\frac{\sin x}{x}\right)}$

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{\frac{k^2}{2} \left(\frac{\sin (\frac{kx}{2})}{\frac{kx}{2}}\right)^2}{\left(\frac{\sin x}{x}\right)}$

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ (with $\theta = \frac{kx}{2}$ and $\theta = x$):

$\lim\limits_{x \to 0} f(x) = \frac{\frac{k^2}{2} \cdot (1)^2}{1}$

$\lim\limits_{x \to 0} f(x) = \frac{k^2}{2}$

Step 3: Set the limit equal to the function value for continuity

For $f(x)$ to be continuous at $x=0$, we must have $\lim\limits_{x \to 0} f(x) = f(0)$.

From Step 1, $f(0) = \frac{1}{2}$.

From Step 2, $\lim\limits_{x \to 0} f(x) = \frac{k^2}{2}$.

Equating these values:

$\frac{k^2}{2} = \frac{1}{2}$

Multiply both sides by 2:

$k^2 = 1$

Taking the square root of both sides:

$k = \pm \sqrt{1}$

$k = \pm 1$

So, $k=1$ or $k=-1$.

Conclusion:

For the function $f(x)$ to be continuous at $x=0$, the value of $k$ must be $1$ or $-1$.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}\frac{x}{|x| + 2x^2},& x≠0 \\ k,& x=0\end{cases}$

To Prove:

The function $f(x)$ is discontinuous at $x=0$ for any value of $k$.

Proof:

For a function $f(x)$ to be continuous at a point $x=a$, the following condition must be satisfied:

$\lim\limits_{x \to a} f(x) = f(a)$

In this specific case, the point of interest is $a=0$. Thus, for $f(x)$ to be continuous at $x=0$, we must have:

$\lim\limits_{x \to 0} f(x) = f(0) = k$

We need to evaluate the limit of $f(x)$ as $x$ approaches $0$. Since the expression for $f(x)$ when $x \neq 0$ involves the absolute value function $|x|$, we need to consider the left-hand limit (LHL) and the right-hand limit (RHL) at $x=0$.

Right-Hand Limit (RHL) at $x=0$:

As $x$ approaches $0$ from the right side ($x \to 0^+$), $x$ is positive ($x > 0$). Therefore, $|x| = x$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} \frac{x}{|x| + 2x^2} = \lim\limits_{x \to 0^+} \frac{x}{x + 2x^2}$

For $x \neq 0$, we can factor out $x$ from the denominator:

$\lim\limits_{x \to 0^+} \frac{x}{x(1 + 2x)} = \lim\limits_{x \to 0^+} \frac{\cancel{x}}{\cancel{x}(1 + 2x)} = \lim\limits_{x \to 0^+} \frac{1}{1 + 2x}$

Now, substituting $x=0$ into the simplified expression:

$\lim\limits_{x \to 0^+} f(x) = \frac{1}{1 + 2(0)} = \frac{1}{1} = 1$

Thus, the right-hand limit at $x=0$ is $1$.

Left-Hand Limit (LHL) at $x=0$:

As $x$ approaches $0$ from the left side ($x \to 0^-$), $x$ is negative ($x < 0$). Therefore, $|x| = -x$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \frac{x}{|x| + 2x^2} = \lim\limits_{x \to 0^-} \frac{x}{-x + 2x^2}$

For $x \neq 0$, we can factor out $x$ from the denominator:

$\lim\limits_{x \to 0^-} \frac{x}{x(-1 + 2x)} = \lim\limits_{x \to 0^-} \frac{\cancel{x}}{\cancel{x}(-1 + 2x)} = \lim\limits_{x \to 0^-} \frac{1}{-1 + 2x}$

Now, substituting $x=0$ into the simplified expression:

$\lim\limits_{x \to 0^-} f(x) = \frac{1}{-1 + 2(0)} = \frac{1}{-1} = -1$

Thus, the left-hand limit at $x=0$ is $-1$.

For the limit $\lim\limits_{x \to 0} f(x)$ to exist, the left-hand limit and the right-hand limit must be equal. In this case, we have:

LHL at $x=0 = -1$

RHL at $x=0 = 1$

Since $-1 \neq 1$, the left-hand limit and the right-hand limit are not equal. Therefore, the limit $\lim\limits_{x \to 0} f(x)$ does not exist.

The condition for continuity at $x=0$ requires that the limit $\lim\limits_{x \to 0} f(x)$ must exist and be equal to $f(0) = k$. Since the limit $\lim\limits_{x \to 0} f(x)$ does not exist, this condition cannot be satisfied for any finite value of $k$.

Hence, the function $f(x)$ is discontinuous at $x=0$, regardless of the value assigned to $k$.

This completes the proof.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}\frac{x−4}{|x−4|}+a,& if\; x<4 \\ a+b,& if\; x=4 \\ \frac{x−4}{|x−4|}+b,& if\; x>4\end{cases}$

To Find:

The values of $a$ and $b$ for which the function $f(x)$ is continuous at $x=4$.

Solution:

For the function $f(x)$ to be continuous at $x=4$, the following condition must be satisfied:

$\lim\limits_{x \to 4^-} f(x) = \lim\limits_{x \to 4^+} f(x) = f(4)$

First, let's find the value of the function at $x=4$:

$f(4) = a+b$

Next, let's evaluate the Left-Hand Limit (LHL) at $x=4$. For $x < 4$, $x-4$ is negative, so $|x-4| = -(x-4)$.

$\lim\limits_{x \to 4^-} f(x) = \lim\limits_{x \to 4^-} \left(\frac{x-4}{|x-4|}+a\right)$

$= \lim\limits_{x \to 4^-} \left(\frac{x-4}{-(x-4)}+a\right)$

For $x \neq 4$, we can cancel out the $(x-4)$ term:

$= \lim\limits_{x \to 4^-} \left(\frac{\cancel{x-4}}{-\cancel{(x-4)}}+a\right)$

$= \lim\limits_{x \to 4^-} (-1+a)$

$= -1+a$

Now, let's evaluate the Right-Hand Limit (RHL) at $x=4$. For $x > 4$, $x-4$ is positive, so $|x-4| = x-4$.

$\lim\limits_{x \to 4^+} f(x) = \lim\limits_{x \to 4^+} \left(\frac{x-4}{|x-4|}+b\right)$

$= \lim\limits_{x \to 4^+} \left(\frac{x-4}{x-4}+b\right)$

For $x \neq 4$, we can cancel out the $(x-4)$ term:

$= \lim\limits_{x \to 4^+} \left(\frac{\cancel{x-4}}{\cancel{x-4}}+b\right)$

$= \lim\limits_{x \to 4^+} (1+b)$

$= 1+b$

For continuity at $x=4$, we must have LHL = RHL = $f(4)$.

$-1+a = 1+b = a+b$

We get two equations from this equality:

$-1+a = 1+b \quad$ (Equation 1)

$1+b = a+b \quad$ (Equation 2)

Let's solve Equation 2:

$1+b = a+b$

Subtract $b$ from both sides:

$1 = a$

Now substitute $a=1$ into Equation 1:

$-1 + (1) = 1 + b$

$0 = 1 + b$

Subtract $1$ from both sides:

$b = -1$

Thus, the values of $a$ and $b$ that make the function continuous at $x=4$ are $a=1$ and $b=-1$.

Let's verify the continuity condition with these values:

LHL = $-1+a = -1+1 = 0$

RHL = $1+b = 1+(-1) = 0$

$f(4) = a+b = 1+(-1) = 0$

Since LHL = RHL = $f(4) = 0$, the function is continuous at $x=4$ for $a=1$ and $b=-1$.

Given:

The function $f(x)$ is defined as:

$f(x) = \frac{1}{x+2}$

To Find:

The points of discontinuity of the composite function $y = f(f(x))$.

Solution:

The given function is $f(x) = \frac{1}{x+2}$.

A rational function is discontinuous where its denominator is zero.

For $f(x)$ to be defined, the denominator $x+2$ must not be equal to zero.

$x+2 \neq 0$

$x \neq -2$

Thus, the function $f(x)$ is discontinuous at $x = -2$.

Now, we need to find the composite function $y = f(f(x))$.

$f(f(x)) = f\left(f(x)\right) = f\left(\frac{1}{x+2}\right)$

To find $f\left(\frac{1}{x+2}\right)$, we substitute $\frac{1}{x+2}$ into the expression for $f(x)$:

$f\left(\frac{1}{x+2}\right) = \frac{1}{\left(\frac{1}{x+2}\right) + 2}$

Simplify the denominator:

$= \frac{1}{\frac{1}{x+2} + \frac{2(x+2)}{x+2}} = \frac{1}{\frac{1 + 2x + 4}{x+2}} = \frac{1}{\frac{2x + 5}{x+2}}$

Invert the fraction in the denominator and multiply:

$= 1 \times \frac{x+2}{2x+5} = \frac{x+2}{2x+5}$

So, the composite function is $f(f(x)) = \frac{x+2}{2x+5}$.

The composite function $y = f(f(x))$ can be discontinuous at two types of points:

1. Points where the inner function $f(x)$ is discontinuous.

As found earlier, $f(x)$ is discontinuous at $x = -2$. Thus, $x = -2$ is a point of discontinuity for $f(f(x))$.

2. Points where the inner function $f(x)$ is defined, but the value of $f(x)$ makes the outer function $f$ undefined.

The outer function is $f(y) = \frac{1}{y+2}$. It is undefined when the argument $y$ is equal to $-2$. So, we need to find the values of $x$ such that $f(x) = -2$.

$f(x) = -2$

$\frac{1}{x+2} = -2$

Multiply both sides by $(x+2)$, assuming $x \neq -2$ (which we already identified as a discontinuity):

$1 = -2(x+2)$

$1 = -2x - 4$

$2x = -4 - 1$

$2x = -5$

$x = -\frac{5}{2}$

Thus, $x = -\frac{5}{2}$ is another point of discontinuity for $f(f(x))$.

Also, the composite function $f(f(x)) = \frac{x+2}{2x+5}$ itself is a rational function and is discontinuous where its denominator is zero.

$2x+5 = 0$

$2x = -5$

$x = -\frac{5}{2}$

This confirms that $x = -\frac{5}{2}$ is a point of discontinuity for $f(f(x))$. Note that the case $x=-2$ is already covered by the first condition regarding the inner function.

Combining both types of discontinuities, the points of discontinuity for the composite function $y = f(f(x))$ are $x = -2$ and $x = -\frac{5}{2}$.

Given:

The function $f(t)$ is defined as $f(t) = \frac{1}{t^2 + t −2}$, where $t = \frac{1}{x − 1}$.

To Find:

The points of discontinuity of the function $f(t)$ in terms of $x$.

Solution:

We are given the function $f(t) = \frac{1}{t^2 + t −2}$ and $t = \frac{1}{x − 1}$. We need to find the points of discontinuity of the composite function $f(t(x))$.

A composite function $h(g(x))$ is discontinuous at points where:

1. The inner function $g(x)$ is discontinuous.

2. The inner function $g(x)$ is defined, but the value $g(x)$ makes the outer function $h$ discontinuous.

In this case, the inner function is $t(x) = \frac{1}{x-1}$ and the outer function is $f(t) = \frac{1}{t^2 + t - 2}$.

Step 1: Find points where the inner function $t(x)$ is discontinuous.

The function $t(x) = \frac{1}{x-1}$ is a rational function. It is discontinuous where its denominator is zero.

$x - 1 = 0$

$x = 1$

So, $t(x)$ is discontinuous at $x=1$. This is a point of discontinuity for $f(t(x))$.

Step 2: Find points where the value of the inner function $t(x)$ makes the outer function $f(t)$ discontinuous.

The outer function $f(t) = \frac{1}{t^2 + t - 2}$ is discontinuous where its denominator is zero.

$t^2 + t - 2 = 0$

Factor the quadratic expression:

$(t+2)(t-1) = 0$

So, $f(t)$ is discontinuous when $t = -2$ or $t = 1$.

We need to find the values of $x$ for which $t(x) = -2$ or $t(x) = 1$.

Case A: $t(x) = -2$

$\frac{1}{x-1} = -2$

Assuming $x \neq 1$, we can multiply both sides by $(x-1)$:

$1 = -2(x-1)$

$1 = -2x + 2$

$2x = 2 - 1$

$2x = 1$

$x = \frac{1}{2}$

So, when $x = \frac{1}{2}$, $t = -2$, which makes $f(t)$ discontinuous. Thus, $x = \frac{1}{2}$ is a point of discontinuity for $f(t(x))$.

Case B: $t(x) = 1$

$\frac{1}{x-1} = 1$

Assuming $x \neq 1$, we can multiply both sides by $(x-1)$:

$1 = 1(x-1)$

$1 = x - 1$

$x = 1 + 1$

$x = 2$

So, when $x = 2$, $t = 1$, which makes $f(t)$ discontinuous. Thus, $x = 2$ is a point of discontinuity for $f(t(x))$.

Combining the points of discontinuity from Step 1 and Step 2, the points of discontinuity for $f(t(x))$ are $x = 1$, $x = \frac{1}{2}$, and $x = 2$.

These can be written in increasing order as $x = \frac{1}{2}, 1, 2$.

Given:

The function $f(x) = |\sin x + \cos x|$.

The point at which continuity is to be shown is $x = \pi$.

To Show:

The function $f(x)$ is continuous at $x = \pi$.

Proof:

For a function $f(x)$ to be continuous at a point $x=a$, the following three conditions must be met:

1. $f(a)$ is defined.

2. $\lim\limits_{x \to a} f(x)$ exists.

3. $\lim\limits_{x \to a} f(x) = f(a)$.

Here, $a = \pi$.

Step 1: Evaluate $f(\pi)$.

Substitute $x = \pi$ into the function $f(x)$:

$f(\pi) = |\sin \pi + \cos \pi|$

We know the standard trigonometric values:

$\sin \pi = 0$

$\cos \pi = -1$

Substitute these values:

$f(\pi) = |0 + (-1)| = |-1| = 1$

Since $f(\pi)$ has a finite value ($1$), $f(\pi)$ is defined.

Step 2: Evaluate the limit $\lim\limits_{x \to \pi} f(x)$.

We need to evaluate $\lim\limits_{x \to \pi} |\sin x + \cos x|$.

Consider the function $g(x) = \sin x + \cos x$. The sine function ($\sin x$) and the cosine function ($\cos x$) are known to be continuous for all real numbers.

The sum of two continuous functions is also continuous. Therefore, $g(x) = \sin x + \cos x$ is continuous at $x = \pi$.

Since $g(x)$ is continuous at $x = \pi$, the limit of $g(x)$ as $x \to \pi$ exists and is equal to $g(\pi)$:

$\lim\limits_{x \to \pi} (\sin x + \cos x) = \sin \pi + \cos \pi = 0 + (-1) = -1$

Now consider the absolute value function, $h(y) = |y|$. The absolute value function is continuous for all real numbers $y$.

The given function $f(x)$ is the composite function $h(g(x)) = |g(x)| = |\sin x + \cos x|$.

For a composite function $h(g(x))$, if $\lim\limits_{x \to a} g(x) = L$ and $h$ is continuous at $L$, then $\lim\limits_{x \to a} h(g(x)) = h(L)$.

In our case, $a=\pi$, $g(x) = \sin x + \cos x$, $h(y) = |y|$, and $\lim\limits_{x \to \pi} g(x) = -1$. The function $h(y) = |y|$ is continuous at $y = -1$.

Therefore, we can write:

$\lim\limits_{x \to \pi} |\sin x + \cos x| = |\lim\limits_{x \to \pi} (\sin x + \cos x)|$

$= |-1|$

$= 1$

Thus, the limit $\lim\limits_{x \to \pi} f(x)$ exists and is equal to $1$.

Step 3: Compare $\lim\limits_{x \to \pi} f(x)$ and $f(\pi)$.

From Step 1, $f(\pi) = 1$.

From Step 2, $\lim\limits_{x \to \pi} f(x) = 1$.

Since $\lim\limits_{x \to \pi} f(x) = f(\pi) = 1$, the third condition for continuity is satisfied.

As all three conditions for continuity at $x=\pi$ are met, the function $f(x) = |\sin x + \cos x|$ is continuous at $x = \pi$.

This completes the proof.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}x[x],& if\; 0≤x<2 \\ (x−1)x,& if\; 2≤x<3\end{cases}$

where $[x]$ denotes the greatest integer function.

To Examine:

The differentiability of the function $f(x)$ at $x=2$.

Solution:

For a function $f(x)$ to be differentiable at a point $x=a$, the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x=a$ must exist and be equal.

The formula for the derivative at a point $a$ is $f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$.

The LHD at $x=a$ is given by $\lim\limits_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$.

The RHD at $x=a$ is given by $\lim\limits_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$.

We need to examine differentiability at $x=2$, so we take $a=2$.

First, let's find the value of the function at $x=2$. According to the definition, for $x=2$ (which falls in the range $2 \le x < 3$), $f(x) = (x-1)x$.

$f(2) = (2-1) \times 2 = 1 \times 2 = 2$

Now, let's calculate the Left-Hand Derivative (LHD) at $x=2$:

LHD $= \lim\limits_{h \to 0^-} \frac{f(2+h) - f(2)}{h}$

As $h \to 0^-$, $h$ is a small negative number, so $2+h$ is slightly less than $2$. For $h$ in the interval $(-1, 0)$, we have $1 < 2+h < 2$. In the interval $0 \le x < 2$, the function is defined as $f(x) = x[x]$.

For $1 < x < 2$, $[x] = 1$. Thus, for $1 < 2+h < 2$ (i.e., $-1 < h < 0$), $f(2+h) = (2+h)[2+h] = (2+h) \times 1 = 2+h$.

Substitute this into the LHD expression:

LHD $= \lim\limits_{h \to 0^-} \frac{(2+h) - 2}{h} = \lim\limits_{h \to 0^-} \frac{h}{h}$

For $h \neq 0$, $\frac{h}{h} = 1$.

LHD $= \lim\limits_{h \to 0^-} 1 = 1$

Next, let's calculate the Right-Hand Derivative (RHD) at $x=2$:

RHD $= \lim\limits_{h \to 0^+} \frac{f(2+h) - f(2)}{h}$

As $h \to 0^+$, $h$ is a small positive number, so $2+h$ is slightly greater than $2$. For $h$ in the interval $(0, 1)$, we have $2 < 2+h < 3$. In the interval $2 \le x < 3$, the function is defined as $f(x) = (x-1)x = x^2 - x$.

Thus, for $2 < 2+h < 3$ (i.e., $0 < h < 1$), $f(2+h) = (2+h-1)(2+h) = (1+h)(2+h)$.

$f(2+h) = 2 + 2h + h + h^2 = h^2 + 3h + 2$

Substitute this into the RHD expression:

RHD $= \lim\limits_{h \to 0^+} \frac{(h^2 + 3h + 2) - 2}{h} = \lim\limits_{h \to 0^+} \frac{h^2 + 3h}{h}$

For $h \neq 0$, we can factor out $h$ from the numerator:

RHD $= \lim\limits_{h \to 0^+} \frac{h(h + 3)}{h} = \lim\limits_{h \to 0^+} (h + 3)$

Substitute $h=0$:

RHD $= 0 + 3 = 3$

Conclusion:

We have calculated the LHD and RHD at $x=2$:

LHD at $x=2 = 1$

RHD at $x=2 = 3$

Since the Left-Hand Derivative is not equal to the Right-Hand Derivative at $x=2$ ($1 \neq 3$), the function $f(x)$ is not differentiable at $x=2$.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}x^2 \sin \frac{1}{x} ,& if\; x≠0 \\ 0,& if\; x=0\end{cases}$

To Examine:

The differentiability of the function $f(x)$ at $x=0$.

Solution:

For a function $f(x)$ to be differentiable at a point $x=a$, the limit of the difference quotient $\lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$ must exist and be finite.

In this case, we need to examine differentiability at $x=0$, so we take $a=0$. We need to evaluate the limit:

$f'(0) = \lim\limits_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim\limits_{h \to 0} \frac{f(h) - f(0)}{h}$

From the definition of the function, we have $f(0) = 0$.

For $h \neq 0$, $f(h) = h^2 \sin \frac{1}{h}$.

Substitute these into the limit expression for $h \neq 0$:

$f'(0) = \lim\limits_{h \to 0} \frac{h^2 \sin \frac{1}{h} - 0}{h}$

$f'(0) = \lim\limits_{h \to 0} \frac{h^2 \sin \frac{1}{h}}{h}$

For $h \neq 0$, we can cancel out one $h$ term from the numerator and the denominator:

$f'(0) = \lim\limits_{h \to 0} \frac{\cancel{h}^2 \sin \frac{1}{h}}{\cancel{h}} = \lim\limits_{h \to 0} h \sin \frac{1}{h}$

To evaluate this limit, we can use the Squeeze Theorem. We know that for any real number $y$, the value of $\sin y$ is bounded between $-1$ and $1$.

$-1 \le \sin \frac{1}{h} \le 1$, for $h \neq 0$

Multiply the inequality by $|h|$. Since $|h| \ge 0$, the direction of the inequality remains the same. Note that for $h \neq 0$, $|h| > 0$.

$-|h| \le h \sin \frac{1}{h} \le |h|$

Now, take the limit as $h \to 0$ for each part of the inequality:

$\lim\limits_{h \to 0} (-|h|) \le \lim\limits_{h \to 0} h \sin \frac{1}{h} \le \lim\limits_{h \to 0} |h|$

We know that $\lim\limits_{h \to 0} |h| = 0$ and $\lim\limits_{h \to 0} (-|h|) = 0$.

By the Squeeze Theorem, since the limits of the functions on either side of $h \sin \frac{1}{h}$ are both $0$, the limit of $h \sin \frac{1}{h}$ as $h \to 0$ must also be $0$.

$\lim\limits_{h \to 0} h \sin \frac{1}{h} = 0$

Thus, $f'(0) = 0$.

Since the limit of the difference quotient exists and is a finite value ($0$), the function $f(x)$ is differentiable at $x=0$.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases}1+x,& if\; x≤2 \\ 5−x ,& if\; x>2\end{cases}$

To Examine:

The differentiability of the function $f(x)$ at $x=2$.

Solution:

For a function $f(x)$ to be differentiable at a point $x=a$, the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x=a$ must exist and be equal.

The formula for the derivative at a point $a$ is $f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$.

The LHD at $x=a$ is given by $\lim\limits_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$.

The RHD at $x=a$ is given by $\lim\limits_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$.

We need to examine differentiability at $x=2$, so we take $a=2$.

First, let's find the value of the function at $x=2$. According to the definition, for $x \le 2$, $f(x) = 1+x$.

$f(2) = 1+2 = 3$

Now, let's calculate the Left-Hand Derivative (LHD) at $x=2$:

LHD $= \lim\limits_{h \to 0^-} \frac{f(2+h) - f(2)}{h}$

As $h \to 0^-$, $h$ is a small negative number, so $2+h$ is slightly less than $2$. Thus, we use the definition $f(x) = 1+x$ for $x=2+h$ (since $2+h < 2$).

$f(2+h) = 1 + (2+h) = 3+h$

Substitute this into the LHD expression:

LHD $= \lim\limits_{h \to 0^-} \frac{(3+h) - 3}{h} = \lim\limits_{h \to 0^-} \frac{h}{h}$

For $h \neq 0$, $\frac{h}{h} = 1$.

LHD $= \lim\limits_{h \to 0^-} 1 = 1$

Next, let's calculate the Right-Hand Derivative (RHD) at $x=2$:

RHD $= \lim\limits_{h \to 0^+} \frac{f(2+h) - f(2)}{h}$

As $h \to 0^+$, $h$ is a small positive number, so $2+h$ is slightly greater than $2$. Thus, we use the definition $f(x) = 5-x$ for $x=2+h$ (since $2+h > 2$).

$f(2+h) = 5 - (2+h) = 5 - 2 - h = 3-h$

Substitute this into the RHD expression:

RHD $= \lim\limits_{h \to 0^+} \frac{(3-h) - 3}{h} = \lim\limits_{h \to 0^+} \frac{-h}{h}$

For $h \neq 0$, $\frac{-h}{h} = -1$.

RHD $= \lim\limits_{h \to 0^+} (-1) = -1$

Conclusion:

We have calculated the LHD and RHD at $x=2$:

LHD at $x=2 = 1$

RHD at $x=2 = -1$

Since the Left-Hand Derivative is not equal to the Right-Hand Derivative at $x=2$ ($1 \neq -1$), the function $f(x)$ is not differentiable at $x=2$.

Note: It is important to check for continuity before checking differentiability. For continuity at $x=2$: $\lim\limits_{x \to 2^-} (1+x) = 3$, $\lim\limits_{x \to 2^+} (5-x) = 3$, and $f(2) = 3$. Since LHL = RHL = $f(2)$, the function is continuous at $x=2$. However, continuity does not imply differentiability, as shown by this example.

Given:

The function $f(x) = |x - 5|$.

To Show:

The function $f(x)$ is continuous but not differentiable at $x = 5$.

Solution:

First, let's express the function $f(x) = |x-5|$ as a piecewise function:

$f(x) = \begin{cases} -(x-5),& if\; x-5 < 0 \\ x-5,& if\; x-5 \ge 0 \end{cases}$

$f(x) = \begin{cases} 5-x,& if\; x < 5 \\ x-5,& if\; x \ge 5 \end{cases}$

Part 1: Continuity at x = 5

For a function $f(x)$ to be continuous at $x=a$, the following condition must be satisfied:

$\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^+} f(x) = f(a)$

Here, $a = 5$.

Evaluate $f(5)$:

Using the definition for $x \ge 5$, we have $f(x) = x-5$.

$f(5) = 5 - 5 = 0$

Evaluate the Left-Hand Limit (LHL) at $x=5$:

As $x \to 5^-$, $x < 5$. We use the definition $f(x) = 5-x$.

$\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^-} (5-x)$

Substitute $x=5$:

$= 5 - 5 = 0$

Evaluate the Right-Hand Limit (RHL) at $x=5$:

As $x \to 5^+$, $x > 5$. We use the definition $f(x) = x-5$.

$\lim\limits_{x \to 5^+} f(x) = \lim\limits_{x \to 5^+} (x-5)$

Substitute $x=5$:

$= 5 - 5 = 0$

Since LHL = RHL = $f(5) = 0$, the function $f(x) = |x-5|$ is continuous at $x=5$.

Part 2: Differentiability at x = 5

For a function $f(x)$ to be differentiable at $x=a$, the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x=a$ must exist and be equal.

The LHD at $x=a$ is $\lim\limits_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$.

The RHD at $x=a$ is $\lim\limits_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$.

Here, $a = 5$ and $f(5) = 0$.

Calculate the Left-Hand Derivative (LHD) at $x=5$:

LHD $= \lim\limits_{h \to 0^-} \frac{f(5+h) - f(5)}{h}$

As $h \to 0^-$, $h$ is a small negative number, so $5+h < 5$. We use $f(x) = 5-x$ for $x=5+h$.

$f(5+h) = 5 - (5+h) = 5 - 5 - h = -h$

Substitute this into the LHD expression:

LHD $= \lim\limits_{h \to 0^-} \frac{-h - 0}{h} = \lim\limits_{h \to 0^-} \frac{-h}{h}$

For $h \neq 0$, $\frac{-h}{h} = -1$.

LHD $= \lim\limits_{h \to 0^-} (-1) = -1$

Calculate the Right-Hand Derivative (RHD) at $x=5$:

RHD $= \lim\limits_{h \to 0^+} \frac{f(5+h) - f(5)}{h}$

As $h \to 0^+$, $h$ is a small positive number, so $5+h > 5$. We use $f(x) = x-5$ for $x=5+h$.

$f(5+h) = (5+h) - 5 = h$

Substitute this into the RHD expression:

RHD $= \lim\limits_{h \to 0^+} \frac{h - 0}{h} = \lim\limits_{h \to 0^+} \frac{h}{h}$

For $h \neq 0$, $\frac{h}{h} = 1$.

RHD $= \lim\limits_{h \to 0^+} 1 = 1$

Since the Left-Hand Derivative ($LHD = -1$) is not equal to the Right-Hand Derivative ($RHD = 1$) at $x=5$, the function $f(x) = |x-5|$ is not differentiable at $x=5$.

Conclusion:

We have shown that $f(x) = |x-5|$ is continuous at $x=5$ (since LHL = RHL = $f(5) = 0$), but it is not differentiable at $x=5$ (since LHD = $-1$ and RHD = $1$, and $-1 \neq 1$).

This completes the demonstration.

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ satisfying the equation $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$.

Also, $f(x) \neq 0$ for all $x \in \mathbb{R}$.

The function is differentiable at $x = 0$, and $f'(0) = 2$.

To Prove:

$f'(x) = 2 f(x)$ for all $x \in \mathbb{R}$.

Proof:

First, let's determine the value of $f(0)$. Using the given functional equation, we set $x=0$ and $y=0$:

$f(0+0) = f(0)f(0)$

$f(0) = [f(0)]^2$

Rearranging this equation, we get $[f(0)]^2 - f(0) = 0$, which factors as $f(0)(f(0) - 1) = 0$.

This implies either $f(0) = 0$ or $f(0) = 1$.

However, the problem statement explicitly says that $f(x) \neq 0$ for all $x \in \mathbb{R}$. This includes $x=0$, so $f(0) \neq 0$.

Therefore, we must have $f(0) = 1$.

Now, let's use the definition of the derivative to find the derivative of $f(x)$ at any point $x \in \mathbb{R}$. The derivative of $f(x)$ is given by:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

We are given the functional equation $f(x+h) = f(x)f(h)$. We can substitute this into the numerator of the derivative expression:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x)f(h) - f(x)}{h}$

We can factor out $f(x)$ from the numerator:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x)(f(h) - 1)}{h}$

Since $f(x)$ does not depend on the limit variable $h$, we can take $f(x)$ outside the limit:

$f'(x) = f(x) \lim\limits_{h \to 0} \frac{f(h) - 1}{h}$

Now, let's look closely at the limit term $\lim\limits_{h \to 0} \frac{f(h) - 1}{h}$. We can rewrite this expression using the value of $f(0)$ that we found ($f(0)=1$):

$\lim\limits_{h \to 0} \frac{f(h) - 1}{h} = \lim\limits_{h \to 0} \frac{f(h) - f(0)}{h}$

By the definition of the derivative, this limit is the derivative of $f$ at $x=0$, which is $f'(0)$.

$f'(0) = \lim\limits_{h \to 0} \frac{f(h) - f(0)}{h} = \lim\limits_{h \to 0} \frac{f(h) - 1}{h}$

We are given that $f'(0) = 2$.

Substituting this value back into the expression for $f'(x)$:

$f'(x) = f(x) \times 2$

$f'(x) = 2f(x)$

This holds for all $x \in \mathbb{R}$ because the derivative $f'(x)$ was evaluated at an arbitrary point $x$.

Thus, we have proved that $f'(x) = 2 f(x)$.

This completes the proof.

Given:

The function to differentiate is $f(x) = 2^{\cos^2x}$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}(2^{\cos^2x})$.

Solution:

We will use the chain rule for differentiation.

Let $y = 2^{\cos^2x}$.

This is of the form $a^u$, where $a=2$ and $u = \cos^2x$. The derivative of $a^u$ is $a^u \ln a \frac{du}{dx}$.

So, we need to find $\frac{du}{dx}$, where $u = \cos^2x$.

Let $u = (\cos x)^2$. This is of the form $v^n$, where $v = \cos x$ and $n=2$. The derivative of $v^n$ is $n v^{n-1} \frac{dv}{dx}$.

Here, $v = \cos x$. The derivative of $\cos x$ with respect to $x$ is $-\sin x$.

Applying the chain rule to $u = (\cos x)^2$:

$\frac{du}{dx} = \frac{d}{dx}(\cos x)^2 = 2 (\cos x)^{2-1} \cdot \frac{d}{dx}(\cos x)$

$= 2 \cos x \cdot (-\sin x)$

$= -2 \sin x \cos x$

Using the identity $\sin(2x) = 2 \sin x \cos x$, we can write:

$\frac{du}{dx} = -\sin(2x)$

Now, we apply the chain rule to the original function $y = 2^u$:

$\frac{dy}{dx} = \frac{d}{du}(2^u) \cdot \frac{du}{dx}$

$= (2^u \ln 2) \cdot (-\sin(2x))$

Substitute $u = \cos^2x$ back into the expression:

$\frac{dy}{dx} = (2^{\cos^2x} \ln 2) \cdot (-\sin(2x))$

$= -2^{\cos^2x} \sin(2x) \ln 2$

Thus, the derivative of $2^{\cos^2x}$ with respect to $x$ is $-2^{\cos^2x} \sin(2x) \ln 2$.

Given:

The function to differentiate is $f(x) = \frac{8^x}{x^8}$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\frac{8^x}{x^8}\right)$.

Solution:

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

Let $u(x) = 8^x$ and $v(x) = x^8$.

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(8^x)$

Using the formula for the derivative of $a^x$, which is $a^x \ln a$:

$u'(x) = 8^x \ln 8$

Next, find the derivative of $v(x)$ with respect to $x$:

$v'(x) = \frac{d}{dx}(x^8)$

Using the power rule for differentiation, which is $\frac{d}{dx}(x^n) = nx^{n-1}$:

$v'(x) = 8x^{8-1} = 8x^7$

Now, apply the quotient rule:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Substitute the expressions for $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$:

$f'(x) = \frac{(8^x \ln 8)(x^8) - (8^x)(8x^7)}{(x^8)^2}$

Simplify the denominator:

$(x^8)^2 = x^{8 \times 2} = x^{16}$

So, the expression becomes:

$f'(x) = \frac{x^8 \cdot 8^x \ln 8 - 8x^7 \cdot 8^x}{x^{16}}$

Factor out the common terms $8^x x^7$ from the numerator:

$f'(x) = \frac{8^x x^7 (x \ln 8 - 8)}{x^{16}}$

Cancel out $x^7$ from the numerator and the denominator:

$f'(x) = \frac{8^x (x \ln 8 - 8)}{x^{16-7}}$

$f'(x) = \frac{8^x (x \ln 8 - 8)}{x^9}$

Thus, the derivative of $\frac{8^x}{x^8}$ with respect to $x$ is $\frac{8^x (x \ln 8 - 8)}{x^9}$.

Given:

The function to differentiate is $f(x) = \log \left( x + \sqrt{x^2 +a} \right)$.

We assume 'log' denotes the natural logarithm, i.e., $\log u = \ln u$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\log \left( x + \sqrt{x^2 +a} \right)\right)$.

Solution:

We will use the chain rule for differentiation. Let $y = \log \left( x + \sqrt{x^2 +a} \right)$.

Let $u = x + \sqrt{x^2 +a}$. Then $y = \log(u)$.

The derivative of $y$ with respect to $x$ is given by the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

First, find $\frac{dy}{du}$. Assuming 'log' is the natural logarithm (ln):

$\frac{dy}{du} = \frac{d}{du}(\log u) = \frac{d}{du}(\ln u) = \frac{1}{u}$

Next, find $\frac{du}{dx}$. We need to differentiate $u = x + \sqrt{x^2 +a}$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x + \sqrt{x^2 +a}) = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{x^2 +a})$

The derivative of $x$ with respect to $x$ is $1$.

$\frac{d}{dx}(x) = 1$

To find the derivative of $\sqrt{x^2 +a}$, we use the chain rule again. Let $v = x^2 + a$. Then $\sqrt{x^2 +a} = \sqrt{v} = v^{1/2}$.

$\frac{d}{dx}(\sqrt{x^2 +a}) = \frac{d}{dv}(v^{1/2}) \cdot \frac{dv}{dx}$

$\frac{d}{dv}(v^{1/2}) = \frac{1}{2}v^{1/2 - 1} = \frac{1}{2}v^{-1/2} = \frac{1}{2\sqrt{v}} = \frac{1}{2\sqrt{x^2 +a}}$

$\frac{dv}{dx} = \frac{d}{dx}(x^2 + a) = \frac{d}{dx}(x^2) + \frac{d}{dx}(a) = 2x + 0 = 2x$

So, the derivative of $\sqrt{x^2 +a}$ with respect to $x$ is:

$\frac{d}{dx}(\sqrt{x^2 +a}) = \frac{1}{2\sqrt{x^2 +a}} \cdot (2x) = \frac{x}{\sqrt{x^2 +a}}$

Now, substitute this back into the expression for $\frac{du}{dx}$:

$\frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2 +a}} = \frac{\sqrt{x^2 +a}}{\sqrt{x^2 +a}} + \frac{x}{\sqrt{x^2 +a}} = \frac{\sqrt{x^2 +a} + x}{\sqrt{x^2 +a}}$

Finally, substitute $\frac{dy}{du}$ and $\frac{du}{dx}$ into the chain rule formula $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:

$\frac{dy}{dx} = \frac{1}{u} \cdot \frac{\sqrt{x^2 +a} + x}{\sqrt{x^2 +a}}$

Substitute $u = x + \sqrt{x^2 +a}$ back into the expression:

$\frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 +a}} \cdot \frac{x + \sqrt{x^2 +a}}{\sqrt{x^2 +a}}$

Cancel the term $(x + \sqrt{x^2 +a})$ from the numerator and the denominator (assuming $x + \sqrt{x^2 +a} \neq 0$, which is required for the logarithm to be defined):

$\frac{dy}{dx} = \frac{1}{\sqrt{x^2 +a}}$

Thus, the derivative of $\log \left( x + \sqrt{x^2 +a} \right)$ with respect to $x$ is $\frac{1}{\sqrt{x^2 +a}}$.

Given:

The function to differentiate is $f(x) = \log \left[ \log (\log x^5) \right]$.

We assume 'log' denotes the natural logarithm, i.e., $\log u = \ln u$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\log \left[ \log (\log x^5) \right]\right)$.

Solution:

We will use the chain rule for differentiation repeatedly. The chain rule states that if $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. Also, the derivative of $\log u$ (natural logarithm) is $\frac{1}{u} \frac{du}{dx}$.

Let $y = \log \left[ \log (\log x^5) \right]$.

Apply the chain rule by considering the outermost logarithm. Let $u = \log (\log x^5)$. Then $y = \log u$.

$\frac{dy}{dx} = \frac{d}{du}(\log u) \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{d}{dx}\left(\log (\log x^5)\right)$

$= \frac{1}{\log (\log x^5)} \cdot \frac{d}{dx}\left(\log (\log x^5)\right)$

Now, differentiate the term $\log (\log x^5)$. Let $v = \log x^5$. Then this term is $\log v$.

$\frac{d}{dx}\left(\log (\log x^5)\right) = \frac{d}{dv}(\log v) \cdot \frac{dv}{dx} = \frac{1}{v} \cdot \frac{d}{dx}(\log x^5)$

$= \frac{1}{\log x^5} \cdot \frac{d}{dx}(\log x^5)$

Finally, differentiate the innermost term $\log x^5$. We can use the logarithm property $\log a^b = b \log a$, so $\log x^5 = 5 \log x$.

$\frac{d}{dx}(\log x^5) = \frac{d}{dx}(5 \log x) = 5 \cdot \frac{d}{dx}(\log x)$

The derivative of $\log x$ is $\frac{1}{x}$.

$= 5 \cdot \frac{1}{x} = \frac{5}{x}$

Now, substitute back the derivatives into the previous steps.

Substitute $\frac{d}{dx}(\log x^5) = \frac{5}{x}$ into the expression for $\frac{d}{dx}\left(\log (\log x^5)\right)$:

$\frac{d}{dx}\left(\log (\log x^5)\right) = \frac{1}{\log x^5} \cdot \frac{5}{x} = \frac{5}{x \log x^5}$

Substitute this result into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{\log (\log x^5)} \cdot \frac{5}{x \log x^5}$

$\frac{dy}{dx} = \frac{5}{x \log x^5 \log (\log x^5)}$

We can further simplify by replacing $\log x^5$ with $5 \log x$ in the denominator:

$\frac{dy}{dx} = \frac{5}{x (5 \log x) \log (5 \log x)}$

$\frac{dy}{dx} = \frac{\cancel{5}}{x (\cancel{5} \log x) \log (5 \log x)}$

$\frac{dy}{dx} = \frac{1}{x \log x \log (5 \log x)}$

Thus, the derivative of $\log \left[ \log (\log x^5) \right]$ with respect to $x$ is $\frac{1}{x \log x \log (5 \log x)}$.

Given:

The function to differentiate is $f(x) = \sin \sqrt{x} + \cos^2 \sqrt{x}$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\sin \sqrt{x} + \cos^2 \sqrt{x}\right)$.

Solution:

We need to find the derivative of the sum of two functions. Using the sum rule, $\frac{d}{dx}(g(x) + h(x)) = \frac{dg}{dx} + \frac{dh}{dx}$.

Let $g(x) = \sin \sqrt{x}$ and $h(x) = \cos^2 \sqrt{x}$.

First, differentiate $g(x) = \sin \sqrt{x}$ with respect to $x$. We use the chain rule. Let $u = \sqrt{x}$. Then $g(x) = \sin u$.

$\frac{dg}{dx} = \frac{dg}{du} \cdot \frac{du}{dx}$

The derivative of $\sin u$ with respect to $u$ is $\cos u$.

$\frac{dg}{du} = \frac{d}{du}(\sin u) = \cos u = \cos \sqrt{x}$

The derivative of $\sqrt{x}$ with respect to $x$ is $\frac{1}{2\sqrt{x}}$.

$\frac{du}{dx} = \frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

So, the derivative of $g(x)$ is:

$\frac{dg}{dx} = (\cos \sqrt{x}) \cdot \left(\frac{1}{2\sqrt{x}}\right) = \frac{\cos \sqrt{x}}{2\sqrt{x}}$

Next, differentiate $h(x) = \cos^2 \sqrt{x} = (\cos \sqrt{x})^2$ with respect to $x$. We use the chain rule. Let $v = \cos \sqrt{x}$. Then $h(x) = v^2$.

$\frac{dh}{dx} = \frac{dh}{dv} \cdot \frac{dv}{dx}$

The derivative of $v^2$ with respect to $v$ is $2v$.

$\frac{dh}{dv} = \frac{d}{dv}(v^2) = 2v = 2 \cos \sqrt{x}$

Now, we need to find the derivative of $v = \cos \sqrt{x}$ with respect to $x$. We use the chain rule again. Let $w = \sqrt{x}$. Then $v = \cos w$.

$\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx}$

The derivative of $\cos w$ with respect to $w$ is $-\sin w$.

$\frac{dv}{dw} = \frac{d}{dw}(\cos w) = -\sin w = -\sin \sqrt{x}$

The derivative of $\sqrt{x}$ with respect to $x$ is $\frac{1}{2\sqrt{x}}$ (as calculated earlier).

$\frac{dw}{dx} = \frac{1}{2\sqrt{x}}$

So, the derivative of $v$ is:

$\frac{dv}{dx} = (-\sin \sqrt{x}) \cdot \left(\frac{1}{2\sqrt{x}}\right) = -\frac{\sin \sqrt{x}}{2\sqrt{x}}$

Now, substitute this back into the expression for $\frac{dh}{dx}$:

$\frac{dh}{dx} = (2 \cos \sqrt{x}) \cdot \left(-\frac{\sin \sqrt{x}}{2\sqrt{x}}\right) = -\frac{2 \cos \sqrt{x} \sin \sqrt{x}}{2\sqrt{x}}$

Using the trigonometric identity $\sin(2\theta) = 2 \sin \theta \cos \theta$ with $\theta = \sqrt{x}$:

$2 \sin \sqrt{x} \cos \sqrt{x} = \sin(2\sqrt{x})$

$\frac{dh}{dx} = -\frac{\sin(2\sqrt{x})}{2\sqrt{x}}$

Finally, add the derivatives of $g(x)$ and $h(x)$ to find the derivative of $f(x)$:

$\frac{df}{dx} = \frac{dg}{dx} + \frac{dh}{dx} = \frac{\cos \sqrt{x}}{2\sqrt{x}} + \left(-\frac{\sin(2\sqrt{x})}{2\sqrt{x}}\right)$

$\frac{df}{dx} = \frac{\cos \sqrt{x} - \sin(2\sqrt{x})}{2\sqrt{x}}$

Thus, the derivative of $\sin \sqrt{x} + \cos^2 \sqrt{x}$ with respect to $x$ is $\frac{\cos \sqrt{x} - \sin(2\sqrt{x})}{2\sqrt{x}}$.

Given:

The function to differentiate is $f(x) = \sin^n (ax^2 + bx + x)$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\sin^n (ax^2 + bx + x)\right)$.

Solution:

We will use the chain rule for differentiation. The function is a composition of several functions: a polynomial inside a sine function, and the result raised to the power $n$.

Let $y = \sin^n (ax^2 + bx + x)$.

We can rewrite the expression inside the sine function as $ax^2 + (b+1)x$.

So, $y = (\sin (ax^2 + (b+1)x))^n$.

Let's apply the chain rule step-by-step from the outermost function inwards.

Let $u = \sin (ax^2 + (b+1)x)$. Then $y = u^n$.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

First, differentiate $y = u^n$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(u^n) = n u^{n-1}$

Substitute $u = \sin (ax^2 + (b+1)x)$ back:

$\frac{dy}{du} = n (\sin (ax^2 + (b+1)x))^{n-1} = n \sin^{n-1} (ax^2 + (b+1)x)$

Next, differentiate $u = \sin (ax^2 + (b+1)x)$ with respect to $x$. This is a composite function itself. Let $v = ax^2 + (b+1)x$. Then $u = \sin v$.

$\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}$

Differentiate $u = \sin v$ with respect to $v$:

$\frac{du}{dv} = \frac{d}{dv}(\sin v) = \cos v$

Substitute $v = ax^2 + (b+1)x$ back:

$\frac{du}{dv} = \cos (ax^2 + (b+1)x)$

Differentiate $v = ax^2 + (b+1)x$ with respect to $x$:

$\frac{dv}{dx} = \frac{d}{dx}(ax^2 + (b+1)x)$

$= \frac{d}{dx}(ax^2) + \frac{d}{dx}((b+1)x)$

$= a(2x) + (b+1)(1)$

$= 2ax + b + 1$

Now, combine the derivatives using the chain rule $\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}$:

$\frac{du}{dx} = \cos (ax^2 + (b+1)x) \cdot (2ax + b + 1)$

$\frac{du}{dx} = (2ax + b + 1) \cos (ax^2 + (b+1)x)$

Substitute $ax^2 + (b+1)x$ back as $ax^2 + bx + x$:

$\frac{du}{dx} = (2ax + b + 1) \cos (ax^2 + bx + x)$

Finally, combine the results using the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:

$\frac{dy}{dx} = \left(n \sin^{n-1} (ax^2 + bx + x)\right) \cdot \left((2ax + b + 1) \cos (ax^2 + bx + x)\right)$

$\frac{dy}{dx} = n (2ax + b + 1) \sin^{n-1} (ax^2 + bx + x) \cos (ax^2 + bx + x)$

Thus, the derivative of $\sin^n (ax^2 + bx + x)$ with respect to $x$ is $n (2ax + b + 1) \sin^{n-1} (ax^2 + bx + x) \cos (ax^2 + bx + x)$.

Given:

The function to differentiate is $f(x) = \cos (\tan \sqrt{x+1})$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\cos (\tan \sqrt{x+1})\right)$.

Solution:

We will use the chain rule for differentiation. The function is a composition of several functions.

Let $y = \cos (\tan \sqrt{x+1})$.

We can view this as $y = \cos(u)$, where $u = \tan(v)$, and $v = \sqrt{w}$, and $w = x+1$.

Using the chain rule, the derivative $\frac{dy}{dx}$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}$

First, differentiate $y = \cos(u)$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\cos u) = -\sin u$

Substituting $u = \tan \sqrt{x+1}$ back, we get:

$\frac{dy}{du} = -\sin (\tan \sqrt{x+1})$

Next, differentiate $u = \tan(v)$ with respect to $v$:

$\frac{du}{dv} = \frac{d}{dv}(\tan v) = \sec^2 v$

Substituting $v = \sqrt{x+1}$ back, we get:

$\frac{du}{dv} = \sec^2 (\sqrt{x+1})$

Next, differentiate $v = \sqrt{w}$ with respect to $w$:

$\frac{dv}{dw} = \frac{d}{dw}(\sqrt{w}) = \frac{d}{dw}(w^{1/2}) = \frac{1}{2} w^{1/2 - 1} = \frac{1}{2} w^{-1/2} = \frac{1}{2\sqrt{w}}$

Substituting $w = x+1$ back, we get:

$\frac{dv}{dw} = \frac{1}{2\sqrt{x+1}}$

Finally, differentiate $w = x+1$ with respect to $x$:

$\frac{dw}{dx} = \frac{d}{dx}(x+1) = \frac{d}{dx}(x) + \frac{d}{dx}(1) = 1 + 0 = 1$

Now, multiply these results together to find $\frac{dy}{dx}$:

$\frac{dy}{dx} = \left(-\sin (\tan \sqrt{x+1})\right) \cdot \left(\sec^2 (\sqrt{x+1})\right) \cdot \left(\frac{1}{2\sqrt{x+1}}\right) \cdot (1)$

$\frac{dy}{dx} = -\frac{\sin (\tan \sqrt{x+1}) \sec^2 (\sqrt{x+1})}{2\sqrt{x+1}}$

Thus, the derivative of $\cos (\tan \sqrt{x+1})$ with respect to $x$ is $-\frac{\sin (\tan \sqrt{x+1}) \sec^2 (\sqrt{x+1})}{2\sqrt{x+1}}$.

Given:

The function to differentiate is $f(x) = \sin x^2 + \sin^2 x + \sin^2 (x^2)$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\sin x^2 + \sin^2 x + \sin^2 (x^2)\right)$.

Solution:

We need to find the derivative of the sum of three functions. Using the sum rule, $\frac{d}{dx}(g(x) + h(x) + k(x)) = \frac{dg}{dx} + \frac{dh}{dx} + \frac{dk}{dx}$.

Let $g(x) = \sin x^2$, $h(x) = \sin^2 x$, and $k(x) = \sin^2 (x^2)$.

First, differentiate $g(x) = \sin x^2$ with respect to $x$. We use the chain rule. Let $u = x^2$. Then $g(x) = \sin u$.

$\frac{dg}{dx} = \frac{dg}{du} \cdot \frac{du}{dx} = \frac{d}{du}(\sin u) \cdot \frac{d}{dx}(x^2)$

$= (\cos u) \cdot (2x)$

Substitute $u = x^2$ back:

$\frac{dg}{dx} = 2x \cos x^2$

Next, differentiate $h(x) = \sin^2 x = (\sin x)^2$ with respect to $x$. We use the chain rule. Let $v = \sin x$. Then $h(x) = v^2$.

$\frac{dh}{dx} = \frac{dh}{dv} \cdot \frac{dv}{dx} = \frac{d}{dv}(v^2) \cdot \frac{d}{dx}(\sin x)$

$= (2v) \cdot (\cos x)$

Substitute $v = \sin x$ back:

$\frac{dh}{dx} = 2 \sin x \cos x$

Using the identity $\sin(2x) = 2 \sin x \cos x$, we can write:

$\frac{dh}{dx} = \sin(2x)$

Finally, differentiate $k(x) = \sin^2 (x^2) = (\sin (x^2))^2$ with respect to $x$. We use the chain rule. Let $w = \sin (x^2)$. Then $k(x) = w^2$.

$\frac{dk}{dx} = \frac{dk}{dw} \cdot \frac{dw}{dx} = \frac{d}{dw}(w^2) \cdot \frac{d}{dx}(\sin (x^2))$

$= (2w) \cdot \frac{d}{dx}(\sin (x^2))$

To differentiate $\sin (x^2)$, we use the chain rule again (this is the same as the derivative of $g(x)$). Let $z = x^2$. Then $\sin(x^2) = \sin z$.

$\frac{d}{dx}(\sin (x^2)) = \frac{d}{dz}(\sin z) \cdot \frac{d}{dx}(x^2) = (\cos z) \cdot (2x) = 2x \cos x^2$

Now, substitute $w = \sin (x^2)$ and $\frac{d}{dx}(\sin (x^2)) = 2x \cos x^2$ back into the expression for $\frac{dk}{dx}$:

$\frac{dk}{dx} = (2 \sin (x^2)) \cdot (2x \cos x^2) = 4x \sin (x^2) \cos (x^2)$

Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$ with $\theta = x^2$, we can write:

$\frac{dk}{dx} = 2x (2 \sin (x^2) \cos (x^2)) = 2x \sin (2x^2)$

Now, add the derivatives of the three terms:

$\frac{df}{dx} = \frac{dg}{dx} + \frac{dh}{dx} + \frac{dk}{dx}$

$\frac{df}{dx} = (2x \cos x^2) + (2 \sin x \cos x) + (4x \sin (x^2) \cos (x^2))$

Or, using the simplified forms for the second and third terms:

$\frac{df}{dx} = 2x \cos x^2 + \sin(2x) + 2x \sin (2x^2)$

Thus, the derivative of $\sin x^2 + \sin^2 x + \sin^2 (x^2)$ with respect to $x$ is $2x \cos x^2 + 2 \sin x \cos x + 4x \sin (x^2) \cos (x^2)$ or $2x \cos x^2 + \sin(2x) + 2x \sin (2x^2)$.

Given:

The function to differentiate is $f(x) = \sin^{-1} \left( \frac{1}{\sqrt{x+1}} \right)$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\sin^{-1} \left( \frac{1}{\sqrt{x+1}} \right)\right)$.

Solution:

We will use the chain rule for differentiation. The derivative of $\sin^{-1}(u)$ with respect to $x$ is $\frac{1}{\sqrt{1-u^2}} \frac{du}{dx}$.

Let $y = \sin^{-1} \left( \frac{1}{\sqrt{x+1}} \right)$.

Let $u = \frac{1}{\sqrt{x+1}} = (x+1)^{-1/2}$.

Then $y = \sin^{-1}(u)$.

By the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

First, find the derivative of $y = \sin^{-1}(u)$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\sin^{-1} u) = \frac{1}{\sqrt{1-u^2}}$

Substitute $u = \frac{1}{\sqrt{x+1}}$ back:

$\sqrt{1-u^2} = \sqrt{1 - \left(\frac{1}{\sqrt{x+1}}\right)^2} = \sqrt{1 - \frac{1}{x+1}}$

$= \sqrt{\frac{x+1 - 1}{x+1}} = \sqrt{\frac{x}{x+1}} = \frac{\sqrt{x}}{\sqrt{x+1}}$

So, $\frac{dy}{du} = \frac{1}{\frac{\sqrt{x}}{\sqrt{x+1}}} = \frac{\sqrt{x+1}}{\sqrt{x}}$.

Next, find the derivative of $u = (x+1)^{-1/2}$ with respect to $x$. We use the chain rule again. Let $v = x+1$. Then $u = v^{-1/2}$.

$\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}$

Differentiate $u = v^{-1/2}$ with respect to $v$:

$\frac{du}{dv} = \frac{d}{dv}(v^{-1/2}) = -\frac{1}{2} v^{-1/2 - 1} = -\frac{1}{2} v^{-3/2}$

Substitute $v = x+1$ back:

$\frac{du}{dv} = -\frac{1}{2} (x+1)^{-3/2}$

Differentiate $v = x+1$ with respect to $x$:

$\frac{dv}{dx} = \frac{d}{dx}(x+1) = 1$

Combine these to get $\frac{du}{dx}$:

$\frac{du}{dx} = \left(-\frac{1}{2} (x+1)^{-3/2}\right) \cdot (1) = -\frac{1}{2} (x+1)^{-3/2} = -\frac{1}{2(x+1)^{3/2}}$

Note that $(x+1)^{3/2} = (x+1)\sqrt{x+1}$.

$\frac{du}{dx} = -\frac{1}{2(x+1)\sqrt{x+1}}$

Finally, combine $\frac{dy}{du}$ and $\frac{du}{dx}$ using the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:

$\frac{dy}{dx} = \left(\frac{\sqrt{x+1}}{\sqrt{x}}\right) \cdot \left(-\frac{1}{2(x+1)\sqrt{x+1}}\right)$

Cancel the $\sqrt{x+1}$ terms:

$\frac{dy}{dx} = \frac{\cancel{\sqrt{x+1}}}{\sqrt{x}} \cdot \left(-\frac{1}{2(x+1)\cancel{\sqrt{x+1}}}\right)$

$\frac{dy}{dx} = -\frac{1}{2\sqrt{x}(x+1)}$

Thus, the derivative of $\sin^{-1} \left( \frac{1}{\sqrt{x+1}} \right)$ with respect to $x$ is $-\frac{1}{2\sqrt{x}(x+1)}$.

Given:

The function to differentiate is $f(x) = (\sin x)^{\cos x}$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left((\sin x)^{\cos x}\right)$.

Solution:

We will use logarithmic differentiation, as the function is of the form $[g(x)]^{h(x)}$.

Let $y = (\sin x)^{\cos x}$.

Take the natural logarithm of both sides:

$\log y = \log \left( (\sin x)^{\cos x} \right)$

Using the logarithm property $\log a^b = b \log a$:

$\log y = \cos x \log (\sin x)$

Now, differentiate both sides with respect to $x$. We will use the chain rule on the left side and the product rule on the right side.

Differentiate the left side:

$\frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx}$

Differentiate the right side using the product rule $\frac{d}{dx}(uv) = u'\v + uv'$, where $u = \cos x$ and $v = \log (\sin x)$.

First, find the derivative of $u = \cos x$:

$u' = \frac{d}{dx}(\cos x) = -\sin x$

Next, find the derivative of $v = \log (\sin x)$ using the chain rule. Let $w = \sin x$. Then $v = \log w$.

$\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx} = \frac{d}{dw}(\log w) \cdot \frac{d}{dx}(\sin x)$

$= \frac{1}{w} \cdot \cos x = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x$

So, $v' = \cot x$.

Now apply the product rule to the right side $\cos x \log (\sin x)$:

$\frac{d}{dx}(\cos x \log (\sin x)) = u'v + uv'$

$= (-\sin x) \log (\sin x) + (\cos x)(\cot x)$

$= -\sin x \log (\sin x) + \cos x \frac{\cos x}{\sin x}$

$= \frac{\cos^2 x}{\sin x} - \sin x \log (\sin x)$

Equate the derivatives of the left and right sides:

$\frac{1}{y} \frac{dy}{dx} = \frac{\cos^2 x}{\sin x} - \sin x \log (\sin x)$

Multiply both sides by $y$ to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = y \left( \frac{\cos^2 x}{\sin x} - \sin x \log (\sin x) \right)$

Substitute $y = (\sin x)^{\cos x}$ back into the expression:

$\frac{dy}{dx} = (\sin x)^{\cos x} \left( \frac{\cos^2 x}{\sin x} - \sin x \log (\sin x) \right)$

We can also write $\frac{\cos^2 x}{\sin x}$ as $\cot x \cos x$:

$\frac{dy}{dx} = (\sin x)^{\cos x} \left( \cot x \cos x - \sin x \log (\sin x) \right)$

Thus, the derivative of $(\sin x)^{\cos x}$ with respect to $x$ is $(\sin x)^{\cos x} \left( \cot x \cos x - \sin x \log (\sin x) \right)$.

Given:

The function to differentiate is $f(x) = \sin^m x \cos^n x$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\sin^m x \cos^n x\right)$.

Solution:

We can use either the product rule directly or logarithmic differentiation. Let's use logarithmic differentiation, which is often helpful for functions of the form $[g(x)]^{h(x)}$ or products/quotients with powers.

Let $y = \sin^m x \cos^n x$.

Take the natural logarithm of both sides:

$\log y = \log (\sin^m x \cos^n x)$

Using the logarithm properties $\log(ab) = \log a + \log b$ and $\log a^b = b \log a$:

$\log y = \log (\sin^m x) + \log (\cos^n x)$

$\log y = m \log (\sin x) + n \log (\cos x)$

Now, differentiate both sides with respect to $x$. On the left side, we use the chain rule for implicit differentiation. On the right side, we use the sum rule and chain rule.

$\frac{d}{dx}(\log y) = \frac{d}{dx}(m \log (\sin x) + n \log (\cos x))$

$\frac{1}{y} \frac{dy}{dx} = m \frac{d}{dx}(\log (\sin x)) + n \frac{d}{dx}(\log (\cos x))$

Differentiate $\log (\sin x)$ using the chain rule. Let $u = \sin x$. $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$.

$\frac{d}{dx}(\log (\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$

Differentiate $\log (\cos x)$ using the chain rule. Let $v = \cos x$. $\frac{d}{dx}(\log v) = \frac{1}{v} \frac{dv}{dx}$.

$\frac{d}{dx}(\log (\cos x)) = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$

Substitute these derivatives back into the equation:

$\frac{1}{y} \frac{dy}{dx} = m (\cot x) + n (-\tan x)$

$\frac{1}{y} \frac{dy}{dx} = m \cot x - n \tan x$

Multiply both sides by $y$ to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = y (m \cot x - n \tan x)$

Substitute $y = \sin^m x \cos^n x$ back into the expression:

$\frac{dy}{dx} = \sin^m x \cos^n x (m \cot x - n \tan x)$

We can rewrite this expression by replacing $\cot x = \frac{\cos x}{\sin x}$ and $\tan x = \frac{\sin x}{\cos x}$:

$\frac{dy}{dx} = \sin^m x \cos^n x \left( m \frac{\cos x}{\sin x} - n \frac{\sin x}{\cos x} \right)$

$= \sin^m x \cos^n x \left( \frac{m \cos^2 x - n \sin^2 x}{\sin x \cos x} \right)$

$= \frac{\sin^m x \cos^n x (m \cos^2 x - n \sin^2 x)}{\sin x \cos x}$

$= \sin^{m-1} x \cos^{n-1} x (m \cos^2 x - n \sin^2 x)$

Thus, the derivative of $\sin^m x \cos^n x$ with respect to $x$ is $\sin^{m-1} x \cos^{n-1} x (m \cos^2 x - n \sin^2 x)$.

Given:

The function to differentiate is $f(x) = (x+1)^2 (x+2)^3 (x+3)^4$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left((x+1)^2 (x+2)^3 (x+3)^4\right)$.

Solution:

We will use logarithmic differentiation.

Let $y = (x+1)^2 (x+2)^3 (x+3)^4$.

Take the natural logarithm of both sides:

$\log y = \log \left[ (x+1)^2 (x+2)^3 (x+3)^4 \right]$

Using the logarithm properties $\log(abc) = \log a + \log b$ and $\log a^b = b \log a$:

$\log y = \log (x+1)^2 + \log (x+2)^3 + \log (x+3)^4$

$\log y = 2 \log (x+1) + 3 \log (x+2) + 4 \log (x+3)$

Now, differentiate both sides with respect to $x$. Use the chain rule on the left side and the sum rule and chain rule on the right side.

$\frac{d}{dx}(\log y) = \frac{d}{dx}[2 \log (x+1) + 3 \log (x+2) + 4 \log (x+3)]$

$\frac{1}{y} \frac{dy}{dx} = 2 \frac{d}{dx}(\log (x+1)) + 3 \frac{d}{dx}(\log (x+2)) + 4 \frac{d}{dx}(\log (x+3))$

Using the chain rule $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$:

$\frac{d}{dx}(\log (x+1)) = \frac{1}{x+1} \cdot \frac{d}{dx}(x+1) = \frac{1}{x+1} \cdot 1 = \frac{1}{x+1}$

$\frac{d}{dx}(\log (x+2)) = \frac{1}{x+2} \cdot \frac{d}{dx}(x+2) = \frac{1}{x+2} \cdot 1 = \frac{1}{x+2}$

$\frac{d}{dx}(\log (x+3)) = \frac{1}{x+3} \cdot \frac{d}{dx}(x+3) = \frac{1}{x+3} \cdot 1 = \frac{1}{x+3}$

Substitute these derivatives back into the equation:

$\frac{1}{y} \frac{dy}{dx} = 2 \left(\frac{1}{x+1}\right) + 3 \left(\frac{1}{x+2}\right) + 4 \left(\frac{1}{x+3}\right)$

$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} + \frac{3}{x+2} + \frac{4}{x+3}$

Multiply both sides by $y$ to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = y \left( \frac{2}{x+1} + \frac{3}{x+2} + \frac{4}{x+3} \right)$

Substitute $y = (x+1)^2 (x+2)^3 (x+3)^4$ back into the expression:

$\frac{dy}{dx} = (x+1)^2 (x+2)^3 (x+3)^4 \left( \frac{2}{x+1} + \frac{3}{x+2} + \frac{4}{x+3} \right)$

Distribute the term $(x+1)^2 (x+2)^3 (x+3)^4$ across the terms in the parenthesis:

$\frac{dy}{dx} = (x+1)^2 (x+2)^3 (x+3)^4 \cdot \frac{2}{x+1} + (x+1)^2 (x+2)^3 (x+3)^4 \cdot \frac{3}{x+2} + (x+1)^2 (x+2)^3 (x+3)^4 \cdot \frac{4}{x+3}$

$= 2 (x+1)^{2-1} (x+2)^3 (x+3)^4 + 3 (x+1)^2 (x+2)^{3-1} (x+3)^4 + 4 (x+1)^2 (x+2)^3 (x+3)^{4-1}$

$= 2 (x+1) (x+2)^3 (x+3)^4 + 3 (x+1)^2 (x+2)^2 (x+3)^4 + 4 (x+1)^2 (x+2)^3 (x+3)^3$

We can factor out the common term $(x+1)(x+2)^2(x+3)^3$ from each term:

$\frac{dy}{dx} = (x+1)(x+2)^2(x+3)^3 \left[ 2(x+2)(x+3) + 3(x+1)(x+3) + 4(x+1)(x+2) \right]$

Expand the terms inside the square brackets:

$2(x^2 + 5x + 6) + 3(x^2 + 4x + 3) + 4(x^2 + 3x + 2)$

$= (2x^2 + 10x + 12) + (3x^2 + 12x + 9) + (4x^2 + 12x + 8)$

$= (2x^2 + 3x^2 + 4x^2) + (10x + 12x + 12x) + (12 + 9 + 8)$

$= 9x^2 + 34x + 29$

Substitute this back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = (x+1)(x+2)^2(x+3)^3 (9x^2 + 34x + 29)$

Thus, the derivative of $(x+1)^2 \;(x+2)^3 \;(x+3)^4$ with respect to $x$ is $(x+1)(x+2)^2(x+3)^3 (9x^2 + 34x + 29)$.

Given:

The function to differentiate is $f(x) = \cos^{-1} \left( \frac{ \sin x + \cos x}{\sqrt{2}} \right)$.

The domain for $x$ is given as $-\frac{\pi}{4} < x < \frac{\pi}{4}$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\cos^{-1} \left( \frac{ \sin x + \cos x}{\sqrt{2}} \right)\right)$.

Solution:

Let $y = \cos^{-1} \left( \frac{ \sin x + \cos x}{\sqrt{2}} \right)$.

We can simplify the expression inside the inverse cosine function using trigonometric identities.

$\frac{ \sin x + \cos x}{\sqrt{2}} = \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x$

We know that $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

Substitute these values into the expression:

$\frac{ \sin x + \cos x}{\sqrt{2}} = \cos \frac{\pi}{4} \sin x + \sin \frac{\pi}{4} \cos x$

Using the angle addition formula for sine, $\sin(A+B) = \sin A \cos B + \cos A \sin B$, with $A = x$ and $B = \frac{\pi}{4}$:

$\cos \frac{\pi}{4} \sin x + \sin \frac{\pi}{4} \cos x = \sin \left( x + \frac{\pi}{4} \right)$

So, the function becomes:

$y = \cos^{-1} \left( \sin \left( x + \frac{\pi}{4} \right) \right)$

Now, we use the identity $\sin \theta = \cos \left( \frac{\pi}{2} - \theta \right)$. Let $\theta = x + \frac{\pi}{4}$.

$\sin \left( x + \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{2} - \left( x + \frac{\pi}{4} \right) \right) = \cos \left( \frac{\pi}{2} - x - \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{4} - x \right)$

So, the function simplifies to:

$y = \cos^{-1} \left( \cos \left( \frac{\pi}{4} - x \right) \right)$

Now, we consider the given domain $-\frac{\pi}{4} < x < \frac{\pi}{4}$. Let $z = \frac{\pi}{4} - x$. We need to find the range of $z$ in this domain.

If $-\frac{\pi}{4} < x < \frac{\pi}{4}$, then multiplying by $-1$ reverses the inequalities:

$\frac{\pi}{4} > -x > -\frac{\pi}{4}$

Rearranging, we get:

$-\frac{\pi}{4} < -x < \frac{\pi}{4}$

Adding $\frac{\pi}{4}$ to all parts:

$\frac{\pi}{4} - \frac{\pi}{4} < \frac{\pi}{4} - x < \frac{\pi}{4} + \frac{\pi}{4}$

$0 < \frac{\pi}{4} - x < \frac{\pi}{2}$

The range of $\frac{\pi}{4} - x$ is the interval $(0, \frac{\pi}{2})$.

The principal range of the inverse cosine function $\cos^{-1}(u)$ is $[0, \pi]$.

For any value $z$ in the interval $[0, \pi]$, $\cos^{-1}(\cos z) = z$.

Since our argument $\frac{\pi}{4} - x$ is in the interval $(0, \frac{\pi}{2})$, which is a subset of $[0, \pi]$, we can use this property directly:

$y = \frac{\pi}{4} - x$

Now, we differentiate this simplified expression with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} - x\right)$

$= \frac{d}{dx}\left(\frac{\pi}{4}\right) - \frac{d}{dx}(x)$

The derivative of a constant ($\frac{\pi}{4}$) is $0$. The derivative of $x$ with respect to $x$ is $1$.

$\frac{dy}{dx} = 0 - 1$

$\frac{dy}{dx} = -1$

Thus, the derivative of $\cos^{-1} \left( \frac{ \sin x + \cos x}{\sqrt{2}} \right)$ with respect to $x$ is $-1$ for the given domain.

Given:

The function to differentiate is $f(x) = \cos^{-1} \left( \frac{ \sin x + \cos x}{\sqrt{2}} \right)$.

The domain for $x$ is given as $-\frac{\pi}{4} < x < \frac{\pi}{4}$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\cos^{-1} \left( \frac{ \sin x + \cos x}{\sqrt{2}} \right)\right)$.

Solution:

Let $y = \cos^{-1} \left( \frac{ \sin x + \cos x}{\sqrt{2}} \right)$.

We simplify the expression inside the inverse cosine function using trigonometric identities.

$\frac{ \sin x + \cos x}{\sqrt{2}} = \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x$

We know that $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$. Substituting these values:

$= \cos \frac{\pi}{4} \sin x + \sin \frac{\pi}{4} \cos x$

Using the angle addition formula for sine, $\sin(A+B) = \sin A \cos B + \cos A \sin B$, with $A = x$ and $B = \frac{\pi}{4}$:

$= \sin \left( x + \frac{\pi}{4} \right)$

So, the function becomes:

$y = \cos^{-1} \left( \sin \left( x + \frac{\pi}{4} \right) \right)$

Now, we use the identity $\sin \theta = \cos \left( \frac{\pi}{2} - \theta \right)$. Let $\theta = x + \frac{\pi}{4}$.

$\sin \left( x + \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{2} - \left( x + \frac{\pi}{4} \right) \right) = \cos \left( \frac{\pi}{2} - x - \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{4} - x \right)$

So, the function simplifies to:

$y = \cos^{-1} \left( \cos \left( \frac{\pi}{4} - x \right) \right)$

Now, we consider the given domain $-\frac{\pi}{4} < x < \frac{\pi}{4}$. We find the range of the argument $\frac{\pi}{4} - x$ in this domain.

Given: $-\frac{\pi}{4} < x < \frac{\pi}{4}$.

Multiplying by $-1$ reverses the inequalities:

$\frac{\pi}{4} > -x > -\frac{\pi}{4}$

Rearranging the inequality:

$-\frac{\pi}{4} < -x < \frac{\pi}{4}$

Adding $\frac{\pi}{4}$ to all parts of the inequality:

$\frac{\pi}{4} - \frac{\pi}{4} < \frac{\pi}{4} - x < \frac{\pi}{4} + \frac{\pi}{4}$

$0 < \frac{\pi}{4} - x < \frac{\pi}{2}$

The range of $\frac{\pi}{4} - x$ is the interval $(0, \frac{\pi}{2})$.

The principal range of the inverse cosine function $\cos^{-1}(u)$ is $[0, \pi]$. For any value $\theta$ in the interval $[0, \pi]$, $\cos^{-1}(\cos \theta) = \theta$.

Since our argument $\frac{\pi}{4} - x$ is in the interval $(0, \frac{\pi}{2})$, which is a subset of $[0, \pi]$, we can use this property directly:

$y = \frac{\pi}{4} - x$

Now, we differentiate this simplified expression with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} - x\right)$

$= \frac{d}{dx}\left(\frac{\pi}{4}\right) - \frac{d}{dx}(x)$

The derivative of a constant ($\frac{\pi}{4}$) is $0$. The derivative of $x$ with respect to $x$ is $1$.

$\frac{dy}{dx} = 0 - 1$

$\frac{dy}{dx} = -1$

Thus, the derivative of $\cos^{-1} \left( \frac{ \sin x + \cos x}{\sqrt{2}} \right)$ with respect to $x$ is $-1$ for the given domain.

Given:

The function to differentiate is $f(x) = \tan^{-1} (\sec x + \tan x)$.

The domain for $x$ is given as $-\frac{\pi}{2} < x < \frac{\pi}{2}$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\tan^{-1} (\sec x + \tan x)\right)$.

Solution:

Let $y = \tan^{-1} (\sec x + \tan x)$.

We simplify the expression inside the inverse tangent function using trigonometric identities:

$\sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x}$

Now, we use the identities $\sin x = \cos\left(\frac{\pi}{2} - x\right)$ and $\cos x = \sin\left(\frac{\pi}{2} - x\right)$.

$\frac{1 + \sin x}{\cos x} = \frac{1 + \cos\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right)}$

Let $\theta = \frac{\pi}{2} - x$. The expression becomes $\frac{1 + \cos \theta}{\sin \theta}$.

Using the half-angle identities $1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right)$ and $\sin \theta = 2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)$:

$\frac{1 + \cos \theta}{\sin \theta} = \frac{2 \cos^2\left(\frac{\theta}{2}\right)}{2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)}$

Assuming $\cos\left(\frac{\theta}{2}\right) \neq 0$, we can cancel a term:

$= \frac{\cos\left(\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} = \cot\left(\frac{\theta}{2}\right)$

Substitute $\theta = \frac{\pi}{2} - x$ back:

$\cot\left(\frac{\frac{\pi}{2} - x}{2}\right) = \cot\left(\frac{\pi}{4} - \frac{x}{2}\right)$

Using the identity $\cot \alpha = \tan\left(\frac{\pi}{2} - \alpha\right)$, with $\alpha = \frac{\pi}{4} - \frac{x}{2}$:

$\cot\left(\frac{\pi}{4} - \frac{x}{2}\right) = \tan\left(\frac{\pi}{2} - \left(\frac{\pi}{4} - \frac{x}{2}\right)\right) = \tan\left(\frac{\pi}{2} - \frac{\pi}{4} + \frac{x}{2}\right) = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$

So, the function becomes:

$y = \tan^{-1} \left( \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right)$

Now, we consider the given domain $-\frac{\pi}{2} < x < \frac{\pi}{2}$. We find the range of the argument $\frac{\pi}{4} + \frac{x}{2}$.

Given: $-\frac{\pi}{2} < x < \frac{\pi}{2}$.

Dividing by 2:

$-\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{4}$

Adding $\frac{\pi}{4}$ to all parts:

$\frac{\pi}{4} - \frac{\pi}{4} < \frac{\pi}{4} + \frac{x}{2} < \frac{\pi}{4} + \frac{\pi}{4}$

$0 < \frac{\pi}{4} + \frac{x}{2} < \frac{\pi}{2}$

The range of $\frac{\pi}{4} + \frac{x}{2}$ is the interval $(0, \frac{\pi}{2})$. The principal range of the inverse tangent function $\tan^{-1}(u)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

For any value $\phi$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\tan^{-1}(\tan \phi) = \phi$.

Since our argument $\frac{\pi}{4} + \frac{x}{2}$ is in the interval $(0, \frac{\pi}{2})$, which is a subset of $(-\frac{\pi}{2}, \frac{\pi}{2})$, we can use this property directly:

$y = \frac{\pi}{4} + \frac{x}{2}$

Now, we differentiate this simplified expression with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + \frac{x}{2}\right)$

$= \frac{d}{dx}\left(\frac{\pi}{4}\right) + \frac{d}{dx}\left(\frac{x}{2}\right)$

The derivative of a constant ($\frac{\pi}{4}$) is $0$. The derivative of $\frac{x}{2} = \frac{1}{2}x$ with respect to $x$ is $\frac{1}{2}$.

$\frac{dy}{dx} = 0 + \frac{1}{2}$

$\frac{dy}{dx} = \frac{1}{2}$

Thus, the derivative of $\tan^{-1} (\sec x + \tan x)$ with respect to $x$ is $\frac{1}{2}$ for the given domain.

Given:

The function to differentiate is $f(x) = \tan^{-1} \left( \frac{a \cos x − b \sin x}{b \cos x + a \sin x} \right)$.

The domain for $x$ is given as $-\frac{\pi}{2} < x < \frac{\pi}{2}$, and the condition $\frac{a}{b} \tan x > -1$ is provided.

We assume $b \neq 0$, otherwise the term $\frac{a}{b}$ is undefined. If $b=0$, the function is $\tan^{-1}\left(\frac{a \cos x}{a \sin x}\right) = \tan^{-1}(\cot x)$ (assuming $a \neq 0$), which has derivative $-1$ for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $x \neq 0$. The condition $\frac{a}{b} \tan x > -1$ is not applicable in this case.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\tan^{-1} \left( \frac{a \cos x − b \sin x}{b \cos x + a \sin x} \right)\right)$.

Solution:

Let $y = \tan^{-1} \left( \frac{a \cos x − b \sin x}{b \cos x + a \sin x} \right)$.

We simplify the expression inside the inverse tangent function. Divide the numerator and the denominator by $b \cos x$ (assuming $b \cos x \neq 0$; since $-\frac{\pi}{2} < x < \frac{\pi}{2}$, $\cos x \neq 0$. Assuming $b \neq 0$):

$y = \tan^{-1} \left( \frac{\frac{a \cos x}{b \cos x} − \frac{b \sin x}{b \cos x}}{\frac{b \cos x}{b \cos x} + \frac{a \sin x}{b \cos x}} \right)$

$y = \tan^{-1} \left( \frac{\frac{a}{b} - \tan x}{1 + \frac{a}{b} \tan x} \right)$

Let $\tan \alpha = \frac{a}{b}$ for some angle $\alpha$. We can choose $\alpha$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

$y = \tan^{-1} \left( \frac{\tan \alpha - \tan x}{1 + \tan \alpha \tan x} \right)$

Using the tangent subtraction formula, $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$, with $A = \alpha$ and $B = x$, we get:

$y = \tan^{-1} (\tan (\alpha - x))$

The given condition is $\frac{a}{b} \tan x > -1$, which is $\tan \alpha \tan x > -1$. This inequality is equivalent to $1 + \tan \alpha \tan x > 0$.

The identity $\tan^{-1}(\tan \phi) = \phi$ holds if and only if $\phi \in (-\frac{\pi}{2}, \frac{\pi}{2})$. If $\phi$ is outside this interval, then $\tan^{-1}(\tan \phi) = \phi - n\pi$ for some integer $n$ such that $\phi - n\pi \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

The condition $1 + \tan \alpha \tan x > 0$ implies that $\tan(\alpha - x)$ has the same sign as $\alpha - x$ in the range of angles for which $\tan$ is defined. More precisely, the condition $1 + \tan \alpha \tan x > 0$ implies that $\alpha - x$ lies in an interval of the form $(k\pi - \frac{\pi}{2}, k\pi + \frac{\pi}{2})$ for some integer $k$.

Since the domain of $x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, the range of $\alpha - x$ is $(\alpha - \frac{\pi}{2}, \alpha + \frac{\pi}{2})$, which is an interval of length $\pi$. Because the condition $\tan \alpha \tan x > -1$ is satisfied on this domain, the argument $\alpha - x$ does not take values $(k + \frac{1}{2})\pi$ where $\tan(\alpha - x)$ is undefined. Consequently, the function $y = \tan^{-1}(\tan(\alpha - x))$ simplifies to a linear expression of the form $\alpha - x + n\pi$ for some fixed integer $n$ over the given domain of $x$.

$y = \alpha - x + n\pi$

Here, $\alpha = \tan^{-1}(a/b)$ is a constant, and $n$ is a constant integer determined by the range of $\alpha - x$ relative to the intervals $(k\pi - \frac{\pi}{2}, k\pi + \frac{\pi}{2})$.

Now, we differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(\alpha - x + n\pi)$

$= \frac{d}{dx}(\alpha) - \frac{d}{dx}(x) + \frac{d}{dx}(n\pi)$

$= 0 - 1 + 0$

$\frac{dy}{dx} = -1$

Thus, the derivative of $\tan^{-1} \left( \frac{a \cos x − b \sin x}{b \cos x + a \sin x} \right)$ with respect to $x$ is $-1$ for the given domain and condition.

Let the given function be $y$.

$y = \sec^{-1} \left( \frac{1}{4x^3 − 3x} \right)$

The domain is given as $0 < x <\frac{1}{\sqrt{2}}$.

We will differentiate $y$ with respect to $x$ using the chain rule.

The derivative of $\sec^{-1}(u)$ with respect to $x$ is given by $\frac{d}{dx}(\sec^{-1}u) = \frac{1}{|u|\sqrt{u^2 - 1}} \frac{du}{dx}$.

Here, let $u = \frac{1}{4x^3 - 3x} = (4x^3 - 3x)^{-1}$.

First, we find $\frac{du}{dx}$:

$\frac{du}{dx} = \frac{d}{dx}(4x^3 - 3x)^{-1}$

Using the chain rule $\frac{d}{dx}(v^n) = nv^{n-1}\frac{dv}{dx}$, where $v = 4x^3 - 3x$ and $n = -1$:

$\frac{du}{dx} = -1 \cdot (4x^3 - 3x)^{-2} \cdot \frac{d}{dx}(4x^3 - 3x)$

$= - (4x^3 - 3x)^{-2} \cdot (12x^2 - 3)$

$= - \frac{12x^2 - 3}{(4x^3 - 3x)^2}$

$= \frac{3 - 12x^2}{(4x^3 - 3x)^2}$

Now, we need to find $\sqrt{u^2 - 1}$.

$u^2 - 1 = \left(\frac{1}{4x^3 - 3x}\right)^2 - 1 = \frac{1}{(4x^3 - 3x)^2} - 1 = \frac{1 - (4x^3 - 3x)^2}{(4x^3 - 3x)^2}$

$\sqrt{u^2 - 1} = \sqrt{\frac{1 - (4x^3 - 3x)^2}{(4x^3 - 3x)^2}} = \frac{\sqrt{1 - (4x^3 - 3x)^2}}{\sqrt{(4x^3 - 3x)^2}} = \frac{\sqrt{1 - (4x^3 - 3x)^2}}{|4x^3 - 3x|}$

Also, we need $|u| = \left|\frac{1}{4x^3 - 3x}\right| = \frac{1}{|4x^3 - 3x|}$.

Now substitute $u$, $\sqrt{u^2 - 1}$, $|u|$ and $\frac{du}{dx}$ into the derivative formula for $\sec^{-1}(u)$:

$\frac{dy}{dx} = \frac{1}{|u|\sqrt{u^2 - 1}} \frac{du}{dx}$

$= \frac{1}{\frac{1}{|4x^3 - 3x|} \cdot \frac{\sqrt{1 - (4x^3 - 3x)^2}}{|4x^3 - 3x|}} \cdot \frac{3 - 12x^2}{(4x^3 - 3x)^2}$

$= \frac{|4x^3 - 3x|^2}{\sqrt{1 - (4x^3 - 3x)^2}} \cdot \frac{3 - 12x^2}{(4x^3 - 3x)^2}$

Since $|A|^2 = A^2$, the term $|4x^3 - 3x|^2$ is equal to $(4x^3 - 3x)^2$. Assuming $(4x^3 - 3x) \neq 0$ (which is true for $0 < x < \frac{1}{\sqrt{2}}$), we can cancel $(4x^3 - 3x)^2$ terms:

$\frac{dy}{dx} = \frac{3 - 12x^2}{\sqrt{1 - (4x^3 - 3x)^2}}$

Let's simplify the denominator using the substitution $x = \cos\theta$. Given $0 < x < \frac{1}{\sqrt{2}}$, we have $\frac{\pi}{4} < \theta < \frac{\pi}{2}$.

The term $(4x^3 - 3x)$ becomes $4\cos^3\theta - 3\cos\theta = \cos(3\theta)$.

So, $\sqrt{1 - (4x^3 - 3x)^2} = \sqrt{1 - \cos^2(3\theta)} = \sqrt{\sin^2(3\theta)} = |\sin(3\theta)|$.

Now we need to express $|\sin(3\theta)|$ in terms of $x$. We know $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$.

Since $x = \cos\theta$ and $\frac{\pi}{4} < \theta < \frac{\pi}{2}$, $\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - x^2}$. Note $\sin\theta > 0$ for this range of $\theta$.

$\sin(3\theta) = 3\sqrt{1 - x^2} - 4(\sqrt{1 - x^2})^3 = \sqrt{1 - x^2}(3 - 4(1 - x^2)) = \sqrt{1 - x^2}(3 - 4 + 4x^2) = \sqrt{1 - x^2}(4x^2 - 1)$.

$|\sin(3\theta)| = |\sqrt{1 - x^2}(4x^2 - 1)| = \sqrt{1 - x^2}|4x^2 - 1|$. (Since $\sqrt{1-x^2} > 0$).

The value of $|4x^2 - 1|$ depends on the range of $x^2$. Since $0 < x < \frac{1}{\sqrt{2}}$, we have $0 < x^2 < \frac{1}{2}$.

The term $4x^2 - 1$ is positive when $4x^2 > 1$, i.e., $x^2 > \frac{1}{4}$, or $x > \frac{1}{2}$ (since $x > 0$).

The term $4x^2 - 1$ is negative when $4x^2 < 1$, i.e., $x^2 < \frac{1}{4}$, or $x < \frac{1}{2}$ (since $x > 0$).

So we need to consider two cases for the given domain $0 < x < \frac{1}{\sqrt{2}}$:

Case 1: $0 < x < \frac{1}{2}$

In this case, $4x^2 - 1 < 0$. So $|4x^2 - 1| = -(4x^2 - 1) = 1 - 4x^2$. Note that $1 - 4x^2 > 0$ for this range.

$|\sin(3\theta)| = \sqrt{1 - x^2}(1 - 4x^2)$.

The derivative is $\frac{dy}{dx} = \frac{3 - 12x^2}{\sqrt{1 - x^2}(1 - 4x^2)} = \frac{3(1 - 4x^2)}{\sqrt{1 - x^2}(1 - 4x^2)}$.

Since $1 - 4x^2 \neq 0$ for $0 < x < \frac{1}{2}$, we can cancel the term $(1 - 4x^2)$.

$\frac{dy}{dx} = \frac{3}{\sqrt{1 - x^2}}$.

Case 2: $\frac{1}{2} < x < \frac{1}{\sqrt{2}}$

In this case, $4x^2 - 1 > 0$. So $|4x^2 - 1| = 4x^2 - 1$. Note that $4x^2 - 1 > 0$ for this range.

$|\sin(3\theta)| = \sqrt{1 - x^2}(4x^2 - 1)$.

The derivative is $\frac{dy}{dx} = \frac{3 - 12x^2}{\sqrt{1 - x^2}(4x^2 - 1)} = \frac{3(1 - 4x^2)}{\sqrt{1 - x^2}(4x^2 - 1)}$.

We can write $(1 - 4x^2) = -(4x^2 - 1)$.

$\frac{dy}{dx} = \frac{-3(4x^2 - 1)}{\sqrt{1 - x^2}(4x^2 - 1)}$.

Since $4x^2 - 1 \neq 0$ for $\frac{1}{2} < x < \frac{1}{\sqrt{2}}$, we can cancel the term $(4x^2 - 1)$.

$\frac{dy}{dx} = -\frac{3}{\sqrt{1 - x^2}}$.

The derivative of the function is piecewise over the given domain.

$\frac{d}{dx} \left( \sec^{-1} \left( \frac{1}{4x^3 − 3x} \right) \right) = \begin{cases} \frac{3}{\sqrt{1 - x^2}} & , & 0 < x < \frac{1}{2} \\ -\frac{3}{\sqrt{1 - x^2}} & , & \frac{1}{2} < x < \frac{1}{\sqrt{2}} \end{cases}$

Given:

The function to differentiate is $f(x) = \tan^{-1} \left( \frac{3a^2x − x^3}{a^3 − 3ax^2} \right)$.

The domain for $x$ is given by the condition $\frac{-1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$.

To Find:

The derivative of the function $f(x)$ with respect to $x$, i.e., $\frac{d}{dx}\left(\tan^{-1} \left( \frac{3a^2x − x^3}{a^3 − 3ax^2} \right)\right)$.

Solution:

Let $y = \tan^{-1} \left( \frac{3a^2x − x^3}{a^3 − 3ax^2} \right)$.

We simplify the expression inside the inverse tangent function by dividing the numerator and the denominator by $a^3$. We assume $a \neq 0$.

$\frac{3a^2x − x^3}{a^3 − 3ax^2} = \frac{\frac{3a^2x}{a^3} − \frac{x^3}{a^3}}{\frac{a^3}{a^3} − \frac{3ax^2}{a^3}} = \frac{3\frac{x}{a} − \left(\frac{x}{a}\right)^3}{1 − 3\left(\frac{x}{a}\right)^2}$

This expression is in the form of the tangent triple angle formula, $\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$.

Let $\frac{x}{a} = \tan \theta$. This implies $\theta = \tan^{-1}\left(\frac{x}{a}\right)$.

Substitute $\frac{x}{a} = \tan \theta$ into the simplified expression:

$\frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} = \tan(3\theta)$

So, the function becomes:

$y = \tan^{-1} (\tan (3\theta))$

Now, we need to consider the range of $3\theta$ based on the given domain $\frac{-1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$.

Since $\frac{x}{a} = \tan \theta$, the domain is $\frac{-1}{\sqrt{3}} < \tan \theta < \frac{1}{\sqrt{3}}$.

The values $\pm \frac{1}{\sqrt{3}}$ correspond to $\tan(\pm \frac{\pi}{6})$. Since the interval $(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ is within the range of $\tan \theta$ for $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, and $\tan \theta$ is strictly increasing in this interval, the inequality implies:

$\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) < \theta < \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$-\frac{\pi}{6} < \theta < \frac{\pi}{6}$

Now, find the range of $3\theta$ by multiplying the inequality by 3:

$3 \times \left(-\frac{\pi}{6}\right) < 3\theta < 3 \times \left(\frac{\pi}{6}\right)$

$-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$

The range of $3\theta$ is the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. This is the principal range of the inverse tangent function. For any value $\phi$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\tan^{-1}(\tan \phi) = \phi$.

Since $3\theta$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$, we can simplify the function:

$y = \tan^{-1} (\tan (3\theta)) = 3\theta$

Substitute back $\theta = \tan^{-1}\left(\frac{x}{a}\right)$:

$y = 3 \tan^{-1}\left(\frac{x}{a}\right)$

Now, differentiate $y$ with respect to $x$. We use the chain rule, where the derivative of $\tan^{-1}(u)$ is $\frac{1}{1+u^2} \frac{du}{dx}$, and $u = \frac{x}{a}$.

$\frac{dy}{dx} = \frac{d}{dx}\left(3 \tan^{-1}\left(\frac{x}{a}\right)\right) = 3 \frac{d}{dx}\left(\tan^{-1}\left(\frac{x}{a}\right)\right)$

$= 3 \cdot \frac{1}{1 + \left(\frac{x}{a}\right)^2} \cdot \frac{d}{dx}\left(\frac{x}{a}\right)$

$= 3 \cdot \frac{1}{1 + \frac{x^2}{a^2}} \cdot \frac{1}{a}$

$= 3 \cdot \frac{1}{\frac{a^2+x^2}{a^2}} \cdot \frac{1}{a}$

$= 3 \cdot \frac{a^2}{a^2+x^2} \cdot \frac{1}{a}$

$= 3 \cdot \frac{\cancel{a}^2}{a^2+x^2} \cdot \frac{1}{\cancel{a}}$

$= \frac{3a}{a^2+x^2}$

Thus, the derivative of $\tan^{-1} \left( \frac{3a^2x − x^3}{a^3 − 3ax^2} \right)$ with respect to $x$ is $\frac{3a}{a^2+x^2}$ for the given domain.

Let the given function be $y$.

$y = \tan^{-1} \left( \frac{\sqrt{1+x^2} + \sqrt{1−x^2}}{\sqrt{1+x^2} − \sqrt{1−x^2}} \right)$

The domain is given as $-1 < x < 1, x \neq 0$. This implies $0 < x^2 < 1$.

We use the substitution $x^2 = \cos(2\theta)$.

Since $0 < x^2 < 1$, we can choose the principal value of $2\theta$ such that $0 < 2\theta < \frac{\pi}{2}$.

This implies $0 < \theta < \frac{\pi}{4}$.

Substitute $x^2 = \cos(2\theta)$ into the expression inside the $\tan^{-1}$ function:

$\frac{\sqrt{1+x^2} + \sqrt{1−x^2}}{\sqrt{1+x^2} − \sqrt{1−x^2}} = \frac{\sqrt{1+\cos(2\theta)} + \sqrt{1−\cos(2\theta)}}{\sqrt{1+\cos(2\theta)} − \sqrt{1−\cos(2\theta)}}$

Using the trigonometric identities $1+\cos(2\theta) = 2\cos^2\theta$ and $1−\cos(2\theta) = 2\sin^2\theta$, we get:

$= \frac{\sqrt{2\cos^2\theta} + \sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta} − \sqrt{2\sin^2\theta}}$

$= \frac{\sqrt{2}|\cos\theta| + \sqrt{2}|\sin\theta|}{\sqrt{2}|\cos\theta| − \sqrt{2}|\sin\theta|}$

Since $0 < \theta < \frac{\pi}{4}$, both $\cos\theta$ and $\sin\theta$ are positive. So, $|\cos\theta| = \cos\theta$ and $|\sin\theta| = \sin\theta$.

$= \frac{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta − \sqrt{2}\sin\theta}$

Divide the numerator and the denominator by $\sqrt{2}\cos\theta$ (since $\cos\theta \neq 0$ for $0 < \theta < \frac{\pi}{4}$):

$= \frac{1 + \frac{\sin\theta}{\cos\theta}}{1 − \frac{\sin\theta}{\cos\theta}} = \frac{1 + \tan\theta}{1 − \tan\theta}$

Using the tangent addition formula, $\tan\left(\frac{\pi}{4} + \theta\right) = \frac{\tan(\pi/4) + \tan\theta}{1 - \tan(\pi/4)\tan\theta} = \frac{1 + \tan\theta}{1 - \tan\theta}$.

So the expression simplifies to $\tan\left(\frac{\pi}{4} + \theta\right)$.

Now, substitute this back into the function for $y$:

$y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \theta\right)\right)$

Since $0 < \theta < \frac{\pi}{4}$, the range of $\frac{\pi}{4} + \theta$ is $\frac{\pi}{4} < \frac{\pi}{4} + \theta < \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.

For any value $A$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\tan^{-1}(\tan A) = A$.

Since $\frac{\pi}{4} < \frac{\pi}{4} + \theta < \frac{\pi}{2}$ lies within the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have:

$y = \frac{\pi}{4} + \theta$ for the given domain.

From the substitution $x^2 = \cos(2\theta)$, we can express $\theta$ in terms of $x^2$.

$2\theta = \cos^{-1}(x^2)$

$\theta = \frac{1}{2}\cos^{-1}(x^2)$

Substitute this back into the expression for $y$:

$y = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}(x^2)$

This simplified form of $y$ is valid for the entire given domain $-1 < x < 1, x \neq 0$.

Now, differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}(x^2)\right)$

Using the linearity of differentiation, $\frac{d}{dx}(c_1f(x) + c_2g(x)) = c_1f'(x) + c_2g'(x)$:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4}\right) + \frac{1}{2}\frac{d}{dx}\left(\cos^{-1}(x^2)\right)$

The derivative of a constant is zero:

$\frac{d}{dx}\left(\frac{\pi}{4}\right) = 0$

To differentiate $\cos^{-1}(x^2)$, we use the chain rule. Let $u = x^2$. Then $\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x$.

The derivative of $\cos^{-1}(u)$ with respect to $u$ is $-\frac{1}{\sqrt{1-u^2}}$.

Using the chain rule, $\frac{d}{dx}(\cos^{-1}(x^2)) = \frac{d}{du}(\cos^{-1}u) \cdot \frac{du}{dx}$:

$= -\frac{1}{\sqrt{1-(x^2)^2}} \cdot (2x)$

$= -\frac{2x}{\sqrt{1-x^4}}$

Substitute this back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = 0 + \frac{1}{2} \cdot \left(-\frac{2x}{\sqrt{1-x^4}}\right)$

$\frac{dy}{dx} = -\frac{x}{\sqrt{1-x^4}}$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{\sqrt{1+x^2} + \sqrt{1−x^2}}{\sqrt{1+x^2} − \sqrt{1−x^2}} \right) \right) = -\frac{x}{\sqrt{1-x^4}}$

We are given the parametric equations:

$x = t + \frac{1}{t}$

$y = t − \frac{1}{t}$

To find $\frac{dy}{dx}$, we use the formula $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

First, find $\frac{dx}{dt}$:

$\frac{dx}{dt} = \frac{d}{dt}\left(t + \frac{1}{t}\right)$

$\frac{dx}{dt} = \frac{d}{dt}(t) + \frac{d}{dt}(t^{-1})$

Using the power rule $\frac{d}{dt}(t^n) = nt^{n-1}$:

$\frac{dx}{dt} = 1 + (-1)t^{-2} = 1 - \frac{1}{t^2}$

Combine the terms:

$\frac{dx}{dt} = \frac{t^2 - 1}{t^2}$

Next, find $\frac{dy}{dt}$:

$\frac{dy}{dt} = \frac{d}{dt}\left(t - \frac{1}{t}\right)$

$\frac{dy}{dt} = \frac{d}{dt}(t) - \frac{d}{dt}(t^{-1})$

Using the power rule:

$\frac{dy}{dt} = 1 - (-1)t^{-2} = 1 + \frac{1}{t^2}$

Combine the terms:

$\frac{dy}{dt} = \frac{t^2 + 1}{t^2}$

Now, calculate $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$:

$\frac{dy}{dx} = \frac{\frac{t^2 + 1}{t^2}}{\frac{t^2 - 1}{t^2}}$

For $t \neq 0$, the $t^2$ terms in the numerator and denominator cancel out:

$\frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1}$

This is the derivative in terms of the parameter $t$.

We can also express the result in terms of $x$ and $y$.

We have $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$.

Consider $x^2 = \left(t + \frac{1}{t}\right)^2 = t^2 + 2(t)\left(\frac{1}{t}\right) + \left(\frac{1}{t}\right)^2 = t^2 + 2 + \frac{1}{t^2}$.

$x^2 - 2 = t^2 + \frac{1}{t^2}$.

Consider $y^2 = \left(t - \frac{1}{t}\right)^2 = t^2 - 2(t)\left(\frac{1}{t}\right) + \left(\frac{1}{t}\right)^2 = t^2 - 2 + \frac{1}{t^2}$.

$y^2 + 2 = t^2 + \frac{1}{t^2}$.

From these, we have $t^2 + \frac{1}{t^2} = x^2 - 2 = y^2 + 2$. (This also shows $y^2 = x^2 - 4$, which is a hyperbola equation relating $x$ and $y$).

Also, we have $t^2 + \frac{1}{t^2} = \frac{t^4 + 1}{t^2}$.

From $x = t + \frac{1}{t}$, squaring both sides gives $x^2 = t^2 + 2 + \frac{1}{t^2}$.

From $y = t - \frac{1}{t}$, squaring both sides gives $y^2 = t^2 - 2 + \frac{1}{t^2}$.

Subtracting the second equation from the first: $x^2 - y^2 = (t^2 + 2 + \frac{1}{t^2}) - (t^2 - 2 + \frac{1}{t^2}) = 4$.

The required derivative is $\frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1}$.

We can express $t^2+1$ and $t^2-1$ in terms of $x$ and $y$.

Adding the original equations: $x+y = (t + \frac{1}{t}) + (t - \frac{1}{t}) = 2t$, so $t = \frac{x+y}{2}$.

Subtracting the second original equation from the first: $x-y = (t + \frac{1}{t}) - (t - \frac{1}{t}) = \frac{2}{t}$, so $t = \frac{2}{x-y}$.

Using $t = \frac{x+y}{2}$:

$t^2 = \left(\frac{x+y}{2}\right)^2 = \frac{(x+y)^2}{4}$.

$\frac{dy}{dx} = \frac{\frac{(x+y)^2}{4} + 1}{\frac{(x+y)^2}{4} - 1} = \frac{(x+y)^2 + 4}{(x+y)^2 - 4}$.

Alternatively, we know $x^2 - y^2 = 4$, so $(x-y)(x+y) = 4$. Since $t \neq \pm 1$ (as $\frac{dx}{dt} \neq 0$), $x = t+1/t \neq \pm 2$ and $y = t-1/t \neq 0$. Also, $x-y = 2/t \neq 0$.

From the original equations: $y = t - \frac{1}{t}$. We want to express $\frac{t^2+1}{t^2-1}$ in terms of $y$. This seems difficult directly.

Let's use the relation $x^2 - y^2 = 4$. Differentiating implicitly with respect to $x$:

$\frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}(4)$

$2x - 2y \frac{dy}{dx} = 0$

$2y \frac{dy}{dx} = 2x$

If $y \neq 0$, we can divide by $2y$:

$\frac{dy}{dx} = \frac{2x}{2y} = \frac{x}{y}$

This form is valid as long as $y \neq 0$. Note that $y=0$ when $t - 1/t = 0$, which means $t^2 = 1$, or $t = \pm 1$. However, $\frac{dx}{dt} = \frac{t^2-1}{t^2}$ is zero for $t = \pm 1$, so the derivative $\frac{dy}{dx}$ is undefined at $t=\pm 1$. Thus $y=0$ is excluded from the points where $\frac{dy}{dx}$ exists.

The derivative $\frac{dy}{dx}$ can be expressed in terms of $t$ or in terms of $x$ and $y$. Both are valid answers.

Using $t$:

$\frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1}$

Using $x$ and $y$:

$\frac{dy}{dx} = \frac{x}{y}$

We are given the parametric equations:

$x = e^\theta \left( \theta+\frac{1}{θ} \right)$

$y = e^{−θ} \left( θ−\frac{1}{θ} \right)$

To find $\frac{dy}{dx}$, we use the formula $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$.

First, find $\frac{dx}{d\theta}$. We use the product rule $\frac{d}{d\theta}(uv) = u'v + uv'$ with $u = e^\theta$ and $v = \theta+\frac{1}{θ} = \theta + \theta^{-1}$.

$\frac{du}{d\theta} = \frac{d}{d\theta}(e^\theta) = e^\theta$

$\frac{dv}{d\theta} = \frac{d}{d\theta}(\theta + \theta^{-1}) = 1 + (-1)\theta^{-2} = 1 - \frac{1}{\theta^2}$

So,

$\frac{dx}{d\theta} = \frac{d}{d\theta}(e^\theta) \left( \theta+\frac{1}{θ} \right) + e^\theta \frac{d}{d\theta}\left( \theta+\frac{1}{θ} \right)$

$\frac{dx}{d\theta} = e^\theta \left( \theta+\frac{1}{θ} \right) + e^\theta \left( 1 - \frac{1}{\theta^2} \right)$

Factor out $e^\theta$:

$\frac{dx}{d\theta} = e^\theta \left( \theta + \frac{1}{\theta} + 1 - \frac{1}{\theta^2} \right)$

Next, find $\frac{dy}{d\theta}$. We use the product rule $\frac{d}{d\theta}(uv) = u'v + uv'$ with $u = e^{-\theta}$ and $v = \theta−\frac{1}{θ} = \theta - \theta^{-1}$.

$\frac{du}{d\theta} = \frac{d}{d\theta}(e^{-\theta}) = -e^{-\theta}$

$\frac{dv}{d\theta} = \frac{d}{d\theta}(\theta - \theta^{-1}) = 1 - (-1)\theta^{-2} = 1 + \frac{1}{\theta^2}$

So,

$\frac{dy}{d\theta} = \frac{d}{d\theta}(e^{-\theta}) \left( θ−\frac{1}{θ} \right) + e^{-\theta} \frac{d}{d\theta}\left( θ−\frac{1}{θ} \right)$

$\frac{dy}{d\theta} = -e^{-\theta} \left( θ−\frac{1}{θ} \right) + e^{-\theta} \left( 1 + \frac{1}{\theta^2} \right)$

Factor out $e^{-\theta}$:

$\frac{dy}{d\theta} = e^{-\theta} \left( - \left( θ−\frac{1}{θ} \right) + \left( 1 + \frac{1}{\theta^2} \right) \right)$

$\frac{dy}{d\theta} = e^{-\theta} \left( -\theta + \frac{1}{\theta} + 1 + \frac{1}{\theta^2} \right)$

Now, calculate $\frac{dy}{dx}$ by dividing $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$:

$\frac{dy}{dx} = \frac{e^{-\theta} \left( -\theta + \frac{1}{\theta} + 1 + \frac{1}{\theta^2} \right)}{e^\theta \left( \theta + \frac{1}{\theta} + 1 - \frac{1}{\theta^2} \right)}$

$\frac{dy}{dx} = e^{-\theta} \cdot e^{-\theta} \cdot \frac{1 + \frac{1}{\theta} - \theta + \frac{1}{\theta^2}}{1 + \frac{1}{\theta} + \theta - \frac{1}{\theta^2}}$

$\frac{dy}{dx} = e^{-2\theta} \frac{1 + \frac{1}{\theta} - \theta + \frac{1}{\theta^2}}{1 + \frac{1}{\theta} + \theta - \frac{1}{\theta^2}}$

To simplify the fraction part, we can multiply the numerator and the denominator by $\theta^2$ (assuming $\theta \neq 0$):

$\frac{1 + \frac{1}{\theta} - \theta + \frac{1}{\theta^2}}{1 + \frac{1}{\theta} + \theta - \frac{1}{\theta^2}} = \frac{\theta^2 \left( 1 + \frac{1}{\theta} - \theta + \frac{1}{\theta^2} \right)}{\theta^2 \left( 1 + \frac{1}{\theta} + \theta - \frac{1}{\theta^2} \right)} = \frac{\theta^2 + \theta - \theta^3 + 1}{\theta^2 + \theta + \theta^3 - 1}$

Rearranging the terms in the numerator and denominator polynomials:

Numerator: $-\theta^3 + \theta^2 + \theta + 1$

Denominator: $\theta^3 + \theta^2 + \theta - 1$ (assuming $\theta^3$ term is positive in the denominator as written)

So the fraction is $\frac{-\theta^3 + \theta^2 + \theta + 1}{\theta^3 + \theta^2 + \theta - 1}$.

Thus, the derivative $\frac{dy}{dx}$ in terms of $\theta$ is:

$\frac{dy}{dx} = e^{-2\theta} \left( \frac{1 + \frac{1}{\theta} - \theta + \frac{1}{\theta^2}}{1 + \frac{1}{\theta} + \theta - \frac{1}{\theta^2}} \right)$

Or, equivalently:

$\frac{dy}{dx} = e^{-2\theta} \left( \frac{-\theta^3 + \theta^2 + \theta + 1}{\theta^3 + \theta^2 + \theta - 1} \right)$

We are given the parametric equations:

$x = 3\cos θ – 2\cos^3 θ$

$y = 3\sin θ – 2\sin^3 θ$

To find $\frac{dy}{dx}$, we use the formula $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$.

First, find $\frac{dx}{d\theta}$ by differentiating $x$ with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}(3\cos θ – 2\cos^3 θ)$

$\frac{dx}{d\theta} = 3\frac{d}{d\theta}(\cos θ) – 2\frac{d}{d\theta}(\cos^3 θ)$

$\frac{dx}{d\theta} = 3(-\sin θ) – 2(3\cos^2 θ) \cdot \frac{d}{d\theta}(\cos θ)$ (using the chain rule for $\cos^3 θ$)

$\frac{dx}{d\theta} = -3\sin θ – 6\cos^2 θ (-\sin θ)$

$\frac{dx}{d\theta} = -3\sin θ + 6\cos^2 θ \sin θ$

Factor out $3\sin θ$:

$\frac{dx}{d\theta} = 3\sin θ (2\cos^2 θ - 1)$

Using the double angle identity $\cos(2\theta) = 2\cos^2\theta - 1$:

$\frac{dx}{d\theta} = 3\sin θ \cos(2θ)$

Next, find $\frac{dy}{d\theta}$ by differentiating $y$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta}(3\sin θ – 2\sin^3 θ)$

$\frac{dy}{d\theta} = 3\frac{d}{d\theta}(\sin θ) – 2\frac{d}{d\theta}(\sin^3 θ)$

$\frac{dy}{d\theta} = 3(\cos θ) – 2(3\sin^2 θ) \cdot \frac{d}{d\theta}(\sin θ)$ (using the chain rule for $\sin^3 θ$)

$\frac{dy}{d\theta} = 3\cos θ – 6\sin^2 θ (\cos θ)$

Factor out $3\cos θ$:

$\frac{dy}{d\theta} = 3\cos θ (1 - 2\sin^2 θ)$

Using the double angle identity $\cos(2\theta) = 1 - 2\sin^2\theta$:

$\frac{dy}{d\theta} = 3\cos θ \cos(2θ)$

Now, calculate $\frac{dy}{dx}$ by dividing $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$:

$\frac{dy}{dx} = \frac{3\cos θ \cos(2θ)}{3\sin θ \cos(2θ)}$

Assuming $3\sin θ \cos(2θ) \neq 0$, we can cancel the common terms $3$ and $\cos(2θ)$:

$\frac{dy}{dx} = \frac{\cos θ}{\sin θ}$

Using the identity $\cot θ = \frac{\cos θ}{\sin θ}$:

$\frac{dy}{dx} = \cot θ$

Thus, the derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \cot θ$

Given:

$x$ is defined by the equation $\sin x = \frac{2t}{1 + t^2}$

$y$ is defined by the equation $\tan y = \frac{2t}{1−t^2}$

To Find:

$\frac{dy}{dx}$

Solution:

We are given the equations implicitly defining $x$ and $y$ in terms of the parameter $t$.

$\sin x = \frac{2t}{1 + t^2}$

$\tan y = \frac{2t}{1−t^2}$

To find $\frac{dy}{dx}$, we will first compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ by differentiating the given equations with respect to $t$. Then we use the formula $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

Consider the first equation: $\sin x = \frac{2t}{1 + t^2}$.

Differentiating both sides with respect to $t$:

$\frac{d}{dt}(\sin x) = \frac{d}{dt}\left(\frac{2t}{1 + t^2}\right)$

Using the chain rule on the left side, $\frac{d}{dt}(\sin x) = \cos x \frac{dx}{dt}$.

Using the quotient rule on the right side, $\frac{d}{dt}\left(\frac{2t}{1 + t^2}\right) = \frac{(1+t^2) \frac{d}{dt}(2t) - (2t) \frac{d}{dt}(1+t^2)}{(1+t^2)^2}$

$= \frac{(1+t^2)(2) - (2t)(2t)}{(1+t^2)^2} = \frac{2 + 2t^2 - 4t^2}{(1+t^2)^2} = \frac{2 - 2t^2}{(1+t^2)^2} = \frac{2(1-t^2)}{(1+t^2)^2}$

Equating the derivatives:

$\cos x \frac{dx}{dt} = \frac{2(1-t^2)}{(1+t^2)^2}$

Assuming $\cos x \neq 0$, we get:

$\frac{dx}{dt} = \frac{2(1-t^2)}{(1+t^2)^2 \cos x}$

Consider the second equation: $\tan y = \frac{2t}{1−t^2}$.

Differentiating both sides with respect to $t$:

$\frac{d}{dt}(\tan y) = \frac{d}{dt}\left(\frac{2t}{1−t^2}\right)$

Using the chain rule on the left side, $\frac{d}{dt}(\tan y) = \sec^2 y \frac{dy}{dt}$.

Using the quotient rule on the right side, $\frac{d}{dt}\left(\frac{2t}{1−t^2}\right) = \frac{(1-t^2) \frac{d}{dt}(2t) - (2t) \frac{d}{dt}(1-t^2)}{(1-t^2)^2}$

$= \frac{(1-t^2)(2) - (2t)(-2t)}{(1-t^2)^2} = \frac{2 - 2t^2 + 4t^2}{(1-t^2)^2} = \frac{2 + 2t^2}{(1-t^2)^2} = \frac{2(1+t^2)}{(1-t^2)^2}$

Equating the derivatives:

$\sec^2 y \frac{dy}{dt} = \frac{2(1+t^2)}{(1-t^2)^2}$

Assuming $\sec^2 y \neq 0$ (which is true for real $y$), we get:

$\frac{dy}{dt} = \frac{2(1+t^2)}{(1-t^2)^2 \sec^2 y}$

Now, we compute $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. Assuming $\frac{dx}{dt} \neq 0$ (i.e., $t \neq \pm 1$ and $\cos x \neq 0$):

$\frac{dy}{dx} = \frac{\frac{2(1+t^2)}{(1-t^2)^2 \sec^2 y}}{\frac{2(1-t^2)}{(1+t^2)^2 \cos x}}$

$\frac{dy}{dx} = \frac{2(1+t^2)}{(1-t^2)^2 \sec^2 y} \cdot \frac{(1+t^2)^2 \cos x}{2(1-t^2)}$

$\frac{dy}{dx} = \frac{(1+t^2)^3}{(1-t^2)^3} \frac{\cos x}{\sec^2 y}$

Alternatively, we can express $x$ and $y$ directly in terms of $t$ using inverse trigonometric functions.

From $\sin x = \frac{2t}{1 + t^2}$, we have $x = \sin^{-1}\left(\frac{2t}{1+t^2}\right)$.

Differentiate $x$ with respect to $t$ using the chain rule $\frac{d}{dt}(\sin^{-1}u) = \frac{1}{\sqrt{1-u^2}}\frac{du}{dt}$, where $u = \frac{2t}{1+t^2}$.

$\frac{du}{dt} = \frac{d}{dt}\left(\frac{2t}{1+t^2}\right) = \frac{2(1-t^2)}{(1+t^2)^2}$.

$\sqrt{1-u^2} = \sqrt{1 - \left(\frac{2t}{1+t^2}\right)^2} = \sqrt{\frac{(1+t^2)^2 - (2t)^2}{(1+t^2)^2}} = \sqrt{\frac{1-2t^2+t^4}{(1+t^2)^2}} = \sqrt{\frac{(1-t^2)^2}{(1+t^2)^2}} = \frac{|1-t^2|}{1+t^2}$.

So, $\frac{dx}{dt} = \frac{1}{\frac{|1-t^2|}{1+t^2}} \cdot \frac{2(1-t^2)}{(1+t^2)^2} = \frac{1+t^2}{|1-t^2|} \cdot \frac{2(1-t^2)}{(1+t^2)^2} = \frac{2(1-t^2)}{(1+t^2)|1-t^2|}$.

From $\tan y = \frac{2t}{1−t^2}$, we have $y = \tan^{-1}\left(\frac{2t}{1−t^2}\right)$.

Differentiate $y$ with respect to $t$ using the chain rule $\frac{d}{dt}(\tan^{-1}v) = \frac{1}{1+v^2}\frac{dv}{dt}$, where $v = \frac{2t}{1−t^2}$.

$\frac{dv}{dt} = \frac{d}{dt}\left(\frac{2t}{1−t^2}\right) = \frac{2(1+t^2)}{(1-t^2)^2}$.

$1+v^2 = 1 + \left(\frac{2t}{1-t^2}\right)^2 = \frac{(1-t^2)^2 + 4t^2}{(1-t^2)^2} = \frac{(1+t^2)^2}{(1-t^2)^2}$.

So, $\frac{dy}{dt} = \frac{1}{\frac{(1+t^2)^2}{(1-t^2)^2}} \cdot \frac{2(1+t^2)}{(1-t^2)^2} = \frac{(1-t^2)^2}{(1+t^2)^2} \cdot \frac{2(1+t^2)}{(1-t^2)^2} = \frac{2}{1+t^2}$.

Now, calculate $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, assuming $\frac{dx}{dt} \neq 0$ (i.e., $t \neq \pm 1$):

$\frac{dy}{dx} = \frac{\frac{2}{1+t^2}}{\frac{2(1-t^2)}{(1+t^2)|1-t^2|}}$

$\frac{dy}{dx} = \frac{2}{1+t^2} \cdot \frac{(1+t^2)|1-t^2|}{2(1-t^2)}$

Assuming $1+t^2 \neq 0$ (always true for real $t$) and $1-t^2 \neq 0$ (i.e., $t \neq \pm 1$):

$\frac{dy}{dx} = \frac{|1-t^2|}{1-t^2}$

The expression $\frac{|A|}{A}$ is equal to $1$ if $A > 0$ and $-1$ if $A < 0$.

Here, $A = 1-t^2$.

If $1-t^2 > 0$, which means $t^2 < 1$ or $-1 < t < 1$, then $|1-t^2| = 1-t^2$.

$\frac{dy}{dx} = \frac{1-t^2}{1-t^2} = 1$ for $-1 < t < 1$.

If $1-t^2 < 0$, which means $t^2 > 1$ or $t < -1$ or $t > 1$, then $|1-t^2| = -(1-t^2) = t^2-1$.

$\frac{dy}{dx} = \frac{-(1-t^2)}{1-t^2} = -1$ for $t < -1$ or $t > 1$.

The derivative $\frac{dy}{dx}$ is undefined at $t = \pm 1$ because $\frac{dx}{dt} = 0$ at these values.

Thus, the derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \begin{cases} 1 & , |t| < 1 \\ -1 & , |t| > 1 \end{cases}$

This can also be written as $\frac{dy}{dx} = \frac{1-t^2}{|1-t^2|}$ for $t \neq \pm 1$.

Given:

$x = \frac{1+ \log t}{t^2}$

$y = \frac{3 + 2 \log t}{t}$

To Find:

$\frac{dy}{dx}$

Solution:

We are given $x$ and $y$ in terms of the parameter $t$. To find $\frac{dy}{dx}$, we will first find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ and then use the formula $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

Note that the presence of $\log t$ implies $t > 0$.

First, differentiate $x$ with respect to $t$ using the quotient rule $\frac{d}{dt}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}$:

$x = \frac{1+ \log t}{t^2}$

Let $u = 1 + \log t$ and $v = t^2$.

Then $\frac{du}{dt} = \frac{d}{dt}(1 + \log t) = \frac{1}{t}$ and $\frac{dv}{dt} = \frac{d}{dt}(t^2) = 2t$.

$\frac{dx}{dt} = \frac{(\frac{1}{t}) \cdot t^2 - (1 + \log t) \cdot (2t)}{(t^2)^2}$

$\frac{dx}{dt} = \frac{t - 2t(1 + \log t)}{t^4}$

Factor out $t$ from the numerator:

$\frac{dx}{dt} = \frac{t(1 - 2(1 + \log t))}{t^4}$

Simplify the numerator and cancel $t$ (since $t>0$, $t \neq 0$):

$\frac{dx}{dt} = \frac{1 - 2 - 2\log t}{t^3}$

$\frac{dx}{dt} = \frac{-1 - 2\log t}{t^3}$

$\frac{dx}{dt} = \frac{-(1 + 2\log t)}{t^3}$

Next, differentiate $y$ with respect to $t$ using the quotient rule:

$y = \frac{3 + 2 \log t}{t}$

Let $u = 3 + 2 \log t$ and $v = t$.

Then $\frac{du}{dt} = \frac{d}{dt}(3 + 2 \log t) = 2 \cdot \frac{1}{t} = \frac{2}{t}$ and $\frac{dv}{dt} = \frac{d}{dt}(t) = 1$.

$\frac{dy}{dt} = \frac{(\frac{2}{t}) \cdot t - (3 + 2 \log t) \cdot (1)}{t^2}$

$\frac{dy}{dt} = \frac{2 - (3 + 2 \log t)}{t^2}$

$\frac{dy}{dt} = \frac{2 - 3 - 2 \log t}{t^2}$

$\frac{dy}{dt} = \frac{-1 - 2 \log t}{t^2}$

Factor out $-1$ from the numerator:

$\frac{dy}{dt} = \frac{-(1 + 2 \log t)}{t^2}$

Now, find $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$, assuming $\frac{dx}{dt} \neq 0$ (i.e., $t > 0$ and $1 + 2\log t \neq 0$):

$\frac{dy}{dx} = \frac{\frac{-(1 + 2 \log t)}{t^2}}{\frac{-(1 + 2 \log t)}{t^3}}$

$\frac{dy}{dx} = \frac{-(1 + 2 \log t)}{t^2} \times \frac{t^3}{-(1 + 2 \log t)}$

Assuming $1 + 2 \log t \neq 0$, cancel the common terms $-(1 + 2 \log t)$. Also, simplify the powers of $t$ (assuming $t \neq 0$):

$\frac{dy}{dx} = \frac{t^3}{t^2}$

$\frac{dy}{dx} = t^{3-2} = t$

Thus, the derivative of the function is:

$\frac{dy}{dx} = t$

Given:

$x = e^{\cos 2t}$

$y = e^{\sin 2t}$

To Prove:

$\frac{dy}{dx} = \frac{−y \log x}{x \log y}$

Solution:

We have the parametric equations:

To find $\frac{dy}{dx}$, we first find $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

Differentiating equation (1) with respect to $t$:

Using the chain rule, $\frac{d}{du}(e^u) = e^u$ and $\frac{d}{dt}(\cos 2t) = -2\sin 2t$, we get:

Differentiating equation (2) with respect to $t$:

Using the chain rule, $\frac{d}{du}(e^u) = e^u$ and $\frac{d}{dt}(\sin 2t) = 2\cos 2t$, we get:

Now, we find $\frac{dy}{dx}$ using the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

From the given equations (1) and (2), taking logarithm (base $e$) on both sides:

From equation (1): $x = e^{\cos 2t}$

Using the property $\log(a^b) = b \log a$ and $\log e = 1$:

From equation (2): $y = e^{\sin 2t}$

Using the property $\log(a^b) = b \log a$ and $\log e = 1$:

Now, substitute equations (6) and (7) into equation (5). Also, note from (1) that $e^{\cos 2t} = x$ and from (2) that $e^{\sin 2t} = y$.

Rearranging the terms, we get:

This matches the expression required to be proved.

Hence Proved.

Given:

$x = a \sin 2t\; (1 + \cos 2t)$

$y = b \cos 2t\; (1 – \cos 2t)$

To Prove:

$\left( \frac{dy}{dx} \right)_{at \;t = \frac{π}{4}} = \frac{b}{a}$

Solution:

We have the parametric equations:

$x = a \sin 2t\; (1 + \cos 2t) = a (\sin 2t + \sin 2t \cos 2t)$

Using the identity $\sin 2A = 2 \sin A \cos A$, so $\sin A \cos A = \frac{1}{2} \sin 2A$, we have $\sin 2t \cos 2t = \frac{1}{2} \sin (2 \cdot 2t) = \frac{1}{2} \sin 4t$.

So, $x = a \left(\sin 2t + \frac{1}{2} \sin 4t\right)$

Differentiating $x$ with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt} \left[a \left(\sin 2t + \frac{1}{2} \sin 4t\right)\right]$

$\frac{dx}{dt} = a \left[\frac{d}{dt}(\sin 2t) + \frac{1}{2} \frac{d}{dt}(\sin 4t)\right]$

$\frac{dx}{dt} = a \left[(\cos 2t \cdot 2) + \frac{1}{2} (\cos 4t \cdot 4)\right]$

$\frac{dx}{dt} = a [2 \cos 2t + 2 \cos 4t]$

$\frac{dx}{dt} = 2a (\cos 2t + \cos 4t)$

Next, consider $y = b \cos 2t\; (1 – \cos 2t) = b (\cos 2t - \cos^2 2t)$

Differentiating $y$ with respect to $t$:

$\frac{dy}{dt} = \frac{d}{dt} [b (\cos 2t - \cos^2 2t)]$

$\frac{dy}{dt} = b \left[\frac{d}{dt}(\cos 2t) - \frac{d}{dt}(\cos^2 2t)\right]$

$\frac{dy}{dt} = b \left[(-\sin 2t \cdot 2) - (2 \cos 2t \cdot (-\sin 2t) \cdot 2)\right]$

$\frac{dy}{dt} = b [-2 \sin 2t + 4 \sin 2t \cos 2t]$

Using the identity $\sin 2A = 2 \sin A \cos A$, we have $4 \sin 2t \cos 2t = 2 (2 \sin 2t \cos 2t) = 2 \sin (2 \cdot 2t) = 2 \sin 4t$.

So, $\frac{dy}{dt} = b [-2 \sin 2t + 2 \sin 4t]$

$\frac{dy}{dt} = 2b (\sin 4t - \sin 2t)$

Now, we find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

$\frac{dy}{dx} = \frac{2b (\sin 4t - \sin 2t)}{2a (\cos 2t + \cos 4t)}$

$\frac{dy}{dx} = \frac{b (\sin 4t - \sin 2t)}{a (\cos 4t + \cos 2t)}$

Using the sum-to-product trigonometric identities:

$\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$

$\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$

Here, $A = 4t$ and $B = 2t$.

$\frac{A+B}{2} = \frac{4t+2t}{2} = 3t$

$\frac{A-B}{2} = \frac{4t-2t}{2} = t$

So, $\sin 4t - \sin 2t = 2 \cos 3t \sin t$

And $\cos 4t + \cos 2t = 2 \cos 3t \cos t$

Substitute these into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{b (2 \cos 3t \sin t)}{a (2 \cos 3t \cos t)}$

Assuming $\cos 3t \neq 0$, we can cancel $2 \cos 3t$ from numerator and denominator.

$\frac{dy}{dx} = \frac{b \sin t}{a \cos t}$

$\frac{dy}{dx} = \frac{b}{a} \tan t$

Now, we need to evaluate $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$:

$\left( \frac{dy}{dx} \right)_{at \;t = \frac{π}{4}} = \frac{b}{a} \tan \left(\frac{\pi}{4}\right)$

We know that $\tan \left(\frac{\pi}{4}\right) = 1$.

Therefore,

$\left( \frac{dy}{dx} \right)_{at \;t = \frac{π}{4}} = \frac{b}{a} \cdot 1 = \frac{b}{a}$

This is the required result.

Hence Shown.

Given:

$x = 3 \sin t – \sin 3t$

$y = 3\cos t – \cos 3t$

To Find:

$\frac{dy}{dx}$ at $t = \frac{\pi}{3}$

Solution:

We have the parametric equations for $x$ and $y$ in terms of $t$:

To find $\frac{dy}{dx}$ for parametric equations, we use the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

First, differentiate equation (1) with respect to $t$ to find $\frac{dx}{dt}$:

Next, differentiate equation (2) with respect to $t$ to find $\frac{dy}{dt}$:

Now, calculate $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ (equation 4) by $\frac{dx}{dt}$ (equation 3):

Using the sum-to-product trigonometric identities:

$\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

$\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

Apply these identities with $A=3t$ and $B=t$ for the numerator:

$\sin 3t - \sin t = 2 \cos \left(\frac{3t+t}{2}\right) \sin \left(\frac{3t-t}{2}\right) = 2 \cos(2t) \sin t$

Apply these identities with $A=t$ and $B=3t$ for the denominator:

$\cos t - \cos 3t = -2 \sin \left(\frac{t+3t}{2}\right) \sin \left(\frac{t-3t}{2}\right) = -2 \sin(2t) \sin(-t)$

Since $\sin(-t) = -\sin t$, the denominator becomes:

$\cos t - \cos 3t = -2 \sin(2t) (-\sin t) = 2 \sin(2t) \sin t$

Substitute these simplified expressions back into the formula for $\frac{dy}{dx}$ (assuming $\sin t \neq 0$):

Now, we need to find the value of $\frac{dy}{dx}$ at $t = \frac{\pi}{3}$. Substitute $t = \frac{\pi}{3}$ into equation (5):

The angle $\frac{2\pi}{3}$ is in the second quadrant, where the cotangent function is negative. The reference angle is $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$.

So, $\cot\left(\frac{2\pi}{3}\right) = -\cot\left(\frac{\pi}{3}\right)$.

We know that $\cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$.

Therefore,

The value of $\frac{dy}{dx}$ at $t = \frac{\pi}{3}$ is $-\frac{1}{\sqrt{3}}$.

Final Answer:

$\left.\frac{dy}{dx}\right|_{t=\frac{\pi}{3}} = -\frac{1}{\sqrt{3}}$

Given:

The expression to be differentiated is $\frac{x}{\sin x}$.

The variable with respect to which we need to differentiate is $\sin x$.

To Find:

The derivative of $\frac{x}{\sin x}$ with respect to $\sin x$.

Solution:

Let $u = \frac{x}{\sin x}$ and $v = \sin x$.

We are asked to find $\frac{du}{dv}$.

We can use the chain rule for differentiation:

First, we find $\frac{du}{dx}$ using the quotient rule $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

Here, $f(x) = x$ and $g(x) = \sin x$.

$f'(x) = \frac{d}{dx}(x) = 1$

$g'(x) = \frac{d}{dx}(\sin x) = \cos x$

Next, we find $\frac{dv}{dx}$.

Now, substitute equations (2) and (3) into equation (1):

This can also be written as:

Thus, the derivative of $\frac{x}{\sin x}$ with respect to $\sin x$ is $\frac{\sin x - x \cos x}{\sin^2 x \cos x}$ or $\text{cosec} x \sec x - x \text{cosec}^2 x$.

Final Answer:

The derivative of $\frac{x}{\sin x}$ with respect to $\sin x$ is $\frac{\sin x - x \cos x}{\sin^2 x \cos x}$.

Given:

The first function is $y = \tan^{-1} \left( \frac{\sqrt{1+ x^2} − 1}{x} \right)$.

The second function is $z = \tan^{-1} x$.

We are given that $x \neq 0$.

To Find:

The derivative of $y$ with respect to $z$, i.e., $\frac{dy}{dz}$.

Solution:

We are asked to find the derivative of $y$ with respect to $z$. We can use the formula:

First, let's simplify the expression for $y$. Let $x = \tan \theta$, where $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ and $\theta \neq 0$ (since $x \neq 0$).

Then $\sqrt{1+x^2} = \sqrt{1+\tan^2 \theta} = \sqrt{\sec^2 \theta}$.

Since $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $\cos \theta > 0$, so $\sec \theta > 0$. Thus, $\sqrt{\sec^2 \theta} = \sec \theta$.

Substitute $x = \tan \theta$ into the expression for $y$:

Rewrite $\sec \theta$ and $\tan \theta$ in terms of $\sin \theta$ and $\cos \theta$:

Use the half-angle trigonometric identities: $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$ and $\sin \theta = 2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)$.

Since $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ and $\theta \neq 0$, we have $\frac{\theta}{2} \in (-\frac{\pi}{4}, \frac{\pi}{4})$ and $\frac{\theta}{2} \neq 0$.

In the interval $(-\frac{\pi}{4}, \frac{\pi}{4})$, $\tan^{-1}(\tan u) = u$.

So, $y = \frac{\theta}{2}$.

Substitute back $\theta = \tan^{-1} x$:

Now, we differentiate $y$ with respect to $x$:

Next, we differentiate the second function $z$ with respect to $x$:

Finally, use equation (1) and substitute equations (3) and (4):

Assuming $1+x^2 \neq 0$, which is true for all real $x$, we can cancel the common term:

Thus, the derivative of $\tan^{-1} \left( \frac{\sqrt{1+ x^2} − 1}{x} \right)$ with respect to $\tan^{-1} x$ is $\frac{1}{2}$.

Final Answer:

The derivative is $\frac{1}{2}$.

Given:

The relation between $x$ and $y$ is $\sin (xy) + \frac{x}{y} = x^2 – y$.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We are given the implicit relation:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation (1) with respect to $x$. Remember to treat $y$ as a function of $x$ and use the chain rule where necessary.

Differentiate each term:

For $\frac{d}{dx}(\sin (xy))$: Using the chain rule with $u = xy$, we have $\frac{d}{dx}(\sin u) = \cos u \frac{du}{dx}$.

And $\frac{d}{dx}(xy) = \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y)$ (Product rule)

So, $\frac{d}{dx}(\sin (xy)) = \cos (xy) \left(y + x\frac{dy}{dx}\right) = y \cos (xy) + x \cos (xy)\frac{dy}{dx}$.

For $\frac{d}{dx}\left(\frac{x}{y}\right)$: Using the quotient rule $\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'\cdot g - f \cdot g'}{g^2}$, where $f=x$ and $g=y$.

For $\frac{d}{dx}(x^2)$: Using the power rule.

For $\frac{d}{dx}(y)$: This is simply $\frac{dy}{dx}$.

Substitute these derivatives back into the differentiated equation:

Now, rearrange the equation to collect all terms involving $\frac{dy}{dx}$ on one side and the other terms on the other side:

Factor out $\frac{dy}{dx}$ from the terms on the left side:

Find a common denominator for the terms inside the parenthesis on the left side and the terms on the right side:

Finally, solve for $\frac{dy}{dx}$ by dividing both sides by the term in the parenthesis:

Invert the denominator fraction and multiply:

This is the expression for $\frac{dy}{dx}$.

Final Answer:

$\frac{dy}{dx} = \frac{y(2xy - y^2 \cos (xy) - 1)}{xy^2 \cos (xy) - x + y^2}$

Given:

The relation between $x$ and $y$ is $\sec (x + y) = xy$.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We are given the implicit relation:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation (1) with respect to $x$. We treat $y$ as a function of $x$ and use the chain rule and product rule.

Differentiate the left side using the chain rule:

So, the derivative of the left side is $\sec (x+y) \tan (x+y) \left(1 + \frac{dy}{dx}\right)$.

Differentiate the right side using the product rule:

Now, equate the derivatives of both sides:

Distribute the term on the left side:

Rearrange the equation to bring terms with $\frac{dy}{dx}$ to one side and other terms to the other side:

Factor out $\frac{dy}{dx}$ from the left side:

Solve for $\frac{dy}{dx}$ by dividing both sides by the term in the parenthesis:

We can also express the result by multiplying the numerator and denominator by -1:

Using the original relation $\sec(x+y) = xy$, we can substitute this into the expression for $\frac{dy}{dx}$ if desired:

Both forms (2) and (3) are valid expressions for $\frac{dy}{dx}$.

Final Answer:

$\frac{dy}{dx} = \frac{y - \sec (x+y) \tan (x+y)}{\sec (x+y) \tan (x+y) - x}$

or

$\frac{dy}{dx} = \frac{y(1 - x \tan(x+y))}{x(y \tan(x+y) - 1)}$

Given:

The relation between $x$ and $y$ is $\tan^{–1} (x^2 + y^2) = a$, where $a$ is a constant.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We are given the implicit relation:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation (1) with respect to $x$. We treat $y$ as a function of $x$ and use the chain rule.

Differentiate the left side using the chain rule. The derivative of $\tan^{-1}(u)$ with respect to $x$ is $\frac{1}{1+u^2} \frac{du}{dx}$. Here, $u = x^2 + y^2$.

So, the derivative of the left side is:

Differentiate the right side. Since $a$ is a constant, its derivative with respect to $x$ is 0.

Equate the derivatives of both sides:

Since $1+(x^2+y^2)^2$ is always positive for real $x$ and $y$, we can multiply both sides by this term:

Now, rearrange the equation to solve for $\frac{dy}{dx}$:

Divide both sides by $2y$ (assuming $y \neq 0$):

Thus, the derivative $\frac{dy}{dx}$ is $-\frac{x}{y}$. This derivative is defined for all $(x, y)$ such that $y \neq 0$ and $(x, y)$ satisfy the original equation.

Final Answer:

$\frac{dy}{dx} = -\frac{x}{y}$

Given:

The relation between $x$ and $y$ is $(x^2 + y^2)^2 = xy$.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We are given the implicit relation:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation (1) with respect to $x$. We treat $y$ as a function of $x$ and use the chain rule and product rule.

Differentiate the left side using the chain rule. Let $u = x^2 + y^2$. The derivative is $\frac{d}{dx}(u^2) = 2u \frac{du}{dx}$.

So, the derivative of the left side is:

Differentiate the right side using the product rule:

Equate the derivatives of both sides:

Expand the left side:

Rearrange the equation to collect all terms involving $\frac{dy}{dx}$ on one side and the other terms on the other side:

Solve for $\frac{dy}{dx}$ by dividing both sides by the term in the parenthesis:

This is the expression for $\frac{dy}{dx}$.

Final Answer:

$\frac{dy}{dx} = \frac{y - 4x^3 - 4xy^2}{4x^2y + 4y^3 - x}$

Given:

The relation between $x$ and $y$ is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$, where $a, h, b, g, f, c$ are constants.

To Show:

$\frac{dy}{dx} \;.\; \frac{dy}{dx} = 1$ (as stated in the question).

Solution:

The question asks to show that $\left(\frac{dy}{dx}\right)^2 = 1$. This would imply that $\frac{dy}{dx} = 1$ or $\frac{dy}{dx} = -1$. This is not generally true for an arbitrary quadratic equation in two variables, which represents a conic section. The phrasing $\frac{dy}{dx} \;.\; \frac{dy}{dx}$ is also unusual for $\left(\frac{dy}{dx}\right)^2$. It is highly probable that the question intends to ask to show the standard reciprocal property of derivatives, which is $\frac{dy}{dx} \cdot \frac{dx}{dy} = 1$. We will proceed by calculating $\frac{dy}{dx}$ and $\frac{dx}{dy}$ and demonstrating this standard property.

We have the implicit relation:

First, to find $\frac{dy}{dx}$, we differentiate equation (1) with respect to $x$, treating $y$ as a function of $x$:

Using the rules of differentiation (product rule for $2hxy$ and chain rule for $by^2$ and $2fy$):

Group the terms containing $\frac{dy}{dx}$:

Divide by $2(hx + by + f)$ (assuming $hx + by + f \neq 0$):

Next, to find $\frac{dx}{dy}$, we differentiate equation (1) with respect to $y$, treating $x$ as a function of $y$:

Using the rules of differentiation (product rule for $2hxy$ and chain rule for $ax^2$ and $2gx$):

Group the terms containing $\frac{dx}{dy}$:

Divide by $2(ax + hy + g)$ (assuming $ax + hy + g \neq 0$):

Now, let's compute the product $\frac{dy}{dx} \cdot \frac{dx}{dy}$ using equations (2) and (3):

Assuming $hx + by + f \neq 0$ and $ax + hy + g \neq 0$, we can cancel the common terms:

The standard property $\frac{dy}{dx} \cdot \frac{dx}{dy} = 1$ is satisfied for the given relation, provided the derivatives are defined and non-zero. The relation $\left(\frac{dy}{dx}\right)^2 = 1$ would only hold for this equation if it represents specific degenerate conics (like pairs of lines $y=x+k$ or $y=-x+k$) where the slope is constantly $1$ or $-1$, which is not true for the general equation.

Conclusion:

While the question asks to show $\left(\frac{dy}{dx}\right)^2 = 1$, which is not generally true for the given equation, the standard property of derivatives, $\frac{dy}{dx} \cdot \frac{dx}{dy} = 1$, is shown to hold for the given relation:

$\frac{dy}{dx} \cdot \frac{dx}{dy} = 1$

Given:

The relation between $x$ and $y$ is $x = e^{\frac{x}{y}}$.

To Prove:

$\frac{dy}{dx} = \frac{x − y}{x \log x}$

Proof:

We are given the equation:

Take the natural logarithm (log) of both sides of equation (1):

Using the property of logarithms $\log(a^b) = b \log a$ and $\log e = 1$:

Rearrange equation (2) to make $y$ the subject (optional, but can simplify differentiation):

Now, differentiate equation (3) implicitly with respect to $x$. Use the product rule on the left side, where $\frac{d}{dx}(\log x) = \frac{1}{x}$ and we treat $y$ as a function of $x$, so $\frac{d}{dx}(y) = \frac{dy}{dx}$.

Isolate the term containing $\frac{dy}{dx}$:

Combine the terms on the right side:

Finally, solve for $\frac{dy}{dx}$ by dividing both sides by $\log x$ (assuming $\log x \neq 0$, which means $x \neq 1$):

This matches the expression we were asked to prove.

Hence Proved.

Given:

The relation between $x$ and $y$ is $y^x = e^{y−x}$.

To Prove:

$\frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y}$

Proof:

We are given the equation:

Take the natural logarithm (log) of both sides of equation (1). We assume $y > 0$ for $\log y$ to be defined, and $x$ is such that $y^x$ is defined.

Using the property of logarithms $\log(a^b) = b \log a$ on the left side and $\log(e^u) = u \log e = u$ on the right side (since $\log e = 1$):

Now, we need to find $\frac{dy}{dx}$. We can differentiate equation (2) implicitly with respect to $x$. We treat $y$ as a function of $x$. The term $x \log y$ requires the product rule $\frac{d}{dx}(uv) = u'v + uv'$. Here $u=x$ and $v=\log y$.

Differentiate the left side:

Differentiate the right side:

Equate the derivatives of both sides:

Rearrange the equation to group terms containing $\frac{dy}{dx}$ on one side:

Solve for $\frac{dy}{dx}$:

The expression (3) is a valid derivative, but it does not match the required form. We need to eliminate the term $(y-x)$ in the denominator using the original relation. From equation (2), we have $x \log y = y - x$.

Substitute $y - x = x \log y$ into the denominator of equation (3):

Equation (4) still does not match the required form, as it contains $x$ in the denominator. Let's use equation (2) again to express $x$ in terms of $y$: $x = \frac{y}{1 + \log y}$.

Substitute this expression for $x$ into equation (4):

Simplify the expression by multiplying the numerator by the reciprocal of the denominator:

Assuming $y \neq 0$ (which must be true for $\log y$ to be defined and for the original equation $y^x=e^{y-x}$ to hold in a non-trivial way), we can cancel $y$ from the numerator and denominator:

This matches the expression we were asked to prove, assuming $\log y \neq 0$, which means $y \neq 1$.

Hence Proved.

Given:

The function is given by the infinite power tower $y = (\cos x)^{(\cos x)^{(\cos x)^{....∞}}}$.

To Show:

$\frac{dy}{dx} = \frac{y^2 \tan x}{y \log \cos x−1}$

Proof:

The given function is an infinite power tower:

Due to the infinite nature of the exponent, the expression in the exponent is the original function $y$. So, we can rewrite the equation as:

To differentiate this equation implicitly, we take the natural logarithm (log) of both sides. We assume $\cos x > 0$ for $\log(\cos x)$ to be defined.

Using the property of logarithms $\log(a^b) = b \log a$:

Now, differentiate equation (3) with respect to $x$. We treat $y$ as a function of $x$ and use the chain rule and product rule.

Differentiate the left side using the chain rule:

Differentiate the right side using the product rule $\frac{d}{dx}(uv) = \frac{du}{dx} v + u \frac{dv}{dx}$, with $u=y$ and $v=\log(\cos x)$.

To find $\frac{d}{dx}(\log (\cos x))$, use the chain rule with inner function $\cos x$:

So, the derivative of the right side is:

Equate the derivatives of both sides:

Rearrange the equation to gather terms with $\frac{dy}{dx}$ on one side:

Factor out $\frac{dy}{dx}$:

Find a common denominator for the terms inside the parenthesis:

Solve for $\frac{dy}{dx}$:

To match the required form, multiply the numerator and the denominator by -1:

This matches the expression we were asked to show, provided $y \log (\cos x) - 1 \neq 0$ and $\cos x > 0$.

Hence Showed.

Given:

The relation between $x$ and $y$ is $x \sin (a + y) + \sin a \cos (a + y) = 0$.

To Prove:

$\frac{dy}{dx} = \frac{\sin^2 (a+y)}{\sin a}$

Proof:

We are given the equation:

Rearrange the equation to express $x$ in terms of $y$. Move the second term to the right side:

Assuming $\sin(a+y) \neq 0$, divide both sides by $\sin(a+y)$:

Using the identity $\frac{\cos \theta}{\sin \theta} = \cot \theta$:

Now, we differentiate $x$ with respect to $y$ ($\frac{dx}{dy}$). Remember that $a$ is a constant, so $\sin a$ is also a constant.

Using the chain rule, the derivative of $\cot u$ with respect to $y$ is $-\text{cosec}^2 u \cdot \frac{du}{dy}$. Here, $u = a+y$.

So, $\frac{d}{dy} (\cot (a + y)) = -\text{cosec}^2 (a+y) \cdot 1 = -\text{cosec}^2 (a+y)$.

Substitute this back into the expression for $\frac{dx}{dy}$:

Using the identity $\text{cosec} \theta = \frac{1}{\sin \theta}$, we have $\text{cosec}^2 (a+y) = \frac{1}{\sin^2 (a+y)}$.

To find $\frac{dy}{dx}$, we take the reciprocal of $\frac{dx}{dy}$, assuming $\frac{dx}{dy} \neq 0$ (i.e., $\sin a \neq 0$ and $\sin(a+y) \neq 0$):

This matches the expression we were asked to prove.

Hence Proved.

Given:

The relation between $x$ and $y$ is $\sqrt{1−x^2} + \sqrt{1−y^2} = a (x - y)$.

(Note: Assuming the equation is $\sqrt{1−x^2} + \sqrt{1−y^2} = a (x - y)$, as the requested proof matches this form. If the equation were $a(x-a)$, the derivative would be different.)

To Prove:

$\frac{dy}{dx} = \sqrt{\frac{1 − y^2}{1 − x^2}}$

Proof:

We are given the equation:

This form suggests a trigonometric substitution. Let $x = \sin A$ and $y = \sin B$.

Assuming the principal values for $\sin^{-1} x$ and $\sin^{-1} y$, we have $A = \sin^{-1} x$ and $B = \sin^{-1} y$, with $A, B \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Then $\sqrt{1-x^2} = \sqrt{1-\sin^2 A} = \sqrt{\cos^2 A}$. Since $A \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\cos A \ge 0$. Thus, $\sqrt{1-x^2} = \cos A$.

Similarly, $\sqrt{1-y^2} = \sqrt{1-\sin^2 B} = \sqrt{\cos^2 B} = \cos B$, since $B \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Substitute these into equation (1):

Use the sum-to-product trigonometric identities:

$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$

$\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$

Substitute these into equation (2):

Assuming $\cos\left(\frac{A+B}{2}\right) \neq 0$, we can divide both sides by $2 \cos\left(\frac{A+B}{2}\right)$:

Assuming $\sin\left(\frac{A-B}{2}\right) \neq 0$, divide both sides of equation (3) by $\sin\left(\frac{A-B}{2}\right)$:

Take the inverse cotangent of both sides:

Substitute back $A = \sin^{-1} x$ and $B = \sin^{-1} y$ into equation (4):

Since $a$ is a constant, $2 \cot^{-1} a$ is also a constant. Let $C = 2 \cot^{-1} a$.

Now, differentiate this equation implicitly with respect to $x$. Remember that $\frac{d}{dx}(\sin^{-1} u) = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx}$ and the derivative of a constant is 0.

Rearrange the equation to solve for $\frac{dy}{dx}$:

Multiply both sides by $\sqrt{1-y^2}$:

Combine the terms under a single square root:

This matches the expression we were asked to prove.

Hence Proved.

Given:

$y = \tan^{-1} x$

To Find:

$\frac{d^2y}{dx^2}$ in terms of $y$ alone.

Solution:

We are given the function:

To find the first derivative $\frac{dy}{dx}$, differentiate equation (1) with respect to $x$:

Now, to find the second derivative $\frac{d^2y}{dx^2}$, differentiate equation (2) with respect to $x$:

We can rewrite $\frac{1}{1+x^2}$ as $(1+x^2)^{-1}$. Using the chain rule:

We need to express $\frac{d^2y}{dx^2}$ in terms of $y$ alone. From the original equation (1), $y = \tan^{-1} x$, which implies:

Also, using the trigonometric identity $1+\tan^2 y = \sec^2 y$, we can write:

Now, substitute equations (4) and (5) into equation (3):

We can express $\tan y$ as $\frac{\sin y}{\cos y}$ and $\sec y$ as $\frac{1}{\cos y}$:

This expression for $\frac{d^2y}{dx^2}$ is in terms of $y$ alone.

Final Answer:

The second derivative of $y = \tan^{-1} x$ in terms of $y$ alone is $-2 \sin y \cos^3 y$.

Given:

The function is $f(x) = x (x – 1)^2$.

The interval is $[0, 1]$.

To Verify:

Rolle’s theorem for the given function in the given interval.

Rolle’s Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$, Rolle’s theorem states that if:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

3. $f(a) = f(b)$.

Then there exists at least one value $c \in (a, b)$ such that $f'(c) = 0$.

Verification of Conditions:

The given function is $f(x) = x (x – 1)^2 = x (x^2 - 2x + 1) = x^3 - 2x^2 + x$.

1. Continuity: $f(x)$ is a polynomial function. Polynomial functions are continuous on the entire real line. Therefore, $f(x)$ is continuous on the closed interval $[0, 1]$.

2. Differentiability: $f(x)$ is a polynomial function. Polynomial functions are differentiable on the entire real line. Therefore, $f(x)$ is differentiable on the open interval $(0, 1)$.

3. Equality of function values at endpoints: We evaluate the function at the endpoints of the interval, $a=0$ and $b=1$.

Since all three conditions of Rolle’s Theorem are satisfied, there must exist at least one value $c \in (0, 1)$ such that $f'(c) = 0$.

Finding the value(s) of c:

First, find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(x^3 - 2x^2 + x)$

Set $f'(x) = 0$ to find the values of $x$ (which will be our $c$ values):

Solve the quadratic equation by factoring:

This gives two possible values for $x$:

The value $x = \frac{1}{3}$ lies in the open interval $(0, 1)$. The value $x = 1$ is an endpoint and not in the open interval.

Thus, we found a value $c = \frac{1}{3}$ in $(0, 1)$ such that $f'(c) = 0$.

Conclusion:

All conditions of Rolle's Theorem are satisfied for the function $f(x) = x(x-1)^2$ on the interval $[0, 1]$, and we have found a value $c = \frac{1}{3} \in (0, 1)$ such that $f'(\frac{1}{3}) = 0$. Therefore, Rolle’s Theorem is verified for the given function and interval.

Given:

The function is $f(x) = \sin^4 x + \cos^4 x$.

The interval is $\left[ 0, \frac{π}{2} \right]$.

To Verify:

Rolle’s theorem for the given function in the given interval.

Rolle’s Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$, Rolle’s theorem states that if:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

3. $f(a) = f(b)$.

Then there exists at least one value $c \in (a, b)$ such that $f'(c) = 0$.

Verification of Conditions:

The given function is $f(x) = \sin^4 x + \cos^4 x$. The interval is $\left[0, \frac{\pi}{2}\right]$, so $a=0$ and $b=\frac{\pi}{2}$.

1. Continuity: The sine and cosine functions are continuous everywhere. Their powers ($\sin^4 x$, $\cos^4 x$) are also continuous. The sum of continuous functions is continuous. Therefore, $f(x) = \sin^4 x + \cos^4 x$ is continuous on the closed interval $\left[0, \frac{\pi}{2}\right]$.

2. Differentiability: The sine and cosine functions are differentiable everywhere. Their powers are also differentiable. The sum of differentiable functions is differentiable. Therefore, $f(x) = \sin^4 x + \cos^4 x$ is differentiable on the open interval $\left(0, \frac{\pi}{2}\right)$.

3. Equality of function values at endpoints: We evaluate the function at the endpoints $x=0$ and $x=\frac{\pi}{2}$.

Since $f(0) = 1$ and $f\left(\frac{\pi}{2}\right) = 1$, we have:

All three conditions of Rolle’s Theorem are satisfied. Therefore, there must exist at least one value $c \in \left(0, \frac{\pi}{2}\right)$ such that $f'(c) = 0$.

Finding the value(s) of c:

First, find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(\sin^4 x + \cos^4 x)$.

Using the chain rule, $\frac{d}{dx}(\sin^n x) = n \sin^{n-1} x \cos x$ and $\frac{d}{dx}(\cos^n x) = n \cos^{n-1} x (-\sin x)$.

Factor out $4 \sin x \cos x$:

Using the trigonometric identities $2 \sin x \cos x = \sin(2x)$ and $\cos^2 x - \sin^2 x = \cos(2x)$:

Using the identity $2 \sin \theta \cos \theta = \sin(2\theta)$ with $\theta = 2x$:

Set $f'(x) = 0$ to find $c$:

The general solution for $\sin \theta = 0$ is $\theta = n\pi$, where $n$ is an integer.

We need to find values of $c$ in the open interval $\left(0, \frac{\pi}{2}\right)$.

• For $n=0$, $c = 0$, which is not in $(0, \frac{\pi}{2})$.
• For $n=1$, $c = \frac{\pi}{4}$. This value is in $\left(0, \frac{\pi}{2}\right)$ since $0 < \frac{\pi}{4} < \frac{\pi}{2}$.
• For $n=2$, $c = \frac{2\pi}{4} = \frac{\pi}{2}$, which is not in $(0, \frac{\pi}{2})$.
• For other integer values of $n$, $c$ will be outside the interval $\left(0, \frac{\pi}{2}\right)$.

Thus, there exists at least one value $c = \frac{\pi}{4} \in \left(0, \frac{\pi}{2}\right)$ such that $f'(c) = 0$.

Conclusion:

All the conditions of Rolle’s Theorem are satisfied for the function $f(x) = \sin^4 x + \cos^4 x$ on the interval $\left[0, \frac{\pi}{2}\right]$, and we have found a value $c = \frac{\pi}{4} \in \left(0, \frac{\pi}{2}\right)$ such that $f'(\frac{\pi}{4}) = 0$. Therefore, Rolle’s Theorem is verified for the given function and interval.

Given:

The function is $f(x) = \log (x^2 + 2) – \log 3$.

The interval is $[-1, 1]$.

To Verify:

Rolle’s theorem for the given function in the given interval.

Rolle’s Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$, Rolle’s theorem states that if:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

3. $f(a) = f(b)$.

Then there exists at least one value $c \in (a, b)$ such that $f'(c) = 0$.

Verification of Conditions:

The given function is $f(x) = \log (x^2 + 2) - \log 3$. The interval is $[-1, 1]$, so $a=-1$ and $b=1$.

1. Continuity: The function $f(x)$ involves a logarithm term $\log(x^2+2)$ and a constant term $\log 3$. The term $x^2+2$ is a polynomial, which is continuous everywhere. For any real $x$, $x^2 \ge 0$, so $x^2+2 \ge 2$. Since $x^2+2$ is always positive, the function $\log(x^2+2)$ is defined and continuous for all real $x$. The constant function $\log 3$ is also continuous. The difference of two continuous functions is continuous. Therefore, $f(x)$ is continuous on the closed interval $[-1, 1]$.

2. Differentiability: To check differentiability, we find the derivative of $f(x)$.

The derivative $f'(x) = \frac{2x}{x^2+2}$ is a rational function. The denominator $x^2+2$ is always positive ($x^2+2 \ge 2$ for all real $x$), so the denominator is never zero. Thus, $f'(x)$ is defined for all real $x$. Therefore, $f(x)$ is differentiable on the open interval $(-1, 1)$.

3. Equality of function values at endpoints: We evaluate $f(x)$ at the endpoints $x=-1$ and $x=1$.

Since all three conditions of Rolle’s Theorem are satisfied, there must exist at least one value $c \in (-1, 1)$ such that $f'(c) = 0$.

Finding the value(s) of c:

We set the derivative $f'(x)$ equal to 0 and solve for $x$ (which will be our $c$ value):

This equation holds if and only if the numerator is zero:

The value $c = 0$ lies in the open interval $(-1, 1)$.

Thus, we have found a value $c = 0 \in (-1, 1)$ such that $f'(0) = 0$.

Conclusion:

All the conditions of Rolle’s Theorem are satisfied for the function $f(x) = \log (x^2 + 2) – \log 3$ on the interval $[-1, 1]$. We have found a value $c = 0 \in (-1, 1)$ such that $f'(0) = 0$. Therefore, Rolle’s Theorem is verified for the given function and interval.

Given:

The function is $f(x) = x (x + 3)e^{–\frac{x}{2}}$.

The interval is $[-3, 0]$.

To Verify:

Rolle’s theorem for the given function in the given interval.

Rolle’s Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$, Rolle’s theorem states that if:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

3. $f(a) = f(b)$.

Then there exists at least one value $c \in (a, b)$ such that $f'(c) = 0$.

Verification of Conditions:

The given function is $f(x) = x(x+3)e^{-\frac{x}{2}} = (x^2 + 3x)e^{-\frac{x}{2}}$. The interval is $[-3, 0]$, so $a=-3$ and $b=0$.

1. Continuity: The function $f(x)$ is the product of a polynomial $p(x) = x^2 + 3x$ and an exponential function $g(x) = e^{-\frac{x}{2}}$. Polynomials are continuous on the entire real line. Exponential functions are continuous on the entire real line. The product of two continuous functions is continuous. Therefore, $f(x)$ is continuous on the closed interval $[-3, 0]$.

2. Differentiability: The polynomial $p(x) = x^2 + 3x$ is differentiable on the entire real line, with $p'(x) = 2x+3$. The exponential function $g(x) = e^{-\frac{x}{2}}$ is differentiable on the entire real line, with $g'(x) = e^{-\frac{x}{2}} \cdot (-\frac{1}{2})$. The product of two differentiable functions is differentiable. Therefore, $f(x)$ is differentiable on the open interval $(-3, 0)$.

3. Equality of function values at endpoints: We evaluate $f(x)$ at the endpoints $x=-3$ and $x=0$.

Since $f(-3) = 0$ and $f(0) = 0$, we have:

All three conditions of Rolle’s Theorem are satisfied. Therefore, there must exist at least one value $c \in (-3, 0)$ such that $f'(c) = 0$.

Finding the value(s) of c:

First, find the derivative of $f(x)$ using the product rule $\frac{d}{dx}(uv) = u'v + uv'$, where $u = x^2 + 3x$ and $v = e^{-\frac{x}{2}}$.

Factor out the common term $e^{-\frac{x}{2}}$:

We can factor out $-\frac{1}{2}$:

Set $f'(x) = 0$ to find the values of $x$ (which will be our $c$ values):

Since $e^{-\frac{x}{2}}$ is never equal to zero for any real $x$, we must have:

Factor the quadratic equation:

This gives two possible values for $x$:

We need to find a value $c$ that lies in the open interval $(-3, 0)$.

• The value $x = 3$ is not in $(-3, 0)$.
• The value $x = -2$ is in $(-3, 0)$, since $-3 < -2 < 0$.

Thus, we found a value $c = -2$ in $(-3, 0)$ such that $f'(c) = 0$.

Conclusion:

All the conditions of Rolle’s Theorem are satisfied for the function $f(x) = x(x+3)e^{-\frac{x}{2}}$ on the interval $[-3, 0]$, and we have found a value $c = -2 \in (-3, 0)$ such that $f'(-2) = 0$. Therefore, Rolle’s Theorem is verified for the given function and interval.

Given:

The function is $f(x) = \sqrt{4−x^2}$.

The interval is $[-2, 2]$.

To Verify:

Rolle’s theorem for the given function in the given interval.

Rolle’s Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$, Rolle’s theorem states that if:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

3. $f(a) = f(b)$.

Then there exists at least one value $c \in (a, b)$ such that $f'(c) = 0$.

Verification of Conditions:

The given function is $f(x) = \sqrt{4-x^2}$. The interval is $[-2, 2]$, so $a=-2$ and $b=2$.

1. Continuity: The function $f(x) = \sqrt{4-x^2}$ is defined when $4-x^2 \ge 0$, which means $x^2 \le 4$, or $-2 \le x \le 2$. The domain of the function is $[-2, 2]$. The square root function is continuous for non-negative values of its argument, and the function $4-x^2$ is a polynomial which is continuous everywhere. Thus, $f(x)$ is continuous on its domain, which is the closed interval $[-2, 2]$. Condition 1 is satisfied.

2. Differentiability: To check differentiability on the open interval $(-2, 2)$, we find the derivative $f'(x)$.

Using the chain rule:

The derivative $f'(x)$ is defined when $4-x^2 \neq 0$, which means $x \neq 2$ and $x \neq -2$. For the open interval $(-2, 2)$, the denominator $\sqrt{4-x^2}$ is always positive, so $f'(x)$ is defined for all $x \in (-2, 2)$. Therefore, $f(x)$ is differentiable on the open interval $(-2, 2)$. Condition 2 is satisfied.

3. Equality of function values at endpoints: We evaluate the function at the endpoints $x=-2$ and $x=2$.

Since $f(-2) = 0$ and $f(2) = 0$, we have:

All three conditions of Rolle’s Theorem are satisfied. Therefore, there must exist at least one value $c \in (-2, 2)$ such that $f'(c) = 0.

Finding the value(s) of c:

We set the derivative $f'(x)$ equal to 0 and solve for $x$ (which will be our $c$ value):

This equation holds if and only if the numerator is zero, provided the denominator is non-zero. For $c \in (-2, 2)$, $\sqrt{4-c^2} \neq 0$.

The value $c = 0$ lies in the open interval $(-2, 2)$, since $-2 < 0 < 2$.

Thus, we have found a value $c = 0 \in (-2, 2)$ such that $f'(0) = 0$.

Conclusion:

All the conditions of Rolle’s Theorem are satisfied for the function $f(x) = \sqrt{4-x^2}$ on the interval $[-2, 2]$. We have found a value $c = 0 \in (-2, 2)$ such that $f'(0) = 0$. Therefore, Rolle’s Theorem is verified for the given function and interval.

Given:

The function is defined as $f(x) = \begin{cases}x^2+1,& if\; 0 ≤ x ≤ 1 \\ 3−x,& if\; 1 ≤ x ≤ 2 \end{cases}$.

The interval is $[0, 2]$.

To Discuss:

The applicability of Rolle’s theorem for the given function on the interval $[0, 2]$.

Rolle’s Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$, Rolle’s theorem is applicable if:

1. $f(x)$ is continuous on the closed interval $[a, b]$.

2. $f(x)$ is differentiable on the open interval $(a, b)$.

3. $f(a) = f(b)$.

If all three conditions are satisfied, then there exists at least one value $c \in (a, b)$ such that $f'(c) = 0$. If any of the conditions are not met, the theorem is not applicable.

Checking the Conditions for $f(x)$ on $[0, 2]$:

Here, $a=0$ and $b=2$.

1. Continuity on $[0, 2]$:

The function is defined by two polynomial pieces: $f(x) = x^2+1$ for $0 \leq x \leq 1$ and $f(x) = 3-x$ for $1 \leq x \leq 2$. Both $x^2+1$ and $3-x$ are polynomial functions and thus continuous on their respective defined intervals.

We need to check the continuity at the point where the definition changes, i.e., at $x=1$.

Left-hand limit at $x=1$:

Right-hand limit at $x=1$:

Function value at $x=1$:

Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2$, the function is continuous at $x=1$.

Thus, $f(x)$ is continuous on the closed interval $[0, 2]$. Condition 1 is satisfied.

2. Differentiability on $(0, 2)$:

We find the derivative of each piece:

For $0 < x < 1$, $f(x) = x^2+1$, so $f'(x) = \frac{d}{dx}(x^2+1) = 2x$.

For $1 < x < 2$, $f(x) = 3-x$, so $f'(x) = \frac{d}{dx}(3-x) = -1$.

We need to check the differentiability at $x=1$. We compare the left-hand derivative and the right-hand derivative at $x=1$.

Left-hand derivative at $x=1$ ($f'(1^-)$):

Since $h \to 0^-$, $1+h < 1$, so $f(1+h) = (1+h)^2+1$. $f(1)=2$.

(Alternatively, evaluate $2x$ at $x=1$: $2(1) = 2$).

Right-hand derivative at $x=1$ ($f'(1^+)$):

Since $h \to 0^+$, $1+h > 1$, so $f(1+h) = 3-(1+h)$. $f(1)=2$.

(Alternatively, evaluate $-1$ for $x > 1$: $-1$).

Since the left-hand derivative at $x=1$ ($2$) is not equal to the right-hand derivative at $x=1$ ($-1$), the function is not differentiable at $x=1$.

Since $x=1$ is a point in the open interval $(0, 2)$, the function is not differentiable on $(0, 2)$. Condition 2 is not satisfied.

3. Equality of function values at endpoints:

Evaluate $f(x)$ at the endpoints $x=0$ and $x=2$.

Conclusion:

The function $f(x)$ satisfies the continuity condition on $[0, 2]$ and the condition $f(0) = f(2)$. However, the function is not differentiable at $x=1$, which is a point within the open interval $(0, 2)$. Since the differentiability condition of Rolle’s Theorem is not satisfied, Rolle’s Theorem is not applicable to the function $f(x)$ on the interval $[0, 2]$.

Given:

The equation of the curve is $y = \cos x - 1$.

The interval for $x$ is $[0, 2π]$.

To Find:

The points on the curve where the tangent is parallel to the x-axis.

Solution:

The tangent to a curve $y = f(x)$ is parallel to the x-axis when its slope is zero. The slope of the tangent is given by the derivative $\frac{dy}{dx}$.

We need to find the derivative of the given function $y = \cos x - 1$ with respect to $x$:

For the tangent to be parallel to the x-axis, the slope $\frac{dy}{dx}$ must be equal to 0:

We need to find the values of $x$ in the interval $[0, 2\pi]$ for which $\sin x = 0$. The general solutions for $\sin x = 0$ are $x = n\pi$, where $n$ is an integer.

Considering the interval $[0, 2\pi]$:

• For $n=0$, $x = 0 \cdot \pi = 0$. This is in the interval.
• For $n=1$, $x = 1 \cdot \pi = \pi$. This is in the interval.
• For $n=2$, $x = 2 \cdot \pi = 2\pi$. This is in the interval.
• For $n < 0$ or $n > 2$, the values of $x$ will be outside the interval $[0, 2\pi]$.

So, the possible values of $x$ are $0$, $\pi$, and $2\pi$.

Now, we find the corresponding $y$-coordinates for these $x$ values by substituting them back into the original equation $y = \cos x - 1$.

• When $x = 0$:

$y = \cos(0) - 1 = 1 - 1 = 0$

The point is $(0, 0)$.

• When $x = \pi$:

$y = \cos(\pi) - 1 = -1 - 1 = -2$

The point is $(\pi, -2)$.

• When $x = 2\pi$:

$y = \cos(2\pi) - 1 = 1 - 1 = 0$

The point is $(2\pi, 0)$.

These are the points on the curve $y = \cos x - 1$ in the interval $[0, 2\pi]$ where the tangent is parallel to the x-axis.

Final Answer:

The points on the curve where the tangent is parallel to the x-axis are $(0, 0)$, $(\pi, -2)$, and $(2\pi, 0)$.

Given:

The equation of the curve is $y = x (x – 4)$.

The interval for $x$ is $[0, 4]$.

To Find:

The point(s) on the curve in the given interval where the tangent is parallel to the x-axis, using Rolle’s theorem.

Solution:

A tangent to a curve $y=f(x)$ is parallel to the x-axis when its slope $\frac{dy}{dx}$ is equal to 0.

Rolle’s theorem states that if a function $f(x)$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one value $c \in (a, b)$ such that $f'(c) = 0$. The value $c$ is the x-coordinate where the tangent to the curve is horizontal (parallel to the x-axis).

The given function is $f(x) = x(x-4) = x^2 - 4x$ on the interval $[0, 4]$. Here, $a=0$ and $b=4$.

Let's verify the conditions of Rolle's theorem:

1. Continuity: $f(x) = x^2 - 4x$ is a polynomial function. Polynomials are continuous everywhere, so $f(x)$ is continuous on the closed interval $[0, 4]$.

2. Differentiability: $f(x) = x^2 - 4x$ is a polynomial function. Polynomials are differentiable everywhere. The derivative is $f'(x) = 2x - 4$, which is defined for all $x$. Thus, $f(x)$ is differentiable on the open interval $(0, 4)$.

3. Equality of function values at endpoints: We evaluate $f(x)$ at the endpoints $x=0$ and $x=4$.

Since all three conditions of Rolle’s theorem are satisfied, there exists at least one value $c \in (0, 4)$ such that $f'(c) = 0$. This value $c$ is the x-coordinate where the tangent is parallel to the x-axis.

Now, we find the derivative $f'(x)$:

Set $f'(c) = 0$ to find the value of $c$:

The value $c=2$ lies in the open interval $(0, 4)$, since $0 < 2 < 4$. This is the x-coordinate of the point where the tangent is parallel to the x-axis.

To find the y-coordinate of this point, substitute $x=2$ back into the original function $y = x(x-4)$:

So, the point on the curve where the tangent is parallel to the x-axis is $(2, -4)$.

Final Answer:

Using Rolle’s theorem, the point on the curve $y = x (x – 4)$ in $[0, 4]$ where the tangent is parallel to the x-axis is $(2, -4)$.

Given:

The function is $f(x) = \frac{1}{4x − 1}$.

The interval is $[1, 4]$.

To Verify:

Mean Value Theorem for the given function in the given interval.

Mean Value Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$, the Mean Value Theorem states that if:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

Then there exists at least one value $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

Verification of Conditions:

The given function is $f(x) = \frac{1}{4x - 1}$ on the interval $[1, 4]$. Here, $a=1$ and $b=4$.

1. Continuity on $[1, 4]$:

The function $f(x)$ is a rational function, which is continuous everywhere its denominator is not equal to zero. The denominator is $4x-1$. $4x-1 = 0$ when $x = \frac{1}{4}$. The point $x = \frac{1}{4}$ is not within the closed interval $[1, 4]$. Therefore, the function $f(x)$ is continuous on $[1, 4]$. Condition 1 is satisfied.

2. Differentiability on $(1, 4)$:

We find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}\left(\frac{1}{4x - 1}\right)$.

Using the chain rule:

The derivative $f'(x)$ is defined everywhere the denominator $(4x-1)^2$ is not zero, i.e., $x \neq \frac{1}{4}$. The point $x = \frac{1}{4}$ is not within the open interval $(1, 4)$. Therefore, the function $f(x)$ is differentiable on $(1, 4)$. Condition 2 is satisfied.

Since both conditions of the Mean Value Theorem are satisfied, the theorem is applicable, and there exists at least one value $c \in (1, 4)$ such that $f'(c) = \frac{f(4) - f(1)}{4 - 1}$.

Finding the value(s) of c:

Calculate the value of the right-hand side of the MVT formula, $\frac{f(b) - f(a)}{b - a}$.

Now, set $f'(c)$ equal to the value from equation (2):

Divide both sides by -4:

Take the reciprocal of both sides:

Take the square root of both sides:

Solve for $c$:

We need to check which of these values lie in the open interval $(1, 4)$.

Approximate value of $\sqrt{5} \approx 2.236$ and $3\sqrt{5} \approx 6.708$.

Case 1: $c = \frac{1 + 3\sqrt{5}}{4} \approx \frac{1 + 6.708}{4} = \frac{7.708}{4} \approx 1.927$

Since $1 < 1.927 < 4$, this value of $c$ is in the interval $(1, 4)$.

Case 2: $c = \frac{1 - 3\sqrt{5}}{4} \approx \frac{1 - 6.708}{4} = \frac{-5.708}{4} \approx -1.427$

Since $-1.427$ is not greater than 1, this value of $c$ is not in the interval $(1, 4)$.

Thus, there exists exactly one value of $c$ in $(1, 4)$ that satisfies the Mean Value Theorem.

Conclusion:

The function $f(x) = \frac{1}{4x - 1}$ is continuous on $[1, 4]$ and differentiable on $(1, 4)$. Thus, the conditions of the Mean Value Theorem are satisfied. We have found a value $c = \frac{1 + 3\sqrt{5}}{4} \in (1, 4)$ such that $f'(c) = \frac{f(4) - f(1)}{4 - 1}$. Therefore, the Mean Value Theorem is verified for the given function and interval.

Given:

The function is $f(x) = x^3 – 2x^2 – x + 3$.

The interval is $[0, 1]$.

To Verify:

Mean Value Theorem for the given function in the given interval.

Mean Value Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$, the Mean Value Theorem states that if:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

Then there exists at least one value $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

Verification of Conditions:

The given function is $f(x) = x^3 – 2x^2 – x + 3$ on the interval $[0, 1]$. Here, $a=0$ and $b=1$.

1. Continuity on $[0, 1]$:

The function $f(x)$ is a polynomial function. Polynomials are continuous on the entire real line. Therefore, $f(x)$ is continuous on the closed interval $[0, 1]$. Condition 1 is satisfied.

2. Differentiability on $(0, 1)$:

The function $f(x)$ is a polynomial function. Polynomials are differentiable on the entire real line. The derivative $f'(x) = 3x^2 - 4x - 1$ is defined for all real $x$. Therefore, $f(x)$ is differentiable on the open interval $(0, 1)$. Condition 2 is satisfied.

Since both conditions of the Mean Value Theorem are satisfied, the theorem is applicable, and there exists at least one value $c \in (0, 1)$ such that $f'(c) = \frac{f(1) - f(0)}{1 - 0}$.

Finding the value(s) of c:

First, calculate the value of the right-hand side of the MVT formula, $\frac{f(b) - f(a)}{b - a}$.

Next, find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(x^3 – 2x^2 – x + 3)$.

Now, set $f'(c)$ equal to the value from equation (1) and solve for $c$:

Rearrange the equation into a standard quadratic form:

Factor the quadratic equation:

This gives two possible values for $c$:

We need to check which of these values lie in the open interval $(0, 1)$.

• The value $c = \frac{1}{3}$ is in the interval $(0, 1)$, since $0 < \frac{1}{3} < 1$.
• The value $c = 1$ is an endpoint and not in the open interval $(0, 1)$.

Thus, there exists exactly one value of $c$ in $(0, 1)$ that satisfies the Mean Value Theorem:

Conclusion:

The function $f(x) = x^3 – 2x^2 – x + 3$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$. Thus, the conditions of the Mean Value Theorem are satisfied. We have found a value $c = \frac{1}{3} \in (0, 1)$ such that $f'(\frac{1}{3}) = \frac{f(1) - f(0)}{1 - 0}$. Therefore, the Mean Value Theorem is verified for the given function and interval.

Given:

The function is $f(x) = \sin x – \sin 2x$.

The interval is $[0, π]$.

To Verify:

Mean Value Theorem for the given function in the given interval.

Mean Value Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$, the Mean Value Theorem states that if:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

Then there exists at least one value $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

Verification of Conditions:

The given function is $f(x) = \sin x – \sin 2x$ on the interval $[0, π]$. Here, $a=0$ and $b=\pi$.

1. Continuity on $[0, π]$:

The sine function ($\sin x$ and $\sin 2x$) is continuous on the entire real line. The difference of two continuous functions is continuous. Therefore, $f(x)$ is continuous on the closed interval $[0, π]$. Condition 1 is satisfied.

2. Differentiability on $(0, π)$:

The sine function is differentiable on the entire real line. The derivative of $\sin x$ is $\cos x$. The derivative of $\sin 2x$ using the chain rule is $\cos 2x \cdot \frac{d}{dx}(2x) = 2 \cos 2x$. The difference of two differentiable functions is differentiable.

The derivative $f'(x)$ is defined for all real $x$. Therefore, $f(x)$ is differentiable on the open interval $(0, π)$. Condition 2 is satisfied.

Since both conditions of the Mean Value Theorem are satisfied, the theorem is applicable, and there exists at least one value $c \in (0, π)$ such that $f'(c) = \frac{f(\pi) - f(0)}{\pi - 0}$.

Finding the value(s) of c:

First, calculate the value of the right-hand side of the MVT formula, $\frac{f(\pi) - f(0)}{\pi - 0}$.

Now, set $f'(c)$ equal to the value from equation (2) and solve for $c$:

Use the double angle identity for cosine: $\cos 2c = 2 \cos^2 c - 1$.

Rearrange the equation into a quadratic form in terms of $\cos c$:

Let $u = \cos c$. We have $4u^2 - u - 2 = 0$. This is a quadratic equation in $u$. Use the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $u$, where $a=4$, $b=-1$, $c=-2$.

So, $\cos c = \frac{1 \pm \sqrt{33}}{8}$.

We need to find values of $c$ in the open interval $(0, \pi)$ such that $\cos c = \frac{1 \pm \sqrt{33}}{8}$.

Approximate value of $\sqrt{33}$. Since $\sqrt{25}=5$ and $\sqrt{36}=6$, $\sqrt{33}$ is between 5 and 6. $\sqrt{33} \approx 5.745$.

Case 1: $\cos c = \frac{1 + \sqrt{33}}{8} \approx \frac{1 + 5.745}{8} = \frac{6.745}{8} \approx 0.843$.

Since $-1 \le 0.843 \le 1$, there exists a value of $c$ such that $\cos c \approx 0.843$. Also, since $0.843 > 0$, $c$ must be in the first quadrant, i.e., $c \in \left(0, \frac{\pi}{2}\right)$. A value in $\left(0, \frac{\pi}{2}\right)$ is certainly in $(0, \pi)$.

Case 2: $\cos c = \frac{1 - \sqrt{33}}{8} \approx \frac{1 - 5.745}{8} = \frac{-4.745}{8} \approx -0.593$.

Since $-1 \le -0.593 \le 1$, there exists a value of $c$ such that $\cos c \approx -0.593$. Since $-0.593 < 0$, $c$ must be in the second quadrant, i.e., $c \in \left(\frac{\pi}{2}, \pi\right)$. A value in $\left(\frac{\pi}{2}, \pi\right)$ is certainly in $(0, \pi)$.

Both $\frac{1 + \sqrt{33}}{8}$ and $\frac{1 - \sqrt{33}}{8}$ are values between -1 and 1, and there exist corresponding values of $c$ in $(0, \pi)$.

For $\cos c = \frac{1 + \sqrt{33}}{8}$, $c = \cos^{-1}\left(\frac{1 + \sqrt{33}}{8}\right)$. Since $\frac{1+\sqrt{33}}{8}$ is positive and less than 1, $c$ is in $(0, \frac{\pi}{2})$, which is in $(0, \pi)$.

For $\cos c = \frac{1 - \sqrt{33}}{8}$, $c = \cos^{-1}\left(\frac{1 - \sqrt{33}}{8}\right)$. Since $\frac{1-\sqrt{33}}{8}$ is negative and between -1 and 0, $c$ is in $(\frac{\pi}{2}, \pi)$, which is in $(0, \pi)$.

Thus, there exist values of $c$ in $(0, \pi)$ that satisfy the Mean Value Theorem. For example, $c = \cos^{-1}\left(\frac{1 + \sqrt{33}}{8}\right)$.

Conclusion:

The function $f(x) = \sin x – \sin 2x$ is continuous on $[0, π]$ and differentiable on $(0, π)$. Thus, the conditions of the Mean Value Theorem are satisfied. We have shown that there exist values $c \in (0, π)$ (specifically, $c = \cos^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)$ that are within the interval) such that $f'(c) = \frac{f(\pi) - f(0)}{\pi - 0}$. Therefore, the Mean Value Theorem is verified for the given function and interval.

Given:

The function is $f(x) = \sqrt{25−x^2}$.

The interval is $[1, 5]$.

To Verify:

Mean Value Theorem for the given function in the given interval.

Mean Value Theorem Conditions:

For a function $f(x)$ on a closed interval $[a, b]$, the Mean Value Theorem states that if:

1. $f(x)$ is continuous on $[a, b]$.

2. $f(x)$ is differentiable on $(a, b)$.

Then there exists at least one value $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

Verification of Conditions:

The given function is $f(x) = \sqrt{25-x^2}$. The interval is $[1, 5]$. Here, $a=1$ and $b=5$.

1. Continuity on $[1, 5]$:

The function $f(x) = \sqrt{25-x^2}$ is defined for values of $x$ such that $25-x^2 \ge 0$, which means $x^2 \le 25$, or $-5 \le x \le 5$. The domain of the function is the closed interval $[-5, 5]$. The given interval $[1, 5]$ is a subset of the domain $[-5, 5]$. The square root function is continuous for non-negative values, and the polynomial $25-x^2$ is continuous everywhere. Therefore, $f(x) = \sqrt{25-x^2}$ is continuous on the closed interval $[1, 5]$. Condition 1 is satisfied.

2. Differentiability on $(1, 5)$:

We find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(\sqrt{25-x^2})$.

Using the chain rule:

The derivative $f'(x)$ is defined when the denominator $\sqrt{25-x^2}$ is not equal to zero, which means $25-x^2 \neq 0$, or $x \neq \pm 5$. The open interval is $(1, 5)$. The points $x=5$ and $x=-5$ are not included in the open interval $(1, 5)$. For all $x \in (1, 5)$, $25-x^2 > 0$, so $\sqrt{25-x^2}$ is a positive real number. Therefore, $f(x)$ is differentiable on the open interval $(1, 5)$. Condition 2 is satisfied.

Since both conditions of the Mean Value Theorem are satisfied, the theorem is applicable, and there exists at least one value $c \in (1, 5)$ such that $f'(c) = \frac{f(5) - f(1)}{5 - 1}$.

Finding the value(s) of c:

First, calculate the value of the right-hand side of the MVT formula, $\frac{f(b) - f(a)}{b - a}$.

Now, set $f'(c)$ equal to the value from equation (2) and solve for $c$. From equation (1), $f'(c) = -\frac{c}{\sqrt{25-c^2}}$.

Multiply both sides by -1:

Square both sides (Since $c \in (1, 5)$, $c$ is positive, so this step is valid):

Cross-multiply:

Add $3c^2$ to both sides:

Divide by 5:

Take the square root of both sides:

We need to check which of these values lie in the open interval $(1, 5)$.

For $c = \sqrt{15}$, we check if $1 < \sqrt{15} < 5$. Squaring the inequality, we get $1^2 < (\sqrt{15})^2 < 5^2$, which is $1 < 15 < 25$. This inequality is true. So, $c = \sqrt{15}$ is in the interval $(1, 5)$.

For $c = -\sqrt{15}$, this value is negative. The interval $(1, 5)$ contains only positive values. So, $c = -\sqrt{15}$ is not in the interval $(1, 5)$.

Thus, there exists exactly one value $c = \sqrt{15}$ in $(1, 5)$ that satisfies the Mean Value Theorem.

Conclusion:

The function $f(x) = \sqrt{25−x^2}$ is continuous on $[1, 5]$ and differentiable on $(1, 5)$. Thus, the conditions of the Mean Value Theorem are satisfied. We have found a value $c = \sqrt{15} \in (1, 5)$ such that $f'(\sqrt{15}) = \frac{f(5) - f(1)}{5 - 1}$. Therefore, the Mean Value Theorem is verified for the given function and interval.

Given:

The curve is $y = (x – 3)^2$.

The chord joins the points $(3, 0)$ and $(4, 1)$.

To Find:

A point on the curve where the tangent is parallel to the chord joining the given points.

Solution:

The problem asks for a point on the curve $y = f(x) = (x-3)^2$ where the tangent is parallel to the chord joining $(3, 0)$ and $(4, 1)$. The coordinates of these points lie on the curve:

This means the problem is asking for a point where the slope of the tangent is equal to the slope of the chord connecting $(3, 0)$ and $(4, 1)$. This is an application of the Mean Value Theorem.

The interval defined by the x-coordinates of the given points is $[3, 4]$. Let $a=3$ and $b=4$.

The function is $f(x) = (x-3)^2 = x^2 - 6x + 9$.

The function $f(x)$ is a polynomial, so it is continuous on the closed interval $[3, 4]$ and differentiable on the open interval $(3, 4)$. Thus, the conditions for the Mean Value Theorem are satisfied.

According to the Mean Value Theorem, there exists at least one value $c \in (3, 4)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

First, calculate the slope of the chord joining $(3, 0)$ and $(4, 1)$:

Next, find the derivative of $f(x) = (x-3)^2$, which represents the slope of the tangent at any point $x$:

Set the slope of the tangent $f'(c)$ equal to the slope of the chord (from equation 1) and solve for $c$:

The value $c = \frac{7}{2} = 3.5$. This value lies in the open interval $(3, 4)$, since $3 < 3.5 < 4$. This is the x-coordinate of the point where the tangent is parallel to the chord.

To find the y-coordinate of this point, substitute $x = c = \frac{7}{2}$ into the original function $y = (x-3)^2$:

The point on the curve where the tangent is parallel to the chord is $\left(\frac{7}{2}, \frac{1}{4}\right)$.

Final Answer:

The point on the curve $y = (x – 3)^2$ where the tangent is parallel to the chord joining the points (3, 0) and (4, 1) is $\left(\frac{7}{2}, \frac{1}{4}\right)$.

Given:

The curve is $y = 2x^2 – 5x + 3$.

The points on the curve are A(1, 0) and B(2, 1).

To Prove:

There is a point on the curve between A and B where the tangent is parallel to the chord AB, using the Mean Value Theorem.

To Find:

That specific point.

Proof/Solution (Using Mean Value Theorem):

The Mean Value Theorem (MVT) states that for a function $f(x)$ that is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, there exists at least one value $c \in (a, b)$ such that the slope of the tangent at $x=c$, $f'(c)$, is equal to the slope of the secant line (chord) joining the points $(a, f(a))$ and $(b, f(b))$. That is, $f'(c) = \frac{f(b) - f(a)}{b - a}$.

The given function is $f(x) = 2x^2 – 5x + 3$. The x-coordinates of the given points A(1, 0) and B(2, 1) define the interval $[a, b] = [1, 2]$. We first verify that points A and B lie on the curve:

Now, we check the conditions of the Mean Value Theorem for $f(x) = 2x^2 – 5x + 3$ on the interval $[1, 2]$.

1. Continuity on $[1, 2]$: $f(x)$ is a polynomial function. Polynomial functions are continuous on the entire real line. Thus, $f(x)$ is continuous on the closed interval $[1, 2]$.

2. Differentiability on $(1, 2)$: $f(x)$ is a polynomial function. Polynomial functions are differentiable on the entire real line. The derivative $f'(x) = 4x - 5$ is defined for all real $x$. Thus, $f(x)$ is differentiable on the open interval $(1, 2)$.

Since both conditions are met, the Mean Value Theorem is applicable. This proves that there exists at least one point $c \in (1, 2)$ such that the tangent at $x=c$ is parallel to the chord AB.

Finding the Point:

The slope of the chord joining points A(1, 0) and B(2, 1) is:

The derivative of the function $f(x)$ is:

According to the MVT, we set $f'(c)$ equal to the slope of the chord:

The value $c = \frac{3}{2} = 1.5$. This value lies in the open interval $(1, 2)$, since $1 < 1.5 < 2$. This is the x-coordinate of the point where the tangent is parallel to the chord AB.

To find the y-coordinate of this point, substitute $x = c = \frac{3}{2}$ into the original function $y = f(x) = 2x^2 – 5x + 3$:

The point on the curve is $\left(\frac{3}{2}, 0\right)$.

Conclusion:

The function $y = 2x^2 – 5x + 3$ is continuous on $[1, 2]$ and differentiable on $(1, 2)$. The Mean Value Theorem is applicable, guaranteeing the existence of a point $c \in (1, 2)$ where $f'(c)$ equals the slope of the chord AB. We found this point to be $\left(\frac{3}{2}, 0\right)$. Thus, the theorem is verified and the point is found.

Given:

The function $f(x) = \begin{cases}x^2+3x+p,& if\; x ≤ 1 \\ qx+2,& if\; x > 1 \end{cases}$.

To Find:

The values of $p$ and $q$ such that $f(x)$ is differentiable at $x = 1$.

Solution:

For a function to be differentiable at a point, it must first be continuous at that point.

Condition for Continuity at $x=1$:

The function $f(x)$ is continuous at $x=1$ if and only if $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x) = f(1)$.

Let's evaluate each part:

The function value at $x=1$ is given by the first case ($x \leq 1$):

$f(1) = (1)^2 + 3(1) + p = 1 + 3 + p = 4 + p$.

The left-hand limit at $x=1$ uses the first case ($x \leq 1$):

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x^2 + 3x + p)$.

Since $x^2 + 3x + p$ is a polynomial, the limit can be found by direct substitution:

$\lim\limits_{x \to 1^-} (x^2 + 3x + p) = (1)^2 + 3(1) + p = 1 + 3 + p = 4 + p$.

The right-hand limit at $x=1$ uses the second case ($x > 1$):

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (qx + 2)$.

Since $qx + 2$ is a polynomial, the limit can be found by direct substitution:

$\lim\limits_{x \to 1^+} (qx + 2) = q(1) + 2 = q + 2$.

For continuity at $x=1$, we must have $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x)$.

Condition for Differentiability at $x=1$:

The function $f(x)$ is differentiable at $x=1$ if and only if the left-hand derivative at $x=1$ equals the right-hand derivative at $x=1$.

The left-hand derivative at $x=1$ is given by $\lim\limits_{h \to 0^-} \frac{f(1+h) - f(1)}{h}$. For $x < 1$, $f'(x) = \frac{d}{dx}(x^2+3x+p) = 2x+3$.

The left-hand derivative at $x=1$ is $\lim\limits_{x \to 1^-} f'(x) = \lim\limits_{x \to 1^-} (2x+3)$.

$\lim\limits_{x \to 1^-} (2x+3) = 2(1) + 3 = 2 + 3 = 5$.

The right-hand derivative at $x=1$ is given by $\lim\limits_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$. For $x > 1$, $f'(x) = \frac{d}{dx}(qx+2) = q$.

The right-hand derivative at $x=1$ is $\lim\limits_{x \to 1^+} f'(x) = \lim\limits_{x \to 1^+} q$.

$\lim\limits_{x \to 1^+} q = q$.

For differentiability at $x=1$, the left-hand derivative must equal the right-hand derivative.

Solving for p and q:

From equation (2), we have the value of $q$ directly:

$q = 5$.

Substitute the value of $q$ into equation (1):

$4 + p = q + 2$

$4 + p = 5 + 2$

$4 + p = 7$

Now, solve for $p$:

$p = 7 - 4$

$p = 3$.

Conclusion:

The values of $p$ and $q$ for which $f(x)$ is differentiable at $x=1$ are $p=3$ and $q=5$.

Given:

The equation $x^m \cdot y^n = (x + y)^{m+n}$.

To Prove:

(i) $\frac{dy}{dx} = \frac{y}{x}$

(ii) $\frac{d^2y}{dx^2} = 0$

Proof:

We are given the equation:

$x^m y^n = (x+y)^{m+n}$

To simplify the differentiation, we take the natural logarithm on both sides of the equation:

$\log(x^m y^n) = \log((x+y)^{m+n})$

Using the logarithm properties $\log(AB) = \log A + \log B$ and $\log(A^p) = p \log A$, we get:

Now, differentiate both sides of equation (1) with respect to $x$. Remember that $y$ is a function of $x$, so we use the chain rule for $\log y$ and $\log(x+y)$.

$\frac{d}{dx}(m \log x) + \frac{d}{dx}(n \log y) = \frac{d}{dx}((m+n) \log(x+y))$

$m \cdot \frac{1}{x} + n \cdot \frac{1}{y} \frac{dy}{dx} = (m+n) \cdot \frac{1}{x+y} \frac{d}{dx}(x+y)$

$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} (1 + \frac{dy}{dx})$

Expand the right side of the equation:

Group the terms containing $\frac{dy}{dx}$ on one side and the other terms on the other side:

$\frac{n}{y} \frac{dy}{dx} - \frac{m+n}{x+y} \frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x}$

Factor out $\frac{dy}{dx}$ on the left side:

$\frac{dy}{dx} \left( \frac{n}{y} - \frac{m+n}{x+y} \right) = \frac{m+n}{x+y} - \frac{m}{x}$

Find a common denominator for the expressions inside the parentheses and on the right side:

$\frac{dy}{dx} \left( \frac{n(x+y) - y(m+n)}{y(x+y)} \right) = \frac{x(m+n) - m(x+y)}{x(x+y)}$

Simplify the numerators:

$\frac{dy}{dx} \left( \frac{nx + ny - my - ny}{y(x+y)} \right) = \frac{mx + nx - mx - my}{x(x+y)}$

$\frac{dy}{dx} \left( \frac{nx - my}{y(x+y)} \right) = \frac{nx - my}{x(x+y)}$

Now, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{\frac{nx - my}{x(x+y)}}{\frac{nx - my}{y(x+y)}}$

Assuming $nx - my \neq 0$ (otherwise $y/x$ is constant which contradicts the equation unless $m=n=0$ or $y \propto x$) and $x+y \neq 0$ (which must be true if $x^m y^n$ is positive for positive $x,y$), we can cancel the common factor $(nx-my)$ and $(x+y)$:

$\frac{dy}{dx} = \frac{nx - my}{x(x+y)} \times \frac{y(x+y)}{nx - my}$

$\frac{dy}{dx} = \frac{y}{x}$

This proves part (i).

Now we need to find the second derivative, $\frac{d^2y}{dx^2}$. We differentiate the result from part (i) with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{y}{x} \right)$

Using the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{\frac{du}{dx}v - u\frac{dv}{dx}}{v^2}$, with $u=y$ and $v=x$:

$\frac{d^2y}{dx^2} = \frac{\frac{dy}{dx} \cdot x - y \cdot \frac{dx}{dx}}{x^2}$

$\frac{d^2y}{dx^2} = \frac{x \frac{dy}{dx} - y \cdot 1}{x^2}$

$\frac{d^2y}{dx^2} = \frac{x \frac{dy}{dx} - y}{x^2}$

Substitute the value of $\frac{dy}{dx} = \frac{y}{x}$ from part (i) into this equation:

$\frac{d^2y}{dx^2} = \frac{x \left(\frac{y}{x}\right) - y}{x^2}$

$\frac{d^2y}{dx^2} = \frac{y - y}{x^2}$

$\frac{d^2y}{dx^2} = \frac{0}{x^2}$

$\frac{d^2y}{dx^2} = 0$

This proves part (ii).

Hence, we have proved that if $x^m \cdot y^n = (x+y)^{m+n}$, then $\frac{dy}{dx} = \frac{y}{x}$ and $\frac{d^2y}{dx^2} = 0$.

Given:

The parametric equations $x = \sin t$ and $y = \sin pt$.

To Prove:

$(1–x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2 y = 0$.

Proof:

We are given $x = \sin t$ and $y = \sin pt$. We need to find the first and second derivatives of $y$ with respect to $x$.

First, find the derivatives with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt}(\sin t) = \cos t$

$\frac{dy}{dt} = \frac{d}{dt}(\sin pt) = p \cos pt$

Now, find the first derivative $\frac{dy}{dx}$ using the chain rule $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

$\frac{dy}{dx} = \frac{p \cos pt}{\cos t}$

To find the second derivative $\frac{d^2y}{dx^2}$, we use the formula $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$.

First, find $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ using the quotient rule:

$\frac{d}{dt}\left(\frac{p \cos pt}{\cos t}\right) = p \cdot \frac{\frac{d}{dt}(\cos pt) \cos t - \cos pt \frac{d}{dt}(\cos t)}{(\cos t)^2}$

$= p \cdot \frac{(-p \sin pt) \cos t - \cos pt (-\sin t)}{\cos^2 t}$

$= p \cdot \frac{-p \sin pt \cos t + \sin t \cos pt}{\cos^2 t}$

Now, substitute this into the formula for $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = \frac{p \cdot \frac{\sin t \cos pt - p \sin pt \cos t}{\cos^2 t}}{\cos t}$

$\frac{d^2y}{dx^2} = \frac{p(\sin t \cos pt - p \sin pt \cos t)}{\cos^3 t}$

Now substitute $x$, $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ into the given differential equation $(1–x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2 y = 0$.

Recall that $x = \sin t$, so $1-x^2 = 1-\sin^2 t = \cos^2 t$. Also, $y = \sin pt$.

Substitute these into the left-hand side (LHS) of the equation:

LHS $= (1-\sin^2 t) \left(\frac{p(\sin t \cos pt - p \sin pt \cos t)}{\cos^3 t}\right) - (\sin t) \left(\frac{p \cos pt}{\cos t}\right) + p^2 (\sin pt)$

LHS $= (\cos^2 t) \left(\frac{p(\sin t \cos pt - p \sin pt \cos t)}{\cos^3 t}\right) - \frac{p \sin t \cos pt}{\cos t} + p^2 \sin pt$

Cancel out $\cos^2 t$ from the first term:

LHS $= \frac{p(\sin t \cos pt - p \sin pt \cos t)}{\cos t} - \frac{p \sin t \cos pt}{\cos t} + p^2 \sin pt$

Combine the first two terms which have a common denominator:

LHS $= \frac{p(\sin t \cos pt - p \sin pt \cos t) - p \sin t \cos pt}{\cos t} + p^2 \sin pt$

LHS $= \frac{p \sin t \cos pt - p^2 \sin pt \cos t - p \sin t \cos pt}{\cos t} + p^2 \sin pt$

Cancel the terms $p \sin t \cos pt$ and $-p \sin t \cos pt$ in the numerator:

LHS $= \frac{-p^2 \sin pt \cos t}{\cos t} + p^2 \sin pt$

Cancel $\cos t$ (assuming $\cos t \neq 0$):

LHS $= -p^2 \sin pt + p^2 \sin pt$

LHS $= 0$

This is equal to the right-hand side (RHS) of the given differential equation.

Hence, $(1–x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2 y = 0$.

The proof is complete.

Given:

The function $y = x^{\tan x} + \sqrt{\frac{x^2 + 1}{2}}$.

To Find:

$\frac{dy}{dx}$.

Solution:

We are asked to find the derivative of the function $y$ with respect to $x$. The function is a sum of two terms. Let's denote the terms as $u$ and $v$, where $u = x^{\tan x}$ and $v = \sqrt{\frac{x^2 + 1}{2}}$.

So, $y = u + v$.

To find $\frac{dy}{dx}$, we can find the derivatives of $u$ and $v$ separately and then add them: $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.

Finding $\frac{du}{dx}$ for $u = x^{\tan x}$:

This is a function of the form $f(x)^{g(x)}$. We use logarithmic differentiation.

Take the natural logarithm on both sides:

$\log u = \log(x^{\tan x})$

Using the property of logarithms $\log(a^b) = b \log a$:

$\log u = \tan x \log x$

Now, differentiate both sides with respect to $x$. On the left side, we use the chain rule, and on the right side, we use the product rule $\frac{d}{dx}(fg) = f'g + fg'$.

$\frac{d}{dx}(\log u) = \frac{d}{dx}(\tan x \log x)$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{d}{dx}(\tan x)\right) (\log x) + (\tan x) \left(\frac{d}{dx}(\log x)\right)$

$\frac{1}{u} \frac{du}{dx} = (\sec^2 x) (\log x) + (\tan x) \left(\frac{1}{x}\right)$

$\frac{1}{u} \frac{du}{dx} = \sec^2 x \log x + \frac{\tan x}{x}$

Now, multiply both sides by $u$ to solve for $\frac{du}{dx}$:

$\frac{du}{dx} = u \left( \sec^2 x \log x + \frac{\tan x}{x} \right)$

Substitute the expression for $u$ back into the equation:

$\frac{du}{dx} = x^{\tan x} \left( \sec^2 x \log x + \frac{\tan x}{x} \right)$

Finding $\frac{dv}{dx}$ for $v = \sqrt{\frac{x^2 + 1}{2}}$:

We can rewrite $v$ as $v = \left(\frac{x^2 + 1}{2}\right)^{1/2}$. Alternatively, $v = \frac{1}{\sqrt{2}} (x^2 + 1)^{1/2}$.

Let's differentiate $v$ with respect to $x$ using the chain rule $\frac{d}{dx}(f(g(x))) = f'(g(x))g'(x)$. Here, the outer function is $(\cdot)^{1/2}$ and the inner function is $\frac{x^2+1}{2}$.

$\frac{dv}{dx} = \frac{d}{dx} \left( \left(\frac{x^2 + 1}{2}\right)^{1/2} \right)$

$\frac{dv}{dx} = \frac{1}{2} \left(\frac{x^2 + 1}{2}\right)^{1/2 - 1} \cdot \frac{d}{dx}\left(\frac{x^2 + 1}{2}\right)$

$\frac{dv}{dx} = \frac{1}{2} \left(\frac{x^2 + 1}{2}\right)^{-1/2} \cdot \frac{1}{2} \frac{d}{dx}(x^2 + 1)$

$\frac{dv}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{x^2 + 1}{2}}} \cdot \frac{1}{2} (2x + 0)$

$\frac{dv}{dx} = \frac{1}{2} \cdot \frac{\sqrt{2}}{\sqrt{x^2 + 1}} \cdot \frac{1}{2} (2x)$

$\frac{dv}{dx} = \frac{1}{2} \cdot \frac{\sqrt{2}}{\sqrt{x^2 + 1}} \cdot x$

$\frac{dv}{dx} = \frac{\sqrt{2} x}{2 \sqrt{x^2 + 1}}$

We can simplify this by writing $2 = \sqrt{2} \cdot \sqrt{2}$:

$\frac{dv}{dx} = \frac{\sqrt{2} x}{\sqrt{2} \sqrt{2} \sqrt{x^2 + 1}}$

Cancel one $\sqrt{2}$:

$\frac{dv}{dx} = \frac{x}{\sqrt{2}\sqrt{x^2 + 1}}$

This can be written under a single square root:

$\frac{dv}{dx} = \frac{x}{\sqrt{2(x^2 + 1)}}$

Combining the derivatives to find $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Substitute the expressions we found for $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$\frac{dy}{dx} = x^{\tan x} \left( \sec^2 x \log x + \frac{\tan x}{x} \right) + \frac{x}{\sqrt{2(x^2 + 1)}}$

Final Answer:

The derivative of the function $y = x^{\tan x} + \sqrt{\frac{x^2 + 1}{2}}$ is:

$\frac{dy}{dx} = x^{\tan x} \left( \sec^2 x \log x + \frac{\tan x}{x} \right) + \frac{x}{\sqrt{2(x^2 + 1)}}$

Given:

$f(x) = 2x$

$g(x) = \frac{x^2}{2} + 1$

Analysis:

The function $f(x) = 2x$ is a polynomial function. Polynomials are continuous for all real numbers.

The function $g(x) = \frac{x^2}{2} + 1$ is also a polynomial function. Polynomials are continuous for all real numbers.

Now let's examine the continuity of the given options:

(A) $f(x) + g(x)$: The sum of two continuous functions is always continuous.

$f(x) + g(x) = 2x + \left(\frac{x^2}{2} + 1\right) = \frac{x^2}{2} + 2x + 1$, which is a polynomial and thus continuous everywhere.

(B) $f(x) – g(x)$: The difference of two continuous functions is always continuous.

$f(x) - g(x) = 2x - \left(\frac{x^2}{2} + 1\right) = -\frac{x^2}{2} + 2x - 1$, which is a polynomial and thus continuous everywhere.

(C) $f(x) \cdot g(x)$: The product of two continuous functions is always continuous.

$f(x) \cdot g(x) = (2x) \left(\frac{x^2}{2} + 1\right) = x^3 + 2x$, which is a polynomial and thus continuous everywhere.

(D) $\frac{g(x)}{f(x)}$: The quotient of two continuous functions is continuous everywhere except where the denominator is zero.

$\frac{g(x)}{f(x)} = \frac{\frac{x^2}{2} + 1}{2x}$

The denominator is $f(x) = 2x$. The denominator is zero when $2x = 0$, which means $x=0$.

Therefore, the function $\frac{g(x)}{f(x)}$ is discontinuous at $x=0$.

Conclusion:

The function that can be discontinuous among the given options is $\frac{g(x)}{f(x)}$.

Answer:

The correct option is (D) $\frac{g(x)}{f(x)}$.

Given:

The function $f(x) = \frac{4 − x^2}{4x − x^3}$.

Analysis:

A rational function is of the form $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials. A rational function is continuous at all points where the denominator $Q(x)$ is not equal to zero. The function is discontinuous at the points where the denominator is equal to zero.

In this case, the numerator is $P(x) = 4 - x^2$ and the denominator is $Q(x) = 4x - x^3$.

To find the points of discontinuity, we need to find the values of $x$ for which the denominator is zero:

$Q(x) = 4x - x^3 = 0$

Factor out $x$ from the expression:

$x(4 - x^2) = 0$

The term $(4 - x^2)$ is a difference of squares, which can be factored as $(2-x)(2+x)$.

So, the equation becomes:

$x(2 - x)(2 + x) = 0$

This equation is satisfied if any of the factors are zero:

1. $x = 0$

2. $2 - x = 0 \implies x = 2$

3. $2 + x = 0 \implies x = -2$

The denominator is zero at $x = 0$, $x = 2$, and $x = -2$. These are the points where the function $f(x)$ is undefined and thus discontinuous.

There are exactly three distinct points where the function is discontinuous.

Conclusion:

The function $f(x) = \frac{4 − x^2}{4x − x^3}$ is discontinuous at $x = -2$, $x = 0$, and $x = 2$. Therefore, it is discontinuous at exactly three points.

Answer:

The correct option is (C) discontinuous at exactly three points.

Given:

The function $f(x) = |2x - 1| \sin x$.

Analysis:

The function $f(x)$ is a product of two functions: $u(x) = |2x - 1|$ and $v(x) = \sin x$.

The function $v(x) = \sin x$ is differentiable for all real numbers $x \in R$.

The function $u(x) = |2x - 1|$ involves an absolute value. The absolute value function $|y|$ is differentiable everywhere except at $y=0$. Here, $y = 2x - 1$. So, $|2x - 1|$ is potentially not differentiable when $2x - 1 = 0$, which means $x = \frac{1}{2}$.

We can write $u(x)$ as a piecewise function:

$|2x - 1| = \begin{cases} 2x - 1 & , & \text{if } 2x - 1 \geq 0 \implies x \geq \frac{1}{2} \\ -(2x - 1) = 1 - 2x & , & \text{if } 2x - 1 < 0 \implies x < \frac{1}{2} \end{cases}$

So, the function $f(x)$ can be written as:

$f(x) = \begin{cases} (2x - 1) \sin x & , & \text{if } x \geq \frac{1}{2} \\ (1 - 2x) \sin x & , & \text{if } x < \frac{1}{2} \end{cases}$

For $x > \frac{1}{2}$, $f(x) = (2x - 1) \sin x$. This is a product of two differentiable functions ($2x-1$ and $\sin x$), so it is differentiable for all $x > \frac{1}{2}$.

For $x < \frac{1}{2}$, $f(x) = (1 - 2x) \sin x$. This is a product of two differentiable functions ($1-2x$ and $\sin x$), so it is differentiable for all $x < \frac{1}{2}$.

The only point where differentiability needs to be explicitly checked is at $x = \frac{1}{2}$.

Differentiability Check at $x = \frac{1}{2}$:

For a function to be differentiable at $x = a$, the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x = a$ must exist and be equal.

The value of the function at $x = \frac{1}{2}$ is:

$f(\frac{1}{2}) = |2(\frac{1}{2}) - 1| \sin(\frac{1}{2}) = |1 - 1| \sin(\frac{1}{2}) = |0| \sin(\frac{1}{2}) = 0$.

The left-hand derivative at $x = \frac{1}{2}$ is:

$\text{LHD} = \lim\limits_{h \to 0^-} \frac{f(\frac{1}{2} + h) - f(\frac{1}{2})}{h}$

Substitute $f(\frac{1}{2})$ and $f(\frac{1}{2} + h)$: For $h < 0$ and small, $\frac{1}{2} + h < \frac{1}{2}$, so $f(\frac{1}{2} + h) = (1 - 2(\frac{1}{2} + h)) \sin(\frac{1}{2} + h) = (1 - 1 - 2h) \sin(\frac{1}{2} + h) = -2h \sin(\frac{1}{2} + h)$.

$\text{LHD} = \lim\limits_{h \to 0^-} \frac{-2h \sin(\frac{1}{2} + h) - 0}{h}$

$\text{LHD} = \lim\limits_{h \to 0^-} \frac{-2h \sin(\frac{1}{2} + h)}{h}$

Cancel $h$ (since $h \neq 0$):

$\text{LHD} = \lim\limits_{h \to 0^-} -2 \sin(\frac{1}{2} + h)$

As $h \to 0^-$, $\frac{1}{2} + h \to \frac{1}{2}$. Since $\sin x$ is continuous, $\lim\limits_{h \to 0^-} \sin(\frac{1}{2} + h) = \sin(\frac{1}{2})$.

The right-hand derivative at $x = \frac{1}{2}$ is:

$\text{RHD} = \lim\limits_{h \to 0^+} \frac{f(\frac{1}{2} + h) - f(\frac{1}{2})}{h}$

Substitute $f(\frac{1}{2})$ and $f(\frac{1}{2} + h)$: For $h > 0$ and small, $\frac{1}{2} + h > \frac{1}{2}$, so $f(\frac{1}{2} + h) = (2(\frac{1}{2} + h) - 1) \sin(\frac{1}{2} + h) = (1 + 2h - 1) \sin(\frac{1}{2} + h) = 2h \sin(\frac{1}{2} + h)$.

$\text{RHD} = \lim\limits_{h \to 0^+} \frac{2h \sin(\frac{1}{2} + h) - 0}{h}$

$\text{RHD} = \lim\limits_{h \to 0^+} \frac{2h \sin(\frac{1}{2} + h)}{h}$

Cancel $h$ (since $h \neq 0$):

$\text{RHD} = \lim\limits_{h \to 0^+} 2 \sin(\frac{1}{2} + h)$

As $h \to 0^+$, $\frac{1}{2} + h \to \frac{1}{2}$. Since $\sin x$ is continuous, $\lim\limits_{h \to 0^+} \sin(\frac{1}{2} + h) = \sin(\frac{1}{2})$.

For differentiability at $x = \frac{1}{2}$, LHD must equal RHD:

$-2 \sin(\frac{1}{2}) = 2 \sin(\frac{1}{2})$

Adding $2 \sin(\frac{1}{2})$ to both sides:

Equation (3) implies $\sin(\frac{1}{2}) = 0$.

The value $\frac{1}{2}$ in $\sin(\frac{1}{2})$ is in radians. The general solution for $\sin \theta = 0$ is $\theta = n\pi$, where $n$ is an integer. Thus, $\sin(\frac{1}{2}) = 0$ if and only if $\frac{1}{2} = n\pi$ for some integer $n$. This is false, as $\frac{1}{2}$ is not an integer multiple of $\pi$.

Therefore, $\sin(\frac{1}{2}) \neq 0$, which means LHD $\neq$ RHD at $x = \frac{1}{2}$.

The function $f(x)$ is not differentiable at $x = \frac{1}{2}$.

Conclusion:

The function $f(x) = |2x - 1| \sin x$ is differentiable for all real numbers except at $x = \frac{1}{2}$. The set of points where the function is differentiable is $R - \left\{ \frac{1}{2} \right\}$.

Answer:

The correct option is (B) R - $\left\{ \frac{1}{2} \right\}$.

Given:

The function $f(x) = \cot x$.

Analysis:

The function $f(x) = \cot x$ can be written as $f(x) = \frac{\cos x}{\sin x}$.

A trigonometric function defined as a ratio of two continuous functions (like $\cos x$ and $\sin x$) is discontinuous at the points where the denominator is equal to zero.

In this case, the denominator is $\sin x$. The function $f(x) = \cot x$ is discontinuous when $\sin x = 0$.

We need to find the values of $x$ for which $\sin x = 0$.

The general solution for $\sin x = 0$ is given by:

The set of points where $\sin x = 0$ is $\{x = n\pi : n \in Z\}$.

These are the points where the function $f(x) = \cot x$ is undefined and therefore discontinuous.

Let's examine the given options:

(A) {x = nπ : n ϵ Z} - This is the set of all integer multiples of $\pi$. This matches our finding.

(B) {x = 2nπ : n ϵ Z} - This is the set of even integer multiples of $\pi$. This is a subset of the points where $\sin x = 0$, but it does not include all of them (e.g., $\pi, 3\pi, -\pi$ are missing).

(C) $\left\{ x = (2n+1) \frac{π}{2} ; n \in Z \right\}$ - This is the set of odd integer multiples of $\frac{\pi}{2}$. These are the points where $\cos x = 0$, not $\sin x = 0$.

(D) $\left\{ x = \frac{nπ}{2} ; n \in Z \right\}$ - This is the set of all integer multiples of $\frac{\pi}{2}$. This includes points where $\cos x = 0$ (like $\frac{\pi}{2}, \frac{3\pi}{2}$) in addition to points where $\sin x = 0$ (like $0, \pi, 2\pi$). The function $\cot x = \frac{\cos x}{\sin x}$ is undefined only when the denominator $\sin x$ is zero, not when the numerator $\cos x$ is zero.

Conclusion:

The function $f(x) = \cot x$ is discontinuous at the points where $\sin x = 0$, which are $x = n\pi$ for any integer $n$. This set is represented by option (A).

Answer:

The correct option is (A) {x = nπ : n ϵ Z}.

Given:

The function $f(x) = e^{|x|}$.

Analysis:

The function $f(x) = e^{|x|}$ can be written as a piecewise function based on the definition of the absolute value function:

$|x| = \begin{cases} x & , & \text{if } x \geq 0 \\ -x & , & \text{if } x < 0 \end{cases}$

So, $f(x)$ is defined as:

$f(x) = \begin{cases} e^x & , & \text{if } x \geq 0 \\ e^{-x} & , & \text{if } x < 0 \end{cases}$

Continuity Analysis:

For $x > 0$, $f(x) = e^x$, which is an exponential function and is continuous for all real numbers. So, $f(x)$ is continuous for $x > 0$.

For $x < 0$, $f(x) = e^{-x}$, which is a composition of exponential and linear functions and is continuous for all real numbers. So, $f(x)$ is continuous for $x < 0$.

We need to check continuity at the transition point $x = 0$. For continuity at $x=0$, we must have $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$.

Let's evaluate each part:

$f(0) = e^{|0|} = e^0 = 1$.

Left-hand limit at $x=0$:

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} e^{-x}$

As $x \to 0^-$, $-x \to 0^+$. Since the exponential function is continuous, $\lim\limits_{x \to 0^-} e^{-x} = e^0 = 1$.

Right-hand limit at $x=0$:

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} e^x$

As $x \to 0^+$, $x \to 0^+$. Since the exponential function is continuous, $\lim\limits_{x \to 0^+} e^x = e^0 = 1$.

Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0) = 1$, the function $f(x)$ is continuous at $x=0$.

Thus, $f(x)$ is continuous everywhere on R.

Differentiability Analysis:

For $x > 0$, $f(x) = e^x$. The derivative is $f'(x) = \frac{d}{dx}(e^x) = e^x$. This derivative exists for all $x > 0$.

For $x < 0$, $f(x) = e^{-x}$. The derivative is $f'(x) = \frac{d}{dx}(e^{-x}) = -e^{-x}$. This derivative exists for all $x < 0$.

We need to check differentiability at the transition point $x = 0$. For a function to be differentiable at $x=0$, the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x=0$ must exist and be equal.

The left-hand derivative at $x=0$ is given by: