Chapter 5 Continuity and Differentiability (Concepts)

To examine the continuity of $f(x) = x^2$ at $x=2$, we check the three conditions for continuity at a point $c=2$.

1. Check if $f(c)$ exists:

Substitute $x=2$ into the function:

$f(2) = (2)^2 = 4$.

Since $f(2)$ is a finite value, $f(c)$ exists at $x=2$.

2. Check if $\lim\limits_{x \to c} f(x)$ exists:

We need to evaluate the limit of $f(x)$ as $x$ approaches 2:

Since $f(x) = x^2$ is a polynomial function, it is continuous everywhere, and the limit as $x \to c$ can be found by direct substitution.

Since the limit is a finite value, $\lim\limits_{x \to 2} f(x)$ exists.

3. Check if $\lim\limits_{x \to c} f(x) = f(c)$:

From step 1, $f(2) = 4$.

From step 2, $\lim\limits_{x \to 2} f(x) = 4$.

Comparing the two values:

Therefore,

Since all three conditions for continuity at a point are satisfied, the function $f(x) = x^2$ is continuous at $x=2$.

To examine the continuity of $f(x)$ at $x=0$, we check the three conditions for continuity at a point $c=0$.

1. Check if $f(c)$ exists:

According to the definition of the function, when $x=0$, $f(x)=0$.

$f(0)$ exists and is equal to 0.

2. Check if $\lim\limits_{x \to c} f(x)$ exists:

Since the function's definition changes at $x=0$ (due to $|x|$ and the piecewise definition), we need to evaluate the left-hand limit (LHL) and the right-hand limit (RHL) at $x=0$.

For the LHL, we consider values of $x$ approaching 0 from the left, i.e., $x < 0$. When $x < 0$, $|x| = -x$.

For the RHL, we consider values of $x$ approaching 0 from the right, i.e., $x > 0$. When $x > 0$, $|x| = x$.

From (ii) and (iii), we see that the left-hand limit and the right-hand limit are not equal:

Since LHL $\neq$ RHL, the limit $\lim\limits_{x \to 0} f(x)$ does not exist.

3. Check if $\lim\limits_{x \to c} f(x) = f(c)$:

Since the limit $\lim\limits_{x \to 0} f(x)$ does not exist (as shown in step 2), the third condition for continuity cannot be satisfied (it requires the limit to exist and be equal to $f(c)$).

Therefore, the function $f(x)$ is discontinuous at $x=0$. Specifically, this is a jump discontinuity because the LHL and RHL exist but are not equal.

We need to show that $f(x) = \sin(x^2)$ is continuous for all real numbers $x$.

The function $f(x) = \sin(x^2)$ can be viewed as a composite function.

Let $g(x) = \sin x$ and $h(x) = x^2$.

Then the composite function $(g \circ h)(x) = g(h(x)) = g(x^2) = \sin(x^2) = f(x)$.

Now we examine the continuity of $g(x)$ and $h(x)$ separately.

1. The function $g(x) = \sin x$ is a sine function. We know that the sine function is continuous for all real numbers $x \in R$.

2. The function $h(x) = x^2$ is a polynomial function of degree 2. We know that all polynomial functions are continuous for all real numbers $x \in R$.

Since $h(x) = x^2$ is continuous for all $x \in R$, the range of $h(x)$ (which is $[0, \infty)$) consists of values that are real numbers.

Since $g(x) = \sin x$ is continuous for all real numbers, it is continuous for all values in the range of $h(x)$.

According to the property of continuity of composite functions, if $h$ is continuous at $c$ and $g$ is continuous at $h(c)$, then $g \circ h$ is continuous at $c$. Since $h$ is continuous for all $x \in R$ and $g$ is continuous for all real numbers (and thus at $h(x)$ for any $x$), their composition $(g \circ h)(x) = \sin(x^2)$ is continuous for all real numbers $x$.

Therefore, the function $f(x) = \sin(x^2)$ is a continuous function.

Given: A function $f$ is differentiable at a point $c$.

By the definition of differentiability, the limit $\lim\limits_{x \to c} \frac{f(x) - f(c)}{x-c}$ exists and is equal to $f'(c)$.

To Prove: $f$ is continuous at $c$.

To prove that $f$ is continuous at $c$, we need to show that $\lim\limits_{x \to c} f(x) = f(c)$. This is equivalent to showing that $\lim\limits_{x \to c} (f(x) - f(c)) = 0$.

Consider the expression $f(x) - f(c)$ for $x \neq c$. We can write this expression by multiplying and dividing by $(x-c)$:

Now, let's take the limit as $x \to c$ on both sides of the equation:

Using the property of limits that the limit of a product is the product of the limits, provided the individual limits exist:

We know from the definition of differentiability that $\lim\limits_{x \to c} \frac{f(x) - f(c)}{x-c} = f'(c)$, which exists and is a finite value.

We also know that $\lim\limits_{x \to c} (x-c) = c-c = 0$.

Substituting these values into the equation:

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

Since $f(c)$ is a constant value (the function evaluated at a fixed point $c$), its limit as $x \to c$ is simply $f(c)$.

Rearranging the equation, we get:

This is the definition of continuity of the function $f$ at the point $c$.

Thus, we have proved that if a function $f$ is differentiable at a point $c$, it is necessarily continuous at $c$.

Given: The function $f(x) = |x|$.

The function can be written as a piecewise function:

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

To Examine: Continuity and differentiability of $f(x)$ at the point $x=0$.

Examination of Continuity at $x=0$

We check the three conditions for continuity at $c=0$:

1. $f(c)$ must exist:

Substitute $x=0$ into the function definition: $f(0) = |0| = 0$.

The function is defined at $x=0$.

2. $\lim\limits_{x \to c} f(x)$ must exist:

This requires the LHL and RHL at $x=0$ to exist and be equal.

For LHL, $x \to 0^-$, meaning $x < 0$. In this case, $f(x) = -x$.

For RHL, $x \to 0^+$, meaning $x > 0$. In this case, $f(x) = x$.

Since LHL = RHL = 0, the limit $\lim\limits_{x \to 0} f(x)$ exists and is equal to 0.

3. $\lim\limits_{x \to c} f(x) = f(c)$:

From step 1, $f(0) = 0$. From step 2, $\lim\limits_{x \to 0} f(x) = 0$.

Since all three conditions are met, the function $f(x) = |x|$ is continuous at $x=0$.

Examination of Differentiability at $x=0$

We check the left-hand derivative (LHD) and right-hand derivative (RHD) at $c=0$. We use the definition $\lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$, considering $h \to 0^+$ for RHD and $h \to 0^+$ on $\frac{f(c-h)-f(c)}{-h}$ for LHD. Here $c=0$.

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

For $h > 0$, $(0-h) = -h < 0$. So, $f(0-h) = |-h| = h$. Also, $f(0) = 0$.

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

For $h > 0$, $(0+h) = h > 0$. So, $f(0+h) = |h| = h$. Also, $f(0) = 0$.

From (viii) and (ix), we have LHD at $0 = -1$ and RHD at $0 = 1$.

Since the left-hand derivative is not equal to the right-hand derivative at $x=0$, the derivative $f'(0) = \lim\limits_{x \to 0} \frac{f(x) - f(0)}{x-0}$ does not exist.

Therefore, the function $f(x) = |x|$ is not differentiable at $x=0$.

This example clearly demonstrates that a function can be continuous at a point (as $f(x)=|x|$ is at $x=0$) but not differentiable at that same point. The point $x=0$ is a point where the graph of $f(x)=|x|$ has a sharp turn or a cusp.

Let the given function be $y = \sin(x^2 + 5)$.

This is a composite function. We can identify the outer function as the sine function and the inner function as the expression inside the sine, which is $x^2 + 5$.

Let $u = x^2 + 5$. Then the function becomes $y = \sin u$.

According to the Chain Rule (Equation i), $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

First, we find the derivative of $y$ with respect to $u$:

Next, we find the derivative of $u$ with respect to $x$:

Using the power rule and the derivative of a constant:

Now, we multiply the two derivatives according to the Chain Rule:

Finally, substitute back $u = x^2 + 5$ to express the derivative in terms of $x$:

Thus, the derivative of $\sin(x^2 + 5)$ with respect to $x$ is $2x \cos(x^2 + 5)$.

Alternatively, using function notation (Equation ii), let $f(v) = \sin v$ and $g(x) = x^2 + 5$. Then $y = f(g(x))$.

The derivative of the outer function $f$ is $f'(v) = \frac{d}{dv}(\sin v) = \cos v$.

The derivative of the inner function $g$ is $g'(x) = \frac{d}{dx}(x^2 + 5) = 2x$.

According to the chain rule, $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$.

Both methods yield the same result.

Let $y = \tan(2x+3)$.

This is a composite function. The outer function is $\tan(u)$ and the inner function is $u = 2x+3$.

Using the Chain Rule (Equation i), $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Let $u = 2x+3$. Then $y = \tan u$.

Derivative of $y$ with respect to $u$:

Derivative of $u$ with respect to $x$:

Applying the Chain Rule:

Substitute back $u = 2x+3$:

The derivative of $\tan(2x+3)$ is $2 \sec^2(2x+3)$.

Let $y = \sin(\cos(x^2))$.

This function is a composition of three functions: the outermost function is sine, the next inner function is cosine, and the innermost function is $x^2$.

We can apply the extended Chain Rule (Equation iii): $\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx}$, where $y = f(v)$, $v = g(u)$, and $u = h(x)$.

Let $v = \cos(x^2)$. Then $y = \sin v$.

Let $u = x^2$. Then $v = \cos u$.

So, we have the chain $y = \sin v$, $v = \cos u$, and $u = x^2$.

Now, we find the derivatives of each link in the chain:

Multiply these derivatives according to the extended Chain Rule:

Finally, substitute back $v = \cos(x^2)$ and $u = x^2$ to express the derivative in terms of $x$:

Rearranging the terms:

The derivative of $\sin(\cos(x^2))$ with respect to $x$ is $-2x \sin(x^2) \cos(\cos(x^2))$.

Given: A function $y = \sin^{-1} x$.

To Find: $\frac{dy}{dx}$.

The definition of the inverse sine function states that if $y = \sin^{-1} x$, then $x = \sin y$.

The domain of $y = \sin^{-1} x$ is $[-1, 1]$, so $-1 \leq x \leq 1$.

The principal value range of $y = \sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$.

Differentiate both sides of the equation $x = \sin y$ with respect to $x$:

The derivative of $x$ with respect to $x$ is 1. For the right side, we use the Chain Rule, treating $y$ as a function of $x$:

Now, solve for $\frac{dy}{dx}$:

We need to express $\cos y$ in terms of $x$. We use the trigonometric identity $\sin^2 y + \cos^2 y = 1$.

$\cos^2 y = 1 - \sin^2 y$

Taking the square root of both sides:

$\cos y = \pm \sqrt{1 - \sin^2 y}$

Since $x = \sin y$, we substitute this into the equation for $\cos y$:

Now, consider the range of $y$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. In this interval, the cosine function is non-negative ($\cos y \ge 0$).

Therefore, we must take the positive square root:

Substitute this expression for $\cos y$ back into equation (ii):

This derivative is valid for $-1 < x < 1$. At $x = \pm 1$, the denominator $\sqrt{1-x^2}$ becomes 0, indicating that the tangent to the graph of $y = \sin^{-1} x$ is vertical at these points, and the function is not differentiable.

Given: A function $y = \cos^{-1} x$.

To Find: $\frac{dy}{dx}$.

The definition states that if $y = \cos^{-1} x$, then $x = \cos y$.

The domain is $[-1, 1]$ ($-1 \leq x \leq 1$). The principal value range is $[0, \pi]$ ($0 \leq y \leq \pi$).

Differentiate $x = \cos y$ with respect to $x$:

$1 = \frac{d}{dy}(\cos y) \cdot \frac{dy}{dx}$ (using Chain Rule)

Solve for $\frac{dy}{dx}$:

We need to express $\sin y$ in terms of $x$. From $\sin^2 y + \cos^2 y = 1$, we get $\sin y = \pm \sqrt{1 - \cos^2 y}$.

Since $x = \cos y$, $\sin y = \pm \sqrt{1 - x^2}$.

Now, consider the range of $y$, which is $[0, \pi]$. In this interval, the sine function is non-negative ($\sin y \ge 0$).

Therefore, we take the positive square root:

Substitute this expression for $\sin y$ back into equation (v):

This derivative is valid for $-1 < x < 1$.

Given: A function $y = \tan^{-1} x$.

To Find: $\frac{dy}{dx}$.

By definition, $x = \tan y$.

The domain is $R$ ($x \in R$). The principal value range is $(-\frac{\pi}{2}, \frac{\pi}{2})$ ($-\frac{\pi}{2} < y < \frac{\pi}{2}$).

Differentiate $x = \tan y$ with respect to $x$:

$1 = \frac{d}{dy}(\tan y) \cdot \frac{dy}{dx}$ (using Chain Rule)

Solve for $\frac{dy}{dx}$:

We need to express $\sec^2 y$ in terms of $x$. Use the identity $1 + \tan^2 y = \sec^2 y$.

Since $x = \tan y$:

Substitute this into equation (viii):

This derivative is valid for all real numbers $x$, as $1+x^2$ is always positive.

Given: A function $y = \cot^{-1} x$.

To Find: $\frac{dy}{dx}$.

By definition, $x = \cot y$.

The domain is $R$ ($x \in R$). The principal value range is $(0, \pi)$ ($0 < y < \pi$).

Differentiate $x = \cot y$ with respect to $x$:

$1 = \frac{d}{dy}(\cot y) \cdot \frac{dy}{dx}$ (using Chain Rule)

Solve for $\frac{dy}{dx}$:

We need to express $\text{cosec}^2 y$ in terms of $x$. Use the identity $1 + \cot^2 y = \text{cosec}^2 y$.

Since $x = \cot y$:

Substitute this into equation (xi):

This derivative is valid for all real numbers $x$.

Given: A function $y = \sec^{-1} x$.

To Find: $\frac{dy}{dx}$.

By definition, $x = \sec y$.

The domain is $R - (-1, 1)$, i.e., $|x| \geq 1$. The principal value range is $[0, \pi] - \{\frac{\pi}{2}\}$.

Differentiate $x = \sec y$ with respect to $x$:

$1 = \frac{d}{dy}(\sec y) \cdot \frac{dy}{dx}$ (using Chain Rule)

Solve for $\frac{dy}{dx}$:

We know $\sec y = x$. We need to express $\tan y$ in terms of $x$. Use the identity $\tan^2 y = \sec^2 y - 1$.

$\tan y = \pm \sqrt{\sec^2 y - 1} = \pm \sqrt{x^2 - 1}$.

Consider the principal value range $[0, \pi] - \{\frac{\pi}{2}\}$:

• If $y \in [0, \frac{\pi}{2})$, then $\sec y \ge 1$ (i.e., $x \ge 1$) and $\tan y \ge 0$. So $\tan y = \sqrt{x^2-1}$. In this case, $\sec y \tan y = x \sqrt{x^2-1}$.
• If $y \in (\frac{\pi}{2}, \pi]$, then $\sec y \le -1$ (i.e., $x \le -1$) and $\tan y \le 0$. So $\tan y = -\sqrt{x^2-1}$. In this case, $\sec y \tan y = x (-\sqrt{x^2-1}) = -x\sqrt{x^2-1}$.

We can express $\sec y \tan y$ compactly using $|x|$. For $x > 1$, $x\sqrt{x^2-1} = |x|\sqrt{x^2-1}$. For $x < -1$, $-x\sqrt{x^2-1} = (-x)\sqrt{x^2-1} = |-x|\sqrt{x^2-1} = |x|\sqrt{x^2-1}$.

So, $\sec y \tan y = |x|\sqrt{x^2-1}$ for $|x| > 1$.

Substitute this into equation (xiv):

This derivative is valid for $|x| > 1$.

Given: A function $y = \text{cosec}^{-1} x$.

To Find: $\frac{dy}{dx}$.

By definition, $x = \text{cosec } y$.

The domain is $R - (-1, 1)$, i.e., $|x| \geq 1$. The principal value range is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

Differentiate $x = \text{cosec } y$ with respect to $x$:

$1 = \frac{d}{dy}(\text{cosec } y) \cdot \frac{dy}{dx}$ (using Chain Rule)

Solve for $\frac{dy}{dx}$:

We know $\text{cosec } y = x$. We need to express $\cot y$ in terms of $x$. Use the identity $\cot^2 y = \text{cosec}^2 y - 1$.

$\cot y = \pm \sqrt{\text{cosec}^2 y - 1} = \pm \sqrt{x^2 - 1}$.

Consider the principal value range $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$:

• If $y \in (0, \frac{\pi}{2}]$, then $\text{cosec } y \ge 1$ (i.e., $x \ge 1$) and $\cot y \ge 0$. So $\cot y = \sqrt{x^2-1}$. In this case, $\text{cosec } y \cot y = x \sqrt{x^2-1}$.
• If $y \in [-\frac{\pi}{2}, 0)$, then $\text{cosec } y \le -1$ (i.e., $x \le -1$) and $\cot y \le 0$. So $\cot y = -\sqrt{x^2-1}$. In this case, $\text{cosec } y \cot y = x (-\sqrt{x^2-1}) = -x\sqrt{x^2-1}$.

Similar to the derivation for $\sec^{-1} x$, we see that $\text{cosec } y \cot y = |x|\sqrt{x^2-1}$ for $|x| > 1$.

Substitute this into equation (xvii):

This derivative is valid for $|x| > 1$.

Given: The function $y = \sin^{-1}(2x)$.

To Find: $\frac{dy}{dx}$.

This is a composite function where the outer function is $\sin^{-1}(u)$ and the inner function is $u = 2x$.

We use the Chain Rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Let $u = 2x$. Then $y = \sin^{-1} u$.

First, find the derivative of $y$ with respect to $u$:

This derivative is valid for $-1 < u < 1$.

Next, find the derivative of $u$ with respect to $x$:

Now, apply the Chain Rule by multiplying the derivatives:

Substitute back $u = 2x$ to express the derivative in terms of $x$:

The condition for the derivative of $\sin^{-1} u$ to be valid is $-1 < u < 1$. Substituting $u=2x$, this becomes $-1 < 2x < 1$, which simplifies to $-\frac{1}{2} < x < \frac{1}{2}$.

Therefore, the derivative of $\sin^{-1}(2x)$ with respect to $x$ is $\frac{2}{\sqrt{1-4x^2}}$, valid for $-\frac{1}{2} < x < \frac{1}{2}$.

Given: The equation $x^2 + y^2 = 25$.

To Find: $\frac{dy}{dx}$.

This is an implicit equation. We differentiate both sides with respect to $x$.

Using the sum rule for derivatives on the left side and the fact that the derivative of a constant is 0 on the right side:

Differentiate $x^2$ with respect to $x$ using the power rule: $\frac{d}{dx}(x^2) = 2x$.

Differentiate $y^2$ with respect to $x$ using the Chain Rule (differentiate $y^2$ with respect to $y$, then multiply by $\frac{dy}{dx}$):

Substitute these derivatives back into the equation:

Now, we need to solve for $\frac{dy}{dx}$. Isolate the term containing $\frac{dy}{dx}$:

Finally, divide by $2y$ (assuming $y \neq 0$) to find $\frac{dy}{dx}$:

The derivative of $y$ with respect to $x$ for the implicit relation $x^2 + y^2 = 25$ is $\frac{dy}{dx} = -\frac{x}{y}$. This derivative is defined for all points on the circle except where $y=0$ (i.e., at $x=\pm 5$), where the tangent is vertical.

Given: The equation $\sin^2 y + \cos(x y) = K$.

To Find: $\frac{dy}{dx}$.

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

Differentiate term by term. The derivative of the constant $K$ on the right side is 0.

For the first term, $\frac{d}{dx}(\sin^2 y)$. This is equivalent to $\frac{d}{dx}((\sin y)^2)$. We use the Chain Rule twice here. First, differentiate the outer square function, then differentiate the sine function, and finally multiply by $\frac{dy}{dx}$:

Using the trigonometric identity $\sin(2y) = 2\sin y \cos y$, the first term's derivative is $\sin(2y) \frac{dy}{dx}$.

For the second term, $\frac{d}{dx}(\cos(x y))$. This is a composite function where the outer function is cosine and the inner function is $xy$. We use the Chain Rule, and the derivative of the inner function $(xy)$ requires the Product Rule.

Now, apply the Product Rule to differentiate $(xy)$ with respect to $x$:

Substitute this back into the derivative of the second term:

Now substitute the derivatives of both terms back into the main differentiated equation:

Collect all terms containing $\frac{dy}{dx}$ on one side (left) and other terms on the other side (right):

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

Solve for $\frac{dy}{dx}$:

This expression for $\frac{dy}{dx}$ is valid provided the denominator $\sin(2y) - x \sin(xy)$ is not equal to zero.

We use the property that any positive number $a$ can be written as $e^{\ln a}$. Therefore, $a^x$ can be rewritten using the property $(b^m)^n = b^{mn}$:

Let $y = a^x = e^{x \ln a}$.

This is a composite function. Let $u = x \ln a$. Then $y = e^u$.

Using the Chain Rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

First, find the derivative of $y$ with respect to $u$:

Next, find the derivative of $u$ with respect to $x$. Note that $\ln a$ is a constant since $a$ is a constant.

Now, apply the Chain Rule:

Substitute back $u = x \ln a$. Since $e^{x \ln a} = a^x$, we get:

Thus, the derivative of $a^x$ with respect to $x$ is $a^x \ln a$.

Given: A function $y = \ln x$.

To Find: $\frac{dy}{dx}$.

From the definition of the natural logarithm, $y = \ln x$ is equivalent to $x = e^y$.

The domain of $\ln x$ is $x > 0$, and the range is $y \in R$.

Differentiate both sides of $x = e^y$ with respect to $x$:

The derivative of $x$ with respect to $x$ is 1. For the right side, we use the Chain Rule, treating $y$ as a function of $x$: $\frac{d}{dx}(e^y) = \frac{d}{dy}(e^y) \cdot \frac{dy}{dx}$.

So, the differentiated equation becomes:

Now, solve for $\frac{dy}{dx}$:

Since we started with $x = e^y$, we can substitute $e^y$ with $x$ in equation (v):

This derivative is valid for $x > 0$, which is the domain of $\ln x$.

Given: A function $y = \log_a x$.

To Find: $\frac{dy}{dx}$.

We use the change of base formula for logarithms, which states that $\log_a x = \frac{\log_b x}{\log_b a}$ for any valid base $b$. Choosing the natural logarithm (base $e$, where $\log_e = \ln$) is convenient for differentiation:

So, let $y = \frac{\ln x}{\ln a}$.

Since $a$ is a constant, $\ln a$ is also a constant. We can rewrite the function as $y = \left(\frac{1}{\ln a}\right) \cdot \ln x$.

Now differentiate $y$ with respect to $x$ using the constant multiple rule:

Using the formula for the derivative of $\ln x$ (Equation iii):

This derivative is valid for $x > 0$.

Given: The function $y = e^{\sin x}$.

To Find: $\frac{dy}{dx}$.

This is a composite function. The outer function is $e^u$ and the inner function is $u = \sin x$.

Using the Chain Rule, $\frac{dy}{dx} = \frac{d}{du}(e^u) \cdot \frac{du}{dx}$.

Let $u = \sin x$. Then $y = e^u$.

Find the derivative of $y$ with respect to $u$:

Find the derivative of $u$ with respect to $x$:

Apply the Chain Rule:

Substitute back $u = \sin x$:

The derivative of $e^{\sin x}$ is $e^{\sin x} \cos x$.

Given: The function $y = \ln(\tan x)$.

To Find: $\frac{dy}{dx}$.

This is a composite function. The outer function is $\ln u$ and the inner function is $u = \tan x$. Note that for $\ln(\tan x)$ to be defined, we must have $\tan x > 0$.

Using the Chain Rule, $\frac{dy}{dx} = \frac{d}{du}(\ln u) \cdot \frac{du}{dx}$.

Let $u = \tan x$. Then $y = \ln u$.

Find the derivative of $y$ with respect to $u$:

Find the derivative of $u$ with respect to $x$:

Apply the Chain Rule:

Substitute back $u = \tan x$:

We can simplify this expression using trigonometric identities:

Further simplification using $\cot x = \frac{\cos x}{\sin x}$ and $\sec x = \frac{1}{\cos x}$:

Multiplying the numerator and denominator by 2, we can use the double angle formula for sine ($\sin(2x) = 2\sin x \cos x$):

Using the reciprocal identity $\frac{1}{\sin \theta} = \text{cosec } \theta$:

So, the derivative can be expressed in various forms.

The derivative of $\ln(\tan x)$ with respect to $x$ is $\frac{\sec^2 x}{\tan x}$, or $\cot x \sec^2 x$, or $\frac{1}{\sin x \cos x}$, or $\frac{2}{\sin(2x)}$, or $2 \text{cosec}(2x)$.

Given: The function $y = x^x$.

To Find: $\frac{dy}{dx}$.

This function is of the form $f(x)^{g(x)}$, where $f(x) = x$ and $g(x) = x$. This is a clear case for logarithmic differentiation. Since the base $x$ must be positive for $x^x$ to be defined for all real exponents $x$, we are given $x > 0$, which also ensures $\ln x$ is defined.

1. Take the natural logarithm of both sides:

2. Simplify using logarithm properties: Use the power rule of logarithms, $\ln(A^p) = p \ln A$.

3. Differentiate both sides with respect to $x$:

On the left side, use the Chain Rule for $\frac{d}{dx}(\ln y)$: $\frac{1}{y} \frac{dy}{dx}$.

On the right side, use the Product Rule for $\frac{d}{dx}(x \ln x)$. The product rule states $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$. Here, let $u=x$ and $v=\ln x$.

Derivative of $u=x$: $\frac{du}{dx} = \frac{d}{dx}(x) = 1$.

Derivative of $v=\ln x$: $\frac{dv}{dx} = \frac{d}{dx}(\ln x) = \frac{1}{x}$.

Applying the Product Rule:

So, the differentiated equation is:

4. Solve for $\frac{dy}{dx}$: Multiply both sides of the equation by $y$:

5. Substitute back $y$: Replace $y$ with its original expression in terms of $x$, which is $y = x^x$.

The derivative of $x^x$ with respect to $x$ is $x^x (1 + \ln x)$.

Given: The function $y = \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}$.

To Find: $\frac{dy}{dx}$.

This function involves a product in the numerator and a product in the denominator. Using the quotient rule directly would be very tedious. Logarithmic differentiation simplifies this. We assume $y > 0$, which implies the factors are such that the fraction is positive.

1. Take the natural logarithm of both sides:

2. Simplify using logarithm properties: Use the quotient rule $\ln(A/B) = \ln A - \ln B$ and the product rule $\ln(AB) = \ln A + \ln B$.

Note: For $\ln(x-c)$ to be defined, $x-c > 0$, i.e., $x > c$. The domain of the original function needs to be considered, and $\ln |f(x)|$ could be used if $f(x)$ can be negative. For simplicity in this example, we proceed assuming the terms inside the logarithm are positive.

3. Differentiate both sides with respect to $x$:

On the left side, $\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}$.

On the right side, differentiate each term using the Chain Rule $\frac{d}{dx}(\ln u) = \frac{1}{u}\frac{du}{dx}$. For $\ln(x-c)$, $u = x-c$, so $\frac{du}{dx} = \frac{d}{dx}(x-c) = 1$. Thus, $\frac{d}{dx}(\ln(x-c)) = \frac{1}{x-c} \cdot 1 = \frac{1}{x-c}$.

4. Solve for $\frac{dy}{dx}$: Multiply both sides by $y$:

5. Substitute back $y$: Replace $y$ with its original expression:

This shows how logarithmic differentiation simplifies the process compared to using the quotient and product rules multiple times.

Given: The parametric equations $x = a \cos t$ and $y = a \sin t$.

To Find: $\frac{dy}{dx}$.

We use the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. First, we find the derivatives of $x$ and $y$ with respect to $t$.

Differentiate $x$ with respect to $t$:

Differentiate $y$ with respect to $t$:

Now, apply the formula (i):

Assuming $a \neq 0$ and $\sin t \neq 0$ (i.e., $t \neq n\pi$ for integer $n$), we can simplify:

The derivative $\frac{dy}{dx}$ is $-\cot t$. This result is expressed in terms of the parameter $t$. The points where $\frac{dx}{dt} = -a \sin t = 0$ (which occurs when $t = n\pi$) correspond to points $( \pm a, 0)$ on the circle, where the tangent is vertical (unless $a=0$).

Alternative Method (Implicit Differentiation after eliminating parameter):

The parametric equations $x = a \cos t$ and $y = a \sin t$ represent a circle centered at the origin with radius $a$. We can see this by squaring both equations and adding them:

$x^2 = (a \cos t)^2 = a^2 \cos^2 t$

$y^2 = (a \sin t)^2 = a^2 \sin^2 t$

$x^2 + y^2 = a^2 \cos^2 t + a^2 \sin^2 t = a^2 (\cos^2 t + \sin^2 t) = a^2(1) = a^2$

So, the equation in Cartesian form is $x^2 + y^2 = a^2$.

Now, we can find $\frac{dy}{dx}$ by implicit differentiation:

To express this in terms of $t$, substitute the parametric expressions for $x$ and $y$:

Both methods give the same result. The parametric method is generally preferred when eliminating the parameter is difficult or impossible.

Given: The parametric equations $x = a(t + \sin t)$ and $y = a(1 - \cos t)$.

To Find: $\frac{dy}{dx}$.

We use the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. First, we find the derivatives of $x$ and $y$ with respect to $t$.

Differentiate $x$ with respect to $t$:

Differentiate $y$ with respect to $t$:

Now, apply the formula (i):

Assuming $a \neq 0$ and $1 + \cos t \neq 0$ (which means $\cos t \neq -1$, so $t \neq (2n+1)\pi$ for integer $n$), we can simplify:

We can further simplify this expression using half-angle trigonometric identities:

$\sin t = 2 \sin \left(\frac{t}{2}\right) \cos \left(\frac{t}{2}\right)$

$1 + \cos t = 2 \cos^2 \left(\frac{t}{2}\right)$

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

Assuming $\cos \frac{t}{2} \neq 0$ (i.e., $\frac{t}{2} \neq (n + \frac{1}{2})\pi$, which means $t \neq (2n+1)\pi$), we can cancel terms:

The derivative $\frac{dy}{dx}$ is $\tan \left(\frac{t}{2}\right)$. The condition $\frac{dx}{dt} = a(1+\cos t) \neq 0$ translates to $1+\cos t \neq 0$, which means $\cos t \neq -1$, or $t \neq (2n+1)\pi$. At these values of $t$, $\frac{dy}{dt} = a \sin((2n+1)\pi) = a \cdot 0 = 0$. So, when $\frac{dx}{dt}=0$, $\frac{dy}{dt}=0$ as well, indicating a cusp point (like in the cycloid curve represented by these parametric equations).

Given: The function $y = x^3 + \sin x$.

To Find: The second order derivative, $\frac{d^2y}{dx^2}$.

First, find the first derivative of $y$ with respect to $x$:

Using the sum rule and standard derivative formulas:

Now, find the second derivative by differentiating the first derivative ($\frac{dy}{dx}$) with respect to $x$:

Using the sum rule and standard derivative formulas again:

The second order derivative of $y = x^3 + \sin x$ is $6x - \sin x$.

Given: The parametric equations $x = a t^2$ and $y = 2 a t$.

To Find: The second order derivative $\frac{d^2y}{dx^2}$.

First, find the derivatives of $x$ and $y$ with respect to the parameter $t$:

Next, find the first derivative $\frac{dy}{dx}$ using the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

Now, find the second derivative $\frac{d^2y}{dx^2}$ using the formula (iv): $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$.

We need to differentiate the expression for $\frac{dy}{dx}$ (which is $\frac{1}{t}$) with respect to $t$:

Using the power rule for differentiation:

Finally, substitute this result (vii) and the expression for $\frac{dx}{dt}$ (v) into the formula for $\frac{d^2y}{dx^2}$ (iv):

Simplify the complex fraction:

The second order derivative is $-\frac{1}{2at^3}$, provided $t \neq 0$ and $a \neq 0$. The parametric equations $x = at^2$, $y = 2at$ represent a parabola $y^2 = 4ax$.

Given: A function $f$ that is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$.

To Prove: There exists $c \in (a, b)$ such that $f'(c) = 0$.

Since $f$ is continuous on the closed interval $[a, b]$, by the Extreme Value Theorem (also known as the Maximum-Minimum Theorem), the function $f$ must attain both an absolute maximum value and an absolute minimum value on $[a, b]$. Let $M$ be the maximum value and $m$ be the minimum value of $f$ on $[a, b]$.

There are two possibilities for the values of $M$ and $m$:

Case 1: The maximum value $M$ and the minimum value $m$ are equal ($M=m$).

If $M = m$, then the function must be constant throughout the interval $[a, b]$. This means $f(x) = k$ for some constant $k$ for all $x \in [a, b]$.

The derivative of a constant function is always zero. So, $f'(x) = 0$ for all $x \in (a, b)$.

In this case, any value $c$ chosen from the open interval $(a, b)$ will satisfy $f'(c) = 0$. Thus, the theorem holds.

Case 2: The maximum value $M$ and the minimum value $m$ are not equal ($M \neq m$).

Since $M \neq m$, and we are given $f(a) = f(b)$, the function is not constant on $[a, b]$. This implies that at least one of the extreme values (either the maximum $M$ or the minimum $m$) must be attained at a point $c$ that is strictly inside the interval, i.e., $c \in (a, b)$. This is because if both the maximum and minimum were attained only at the endpoints, and $M \neq m$, then we would have $f(a)$ and $f(b)$ taking on different values, which contradicts the condition $f(a)=f(b)$.

Let's assume, without loss of generality, that the maximum value $M$ is attained at a point $c \in (a, b)$. So, $f(c) = M$. By the definition of a maximum value, for any small positive number $h$ such that $c+h$ and $c-h$ are within the interval $[a, b]$ (and assuming $h$ is small enough that $c \pm h$ are within $(a, b)$), we have:

$f(c) \geq f(c+h)$ and $f(c) \geq f(c-h)$.

Consider the right-hand derivative at $c$. For $h > 0$:

$f(c+h) - f(c) \leq 0$

Dividing by $h$ (which is positive):

$\frac{f(c+h) - f(c)}{h} \leq 0$

Taking the limit as $h \to 0^+$:

Consider the left-hand derivative at $c$. For $h > 0$, $c-h$ is less than $c$:

$f(c-h) - f(c) \leq 0$

Dividing by $-h$ (which is negative), the inequality reverses:

$\frac{f(c-h) - f(c)}{-h} \geq 0$

Taking the limit as $h \to 0^+$ (which corresponds to $x \to c^-$):

We are given that $f$ is differentiable on the open interval $(a, b)$, and $c \in (a, b)$. Therefore, $f$ is differentiable at $c$. This means the left-hand derivative and the right-hand derivative at $c$ exist and are equal to $f'(c)$.

From the limits above, we have:

The only way for both $f'(c) \leq 0$ and $f'(c) \geq 0$ to be true simultaneously is if $f'(c) = 0$.

Thus, we have shown that there exists at least one point $c \in (a, b)$ such that $f'(c) = 0$. A similar argument applies if the minimum value $m$ is attained at a point $c \in (a, b)$.

This completes the proof idea for Rolle's Theorem.

To verify Rolle's Theorem for $f(x) = x^2 - 4x + 3$ on the interval $[1, 3]$, we need to check if the three conditions of the theorem are satisfied and then find a value $c$ in the open interval $(1, 3)$ such that $f'(c) = 0$.

1. Continuity: The function $f(x) = x^2 - 4x + 3$ is a polynomial function. We know that all polynomial functions are continuous for all real numbers $x$. Therefore, $f(x)$ is continuous on the closed interval $[1, 3]$.

2. Differentiability: The function $f(x) = x^2 - 4x + 3$ is a polynomial function. We know that all polynomial functions are differentiable for all real numbers $x$. Its derivative is $f'(x) = \frac{d}{dx}(x^2 - 4x + 3) = 2x - 4$. Therefore, $f(x)$ is differentiable on the open interval $(1, 3)$.

3. Function values at endpoints: Check if $f(a) = f(b)$ for $a=1$ and $b=3$.

$f(1) = (1)^2 - 4(1) + 3 = 1 - 4 + 3 = 0$

$f(3) = (3)^2 - 4(3) + 3 = 9 - 12 + 3 = 0$

Since $f(1) = 0$ and $f(3) = 0$, the condition $f(a) = f(b)$ is satisfied.

All three conditions of Rolle's Theorem are satisfied. Therefore, according to the theorem, there must exist at least one value $c$ in the open interval $(1, 3)$ such that $f'(c) = 0$.

Let's find the value(s) of $c$. We set the first derivative $f'(x)$ equal to zero:

Set $f'(c) = 0$:

Solving for $c$:

The value $c=2$ lies in the open interval $(1, 3)$. This is the point guaranteed by Rolle's Theorem.

Thus, Rolle's Theorem is verified for the function $f(x) = x^2 - 4x + 3$ on the interval $[1, 3]$.

To examine if Rolle's Theorem is applicable to $f(x) = |x|$ on $[-1, 1]$, we check the three conditions of the theorem for the interval $[a, b] = [-1, 1]$.

1. Continuity: The function $f(x) = |x|$ is the modulus function. The modulus function is continuous for all real numbers $x$. Therefore, $f(x) = |x|$ is continuous on the closed interval $[-1, 1]$.

The first condition is satisfied.

2. Differentiability: The function $f(x) = |x|$ is differentiable for all $x \neq 0$. However, it is not differentiable at $x=0$ because of the sharp corner (cusp) in its graph at the origin.

The open interval $(a, b)$ is $(-1, 1)$. The point $x=0$ is within this open interval $(-1, 1)$. Since $f(x) = |x|$ is not differentiable at $x=0$, the function is not differentiable on the entire open interval $(-1, 1)$.

The second condition of Rolle's Theorem requires differentiability on the *open* interval $(a, b)$. Since this condition is not met, Rolle's Theorem is not applicable to the function $f(x) = |x|$ on the interval $[-1, 1]$.

(Note: Although not necessary to check once a condition fails, let's examine the third condition for completeness):

3. Function values at endpoints: Check $f(a) = f(b)$ for $a=-1$ and $b=1$.

$f(-1) = |-1| = 1$

$f(1) = |1| = 1$

So, $f(-1) = f(1) = 1$. The third condition is satisfied.

Even though the continuity and endpoint conditions are met, the differentiability condition fails at $x=0$ within the open interval $(-1, 1)$. Thus, Rolle's Theorem cannot be applied.

Indeed, if we look at the derivative $f'(x)$ for $x \neq 0$:

$f'(x) = \begin{cases} 1 & , & x > 0 \\ -1 & , & x < 0 \end{cases}$

The derivative is either $1$ or $-1$; it is never zero for any $x \in (-1, 1), x \neq 0$. This is consistent with the fact that Rolle's Theorem does not apply because the required conditions are not fully met.

Given: A function $f$ that is continuous on $[a, b]$ and differentiable on $(a, b)$.

To Prove: There exists $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}$.

Consider the secant line passing through the points $A(a, f(a))$ and $B(b, f(b))$. The equation of this line can be written as:

Let $L(x)$ be the $y$-coordinate on this secant line for a given $x$:

Now, define an auxiliary function $g(x)$ as the difference between the function $f(x)$ and the value on the secant line $L(x)$ at each point $x$ in the interval $[a, b]$:

We will now check if this function $g(x)$ satisfies the three conditions of Rolle's Theorem on the interval $[a, b]$.

1. Continuity of $g(x)$ on $[a, b]$:

• $f(x)$ is continuous on $[a, b]$ (given).
• The function $L(x) = f(a) + \frac{f(b) - f(a)}{b-a} (x - a)$ is a linear function of $x$ (since $f(a)$, $f(b)$, $a$, and $b$ are constants). Linear functions are continuous everywhere, so $L(x)$ is continuous on $[a, b]$.

2. Differentiability of $g(x)$ on $(a, b)$:

• $f(x)$ is differentiable on $(a, b)$ (given).
• The function $L(x)$ is a linear function, $L(x) = (\frac{f(b) - f(a)}{b-a}) x + (f(a) - a \frac{f(b) - f(a)}{b-a})$. The derivative of a linear function $mx+c$ is $m$. So, the derivative of $L(x)$ with respect to $x$ is $L'(x) = \frac{f(b) - f(a)}{b-a}$. This derivative exists for all $x \in (a, b)$ (in fact, for all $x \in R$).

The derivative of $g(x)$ is:

3. Function values at endpoints of $g(x)$: Check if $g(a) = g(b)$.

So, $g(a) = g(b) = 0$. (Condition 3 of Rolle's Theorem satisfied for g(x)).

Since the function $g(x)$ satisfies all three conditions of Rolle's Theorem on the interval $[a, b]$, Rolle's Theorem guarantees that there exists at least one real number $c$ in the open interval $(a, b)$ such that $g'(c) = 0$.

Substitute $x=c$ into the expression we found for $g'(x)$:

Since $g'(c) = 0$:

Rearranging this equation, we get the conclusion of the Mean Value Theorem:

Thus, there exists at least one $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}$. This completes the proof idea for Lagrange's Mean Value Theorem.

To verify Lagrange's Mean Value Theorem for $f(x) = x^2 - 4x - 3$ on $[1, 4]$, we check the conditions and find a value $c \in (1, 4)$ that satisfies the theorem's conclusion.

1. Continuity: The function $f(x) = x^2 - 4x - 3$ is a polynomial function. Polynomials are continuous for all real numbers. Thus, $f(x)$ is continuous on the closed interval $[1, 4]$.

2. Differentiability: The function $f(x) = x^2 - 4x - 3$ is a polynomial function. Polynomials are differentiable for all real numbers. The derivative is $f'(x) = \frac{d}{dx}(x^2 - 4x - 3) = 2x - 4$. Thus, $f(x)$ is differentiable on the open interval $(1, 4)$.

Both conditions of LMVT are satisfied. Therefore, the theorem guarantees the existence of at least one value $c \in (1, 4)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}$.

First, calculate the function values at the endpoints $a=1$ and $b=4$:

Now, calculate the slope of the secant line connecting $(1, f(1))$ and $(4, f(4))$:

Next, find the derivative of $f(x)$ and evaluate it at $x=c$:

According to LMVT, $f'(c)$ must be equal to the slope of the secant line:

Now, solve this equation for $c$:

The value $c = 2.5$ lies in the open interval $(1, 4)$. This value of $c$ is the point guaranteed by Lagrange's Mean Value Theorem for the given function and interval.

Since we found a value $c \in (1, 4)$ such that $f'(c) = \frac{f(4) - f(1)}{4-1}$, Lagrange's Mean Value Theorem is verified for the function $f(x) = x^2 - 4x - 3$ on the interval $[1, 4]$.