Chapter 5 Continuity And Differentiability

Given:

The function is $f(x) = 2x + 3$.

We need to check the continuity at the point $x = 1$.

To Check:

If the function $f(x)$ is continuous at $x = 1$.

Solution:

A function $f(x)$ is continuous at a point $x = c$ if the following three conditions are met:

1. $f(c)$ is defined.

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

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

For the given function $f(x) = 2x + 3$ and the point $x = 1$ (so $c=1$):

Step 1: Check if $f(1)$ is defined.

Substitute $x=1$ into the function:

$f(1) = 2(1) + 3 = 2 + 3 = 5$

$f(1) = 5$ is a finite and well-defined value.

Step 2: Check if $\lim\limits_{x \to 1} f(x)$ exists.

We need to evaluate the limit of $f(x)$ as $x$ approaches 1:

$\lim\limits_{x \to 1} f(x) = \lim\limits_{x \to 1} (2x + 3)$

Since $f(x) = 2x+3$ is a polynomial function, the limit as $x$ approaches a value can be found by direct substitution.

$\lim\limits_{x \to 1} (2x + 3) = 2(1) + 3 = 2 + 3 = 5$

The limit exists and its value is 5.

Step 3: Check if $\lim\limits_{x \to 1} f(x) = f(1)$.

From Step 1, $f(1) = 5$.

From Step 2, $\lim\limits_{x \to 1} f(x) = 5$.

We compare the two values:

$\lim\limits_{x \to 1} f(x) = 5$

$f(1) = 5$

Since the limit of the function as $x$ approaches 1 is equal to the value of the function at $x=1$ ($5=5$), the function $f(x)$ is continuous at $x=1$.

Conclusion:

All three conditions for continuity at $x=1$ are satisfied.

Therefore, the function $f(x) = 2x + 3$ is continuous at $x = 1$.

Given:

The function is $f(x) = x^2$.

We need to examine the continuity at the point $x = 0$.

To Examine:

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

Solution:

A function $f(x)$ is continuous at a point $x = c$ if the following three conditions are met:

1. $f(c)$ is defined.

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

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

For the given function $f(x) = x^2$ and the point $x = 0$ (so $c=0$):

Step 1: Check if $f(0)$ is defined.

Substitute $x=0$ into the function:

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

$f(0) = 0$ is a finite and well-defined value.

Step 2: Check if $\lim\limits_{x \to 0} f(x)$ exists.

We need to evaluate the limit of $f(x)$ as $x$ approaches 0:

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

Since $f(x) = x^2$ is a polynomial function, the limit as $x$ approaches a value can be found by direct substitution.

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

The limit exists and its value is 0.

Alternatively, we can check the left-hand limit (LHL) and the right-hand limit (RHL):

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

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

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

Step 3: Check if $\lim\limits_{x \to 0} f(x) = f(0)$.

From Step 1, $f(0) = 0$.

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

We compare the two values:

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

$f(0) = 0$

Since the limit of the function as $x$ approaches 0 is equal to the value of the function at $x=0$ ($0=0$), the function $f(x)$ is continuous at $x=0$.

Conclusion:

All three conditions for continuity at $x=0$ are satisfied.

Therefore, the function $f(x) = x^2$ is continuous at $x = 0$.

Given:

The function is $f(x) = |x|$.

We need to discuss the continuity at the point $x = 0$.

To Discuss:

The continuity of the function $f(x) = |x|$ at $x = 0$.

Solution:

A function $f(x)$ is continuous at a point $x = c$ if the following three conditions are met:

1. $f(c)$ is defined.

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

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

The function $f(x) = |x|$ can be defined piecewise as:

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

We check the continuity at $x = 0$ (so $c=0$).

Step 1: Check if $f(0)$ is defined.

Substitute $x=0$ into the definition of the function for $x \geq 0$:

$f(0) = |0| = 0$

$f(0) = 0$ is a finite and well-defined value.

Step 2: Check if $\lim\limits_{x \to 0} f(x)$ exists.

For the limit to exist at $x=0$, the left-hand limit (LHL) and the right-hand limit (RHL) must exist and be equal.

The LHL is evaluated as $x$ approaches 0 from the left ($x < 0$). We use the definition $f(x) = -x$ for $x < 0$:

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

As $x$ approaches 0 from the left, $-x$ approaches 0.

LHL = $-(0) = 0$

The RHL is evaluated as $x$ approaches 0 from the right ($x > 0$). We use the definition $f(x) = x$ for $x \geq 0$:

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

As $x$ approaches 0 from the right, $x$ approaches 0.

RHL = $0$

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

Step 3: Check if $\lim\limits_{x \to 0} f(x) = f(0)$.

From Step 1, $f(0) = 0$.

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

We compare the two values:

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

$f(0) = 0$

Since the limit of the function as $x$ approaches 0 is equal to the value of the function at $x=0$ ($0=0$), the function $f(x)$ is continuous at $x=0$.

Conclusion:

All three conditions for continuity at $x=0$ are satisfied.

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

Given:

The function $f$ is defined as:

$f(x) = \begin{cases} x^3+3, & if \;x \neq 0\\1,&if\; x = 0 \end{cases}$

We need to check the continuity of the function at $x = 0$.

To Show:

The function $f$ is not continuous at $x = 0$.

Solution:

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

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

In this case, we need to check continuity at $x = 0$. So, we need to evaluate $\lim\limits_{x \to 0} f(x)$ and $f(0)$.

First, let's find the value of the function at $x = 0$. From the definition of $f(x)$, when $x = 0$, $f(x) = 1$.

$f(0) = 1$

Next, let's find the limit of the function as $x$ approaches $0$. For $x \neq 0$, the function is defined as $f(x) = x^3 + 3$. Therefore, the limit as $x$ approaches $0$ is calculated using this expression:

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} (x^3 + 3)$

As $x$ approaches $0$, $x^3$ approaches $0^3 = 0$.

$\lim\limits_{x \to 0} (x^3 + 3) = 0^3 + 3 = 0 + 3 = 3$

So, we have:

$\lim\limits_{x \to 0} f(x) = 3$

Now we compare the limit of the function as $x$ approaches $0$ with the value of the function at $x = 0$.

$\lim\limits_{x \to 0} f(x) = 3$

$f(0) = 1$

Since $\lim\limits_{x \to 0} f(x) \neq f(0)$, the function $f$ is not continuous at $x = 0$.

Conclusion:

The limit of the function as $x$ approaches $0$ is $3$, while the value of the function at $x=0$ is $1$. Because the limit does not equal the function value at the point, the function $f$ is not continuous at $x = 0$.

Given:

The function $f(x) = k$, where $k$ is a constant.

To Check:

The points where the function $f(x) = k$ is continuous.

Solution:

To determine the points where the constant function $f(x) = k$ is continuous, we consider an arbitrary real number, let's call it $c$.

A function $f$ is continuous at a point $x = c$ if and only if the following condition is met:

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

First, let's evaluate the function at the point $x = c$. Since $f(x) = k$ for all values of $x$, including $x=c$, we have:

$f(c) = k$

Next, let's evaluate the limit of the function as $x$ approaches $c$. Since $f(x) = k$ is a constant function, its value does not change as $x$ gets closer to $c$. The limit of a constant function is the constant itself.

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

Now, we compare the limit of the function as $x$ approaches $c$ with the value of the function at $x = c$.

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

$f(c) = k$

Since $\lim\limits_{x \to c} f(x) = f(c) = k$ for any real number $c$, the constant function $f(x) = k$ satisfies the condition for continuity at every real number.

Conclusion:

The constant function $f(x) = k$ is continuous at every point in its domain, which is the set of all real numbers ($\mathbb{R}$).

Given:

The identity function $f(x) = x$ defined on the set of real numbers ($\mathbb{R}$).

To Prove:

The function $f(x) = x$ is continuous at every real number.

Proof:

To prove that the function $f(x) = x$ is continuous at every real number, we need to show that it is continuous at an arbitrary point $c \in \mathbb{R}$.

A function $f$ is continuous at a point $x = c$ if and only if the following three conditions are met:

1. $f(c)$ is defined.

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

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

Let $c$ be any arbitrary real number.

First, let's find the value of the function at $x = c$. Since $f(x) = x$, we have:

$f(c) = c$

Since $c$ is a real number, $f(c)$ is defined.

Next, let's find the limit of the function as $x$ approaches $c$. For the function $f(x) = x$, the limit as $x$ approaches $c$ is:

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

By the properties of limits, the limit of $x$ as $x$ approaches $c$ is simply $c$.

$\lim\limits_{x \to c} x = c$

So, the limit exists and is equal to $c$.

Now, we compare the limit of the function as $x$ approaches $c$ with the value of the function at $x = c$.

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

$f(c) = c$

We see that $\lim\limits_{x \to c} f(x) = f(c)$.

Since this equality holds for any arbitrary real number $c$, the identity function $f(x) = x$ is continuous at every real number.

Conclusion:

The identity function $f(x) = x$ is continuous at every real number.

Given:

The function $f(x) = |x|$ defined for all real numbers.

To Check:

Whether the function $f(x) = |x|$ is a continuous function.

Solution:

The absolute value function $f(x) = |x|$ can be defined piecewise as:

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

To check if $f(x)$ is continuous, we need to examine its continuity at all points in its domain, which is the set of all real numbers ($\mathbb{R}$). We consider three cases:

Case 1: Continuity at $c > 0$

Let $c$ be any real number such that $c > 0$.

The function value at $x = c$ is $f(c) = |c| = c$ (since $c > 0$).

For values of $x$ close to $c$ where $x > 0$, $f(x) = x$.

The limit of $f(x)$ as $x$ approaches $c$ is:

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

Since $\lim\limits_{x \to c} f(x) = f(c) = c$, the function $f(x)$ is continuous for all $c > 0$.

Case 2: Continuity at $c < 0$

Let $c$ be any real number such that $c < 0$.

The function value at $x = c$ is $f(c) = |c| = -c$ (since $c < 0$, $-c > 0$).

For values of $x$ close to $c$ where $x < 0$, $f(x) = -x$.

The limit of $f(x)$ as $x$ approaches $c$ is:

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

Since $\lim\limits_{x \to c} f(x) = f(c) = -c$, the function $f(x)$ is continuous for all $c < 0$.

Case 3: Continuity at $x = 0$

We need to check if $\lim\limits_{x \to 0} f(x) = f(0)$.

First, find the function value at $x = 0$. From the definition, $f(0) = |0| = 0$.

$f(0) = 0$

Next, find the limit of $f(x)$ as $x$ approaches $0$. We need to check the left-hand limit (LHL) and the right-hand limit (RHL).

LHL: $\lim\limits_{x \to 0^-} f(x)$. For $x < 0$, $f(x) = -x$.

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

RHL: $\lim\limits_{x \to 0^+} f(x)$. For $x > 0$, $f(x) = x$.

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

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

Now, compare the limit with the function value at $x=0$.

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

$f(0) = 0$

Since $\lim\limits_{x \to 0} f(x) = f(0) = 0$, the function $f(x)$ is continuous at $x = 0$.

Since the function is continuous for $x > 0$, $x < 0$, and at $x = 0$, it is continuous at every real number.

Conclusion:

The function $f(x) = |x|$ is continuous at every real number.

Given:

The function $f(x) = x^3 + x^2 - 1$.

To Discuss:

The continuity of the function $f(x)$ at all points in its domain.

Solution:

The function $f(x) = x^3 + x^2 - 1$ is a polynomial function. The domain of a polynomial function is the set of all real numbers ($\mathbb{R}$).

To discuss the continuity of $f(x)$, we need to check if it is continuous at every point in its domain. Let $c$ be an arbitrary real number.

For the function $f$ to be continuous at $x=c$, we must verify that $\lim\limits_{x \to c} f(x) = f(c)$.

First, let's evaluate the function at $x = c$:

$f(c) = c^3 + c^2 - 1$

Next, let's evaluate the limit of the function as $x$ approaches $c$:

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (x^3 + x^2 - 1)$

Using the properties of limits (limit of a sum is the sum of limits, limit of a difference is the difference of limits, limit of a power is the power of the limit, and limit of a constant is the constant), we can evaluate this limit:

$\lim\limits_{x \to c} (x^3 + x^2 - 1) = \lim\limits_{x \to c} x^3 + \lim\limits_{x \to c} x^2 - \lim\limits_{x \to c} 1$

$= (\lim\limits_{x \to c} x)^3 + (\lim\limits_{x \to c} x)^2 - 1$

Since $\lim\limits_{x \to c} x = c$, we substitute $c$ into the expression:

$= c^3 + c^2 - 1$

So, we have found that:

$\lim\limits_{x \to c} f(x) = c^3 + c^2 - 1$

Now, we compare the limit of the function as $x$ approaches $c$ with the value of the function at $x = c$:

$\lim\limits_{x \to c} f(x) = c^3 + c^2 - 1$

$f(c) = c^3 + c^2 - 1$

Since $\lim\limits_{x \to c} f(x) = f(c)$, the function $f(x)$ is continuous at the arbitrary point $c$.

Since $c$ was any arbitrary real number, this holds true for all real numbers.

Alternatively, we know that polynomial functions are continuous everywhere. The function $f(x) = x^3 + x^2 - 1$ is a polynomial of degree 3, so it is continuous on $\mathbb{R}$.

Conclusion:

The function $f(x) = x^3 + x^2 - 1$ is a polynomial function, and polynomial functions are continuous at every real number. Therefore, the function $f(x)$ is continuous for all real numbers.

Given:

The function $f(x) = \frac{1}{x}$, defined for all real numbers $x$ such that $x \neq 0$.

To Discuss:

The continuity of the function $f(x)$ at all points in its domain.

Solution:

The domain of the function $f(x) = \frac{1}{x}$ is the set of all real numbers except $0$. This can be written as $\mathbb{R} \setminus \{0\}$ or $(-\infty, 0) \cup (0, \infty)$.

To discuss the continuity of $f(x)$, we need to check if it is continuous at every point within its domain.

Let $c$ be an arbitrary real number in the domain of $f$, meaning $c \neq 0$.

For the function $f$ to be continuous at $x=c$, we must verify that $\lim\limits_{x \to c} f(x) = f(c)$.

First, let's evaluate the function at $x = c$. Since $c \neq 0$, $f(c)$ is defined:

$f(c) = \frac{1}{c}$

Next, let's evaluate the limit of the function as $x$ approaches $c$. For $c \neq 0$, the limit of $\frac{1}{x}$ as $x \to c$ is given by substituting $c$ into the expression:

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

Using the properties of limits, specifically the limit of a quotient, if the limit of the denominator is not zero (which is true since $c \neq 0$), we have:

$\lim\limits_{x \to c} \frac{1}{x} = \frac{\lim\limits_{x \to c} 1}{\lim\limits_{x \to c} x} = \frac{1}{c}$

So, we have found that:

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

Now, we compare the limit of the function as $x$ approaches $c$ with the value of the function at $x = c$:

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

$f(c) = \frac{1}{c}$

Since $\lim\limits_{x \to c} f(x) = f(c)$, the function $f(x)$ is continuous at the arbitrary point $c$, provided $c \neq 0$.

Since $c$ was any arbitrary real number in the domain of the function, this holds true for all points in the domain.

The function is not defined at $x=0$, so we do not check for continuity at $x=0$ as it is not in the domain.

Conclusion:

The function $f(x) = \frac{1}{x}$ is continuous at every point in its domain, which is the set of all real numbers except $0$ ($\mathbb{R} \setminus \{0\}$).

Given:

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

$f(x) = \begin{cases} x+2,& if \;x≤ 1\\x−2,& if \;x > 1 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Discuss:

The continuity of the function $f(x)$ at all points in its domain.

Solution:

To discuss the continuity of the function $f(x)$, we need to check for continuity in the intervals defined by the piecewise function and at the point where the definition changes, which is $x=1$.

We consider three cases:

Case 1: For $x < 1$

Let $c$ be any real number such that $c < 1$. For all values of $x$ in an open interval around $c$, $x < 1$, so $f(x) = x+2$.

The function value at $x = c$ is $f(c) = c+2$.

The limit of $f(x)$ as $x$ approaches $c$ is:

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

Since $x+2$ is a polynomial expression, the limit can be found by direct substitution:

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

Since $\lim\limits_{x \to c} f(x) = f(c) = c+2$, the function $f(x)$ is continuous for all $x < 1$.

Case 2: For $x > 1$

Let $c$ be any real number such that $c > 1$. For all values of $x$ in an open interval around $c$, $x > 1$, so $f(x) = x-2$.

The function value at $x = c$ is $f(c) = c-2$.

The limit of $f(x)$ as $x$ approaches $c$ is:

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

Since $x-2$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to c} (x-2) = c-2$

Since $\lim\limits_{x \to c} f(x) = f(c) = c-2$, the function $f(x)$ is continuous for all $x > 1$.

Case 3: At $x = 1$

To check for continuity at $x=1$, we need to verify if $\lim\limits_{x \to 1} f(x) = f(1)$. We need to find $f(1)$, the left-hand limit (LHL), and the right-hand limit (RHL) at $x=1$.

First, find the function value at $x = 1$. From the definition, when $x \leq 1$, $f(x) = x+2$. So, at $x=1$, $f(1) = 1+2 = 3$.

$f(1) = 3$

Next, find the left-hand limit at $x=1$. For $x < 1$, $f(x) = x+2$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x+2) = 1+2 = 3$

Finally, find the right-hand limit at $x=1$. For $x > 1$, $f(x) = x-2$.

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

Now, we compare the LHL and RHL:

LHL at $x=1$ is $3$

RHL at $x=1$ is $-1$

Since $\lim\limits_{x \to 1^-} f(x) \neq \lim\limits_{x \to 1^+} f(x)$ ($3 \neq -1$), the limit $\lim\limits_{x \to 1} f(x)$ does not exist.

For a function to be continuous at a point, the limit must exist at that point and be equal to the function value. Since the limit does not exist at $x=1$, the function is not continuous at $x=1$.

Conclusion:

The function $f(x)$ is continuous for all $x < 1$ and for all $x > 1$. However, it is not continuous at $x = 1$.

Given:

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

$f(x) = \begin{cases} x+2,& if\; x < 1\\0,&if \;x = 1\\x−2,& if\; x > 1 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All the points of discontinuity of the function $f(x)$.

Solution:

To find the points of discontinuity, we need to examine the function's continuity at points where its definition changes, and in the intervals where it is defined by a single expression.

We consider three cases:

Case 1: For $x < 1$

Let $c$ be any real number such that $c < 1$. In this interval, $f(x) = x+2$, which is a polynomial function.

We know that polynomial functions are continuous for all real numbers.

Alternatively, we can check the condition $\lim\limits_{x \to c} f(x) = f(c)$.

$f(c) = c+2$

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

Since $\lim\limits_{x \to c} f(x) = f(c)$, the function is continuous for all $x < 1$.

Case 2: For $x > 1$

Let $c$ be any real number such that $c > 1$. In this interval, $f(x) = x-2$, which is a polynomial function.

Polynomial functions are continuous for all real numbers.

Alternatively, we can check the condition $\lim\limits_{x \to c} f(x) = f(c)$.

$f(c) = c-2$

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

Since $\lim\limits_{x \to c} f(x) = f(c)$, the function is continuous for all $x > 1$.

Case 3: At $x = 1$

This is the point where the definition of the function changes. We need to check if $f(1)$, the left-hand limit (LHL) at $x=1$, and the right-hand limit (RHL) at $x=1$ exist and are equal.

Function value at $x = 1$:

From the definition, when $x = 1$, $f(x) = 0$.

$f(1) = 0$

Left-hand limit at $x = 1$:

For $x < 1$, $f(x) = x+2$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x+2) = 1+2 = 3$

Right-hand limit at $x = 1$:

For $x > 1$, $f(x) = x-2$.

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

Now, we compare the LHL and RHL at $x=1$.

$\lim\limits_{x \to 1^-} f(x) = 3$

$\lim\limits_{x \to 1^+} f(x) = -1$

Since the left-hand limit is not equal to the right-hand limit ($\lim\limits_{x \to 1^-} f(x) \neq \lim\limits_{x \to 1^+} f(x)$), the limit of the function as $x$ approaches $1$, i.e., $\lim\limits_{x \to 1} f(x)$, does not exist.

For a function to be continuous at a point, the limit must exist at that point and be equal to the function value. Since the limit does not exist at $x=1$, the function is not continuous at $x=1$.

Combining the cases, the function is continuous for all $x < 1$ and all $x > 1$. The only point where it is not continuous is $x=1$.

Conclusion:

The function $f(x)$ is continuous for all real numbers except at $x=1$. The only point of discontinuity is $x=1$.

Given:

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

$f(x) = \begin{cases} x+2,& if\; x < 0\\−x+2,& if\; x >0 \end{cases}$

From the definition, the function is defined for $x < 0$ and $x > 0$. It is not defined at $x=0$. Thus, the domain of the function is $\mathbb{R} \setminus \{0\}$.

To Discuss:

The continuity of the function $f(x)$ at all points in its domain.

Solution:

To discuss the continuity of the function $f(x)$, we examine the intervals where the function is defined by a single expression.

We consider two cases based on the definition of the function:

Case 1: For $x < 0$

Let $c$ be any real number such that $c < 0$. For all values of $x$ in an open interval around $c$ that lies within $(-\infty, 0)$, the function is defined as $f(x) = x+2$.

The function value at $x = c$ is $f(c) = c+2$.

The limit of $f(x)$ as $x$ approaches $c$ is:

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

Since $x+2$ is a polynomial expression, the limit can be found by direct substitution:

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

Since $\lim\limits_{x \to c} f(x) = f(c) = c+2$, the function $f(x)$ is continuous for all $x < 0$.

Case 2: For $x > 0$

Let $c$ be any real number such that $c > 0$. For all values of $x$ in an open interval around $c$ that lies within $(0, \infty)$, the function is defined as $f(x) = -x+2.

The function value at $x = c$ is $f(c) = -c+2.

The limit of $f(x)$ as $x$ approaches $c$ is:

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (-x+2)$

Since $-x+2$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to c} (-x+2) = -c+2$

Since $\lim\limits_{x \to c} f(x) = f(c) = -c+2$, the function $f(x)$ is continuous for all $x > 0$.

Discussion at $x=0$:

The point $x=0$ is not included in the domain of the function as defined. For a function to be continuous at a point, it must be defined at that point. Since $f(0)$ is not defined, the function is not continuous at $x=0$. However, as per the standard definition of continuity on an interval, we check continuity only at points within the domain of the function.

Based on the definition provided, the function is continuous on the intervals $(-\infty, 0)$ and $(0, \infty)$, which constitute its domain.

Conclusion:

The function $f(x)$ is defined and continuous for all real numbers except $x=0$. Therefore, the function $f(x)$ is continuous at every point in its domain, which is $\mathbb{R} \setminus \{0\}$. The function is not continuous on the set of all real numbers $\mathbb{R}$ because it has a discontinuity at $x=0$ (a removable discontinuity if we consider the limits, but primarily because it is not defined at $x=0$).

Given:

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

$f(x) = \begin{cases} x,& if\; x ≥ 0\\x^2,& if\; x < 0 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Discuss:

The continuity of the function $f(x)$ at all points in its domain.

Solution:

To discuss the continuity of the function $f(x)$, we need to examine its continuity in the intervals defined by the piecewise function and at the point where the definition changes, which is $x=0$.

We consider three cases:

Case 1: For $x < 0$

Let $c$ be any real number such that $c < 0$. For all values of $x$ in an open interval around $c$ that lies within $(-\infty, 0)$, the function is defined as $f(x) = x^2$.

The function value at $x = c$ is $f(c) = c^2$.

The limit of $f(x)$ as $x$ approaches $c$ is:

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

Since $x^2$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to c} x^2 = c^2$

Since $\lim\limits_{x \to c} f(x) = f(c) = c^2$, the function $f(x)$ is continuous for all $x < 0$.

Case 2: For $x > 0$

Let $c$ be any real number such that $c > 0$. For all values of $x$ in an open interval around $c$ that lies within $(0, \infty)$, the function is defined as $f(x) = x.

The function value at $x = c$ is $f(c) = c.$

The limit of $f(x)$ as $x$ approaches $c$ is:

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

Since $x$ is a polynomial expression (identity function), the limit can be found by direct substitution:

$\lim\limits_{x \to c} x = c$

Since $\lim\limits_{x \to c} f(x) = f(c) = c$, the function $f(x)$ is continuous for all $x > 0$.

Case 3: At $x = 0$

This is the point where the definition of the function changes. To check for continuity at $x=0$, we need to verify if $\lim\limits_{x \to 0} f(x) = f(0)$. We need to find $f(0)$, the left-hand limit (LHL) at $x=0$, and the right-hand limit (RHL) at $x=0$.

Function value at $x = 0$:

From the definition, when $x \geq 0$, $f(x) = x$. So, at $x=0$, $f(0) = 0$.

$f(0) = 0$

Left-hand limit at $x = 0$:

For $x < 0$, $f(x) = x^2$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} x^2 = (0)^2 = 0$

Right-hand limit at $x = 0$:

For $x > 0$, $f(x) = x$.

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

Now, we compare the LHL, RHL, and $f(0)$:

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

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

$f(0) = 0$

Since LHL = RHL = $f(0) = 0$, the limit of the function as $x$ approaches $0$ exists and is equal to the function value at $x=0$. Therefore, the function is continuous at $x=0$.

Combining all three cases, the function is continuous for all $x < 0$, for all $x > 0$, and at $x = 0$. Thus, the function is continuous for all real numbers.

Conclusion:

The function $f(x)$ is continuous at every real number.

Given:

A polynomial function $P(x)$.

A general polynomial function of degree $n$ is given by:

$P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$

where $a_0, a_1, \dots, a_n$ are real coefficients and $a_n \neq 0$ (for degree $n \ge 1$). For $n=0$, $P(x) = a_0$, which is a constant function.

The domain of any polynomial function is the set of all real numbers ($\mathbb{R}$).

To Prove:

Every polynomial function is continuous at every real number.

Proof:

To show that a polynomial function $P(x)$ is continuous, we need to prove that it is continuous at every point in its domain. Let $c$ be an arbitrary real number.

For $P(x)$ to be continuous at $x=c$, the following condition must be satisfied:

$\lim\limits_{x \to c} P(x) = P(c)$

First, let's evaluate the function at $x=c$. Substituting $x=c$ into the expression for $P(x)$, we get:

$P(c) = a_n c^n + a_{n-1} c^{n-1} + \dots + a_1 c + a_0$

Since $c$ is a real number and $a_i$ are real coefficients, $P(c)$ is a well-defined real number.

Next, let's evaluate the limit of the function as $x$ approaches $c$:

$\lim\limits_{x \to c} P(x) = \lim\limits_{x \to c} (a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0)$

Using the properties of limits, which state that the limit of a sum (or difference) is the sum (or difference) of the limits, and the limit of a constant times a function is the constant times the limit of the function, we can write:

$\lim\limits_{x \to c} P(x) = \lim\limits_{x \to c} (a_n x^n) + \lim\limits_{x \to c} (a_{n-1} x^{n-1}) + \dots + \lim\limits_{x \to c} (a_1 x) + \lim\limits_{x \to c} a_0$

$= a_n \lim\limits_{x \to c} x^n + a_{n-1} \lim\limits_{x \to c} x^{n-1} + \dots + a_1 \lim\limits_{x \to c} x + \lim\limits_{x \to c} a_0$

We know that for any integer $k \ge 0$, $\lim\limits_{x \to c} x^k = c^k$, and $\lim\limits_{x \to c} a_0 = a_0$ (limit of a constant).

Substituting these into the expression:

$\lim\limits_{x \to c} P(x) = a_n c^n + a_{n-1} c^{n-1} + \dots + a_1 c + a_0$

By comparing the limit value with the function value at $x=c$, we see that:

$\lim\limits_{x \to c} P(x) = P(c)$

This condition for continuity is satisfied for any arbitrary real number $c$. Therefore, the polynomial function $P(x)$ is continuous at every real number.

Conclusion:

Since the continuity condition $\lim\limits_{x \to c} P(x) = P(c)$ holds for any real number $c$, every polynomial function is continuous on the set of all real numbers ($\mathbb{R}$).

Given:

The greatest integer function $f(x) = [x]$, where $[x]$ denotes the greatest integer less than or equal to $x$.

The domain of the greatest integer function is the set of all real numbers ($\mathbb{R}$).

To Find:

All the points of discontinuity of the function $f(x) = [x]$.

Solution:

To find the points of discontinuity of $f(x) = [x]$, we consider points in its domain and check the continuity condition $\lim\limits_{x \to c} f(x) = f(c)$. We consider two cases for any real number $c$:

Case 1: $c$ is an integer.

Let $c = n$, where $n \in \mathbb{Z}$.

The function value at $x = n$ is $f(n) = [n] = n$.

Next, we evaluate the left-hand limit (LHL) and the right-hand limit (RHL) at $x = n$.

LHL at $x = n$: $\lim\limits_{x \to n^-} f(x) = \lim\limits_{x \to n^-} [x]$.

As $x$ approaches $n$ from the left side, $x$ is slightly less than $n$. For example, if $n=3$, as $x \to 3^-$, $x$ could be $2.9, 2.99, 2.999, \dots$. The greatest integer less than or equal to such $x$ is $n-1$.

$\lim\limits_{x \to n^-} [x] = n-1$

RHL at $x = n$: $\lim\limits_{x \to n^+} f(x) = \lim\limits_{x \to n^+} [x]$.

As $x$ approaches $n$ from the right side, $x$ is slightly greater than $n$. For example, if $n=3$, as $x \to 3^+$, $x$ could be $3.1, 3.01, 3.001, \dots$. The greatest integer less than or equal to such $x$ is $n$.

$\lim\limits_{x \to n^+} [x] = n$

Comparing the LHL and RHL at $x=n$:

$n-1 \neq n$

Since the left-hand limit and the right-hand limit are not equal, the limit $\lim\limits_{x \to n} [x]$ does not exist at any integer point $n$.

For a function to be continuous at a point, the limit must exist at that point and be equal to the function value. Since the limit does not exist at any integer point, the function is not continuous at any integer point.

Case 2: $c$ is not an integer.

Let $c$ be any real number that is not an integer. This means there exists an integer $n$ such that $n < c < n+1$.

The function value at $x = c$ is $f(c) = [c]$. Since $n < c < n+1$, the greatest integer less than or equal to $c$ is $n$.

$f(c) = n$

Now, consider the limit of $f(x)$ as $x$ approaches $c$. Since $c$ is not an integer, we can find a small open interval around $c$, say $(c-\delta, c+\delta)$ for some $\delta > 0$, such that for all $x$ in this interval, $n < x < n+1$. (For example, choose $\delta = \min(c-n, n+1-c)$). In this interval, the value of $[x]$ is always $n$.

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} [x]$

For $x$ sufficiently close to $c$ (within the interval where $[x]=n$), the limit is:

$\lim\limits_{x \to c} n = n$

Comparing the limit with the function value at $x=c$:

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

$f(c) = n$

Since $\lim\limits_{x \to c} f(x) = f(c)$, the function $f(x)$ is continuous at any real number $c$ that is not an integer.

Combining both cases, the function is continuous at all non-integer points and discontinuous at all integer points.

Conclusion:

The points of discontinuity of the greatest integer function $f(x) = [x]$ are all the integer points. That is, the function is discontinuous at $x = n$ for all $n \in \mathbb{Z}$.

Given:

A rational function $f(x)$.

A rational function is defined as the ratio of two polynomial functions.

Let $P(x)$ and $Q(x)$ be two polynomial functions, where $Q(x)$ is not the zero polynomial.

A rational function $f(x)$ can be expressed in the form:

$f(x) = \frac{P(x)}{Q(x)}$

The domain of the rational function $f(x)$ is the set of all real numbers $x$ for which $Q(x) \neq 0$.

To Prove:

Every rational function is continuous at every point in its domain.

Proof:

We need to prove that the rational function $f(x) = \frac{P(x)}{Q(x)}$ is continuous at every point $c$ such that $c$ is in the domain of $f(x)$. This means $Q(c) \neq 0$.

We recall a property regarding the continuity of polynomial functions:

A polynomial function is continuous at every real number.

From Example 14 (or a known theorem), we know that both $P(x)$ and $Q(x)$ are polynomial functions, and thus they are continuous everywhere on $\mathbb{R}$.

We also recall a fundamental property of continuous functions regarding their quotients:

If $g(x)$ and $h(x)$ are two continuous functions at a point $c$, and $h(c) \neq 0$, then the function given by their quotient, $\frac{g(x)}{h(x)}$, is also continuous at $c$.

In our case, let $c$ be a point in the domain of $f(x)$. By definition of the domain, $Q(c) \neq 0$.

Since $P(x)$ is a polynomial, it is continuous at $c$.

Since $Q(x)$ is a polynomial, it is continuous at $c$.

At the point $c$, we have $Q(c) \neq 0$.

Applying the property of the quotient of continuous functions, with $g(x) = P(x)$ and $h(x) = Q(x)$, the function $f(x) = \frac{P(x)}{Q(x)}$ is continuous at $c$.

Since $c$ was an arbitrary point in the domain of $f(x)$, this proves that $f(x)$ is continuous at every point in its domain.

Conclusion:

Since a rational function is the quotient of two polynomial functions, and the quotient of continuous functions is continuous wherever the denominator is non-zero, every rational function is continuous at every point in its domain.

Given:

The sine function $f(x) = \sin(x)$.

The domain of the sine function is the set of all real numbers ($\mathbb{R}$).

To Discuss:

The continuity of the sine function at all points in its domain.

Solution:

To discuss the continuity of the function $f(x) = \sin(x)$, we need to check if it is continuous at every point in its domain. Let $c$ be an arbitrary real number.

For the function $f$ to be continuous at $x=c$, we must verify that $\lim\limits_{x \to c} f(x) = f(c)$.

First, let's evaluate the function at $x = c$:

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

Next, let's evaluate the limit of the function as $x$ approaches $c$. We need to find $\lim\limits_{x \to c} \sin(x)$.

Let $x = c + h$. As $x \to c$, $h \to 0$. Substituting this into the limit expression:

$\lim\limits_{x \to c} \sin(x) = \lim\limits_{h \to 0} \sin(c+h)$

Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, we have:

$\lim\limits_{h \to 0} \sin(c+h) = \lim\limits_{h \to 0} (\sin c \cos h + \cos c \sin h)$

Using the properties of limits (limit of a sum is the sum of limits, limit of a product is the product of limits, constants can be factored out of limits):

$= \lim\limits_{h \to 0} (\sin c \cos h) + \lim\limits_{h \to 0} (\cos c \sin h)$

$= \sin c \cdot \lim\limits_{h \to 0} \cos h + \cos c \cdot \lim\limits_{h \to 0} \sin h$

We use the standard limits $\lim\limits_{h \to 0} \cos h = 1$ and $\lim\limits_{h \to 0} \sin h = 0$.

$= \sin c \cdot 1 + \cos c \cdot 0$

$= \sin c + 0$

$= \sin c$

So, we have found that:

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

Now, we compare the limit of the function as $x$ approaches $c$ with the value of the function at $x = c$:

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

$f(c) = \sin c$

Since $\lim\limits_{x \to c} f(x) = f(c)$, the function $f(x) = \sin(x)$ is continuous at the arbitrary point $c$.

Since $c$ was any arbitrary real number, this holds true for all real numbers.

Conclusion:

The sine function $f(x) = \sin(x)$ is continuous at every real number.

Given:

The function $f(x) = \tan x$.

The tangent function is defined as $f(x) = \frac{\sin x}{\cos x}$.

The domain of $f(x) = \tan x$ is the set of all real numbers $x$ for which $\cos x \neq 0$. This means $x \neq \frac{\pi}{2} + n\pi$, where $n$ is any integer ($n \in \mathbb{Z}$).

To Prove:

The function $f(x) = \tan x$ is continuous at every point in its domain.

Proof:

To prove that $f(x) = \tan x$ is continuous, we need to show that it is continuous at every point $c$ in its domain. By definition of the domain, this means we consider any real number $c$ such that $\cos c \neq 0$.

We know from previous discussions (Example 17 for sine, and a similar proof can be constructed for cosine) that the sine function, $g(x) = \sin x$, is continuous for all real numbers, and the cosine function, $h(x) = \cos x$, is continuous for all real numbers.

A property of continuous functions states that if two functions $g(x)$ and $h(x)$ are continuous at a point $c$, then their quotient $\frac{g(x)}{h(x)}$ is continuous at $c$, provided that $h(c) \neq 0$.

In this case, we have $f(x) = \frac{\sin x}{\cos x}$. Let $g(x) = \sin x$ and $h(x) = \cos x$.

Let $c$ be any point in the domain of $f(x) = \tan x$. This implies that $c$ is a real number and $\cos c \neq 0$.

Since $g(x) = \sin x$ is continuous for all real numbers, it is continuous at $c$.

Since $h(x) = \cos x$ is continuous for all real numbers, it is continuous at $c$.

At the point $c$, we have $h(c) = \cos c \neq 0$.

Applying the quotient rule for continuous functions, the function $f(x) = \frac{g(x)}{h(x)} = \frac{\sin x}{\cos x} = \tan x$ is continuous at $c$.

Since $c$ was an arbitrary point in the domain of $f(x) = \tan x$, this proves that $f(x) = \tan x$ is continuous at every point in its domain.

Conclusion:

The function $f(x) = \tan x$ is continuous at every point in its domain, which is $\{x \in \mathbb{R} \mid x \neq \frac{\pi}{2} + n\pi, n \in \mathbb{Z}\}$. Therefore, the function $f(x) = \tan x$ is a continuous function on its domain.

Given:

The function $f(x) = \sin(x^2)$.

The domain of the function $f(x) = \sin(x^2)$ is the set of all real numbers ($\mathbb{R}$).

To Show:

The function $f(x) = \sin(x^2)$ is a continuous function on its domain.

Proof:

We can express the function $f(x) = \sin(x^2)$ as a composition of two simpler functions.

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

Then, $f(x) = h(g(x)) = h(x^2) = \sin(x^2)$.

To show that $f(x)$ is continuous, we can use the theorem on the continuity of composite functions.

The theorem states that if a function $g$ is continuous at a point $c$, and a function $h$ is continuous at $g(c)$, then the composite function $h \circ g$, defined by $(h \circ g)(x) = h(g(x))$, is continuous at $c$.

Let's examine the continuity of the individual functions $g(x)$ and $h(u)$.

1. Continuity of $g(x) = x^2$:

The function $g(x) = x^2$ is a polynomial function. We know that every polynomial function is continuous at every real number (as shown in Example 14). Therefore, $g(x)$ is continuous for all $x \in \mathbb{R}$.

2. Continuity of $h(u) = \sin(u)$:

The function $h(u) = \sin(u)$ is the sine function. We know that the sine function is continuous at every real number (as shown in Example 17). Therefore, $h(u)$ is continuous for all $u \in \mathbb{R}$.

Now, consider an arbitrary real number $c \in \mathbb{R}$.

Since $g(x) = x^2$ is continuous at $c$, the value $g(c) = c^2$ is a real number.

Since $h(u) = \sin(u)$ is continuous for all real numbers, it is continuous at the real number $g(c) = c^2$.

By the theorem on the continuity of composite functions, since $g(x)$ is continuous at $c$ and $h(u)$ is continuous at $g(c)$, the composite function $f(x) = h(g(x))$ is continuous at $c$.

Since $c$ was an arbitrary real number, this proves that $f(x) = \sin(x^2)$ is continuous at every point in its domain, which is $\mathbb{R}$.

Conclusion:

The function $f(x) = \sin(x^2)$ is a composition of the polynomial function $g(x) = x^2$ and the sine function $h(u) = \sin(u)$, both of which are continuous on $\mathbb{R}$. By the theorem on continuity of composite functions, $f(x)$ is continuous at every real number.

Given:

The function $f(x) = |1 - x + |x||$ defined for all real numbers.

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Show:

The function $f(x) = |1 - x + |x||$ is a continuous function on its domain.

Proof:

We can analyze the continuity of $f(x)$ by considering it as a combination and composition of elementary continuous functions.

We know the following functions are continuous for all real numbers:

1. The constant function, say $g_1(x) = 1$.

2. The identity function, say $g_2(x) = x$.

3. The absolute value function, say $g_3(x) = |x|$. (As discussed in Example 7)

We also know that the sum, difference, and composition of continuous functions are continuous on their respective domains.

Let's consider the function inside the outer absolute value sign: $h(x) = 1 - x + |x|$.

This function $h(x)$ can be written as the sum and difference of the continuous functions $g_1(x)$, $g_2(x)$, and $g_3(x)$: $h(x) = g_1(x) - g_2(x) + g_3(x)$.

Since $g_1(x)$, $g_2(x)$, and $g_3(x)$ are continuous for all real numbers, their combination $h(x) = 1 - x + |x|$ is also continuous for all real numbers.

Now, the given function is $f(x) = |h(x)|$. This is the composition of the absolute value function with $h(x)$.

Let $k(u) = |u|$. The function $k(u)$ is continuous for all real numbers.

Then $f(x) = k(h(x))$.

Since $h(x)$ is continuous for all real numbers, and $k(u)$ is continuous for all real numbers, the composite function $f(x) = k(h(x))$ is continuous for all real numbers by the theorem on the continuity of composite functions.

Alternatively, we can examine the function piecewise:

$f(x) = |1 - x + |x|| = \begin{cases} |1 - x + x| & , & \text{if } x \geq 0 \\ |1 - x - x| & , & \text{if } x < 0 \end{cases}$

$f(x) = \begin{cases} |1| & , & \text{if } x \geq 0 \\ |1 - 2x| & , & \text{if } x < 0 \end{cases}$

$f(x) = \begin{cases} 1 & , & \text{if } x \geq 0 \\ |1 - 2x| & , & \text{if } x < 0 \end{cases}$

For $x < 0$, $f(x) = |1 - 2x|$. Since $1-2x$ is a polynomial (hence continuous) and the absolute value function is continuous, the composition $|1-2x|$ is continuous for all $x < 0$.

For $x > 0$, $f(x) = 1$, which is a constant function, hence continuous for all $x > 0$.

Now we check continuity at the point $x=0$, where the definition changes.

Function value at $x=0$:

For $x \geq 0$, $f(x) = 1$. So $f(0) = 1$.

$f(0) = 1$

Left-hand limit at $x=0$:

For $x < 0$, $f(x) = |1 - 2x|$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} |1 - 2x| = |1 - 2(0)| = |1| = 1$

Right-hand limit at $x=0$:

For $x > 0$, $f(x) = 1$.

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

Comparing the LHL, RHL, and $f(0)$:

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

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

$f(0) = 1$

Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0) = 1$, the function is continuous at $x=0$.

Combining all cases, the function is continuous for $x < 0$, $x > 0$, and at $x = 0$. Thus, the function is continuous for all real numbers.

Conclusion:

The function $f(x) = |1 - x + |x||$ is a composition and combination of elementary continuous functions (constant, identity, absolute value). Applying the properties of continuity for sums, differences, and compositions, or by piecewise analysis, we find that the function is continuous at every real number.

Given:

The function $f(x) = 5x - 3$.

The points to check for continuity are $x = 0$, $x = -3$, and $x = 5$.

To Prove:

The function $f(x) = 5x - 3$ is continuous at $x = 0$, at $x = -3$, and at $x = 5$.

Proof:

A function $f$ is continuous at a point $x = c$ if $\lim\limits_{x \to c} f(x) = f(c)$. We will check this condition for each of the given points.

Continuity at $x = 0$:

First, find the function value at $x = 0$:

$f(0) = 5(0) - 3 = 0 - 3 = -3$

Next, find the limit of the function as $x$ approaches $0$:

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} (5x - 3)$

Since $f(x) = 5x - 3$ is a polynomial function, the limit can be evaluated by direct substitution:

$\lim\limits_{x \to 0} (5x - 3) = 5(0) - 3 = -3$

Comparing the limit and the function value at $x=0$:

$\lim\limits_{x \to 0} f(x) = -3$

$f(0) = -3$

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the function $f$ is continuous at $x = 0$.

Continuity at $x = -3$:

First, find the function value at $x = -3$:

$f(-3) = 5(-3) - 3 = -15 - 3 = -18$

Next, find the limit of the function as $x$ approaches $-3$:

$\lim\limits_{x \to -3} f(x) = \lim\limits_{x \to -3} (5x - 3)$

Using direct substitution for the polynomial function:

$\lim\limits_{x \to -3} (5x - 3) = 5(-3) - 3 = -15 - 3 = -18$

Comparing the limit and the function value at $x=-3$:

$\lim\limits_{x \to -3} f(x) = -18$

$f(-3) = -18$

Since $\lim\limits_{x \to -3} f(x) = f(-3)$, the function $f$ is continuous at $x = -3$.

Continuity at $x = 5$:

First, find the function value at $x = 5$:

$f(5) = 5(5) - 3 = 25 - 3 = 22$

Next, find the limit of the function as $x$ approaches $5$:

$\lim\limits_{x \to 5} f(x) = \lim\limits_{x \to 5} (5x - 3)$

Using direct substitution for the polynomial function:

$\lim\limits_{x \to 5} (5x - 3) = 5(5) - 3 = 25 - 3 = 22$

Comparing the limit and the function value at $x=5$:

$\lim\limits_{x \to 5} f(x) = 22$

$f(5) = 22$

Since $\lim\limits_{x \to 5} f(x) = f(5)$, the function $f$ is continuous at $x = 5$.

Conclusion:

In all three cases ($x=0$, $x=-3$, and $x=5$), the limit of the function as $x$ approaches the point is equal to the value of the function at that point. Therefore, the function $f(x) = 5x - 3$ is continuous at $x = 0$, at $x = -3$, and at $x = 5$.

Given:

The function $f(x) = 2x^2 - 1$.

We need to examine the continuity of the function at $x = 3$.

To Examine:

The continuity of $f(x)$ at $x = 3$.

Solution:

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

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

In this problem, we need to check the continuity at $x = 3$. So, we will verify if $\lim\limits_{x \to 3} f(x) = f(3)$.

First, let's find the value of the function at $x = 3$:

$f(3) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$

So, $f(3) = 17$.

Next, let's find the limit of the function as $x$ approaches $3$:

$\lim\limits_{x \to 3} f(x) = \lim\limits_{x \to 3} (2x^2 - 1)$

The function $f(x) = 2x^2 - 1$ is a polynomial function. For polynomial functions, the limit as $x$ approaches a point can be found by direct substitution:

$\lim\limits_{x \to 3} (2x^2 - 1) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$

So, $\lim\limits_{x \to 3} f(x) = 17$.

Now, we compare the limit of the function as $x$ approaches $3$ with the value of the function at $x = 3$:

$\lim\limits_{x \to 3} f(x) = 17$

$f(3) = 17$

Since $\lim\limits_{x \to 3} f(x) = f(3)$, the function $f(x) = 2x^2 - 1$ is continuous at $x = 3$.

Conclusion:

Since the limit of the function at $x = 3$ exists and is equal to the function value at $x = 3$, the function $f(x) = 2x^2 - 1$ is continuous at x = 3.

We examine the continuity of each function separately.

(a) f(x) = x – 5

Given: The function $f(x) = x - 5$.

This is a polynomial function of degree 1. The domain of a polynomial function is the set of all real numbers ($\mathbb{R}$).

To Examine: The continuity of $f(x)$ at all points in its domain.

Solution:

To examine the continuity of $f(x)$ on $\mathbb{R}$, we choose an arbitrary real number $c$ and check if $\lim\limits_{x \to c} f(x) = f(c)$.

First, evaluate the function at $x=c$:

$f(c) = c - 5$

Next, evaluate the limit of the function as $x$ approaches $c$. Since $f(x)$ is a polynomial, the limit can be found by direct substitution:

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (x - 5) = c - 5$

Comparing the limit and the function value:

$\lim\limits_{x \to c} f(x) = c - 5$

$f(c) = c - 5$

Since $\lim\limits_{x \to c} f(x) = f(c)$ for any real number $c$, the function $f(x) = x - 5$ is continuous at every real number.

Conclusion:

The function $f(x) = x - 5$ is a polynomial and is continuous on its domain, which is all real numbers ($\mathbb{R}$).

(b) f(x) = $\frac{1}{x-5}$ , x ≠ 5

Given: The function $f(x) = \frac{1}{x-5}$, with the domain $x \neq 5$, which is $\mathbb{R} \setminus \{5\}$.

This is a rational function, as it is the ratio of two polynomial functions, $P(x) = 1$ and $Q(x) = x-5$.

To Examine: The continuity of $f(x)$ at all points in its domain.

Solution:

To examine the continuity of $f(x)$ on its domain, we choose an arbitrary real number $c$ such that $c \neq 5$. We check if $\lim\limits_{x \to c} f(x) = f(c)$.

First, evaluate the function at $x=c$. Since $c \neq 5$, the denominator is non-zero, and $f(c)$ is defined:

$f(c) = \frac{1}{c-5}$

Next, evaluate the limit of the function as $x$ approaches $c$. Since $f(x)$ is a rational function and the denominator $x-5$ is non-zero at $x=c$ (because $c \neq 5$), the limit can be evaluated by direct substitution:

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} \frac{1}{x-5} = \frac{\lim\limits_{x \to c} 1}{\lim\limits_{x \to c} (x-5)} = \frac{1}{c-5}$

Comparing the limit and the function value:

$\lim\limits_{x \to c} f(x) = \frac{1}{c-5}$

$f(c) = \frac{1}{c-5}$

Since $\lim\limits_{x \to c} f(x) = f(c)$ for any real number $c$ in the domain ($c \neq 5$), the function $f(x) = \frac{1}{x-5}$ is continuous at every point in its domain.

The function is not defined at $x=5$, so it cannot be continuous at $x=5$.

Conclusion:

The function $f(x) = \frac{1}{x-5}$ is a rational function and is continuous on its domain, which is all real numbers except $5$ ($\mathbb{R} \setminus \{5\}$).

(c) f(x) = $\frac{x^2 − 25}{x + 5}$ , x ≠ 5

Given: The function $f(x) = \frac{x^2 - 25}{x + 5}$, with the condition $x \neq 5$.

The expression $\frac{x^2 - 25}{x + 5}$ is a rational expression. It is undefined when the denominator is zero, i.e., $x+5 = 0$, which means $x = -5$. Therefore, the natural domain of the expression $\frac{x^2 - 25}{x + 5}$ is $\mathbb{R} \setminus \{-5\}$. The given condition $x \neq 5$ is also part of the domain definition for this specific function as stated.

Let's simplify the expression for $x \neq -5$:

$x^2 - 25 = (x-5)(x+5)$

So, for $x \neq -5$, we have:

$f(x) = \frac{(x-5)(x+5)}{x+5} = x-5$

Thus, for $x \neq -5$, $f(x)$ behaves like the polynomial $g(x) = x-5$. The question states the function is defined for $x \neq 5$. The function as given by the expression $\frac{x^2 - 25}{x + 5}$ is defined at $x=5$ (where $f(5) = \frac{5^2-25}{5+5} = \frac{0}{10} = 0$) but not at $x=-5$. If the function is strictly defined as $f(x) = \frac{x^2 − 25}{x + 5}$ for $x \neq 5$, and no value is specified at $x=5$, then its domain is $\mathbb{R} \setminus \{-5\}$. Let's assume the question intends the function $f(x) = x-5$ for $x \neq -5$, and the $x \neq 5$ is just part of the original problem statement copy.

Let's discuss continuity on the domain $\mathbb{R} \setminus \{-5\}$. For $x \neq -5$, $f(x) = x-5$, which is a polynomial function.

To Examine: The continuity of $f(x)$ at all points in its domain, which is $\mathbb{R} \setminus \{-5\}$.

Solution:

Let $c$ be any real number such that $c \neq -5$. For such $c$, $f(x) = x-5$ in an interval around $c$. We check if $\lim\limits_{x \to c} f(x) = f(c)$.

First, evaluate the function at $x=c$ (where $c \neq -5$):

$f(c) = c - 5$

Next, evaluate the limit of the function as $x$ approaches $c$ (where $c \neq -5$):

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (x - 5) = c - 5$

Comparing the limit and the function value:

$\lim\limits_{x \to c} f(x) = c - 5$

$f(c) = c - 5$

Since $\lim\limits_{x \to c} f(x) = f(c)$ for any real number $c \neq -5$, the function $f(x)$ is continuous at every point in its domain $\mathbb{R} \setminus \{-5\}$. The function is not defined at $x=-5$, so it is discontinuous at $x=-5$.

Regarding the condition $x \neq 5$ in the prompt: The expression $\frac{x^2 - 25}{x + 5}$ is defined at $x=5$ and its value is 0. If the function was defined piece-wise, e.g., $f(x) = \frac{x^2-25}{x+5}$ for $x \neq 5, x \neq -5$ and $f(5)=k$ for some $k$, we would check continuity at $x=5$. As it is, the function is $f(x) = x-5$ for $x \neq -5$, and this function is continuous everywhere except where it's not defined, which is $x=-5$.

Conclusion:

The function $f(x) = \frac{x^2 - 25}{x + 5}$ is equivalent to $f(x) = x-5$ for $x \neq -5$. Thus, the function is continuous on its domain, which is all real numbers except $-5$ ($\mathbb{R} \setminus \{-5\}$). The point of discontinuity is $x=-5$ (a removable discontinuity).

(d) f(x) = | x – 5 |

Given: The function $f(x) = |x - 5|$.

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Examine: The continuity of $f(x)$ at all points in its domain.

Solution:

We can write $f(x)$ using a piecewise definition:

$f(x) = |x - 5| = \begin{cases} x - 5 & , & \text{if } x - 5 \geq 0 \implies x \geq 5 \\ -(x - 5) = 5 - x & , & \text{if } x - 5 < 0 \implies x < 5 \end{cases}$

We examine continuity in the intervals and at the point where the definition changes, $x=5$.

Case 1: For $x < 5$

Let $c < 5$. For $x$ near $c$, $f(x) = 5 - x$, which is a polynomial (hence continuous). $f(c) = 5-c$, and $\lim\limits_{x \to c} (5-x) = 5-c$. So $f$ is continuous for $x < 5$.

Case 2: For $x > 5$

Let $c > 5$. For $x$ near $c$, $f(x) = x - 5$, which is a polynomial (hence continuous). $f(c) = c-5$, and $\lim\limits_{x \to c} (x-5) = c-5$. So $f$ is continuous for $x > 5$.

Case 3: At $x = 5$

We check if $\lim\limits_{x \to 5} f(x) = f(5)$.

Function value at $x=5$:

Using the definition for $x \geq 5$, $f(5) = 5 - 5 = 0$.

$f(5) = 0$

Left-hand limit at $x=5$:

For $x < 5$, $f(x) = 5 - x$.

$\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^-} (5 - x) = 5 - 5 = 0$

Right-hand limit at $x=5$:

For $x > 5$, $f(x) = x - 5$.

$\lim\limits_{x \to 5^+} f(x) = \lim\limits_{x \to 5^+} (x - 5) = 5 - 5 = 0$

Comparing the LHL, RHL, and $f(5)$:

$\lim\limits_{x \to 5^-} f(x) = 0$

$\lim\limits_{x \to 5^+} f(x) = 0$

$f(5) = 0$

Since LHL = RHL = $f(5) = 0$, the function is continuous at $x = 5$.

Alternatively, we can use the property of composite functions. Let $g(x) = x - 5$ and $h(u) = |u|$. Then $f(x) = h(g(x))$. $g(x)$ is a polynomial, continuous on $\mathbb{R}$. $h(u)$ is the absolute value function, continuous on $\mathbb{R}$. The composition of two continuous functions is continuous. Therefore, $f(x) = |x-5|$ is continuous on $\mathbb{R}$.

Conclusion:

The function $f(x) = |x - 5|$ is continuous for all $x < 5$, for all $x > 5$, and at $x = 5$. Thus, the function $f(x) = |x - 5|$ is continuous at every real number.

Given:

The function $f(x) = x^n$, where $n$ is a positive integer.

The domain of the function $f(x) = x^n$ is the set of all real numbers ($\mathbb{R}$). We are asked to prove its continuity at the specific point $x = n$.

To Prove:

The function $f(x) = x^n$ is continuous at $x = n$, where $n$ is a positive integer.

Proof:

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

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

In this problem, we need to prove the continuity of $f(x) = x^n$ at the point $x = n$. Thus, we need to verify if $\lim\limits_{x \to n} f(x) = f(n)$.

First, let's evaluate the function at $x = n$:

$f(n) = n^n$

Since $n$ is a positive integer, $n^n$ is a well-defined real number.

Next, let's evaluate the limit of the function as $x$ approaches $n$:

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

The function $f(x) = x^n$ is a polynomial function (specifically, a monomial) since $n$ is a positive integer. A fundamental property of limits states that for a polynomial function $P(x)$, the limit as $x$ approaches any real number $c$ is simply the function evaluated at that point, i.e., $\lim\limits_{x \to c} P(x) = P(c)$.

Using this property for $f(x) = x^n$ and the point $c = n$, we have:

$\lim\limits_{x \to n} x^n = n^n$

So, we have found that:

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

Now, we compare the limit of the function as $x$ approaches $n$ with the value of the function at $x = n$:

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

$f(n) = n^n$

Since $\lim\limits_{x \to n} f(x) = f(n)$, the function $f(x) = x^n$ satisfies the condition for continuity at $x = n$.

Conclusion:

Because $\lim\limits_{x \to n} f(x) = f(n)$, the function $f(x) = x^n$ is continuous at x = n for any positive integer $n$. (Note: In fact, polynomial functions $x^n$ are continuous at every real number $x=c$, not just when $c=n$).

Given:

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

$f(x) = \begin{cases} x,& if \;x ≤ 1\\5,& if\; x > 1 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Check:

The continuity of the function $f(x)$ at $x = 0$, $x = 1$, and $x = 2$.

Solution:

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

Let's check the continuity at the specified points:

Continuity at $x = 0$:

At $x=0$, the function definition is $f(x) = x$ since $0 \leq 1$.

Function value at $x = 0$:

$f(0) = 0$

Limit at $x = 0$:

For values of $x$ near $0$ (e.g., in $(-1, 1)$), $f(x) = x$.

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

Comparing the limit and the function value:

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

$f(0) = 0$

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the function is continuous at $x = 0$.

Continuity at $x = 1$:

At $x=1$, the function definition changes. We need to check the function value, the left-hand limit (LHL), and the right-hand limit (RHL).

Function value at $x = 1$:

For $x \leq 1$, $f(x) = x$. So, $f(1) = 1$.

$f(1) = 1$

Left-hand limit at $x = 1$:

For $x < 1$, $f(x) = x$.

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

Right-hand limit at $x = 1$:

For $x > 1$, $f(x) = 5$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} 5 = 5$

Comparing the LHL and RHL:

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

$\lim\limits_{x \to 1^+} f(x) = 5$

Since $\lim\limits_{x \to 1^-} f(x) \neq \lim\limits_{x \to 1^+} f(x)$, the limit $\lim\limits_{x \to 1} f(x)$ does not exist. Therefore, the function is not continuous at $x = 1$.

Continuity at $x = 2$:

At $x=2$, the function definition is $f(x) = 5$ since $2 > 1$.

Function value at $x = 2$:

$f(2) = 5$

Limit at $x = 2$:

For values of $x$ near $2$ (e.g., in $(1, 3)$), $f(x) = 5$.

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

Comparing the limit and the function value:

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

$f(2) = 5$

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

Conclusion:

Based on the examination:

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

The function $f(x)$ is not continuous at x = 1.

The function $f(x)$ is continuous at x = 2.

Given:

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

$f(x) = \begin{cases} 2x+3,& if\; x ≤ 2\\2x−3,& if\; x > 2 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

To find the points of discontinuity, we examine the function's continuity in the intervals defined by the piecewise function and at the point where the definition changes, which is $x = 2$.

We consider three cases:

Case 1: For $x < 2$

Let $c$ be any real number such that $c < 2$. For all values of $x$ in an open interval around $c$ that lies within $(-\infty, 2)$, the function is defined as $f(x) = 2x+3$.

The function $g(x) = 2x+3$ is a polynomial function. Polynomial functions are continuous everywhere.

Alternatively, we can check the continuity condition $\lim\limits_{x \to c} f(x) = f(c)$ for $c < 2$.

$f(c) = 2c+3$

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (2x+3) = 2c+3$

Since $\lim\limits_{x \to c} f(x) = f(c)$, the function $f(x)$ is continuous for all $x < 2$.

Case 2: For $x > 2$

Let $c$ be any real number such that $c > 2$. For all values of $x$ in an open interval around $c$ that lies within $(2, \infty)$, the function is defined as $f(x) = 2x-3.

The function $h(x) = 2x-3$ is a polynomial function. Polynomial functions are continuous everywhere.

Alternatively, we can check the continuity condition $\lim\limits_{x \to c} f(x) = f(c)$ for $c > 2$.

$f(c) = 2c-3$

$\lim\limits_{x \to c} f(x) = \lim\limits_{x \to c} (2x-3) = 2c-3$

Since $\lim\limits_{x \to c} f(x) = f(c)$, the function $f(x)$ is continuous for all $x > 2$.

Case 3: At $x = 2$

This is the critical point where the definition of the function changes. To check for continuity at $x=2$, we need to verify if $\lim\limits_{x \to 2} f(x) = f(2)$. We need to find $f(2)$, the left-hand limit (LHL) at $x=2$, and the right-hand limit (RHL) at $x=2$.

Function value at $x = 2$:

From the definition, when $x \leq 2$, $f(x) = 2x+3$. So, at $x=2$, $f(2) = 2(2)+3 = 4+3 = 7$.

$f(2) = 7$

Left-hand limit at $x = 2$:

For $x < 2$, $f(x) = 2x+3$.

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (2x+3) = 2(2)+3 = 4+3 = 7$

Right-hand limit at $x = 2$:

For $x > 2$, $f(x) = 2x-3$.

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (2x-3) = 2(2)-3 = 4-3 = 1$

Now, we compare the LHL and RHL at $x=2$:

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

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

Since the left-hand limit is not equal to the right-hand limit ($\lim\limits_{x \to 2^-} f(x) \neq \lim\limits_{x \to 2^+} f(x)$), the limit $\lim\limits_{x \to 2} f(x)$ does not exist.

For a function to be continuous at a point, the limit must exist at that point and be equal to the function value. Since the limit does not exist at $x=2$, the function is not continuous at $x=2$.

Combining the cases, the function is continuous for all $x < 2$ and all $x > 2$. The only point where it is not continuous is $x=2$.

Conclusion:

The function $f(x)$ is continuous for all real numbers except at $x=2$. The only point of discontinuity is $x=2$.

Given:

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

$f(x) = \begin{cases} |x|+3,& if\; x ≤ −3\\−2x,& if\; −3 < x < 3\\6x + 2,& if\; x ≥ 3 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

To find the points of discontinuity, we examine the function's continuity in the open intervals defined by the piecewise function and at the points where the definition changes, which are $x = -3$ and $x = 3$.

We consider the continuity in the intervals $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$, and at the boundary points $x = -3$ and $x = 3$.

Case 1: For $x < -3$

Let $c$ be any real number such that $c < -3$. In this interval, $x < 0$, so $|x| = -x$. Thus, for $x < -3$, $f(x) = -x+3$. This is a polynomial function, which is continuous for all real numbers. Hence, $f(x)$ is continuous for all $x < -3$.

Case 2: For $-3 < x < 3$

Let $c$ be any real number such that $-3 < c < 3$. In this interval, $f(x) = -2x$. This is a polynomial function, which is continuous for all real numbers. Hence, $f(x)$ is continuous for all $-3 < x < 3$.

Case 3: For $x > 3$

Let $c$ be any real number such that $c > 3$. In this interval, $f(x) = 6x+2$. This is a polynomial function, which is continuous for all real numbers. Hence, $f(x)$ is continuous for all $x > 3$.

Now, we need to check the continuity at the points where the definition changes, i.e., at $x = -3$ and $x = 3$.

Continuity at $x = -3$

We need to check if $\lim\limits_{x \to -3} f(x) = f(-3)$. This requires evaluating $f(-3)$, the left-hand limit (LHL), and the right-hand limit (RHL) at $x = -3$.

Function value at $x = -3$:

Using the definition for $x \leq -3$, $f(x) = |x|+3$. So, $f(-3) = |-3|+3 = 3+3 = 6$.

$f(-3) = 6$

Left-hand limit at $x = -3$:

For $x < -3$, $f(x) = |x|+3 = -x+3$.

$\lim\limits_{x \to -3^-} f(x) = \lim\limits_{x \to -3^-} (-x+3) = -(-3)+3 = 3+3 = 6$

Right-hand limit at $x = -3$:

For $x > -3$ and $x < 3$, $f(x) = -2x$.

$\lim\limits_{x \to -3^+} f(x) = \lim\limits_{x \to -3^+} (-2x) = -2(-3) = 6$

Comparing the LHL, RHL, and $f(-3)$:

$\lim\limits_{x \to -3^-} f(x) = 6$

$\lim\limits_{x \to -3^+} f(x) = 6$

$f(-3) = 6$

Since $\lim\limits_{x \to -3^-} f(x) = \lim\limits_{x \to -3^+} f(x) = f(-3) = 6$, the function is continuous at $x = -3$.

Continuity at $x = 3$

We need to check if $\lim\limits_{x \to 3} f(x) = f(3)$. This requires evaluating $f(3)$, the left-hand limit (LHL), and the right-hand limit (RHL) at $x = 3$.

Function value at $x = 3$:

Using the definition for $x \geq 3$, $f(x) = 6x+2$. So, $f(3) = 6(3)+2 = 18+2 = 20$.

$f(3) = 20$

Left-hand limit at $x = 3$:

For $x < 3$ and $x > -3$, $f(x) = -2x$.

$\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^-} (-2x) = -2(3) = -6$

Right-hand limit at $x = 3$:

For $x > 3$, $f(x) = 6x+2$.

$\lim\limits_{x \to 3^+} f(x) = \lim\limits_{x \to 3^+} (6x+2) = 6(3)+2 = 18+2 = 20$

Comparing the LHL, RHL, and $f(3)$:

$\lim\limits_{x \to 3^-} f(x) = -6$

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

$f(3) = 20$

Since the left-hand limit is not equal to the right-hand limit ($\lim\limits_{x \to 3^-} f(x) \neq \lim\limits_{x \to 3^+} f(x)$), the limit $\lim\limits_{x \to 3} f(x)$ does not exist. Therefore, the function is not continuous at $x = 3$.

Combining all cases, the function is continuous on $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$, and continuous at $x=-3$. The only point where it is not continuous is $x=3$.

Conclusion:

The function $f(x)$ is continuous everywhere except at $x=3$. The only point of discontinuity is $x=3$.

Given:

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

$f(x) = \begin{cases} \frac{|x|}{x},& if\; x \neq 0\\0,& if\; x = 0 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

We can rewrite the function $f(x)$ for $x \neq 0$ using the definition of $|x|$:

If $x > 0$, then $|x| = x$, so $\frac{|x|}{x} = \frac{x}{x} = 1$.

If $x < 0$, then $|x| = -x$, so $\frac{|x|}{x} = \frac{-x}{x} = -1$.

Thus, the function can be written as:

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

To find the points of discontinuity, we examine the function's continuity in the open intervals defined by the piecewise function and at the point where the definition changes, which is $x = 0$.

We consider the continuity in the intervals $(-\infty, 0)$ and $(0, \infty)$, and at the boundary point $x = 0$.

Case 1: For $x < 0$

Let $c$ be any real number such that $c < 0$. In this interval, $f(x) = -1$. This is a constant function. Constant functions are continuous for all real numbers. Thus, $f(x)$ is continuous for all $x < 0$.

Case 2: For $x > 0$

Let $c$ be any real number such that $c > 0$. In this interval, $f(x) = 1$. This is a constant function. Constant functions are continuous for all real numbers. Thus, $f(x)$ is continuous for all $x > 0$.

Case 3: At $x = 0$

This is the critical point where the definition of the function changes. To check for continuity at $x=0$, we need to verify if $\lim\limits_{x \to 0} f(x) = f(0)$. We need to find $f(0)$, the left-hand limit (LHL) at $x=0$, and the right-hand limit (RHL) at $x=0$.

Function value at $x = 0$:

From the definition, when $x = 0$, $f(x) = 0$.

$f(0) = 0$

Left-hand limit at $x = 0$:

For $x < 0$, $f(x) = -1$.

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

Right-hand limit at $x = 0$:

For $x > 0$, $f(x) = 1$.

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

Now, we compare the LHL and RHL at $x=0$:

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

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

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

For a function to be continuous at a point, the limit must exist at that point and be equal to the function value. Since the limit does not exist at $x=0$, the function is not continuous at $x=0$.

Combining the cases, the function is continuous for all $x < 0$ and all $x > 0$. The only point where it is not continuous is $x=0$.

Conclusion:

The function $f(x)$ is continuous everywhere except at $x=0$. The only point of discontinuity is $x=0$.

Given:

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

$f(x) = \begin{cases} \frac{x}{|x|},& if\; x < 0\\−1,& if\; x ≥ 0 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

We can simplify the expression for $f(x)$ when $x < 0$. If $x < 0$, then $|x| = -x$.

So, for $x < 0$, $\frac{x}{|x|} = \frac{x}{-x} = -1$.

Thus, the function can be written as:

$f(x) = \begin{cases} -1,& if\; x < 0\\−1,& if\; x ≥ 0 \end{cases}$

We can observe from this simplified form that $f(x)$ is always equal to $-1$ for all real numbers $x$.

$f(x) = -1$ for all $x \in \mathbb{R}$.

This means $f(x)$ is a constant function.

To Examine: The continuity of $f(x)$ at all points in its domain.

Solution using simplified form:

The function $f(x) = -1$ is a constant function. We know that constant functions are continuous for all real numbers.

Alternatively, we can check the continuity at the point where the original definition changes, which is $x=0$.

Case 1: For $x < 0$

Let $c < 0$. For $x$ near $c$, $f(x) = -1$, which is a constant function, hence continuous. $f(c) = -1$, and $\lim\limits_{x \to c} (-1) = -1$. So $f$ is continuous for $x < 0$.

Case 2: For $x > 0$

Let $c > 0$. For $x$ near $c$, $f(x) = -1$, which is a constant function, hence continuous. $f(c) = -1$, and $\lim\limits_{x \to c} (-1) = -1$. So $f$ is continuous for $x > 0$.

Case 3: At $x = 0$

We check if $\lim\limits_{x \to 0} f(x) = f(0)$.

Function value at $x=0$:

Using the definition for $x \geq 0$, $f(0) = -1$.

$f(0) = -1$

Left-hand limit at $x=0$:

For $x < 0$, $f(x) = \frac{x}{|x|} = -1$.

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

Right-hand limit at $x=0$:

For $x > 0$, $f(x) = -1$.

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

Comparing the LHL, RHL, and $f(0)$:

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

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

$f(0) = -1$

Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0) = -1$, the function is continuous at $x = 0$.

Combining all cases, the function is continuous for all $x < 0$, for all $x > 0$, and at $x = 0$. Thus, the function is continuous for all real numbers.

Conclusion:

The function $f(x)$ is a constant function equal to $-1$ for all real numbers. Constant functions are continuous everywhere. Therefore, there are no points of discontinuity for this function.

Given:

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

$f(x) = \begin{cases} x+1,& if\; x ≥1\\x^2+1,& if \;x < 1 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

To find the points of discontinuity, we examine the function's continuity in the open intervals defined by the piecewise function and at the point where the definition changes, which is $x = 1$.

We consider the continuity in the intervals $(-\infty, 1)$ and $(1, \infty)$, and at the boundary point $x = 1$.

Case 1: For $x < 1$

Let $c$ be any real number such that $c < 1$. In this interval, $f(x) = x^2+1$. This is a polynomial function. Polynomial functions are continuous everywhere. Thus, $f(x)$ is continuous for all $x < 1$.

Case 2: For $x > 1$

Let $c$ be any real number such that $c > 1$. In this interval, $f(x) = x+1$. This is a polynomial function. Polynomial functions are continuous everywhere. Thus, $f(x)$ is continuous for all $x > 1$.

Case 3: At $x = 1$

This is the critical point where the definition of the function changes. To check for continuity at $x=1$, we need to verify if $\lim\limits_{x \to 1} f(x) = f(1)$. We need to find $f(1)$, the left-hand limit (LHL) at $x=1$, and the right-hand limit (RHL) at $x=1$.

Function value at $x = 1$:

Using the definition for $x \geq 1$, $f(x) = x+1$. So, at $x=1$, $f(1) = 1+1 = 2$.

$f(1) = 2$

Left-hand limit at $x = 1$:

For $x < 1$, $f(x) = x^2+1$.

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

Right-hand limit at $x = 1$:

For $x > 1$, $f(x) = x+1$.

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

Comparing the LHL, RHL, and $f(1)$:

$\lim\limits_{x \to 1^-} f(x) = 2$

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

$f(1) = 2$

Since the left-hand limit is equal to the right-hand limit ($\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x)$), the limit $\lim\limits_{x \to 1} f(x)$ exists and is equal to 2.

Furthermore, the limit of the function at $x=1$ is equal to the function value at $x=1$ ($\lim\limits_{x \to 1} f(x) = f(1) = 2$).

Therefore, the function is continuous at $x = 1$.

Combining all cases, the function is continuous on $(-\infty, 1)$, $(1, \infty)$, and at $x=1$. Thus, the function is continuous for all real numbers.

Conclusion:

The function $f(x)$ is continuous at every real number. Therefore, there are no points of discontinuity for this function.

Given:

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

$f(x) = \begin{cases} x^3−3,& if\; x ≤ 2\\x^2+1,& if\; x > 2 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

To find the points of discontinuity, we examine the function's continuity in the open intervals defined by the piecewise function and at the point where the definition changes, which is $x = 2$.

We consider the continuity in the intervals $(-\infty, 2)$ and $(2, \infty)$, and at the boundary point $x = 2$.

Case 1: For $x < 2$

Let $c$ be any real number such that $c < 2$. In this interval, $f(x) = x^3-3$. This is a polynomial function. Polynomial functions are continuous everywhere. Thus, $f(x)$ is continuous for all $x < 2$.

Case 2: For $x > 2$

Let $c$ be any real number such that $c > 2$. In this interval, $f(x) = x^2+1$. This is a polynomial function. Polynomial functions are continuous everywhere. Thus, $f(x)$ is continuous for all $x > 2$.

Case 3: At $x = 2$

This is the critical point where the definition of the function changes. To check for continuity at $x=2$, we need to verify if $\lim\limits_{x \to 2} f(x) = f(2)$. We need to find $f(2)$, the left-hand limit (LHL) at $x=2$, and the right-hand limit (RHL) at $x=2$.

Function value at $x = 2$:

Using the definition for $x \leq 2$, $f(x) = x^3-3$. So, at $x=2$, $f(2) = (2)^3-3 = 8-3 = 5$.

$f(2) = 5$

Left-hand limit at $x = 2$:

For $x < 2$, $f(x) = x^3-3$.

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (x^3-3) = (2)^3-3 = 8-3 = 5$

Right-hand limit at $x = 2$:

For $x > 2$, $f(x) = x^2+1$.

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (x^2+1) = (2)^2+1 = 4+1 = 5$

Comparing the LHL, RHL, and $f(2)$:

$\lim\limits_{x \to 2^-} f(x) = 5$

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

$f(2) = 5$

Since the left-hand limit is equal to the right-hand limit ($\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^+} f(x)$), the limit $\lim\limits_{x \to 2} f(x)$ exists and is equal to 5.

Furthermore, the limit of the function at $x=2$ is equal to the function value at $x=2$ ($\lim\limits_{x \to 2} f(x) = f(2) = 5$).

Therefore, the function is continuous at $x = 2$.

Combining all cases, the function is continuous on $(-\infty, 2)$, $(2, \infty)$, and at $x=2$. Thus, the function is continuous for all real numbers.

Conclusion:

The function $f(x)$ is continuous at every real number. Therefore, there are no points of discontinuity for this function.

Given:

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

$f(x) = \begin{cases} x^{10}−1,& if\; x ≤ 1\\x^2,& if\; x > 1 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

To find the points of discontinuity, we examine the function's continuity in the open intervals defined by the piecewise function and at the point where the definition changes, which is $x = 1$.

We consider the continuity in the intervals $(-\infty, 1)$ and $(1, \infty)$, and at the boundary point $x = 1$.

Case 1: For $x < 1$

Let $c$ be any real number such that $c < 1$. In this interval, $f(x) = x^{10}-1$. This is a polynomial function. Polynomial functions are continuous everywhere. Thus, $f(x)$ is continuous for all $x < 1$.

Case 2: For $x > 1$

Let $c$ be any real number such that $c > 1$. In this interval, $f(x) = x^2$. This is a polynomial function. Polynomial functions are continuous everywhere. Thus, $f(x)$ is continuous for all $x > 1$.

Case 3: At $x = 1$

This is the critical point where the definition of the function changes. To check for continuity at $x=1$, we need to verify if $\lim\limits_{x \to 1} f(x) = f(1)$. We need to find $f(1)$, the left-hand limit (LHL) at $x=1$, and the right-hand limit (RHL) at $x=1$.

Function value at $x = 1$:

Using the definition for $x \leq 1$, $f(x) = x^{10}-1$. So, at $x=1$, $f(1) = (1)^{10}-1 = 1-1 = 0$.

$f(1) = 0$

Left-hand limit at $x = 1$:

For $x < 1$, $f(x) = x^{10}-1$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x^{10}-1) = (1)^{10}-1 = 1-1 = 0$

Right-hand limit at $x = 1$:

For $x > 1$, $f(x) = x^2$.

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

Comparing the LHL, RHL, and $f(1)$:

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

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

$f(1) = 0$

Since the left-hand limit is not equal to the right-hand limit ($\lim\limits_{x \to 1^-} f(x) \neq \lim\limits_{x \to 1^+} f(x)$), the limit $\lim\limits_{x \to 1} f(x)$ does not exist.

For a function to be continuous at a point, the limit must exist at that point and be equal to the function value. Since the limit does not exist at $x=1$, the function is not continuous at $x=1$.

Combining the cases, the function is continuous for all $x < 1$ and all $x > 1$. The only point where it is not continuous is $x=1$.

Conclusion:

The function $f(x)$ is continuous everywhere except at $x=1$. The only point of discontinuity is $x=1$.

Given:

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

$f(x) = \begin{cases} x+5,& if\; x ≤ 1\\x−5,& if\; x > 1 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Determine:

Whether the function $f(x)$ is a continuous function (i.e., continuous at every point in its domain).

Solution:

To determine if $f(x)$ is a continuous function, we need to check its continuity at all points in its domain. We examine the continuity in the open intervals defined by the piecewise function and at the point where the definition changes, which is $x = 1$.

We consider the continuity in the intervals $(-\infty, 1)$ and $(1, \infty)$, and at the boundary point $x = 1$.

Case 1: For $x < 1$

Let $c$ be any real number such that $c < 1$. In this interval, $f(x) = x+5$. This is a polynomial function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $x < 1$.

Case 2: For $x > 1$

Let $c$ be any real number such that $c > 1$. In this interval, $f(x) = x-5$. This is a polynomial function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $x > 1$.

Case 3: At $x = 1$

This is the critical point where the definition of the function changes. To check for continuity at $x=1$, we need to verify if $\lim\limits_{x \to 1} f(x) = f(1)$. We need to find $f(1)$, the left-hand limit (LHL) at $x=1$, and the right-hand limit (RHL) at $x=1$.

Function value at $x = 1$:

Using the definition for $x \leq 1$, $f(x) = x+5$. So, at $x=1$, $f(1) = 1+5 = 6$.

$f(1) = 6$

Left-hand limit at $x = 1$:

For $x < 1$, $f(x) = x+5$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x+5) = 1+5 = 6$

Right-hand limit at $x = 1$:

For $x > 1$, $f(x) = x-5$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (x-5) = 1-5 = -4$

Comparing the LHL and RHL at $x=1$:

$\lim\limits_{x \to 1^-} f(x) = 6$

$\lim\limits_{x \to 1^+} f(x) = -4$

Since the left-hand limit is not equal to the right-hand limit ($\lim\limits_{x \to 1^-} f(x) \neq \lim\limits_{x \to 1^+} f(x)$), the limit $\lim\limits_{x \to 1} f(x)$ does not exist.

For a function to be continuous at a point, the limit must exist at that point and be equal to the function value. Since the limit does not exist at $x=1$, the function is not continuous at $x=1$.

For a function to be a continuous function, it must be continuous at every point in its domain. Since $f(x)$ is not continuous at $x=1$, it is not a continuous function.

Conclusion:

The function $f(x)$ is continuous for all $x < 1$ and for all $x > 1$. However, it has a discontinuity at $x=1$. Therefore, the function defined by $f(x)$ is not a continuous function.

Given:

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

$f(x) = \begin{cases} 3,& if\; 0 ≤ x ≤ 1\\4,& if\; 1 < x < 3\\5,& if\; 3 ≤ x ≤ 10 \end{cases}$

The domain of the function is the union of the intervals $[0, 1]$, $(1, 3)$, and $[3, 10]$, which is the closed interval $[0, 10]$.

To Discuss:

The continuity of the function $f(x)$ at all points in its domain $[0, 10]$.

Solution:

To discuss the continuity of the function $f(x)$, we examine its continuity in the open intervals where it is defined by a single expression and at the points where the definition changes ($x=1$ and $x=3$) and the endpoints of the domain ($x=0$ and $x=10$).

Case 1: For $0 < x < 1$

Let $c$ be any real number such that $0 < c < 1$. In this interval, $f(x) = 3$. This is a constant function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $x \in (0, 1)$.

Case 2: For $1 < x < 3$

Let $c$ be any real number such that $1 < c < 3$. In this interval, $f(x) = 4$. This is a constant function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $x \in (1, 3)$.

Case 3: For $3 < x < 10$

Let $c$ be any real number such that $3 < c < 10$. In this interval, $f(x) = 5$. This is a constant function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $x \in (3, 10)$.

Now, we check continuity at the points where the definition changes ($x=1$ and $x=3$) and at the endpoints of the domain ($x=0$ and $x=10$).

Continuity at $x = 0$ (Left Endpoint):

We need to check if the right-hand limit at $x=0$ equals $f(0)$.

Function value at $x = 0$:

Using the definition for $0 \leq x \leq 1$, $f(0) = 3$.

$f(0) = 3$

Right-hand limit at $x = 0$:

For $0 < x \leq 1$, $f(x) = 3$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} 3 = 3$

Comparing the RHL and $f(0)$: $\lim\limits_{x \to 0^+} f(x) = f(0) = 3$. The function is continuous from the right at $x=0$. Since it is the left endpoint of the domain, the function is continuous at $x=0$.

Continuity at $x = 1$:

This is a critical point. We check $f(1)$, LHL, and RHL.

Function value at $x = 1$:

Using the definition for $0 \leq x \leq 1$, $f(1) = 3$.

$f(1) = 3$

Left-hand limit at $x = 1$:

For $x < 1$ and $x \geq 0$, $f(x) = 3$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} 3 = 3$

Right-hand limit at $x = 1$:

For $x > 1$ and $x < 3$, $f(x) = 4$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} 4 = 4$

Comparing the LHL and RHL: $\lim\limits_{x \to 1^-} f(x) = 3$ and $\lim\limits_{x \to 1^+} f(x) = 4$. Since $\lim\limits_{x \to 1^-} f(x) \neq \lim\limits_{x \to 1^+} f(x)$, the limit $\lim\limits_{x \to 1} f(x)$ does not exist. Therefore, the function is not continuous at $x = 1$.

Continuity at $x = 3$:

This is a critical point. We check $f(3)$, LHL, and RHL.

Function value at $x = 3$:

Using the definition for $3 \leq x \leq 10$, $f(3) = 5$.

$f(3) = 5$

Left-hand limit at $x = 3$:

For $x < 3$ and $x > 1$, $f(x) = 4$.

$\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^-} 4 = 4$

Right-hand limit at $x = 3$:

For $x > 3$ and $x \leq 10$, $f(x) = 5$.

$\lim\limits_{x \to 3^+} f(x) = \lim\limits_{x \to 3^+} 5 = 5$

Comparing the LHL and RHL: $\lim\limits_{x \to 3^-} f(x) = 4$ and $\lim\limits_{x \to 3^+} f(x) = 5$. Since $\lim\limits_{x \to 3^-} f(x) \neq \lim\limits_{x \to 3^+} f(x)$, the limit $\lim\limits_{x \to 3} f(x)$ does not exist. Therefore, the function is not continuous at $x = 3$.

Continuity at $x = 10$ (Right Endpoint):

We need to check if the left-hand limit at $x=10$ equals $f(10)$.

Function value at $x = 10$:

Using the definition for $3 \leq x \leq 10$, $f(10) = 5$.

$f(10) = 5$

Left-hand limit at $x = 10$:

For $x < 10$ and $x \geq 3$, $f(x) = 5$.

$\lim\limits_{x \to 10^-} f(x) = \lim\limits_{x \to 10^-} 5 = 5$

Comparing the LHL and $f(10)$: $\lim\limits_{x \to 10^-} f(x) = f(10) = 5$. The function is continuous from the left at $x=10$. Since it is the right endpoint of the domain, the function is continuous at $x=10$.

Combining all points in the domain $[0, 10]$: The function is continuous on $(0, 1)$, $(1, 3)$, $(3, 10)$, at $x=0$, and at $x=10$. It is discontinuous at $x=1$ and $x=3$.

Conclusion:

The function $f(x)$ is continuous on the intervals $[0, 1)$, $(1, 3)$, and $(3, 10]$. The points where the function is discontinuous are $x = 1$ and $x = 3$.

Given:

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

$f(x) = \begin{cases} 2x,& if\; x < 0\\0,& if\; 0 ≤ x ≤ 1\\4x,& if\; x > 1 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

To find the points of discontinuity, we examine the function's continuity in the open intervals defined by the piecewise function and at the points where the definition changes, which are $x = 0$ and $x = 1$.

We consider the continuity in the intervals $(-\infty, 0)$, $(0, 1)$, and $(1, \infty)$, and at the boundary points $x = 0$ and $x = 1$.

Case 1: For $x < 0$

Let $c$ be any real number such that $c < 0$. In this interval, $f(x) = 2x$. This is a polynomial function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $x < 0$.

Case 2: For $0 < x < 1$

Let $c$ be any real number such that $0 < c < 1$. In this interval, $f(x) = 0$. This is a constant function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $0 < x < 1$.

Case 3: For $x > 1$

Let $c$ be any real number such that $c > 1$. In this interval, $f(x) = 4x$. This is a polynomial function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $x > 1$.

Now, we need to check the continuity at the points where the definition changes, i.e., at $x = 0$ and $x = 1$.

Continuity at $x = 0$

We need to check if $\lim\limits_{x \to 0} f(x) = f(0)$. This requires evaluating $f(0)$, the left-hand limit (LHL) at $x=0$, and the right-hand limit (RHL) at $x=0$.

Function value at $x = 0$:

Using the definition for $0 \leq x \leq 1$, $f(0) = 0$.

$f(0) = 0$

Left-hand limit at $x = 0$:

For $x < 0$, $f(x) = 2x$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (2x) = 2(0) = 0$

Right-hand limit at $x = 0$:

For $x > 0$ and $x \leq 1$, $f(x) = 0$.

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

Comparing the LHL, RHL, and $f(0)$:

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

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

$f(0) = 0$

Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0) = 0$, the function is continuous at $x = 0$.

Continuity at $x = 1$

We need to check if $\lim\limits_{x \to 1} f(x) = f(1)$. This requires evaluating $f(1)$, the left-hand limit (LHL) at $x=1$, and the right-hand limit (RHL) at $x=1$.

Function value at $x = 1$:

Using the definition for $0 \leq x \leq 1$, $f(1) = 0$.

$f(1) = 0$

Left-hand limit at $x = 1$:

For $x < 1$ and $x \geq 0$, $f(x) = 0$.

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

Right-hand limit at $x = 1$:

For $x > 1$, $f(x) = 4x$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (4x) = 4(1) = 4$

Comparing the LHL, RHL, and $f(1)$:

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

$\lim\limits_{x \to 1^+} f(x) = 4$

$f(1) = 0$

Since the left-hand limit is not equal to the right-hand limit ($\lim\limits_{x \to 1^-} f(x) \neq \lim\limits_{x \to 1^+} f(x)$), the limit $\lim\limits_{x \to 1} f(x)$ does not exist. Therefore, the function is not continuous at $x = 1$.

Combining all cases, the function is continuous on $(-\infty, 0)$, $(0, 1)$, and $(1, \infty)$, and at $x=0$. The only point where it is not continuous is $x=1$.

Conclusion:

The function $f(x)$ is continuous everywhere except at $x=1$. The only point of discontinuity is $x=1$.

Given:

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

$f(x) = \begin{cases} −2,& if\; x ≤ −1\\2x,& if\; −1 < x ≤ 1\\2,& if\; x > 1 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

To find the points of discontinuity, we examine the function's continuity in the open intervals defined by the piecewise function and at the points where the definition changes, which are $x = -1$ and $x = 1$.

We consider the continuity in the intervals $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$, and at the boundary points $x = -1$ and $x = 1$.

Case 1: For $x < -1$

Let $c$ be any real number such that $c < -1$. In this interval, $f(x) = -2$. This is a constant function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $x < -1$.

Case 2: For $-1 < x < 1$

Let $c$ be any real number such that $-1 < c < 1$. In this interval, $f(x) = 2x$. This is a polynomial function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $-1 < x < 1$.

Case 3: For $x > 1$

Let $c$ be any real number such that $c > 1$. In this interval, $f(x) = 2$. This is a constant function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $x > 1$.

Now, we need to check the continuity at the points where the definition changes, i.e., at $x = -1$ and $x = 1$.

Continuity at $x = -1$

We need to check if $\lim\limits_{x \to -1} f(x) = f(-1)$. This requires evaluating $f(-1)$, the left-hand limit (LHL) at $x=-1$, and the right-hand limit (RHL) at $x=-1$.

Function value at $x = -1$:

Using the definition for $x \leq -1$, $f(-1) = -2$.

$f(-1) = -2$

Left-hand limit at $x = -1$:

For $x < -1$, $f(x) = -2$.

$\lim\limits_{x \to -1^-} f(x) = \lim\limits_{x \to -1^-} (-2) = -2$

Right-hand limit at $x = -1$:

For $x > -1$ and $x \leq 1$, $f(x) = 2x$.

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

Comparing the LHL, RHL, and $f(-1)$:

$\lim\limits_{x \to -1^-} f(x) = -2$

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

$f(-1) = -2$

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

Continuity at $x = 1$

We need to check if $\lim\limits_{x \to 1} f(x) = f(1)$. This requires evaluating $f(1)$, the left-hand limit (LHL) at $x=1$, and the right-hand limit (RHL) at $x=1$.

Function value at $x = 1$:

Using the definition for $-1 < x \leq 1$, $f(1) = 2(1) = 2$.

$f(1) = 2$

Left-hand limit at $x = 1$:

For $x < 1$ and $x > -1$, $f(x) = 2x$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (2x) = 2(1) = 2$

Right-hand limit at $x = 1$:

For $x > 1$, $f(x) = 2$.

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

Comparing the LHL, RHL, and $f(1)$:

$\lim\limits_{x \to 1^-} f(x) = 2$

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

$f(1) = 2$

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

Combining all cases, the function is continuous on $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$, and at $x=-1$ and $x=1$. Thus, the function is continuous for all real numbers.

Conclusion:

The function $f(x)$ is continuous at every real number. Therefore, there are no points of discontinuity for this function.

Given:

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

$f(x) = \begin{cases} ax+1,& if\; x ≤ 3\\bx+3,& if\; x > 3 \end{cases}$

The function is continuous at $x = 3$.

To Find:

The relationship between $a$ and $b$ such that $f(x)$ is continuous at $x=3$.

Solution:

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

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

This implies that the left-hand limit (LHL), the right-hand limit (RHL), and the function value at $x=c$ must all be equal:

$\lim\limits_{x \to c^-} f(x) = \lim\limits_{x \to c^+} f(x) = f(c)$

In this problem, the point of interest is $x = 3$. We are given that $f(x)$ is continuous at $x=3$. Therefore, the continuity condition must hold at $x=3$:

$\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^+} f(x) = f(3)$

Let's evaluate each part:

1. Function value at $x = 3$:

Using the definition for $x \leq 3$, $f(x) = ax+1$. So, $f(3) = a(3)+1 = 3a+1$.

$f(3) = 3a+1$

2. Left-hand limit at $x = 3$:

For $x < 3$, $f(x) = ax+1$.

$\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^-} (ax+1)$

Since $ax+1$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to 3^-} (ax+1) = a(3)+1 = 3a+1$

LHL at $x=3$ is $3a+1$

3. Right-hand limit at $x = 3$:

For $x > 3$, $f(x) = bx+3$.

$\lim\limits_{x \to 3^+} f(x) = \lim\limits_{x \to 3^+} (bx+3)$

Since $bx+3$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to 3^+} (bx+3) = b(3)+3 = 3b+3$

RHL at $x=3$ is $3b+3$

For $f(x)$ to be continuous at $x=3$, we must have LHL = RHL = $f(3)$.

This means:

$3a+1 = 3b+3 = 3a+1$

We only need to equate LHL and RHL (or LHL and $f(3)$, or RHL and $f(3)$).

Equating the LHL and RHL:

$3a+1 = 3b+3$

Now, we find the relationship between $a$ and $b$ by rearranging the equation:

$3a - 3b = 3 - 1$

$3a - 3b = 2$

We can divide the entire equation by 3:

$a - b = \frac{2}{3}$

Or express $a$ in terms of $b$, or $b$ in terms of $a$:

$a = b + \frac{2}{3}$

$b = a - \frac{2}{3}$

Any of these forms represents the relationship between $a$ and $b$ for the function to be continuous at $x=3$.

Conclusion:

For the function $f(x)$ to be continuous at $x=3$, the relationship between $a$ and $b$ must be $3a - 3b = 2$ (or equivalently, $a = b + \frac{2}{3}$).

Given:

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

$f(x) = \begin{cases} λ(x^2−2x),& if\; x ≤ 0\\4x+1,& if\; x > 0 \end{cases}$

where λ is a constant.

To Find:

1. The value of λ for which $f(x)$ is continuous at $x = 0$.

2. The continuity of $f(x)$ at $x = 1$.

Solution:

Part 1: Continuity at $x = 0$

For $f(x)$ to be continuous at $x = 0$, the following condition must hold:

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

Let's evaluate each part:

1. Function value at $x = 0$:

Using the definition for $x \leq 0$, $f(x) = λ(x^2-2x)$. So, $f(0) = λ((0)^2 - 2(0)) = λ(0 - 0) = λ(0) = 0$.

$f(0) = 0$

2. Left-hand limit at $x = 0$:

For $x < 0$, $f(x) = λ(x^2-2x)$.

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} λ(x^2-2x)$

Since $λ(x^2-2x)$ is a polynomial expression (in $x$), the limit can be found by direct substitution:

$\lim\limits_{x \to 0^-} λ(x^2-2x) = λ((0)^2 - 2(0)) = λ(0 - 0) = λ(0) = 0$

LHL at $x=0$ is $0$

3. Right-hand limit at $x = 0$:

For $x > 0$, $f(x) = 4x+1$.

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (4x+1)$

Since $4x+1$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to 0^+} (4x+1) = 4(0)+1 = 0+1 = 1$

RHL at $x=0$ is $1$

For continuity at $x=0$, we need LHL = RHL = $f(0)$.

$0 = 1 = 0$

This equality $0 = 1$ is false, regardless of the value of λ.

This means that for the given definition of $f(x)$, the left-hand limit and the right-hand limit at $x=0$ are never equal, regardless of the value of λ.

Therefore, the limit $\lim\limits_{x \to 0} f(x)$ does not exist for any value of λ. For continuity at $x=0$, the limit must exist and be equal to $f(0)$. Since the limit does not exist, the function cannot be continuous at $x=0$ for any value of λ.

Let's double-check the problem statement to ensure there isn't a typo or misunderstanding of the question. Assuming the function is stated correctly:

The condition for continuity at $x=0$ is $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$.

We have $\lim\limits_{x \to 0^-} f(x) = 0$, $\lim\limits_{x \to 0^+} f(x) = 1$, and $f(0) = 0$.

For continuity, we need $0 = 1 = 0$. This is not possible.

It seems there might be an issue with the question setup if the intention was to find a λ for continuity at $x=0$. However, based on the definition provided:

$f(x) = \begin{cases} λ(x^2−2x),& if\; x ≤ 0\\4x+1,& if\; x > 0 \end{cases}$

the left-hand limit as $x \to 0$ is $\lim_{x \to 0^-} \lambda(x^2 - 2x) = \lambda(0^2 - 2(0)) = \lambda \cdot 0 = 0$. The right-hand limit as $x \to 0$ is $\lim_{x \to 0^+} (4x+1) = 4(0)+1 = 1$. Since $0 \neq 1$, the limit at $x=0$ does not exist, and the function is not continuous at $x=0$ for any value of $\lambda$.

Let's assume there is no value of λ for which the function is continuous at $x=0$.

Part 2: Continuity at $x = 1$

Now, we examine the continuity of $f(x)$ at $x = 1$.

At $x=1$, the function definition is $f(x) = 4x+1$, since $1 > 0$.

For a function to be continuous at $x=1$, we need $\lim\limits_{x \to 1} f(x) = f(1)$.

Function value at $x = 1$:

Using the definition for $x > 0$ (which includes $x=1$ and values near $1$), $f(x) = 4x+1$. So, $f(1) = 4(1)+1 = 4+1 = 5$.

$f(1) = 5$

Limit at $x = 1$:

For values of $x$ near $1$ (e.g., in $(0.5, 1.5)$), the definition $f(x) = 4x+1$ applies because $x > 0$.

$\lim\limits_{x \to 1} f(x) = \lim\limits_{x \to 1} (4x+1)$

Since $4x+1$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to 1} (4x+1) = 4(1)+1 = 4+1 = 5$

$\lim\limits_{x \to 1} f(x) = 5$

Comparing the limit and the function value:

$\lim\limits_{x \to 1} f(x) = 5$

$f(1) = 5$

Since $\lim\limits_{x \to 1} f(x) = f(1)$, the function is continuous at $x = 1$. The continuity at $x=1$ does not depend on the value of λ, as the definition $f(x) = 4x+1$ applies in an interval around $x=1$.

Conclusion:

Based on the given function definition, there is no value of λ for which the function $f(x)$ is continuous at $x = 0$, because the left-hand limit ($0$) and the right-hand limit ($1$) at $x=0$ are not equal for any λ. The function $f(x)$ is continuous at x = 1, and its continuity at $x=1$ is independent of the value of λ.

Given:

The function $g(x) = x - [x]$, where $[x]$ is the greatest integer function.

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Show:

The function $g(x) = x - [x]$ is discontinuous at all integral points.

Proof:

To show that the function $g(x)$ is discontinuous at all integral points, we need to examine its continuity at an arbitrary integer point. Let $c = n$, where $n$ is any integer ($n \in \mathbb{Z}$).

For $g(x)$ to be continuous at $x = n$, the following condition must hold:

$\lim\limits_{x \to n} g(x) = g(n)$

This implies that the left-hand limit (LHL), the right-hand limit (RHL), and the function value at $x=n$ must all be equal:

$\lim\limits_{x \to n^-} g(x) = \lim\limits_{x \to n^+} g(x) = g(n)$

Let's evaluate each part at the integer point $x = n$:

1. Function value at $x = n$:

$g(n) = n - [n]$

By the definition of the greatest integer function, $[n] = n$ for any integer $n$.

$g(n) = n - n = 0$

2. Left-hand limit at $x = n$:

$\lim\limits_{x \to n^-} g(x) = \lim\limits_{x \to n^-} (x - [x])$

Using properties of limits, this is $\lim\limits_{x \to n^-} x - \lim\limits_{x \to n^-} [x]$.

As $x$ approaches $n$ from the left side ($x < n$), $x$ is slightly less than $n$. For such $x$, the greatest integer less than or equal to $x$ is $n-1$. For example, if $n=3$, as $x \to 3^-$, $x$ is like $2.99$, and $[x] = 2 = 3-1$.

$\lim\limits_{x \to n^-} x = n$

$\lim\limits_{x \to n^-} [x] = n-1$

So, the LHL is:

$\lim\limits_{x \to n^-} (x - [x]) = n - (n-1) = n - n + 1 = 1$

LHL at $x=n$ is $1$

3. Right-hand limit at $x = n$:

$\lim\limits_{x \to n^+} g(x) = \lim\limits_{x \to n^+} (x - [x])$

Using properties of limits, this is $\lim\limits_{x \to n^+} x - \lim\limits_{x \to n^+} [x]$.

As $x$ approaches $n$ from the right side ($x > n$), $x$ is slightly greater than $n$. For such $x$, the greatest integer less than or equal to $x$ is $n$. For example, if $n=3$, as $x \to 3^+$, $x$ is like $3.01$, and $[x] = 3 = n$.

$\lim\limits_{x \to n^+} x = n$

$\lim\limits_{x \to n^+} [x] = n$

So, the RHL is:

$\lim\limits_{x \to n^+} (x - [x]) = n - n = 0$

RHL at $x=n$ is $0$

Now, we compare the LHL, RHL, and $g(n)$ at $x=n$:

LHL = 1

RHL = 0

$g(n) = 0$

Since the left-hand limit (1) is not equal to the right-hand limit (0), the limit $\lim\limits_{x \to n} g(x)$ does not exist at the integer point $n$.

For a function to be continuous at a point, the limit must exist at that point and be equal to the function value. Since the limit does not exist at any integer point $n$, the function $g(x) = x - [x]$ is not continuous at any integer point.

Conclusion:

For any integer $n$, the left-hand limit of $g(x)$ at $x=n$ is $1$, while the right-hand limit and the function value are $0$. Since the left-hand limit does not equal the right-hand limit, the limit at $x=n$ does not exist. Therefore, the function $g(x) = x - [x]$ is discontinuous at all integral points.

Given:

The function $f(x) = x^2 - \sin x + 5$.

We need to check the continuity of the function at $x = \pi$.

To Check:

Whether the function $f(x) = x^2 - \sin x + 5$ is continuous at $x = \pi$.

Solution:

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

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

In this problem, we need to check the continuity at $x = \pi$. So, we will verify if $\lim\limits_{x \to \pi} f(x) = f(\pi)$.

First, let's find the value of the function at $x = \pi$:

$f(\pi) = (\pi)^2 - \sin(\pi) + 5$

We know that $\sin(\pi) = 0$.

$f(\pi) = \pi^2 - 0 + 5 = \pi^2 + 5$

So, $f(\pi) = \pi^2 + 5$.

Next, let's find the limit of the function as $x$ approaches $\pi$:

$\lim\limits_{x \to \pi} f(x) = \lim\limits_{x \to \pi} (x^2 - \sin x + 5)$

Using the properties of limits (limit of a sum/difference is the sum/difference of limits):

$\lim\limits_{x \to \pi} (x^2 - \sin x + 5) = \lim\limits_{x \to \pi} x^2 - \lim\limits_{x \to \pi} \sin x + \lim\limits_{x \to \pi} 5$

We evaluate each limit:

$\lim\limits_{x \to \pi} x^2 = \pi^2$ (since $x^2$ is a polynomial)

$\lim\limits_{x \to \pi} \sin x = \sin \pi = 0$ (since $\sin x$ is continuous)

$\lim\limits_{x \to \pi} 5 = 5$ (limit of a constant)

Substituting these values:

$\lim\limits_{x \to \pi} f(x) = \pi^2 - 0 + 5 = \pi^2 + 5$

So, $\lim\limits_{x \to \pi} f(x) = \pi^2 + 5$.

Now, we compare the limit of the function as $x$ approaches $\pi$ with the value of the function at $x = \pi$:

$\lim\limits_{x \to \pi} f(x) = \pi^2 + 5$

$f(\pi) = \pi^2 + 5$

Since $\lim\limits_{x \to \pi} f(x) = f(\pi)$, the function $f(x) = x^2 - \sin x + 5$ is continuous at $x = \pi$.

Alternatively, we can consider $f(x)$ as a combination of elementary continuous functions. The function $g(x) = x^2$ is a polynomial, so it is continuous on $\mathbb{R}$. The function $h(x) = \sin x$ is the sine function, so it is continuous on $\mathbb{R}$. The constant function $k(x) = 5$ is continuous on $\mathbb{R}$. Since the sum and difference of continuous functions are continuous, the function $f(x) = g(x) - h(x) + k(x) = x^2 - \sin x + 5$ is continuous on $\mathbb{R}$. Since it is continuous on $\mathbb{R}$, it is continuous at any specific real number, including $x = \pi$.

Conclusion:

Since $\lim\limits_{x \to \pi} f(x) = f(\pi)$, the function $f(x) = x^2 - \sin x + 5$ is continuous at x = π.

We discuss the continuity of each function separately.

To discuss the continuity of these functions, we rely on the known continuity of the sine function and the cosine function.

We know that the function $g(x) = \sin x$ is continuous for all real numbers ($\mathbb{R}$). (Discussed in Example 17)

We also know that the function $h(x) = \cos x$ is continuous for all real numbers ($\mathbb{R}$). (This can be proven similarly to the sine function, using the limit $\lim\limits_{h \to 0} \frac{\cos h - 1}{h} = 0$ and $\lim\limits_{h \to 0} \frac{\sin h}{h} = 1$, and the identity $\cos(c+h) = \cos c \cos h - \sin c \sin h$).

We also use the properties of continuous functions:

If $g(x)$ and $h(x)$ are continuous at a point $c$, then:

• $g(x) + h(x)$ is continuous at $c$.
• $g(x) - h(x)$ is continuous at $c$.
• $g(x) \cdot h(x)$ is continuous at $c$.

Since $g(x) = \sin x$ and $h(x) = \cos x$ are continuous at every real number, the combinations below are also continuous at every real number.

(a) f(x) = sin x + cos x

Given: The function $f(x) = \sin x + \cos x$.

This function is the sum of $g(x) = \sin x$ and $h(x) = \cos x$.

Discussion of Continuity:

Since $\sin x$ is continuous on $\mathbb{R}$ and $\cos x$ is continuous on $\mathbb{R}$, their sum $f(x) = \sin x + \cos x$ is also continuous on $\mathbb{R}$.

For any real number $c$, $\lim\limits_{x \to c} (\sin x + \cos x) = \lim\limits_{x \to c} \sin x + \lim\limits_{x \to c} \cos x = \sin c + \cos c = f(c)$.

Conclusion:

The function $f(x) = \sin x + \cos x$ is continuous for all real numbers ($\mathbb{R}$).

(b) f(x) = sin x – cos x

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

This function is the difference of $g(x) = \sin x$ and $h(x) = \cos x$.

Discussion of Continuity:

Since $\sin x$ is continuous on $\mathbb{R}$ and $\cos x$ is continuous on $\mathbb{R}$, their difference $f(x) = \sin x - \cos x$ is also continuous on $\mathbb{R}$.

For any real number $c$, $\lim\limits_{x \to c} (\sin x - \cos x) = \lim\limits_{x \to c} \sin x - \lim\limits_{x \to c} \cos x = \sin c - \cos c = f(c)$.

Conclusion:

The function $f(x) = \sin x - \cos x$ is continuous for all real numbers ($\mathbb{R}$).

(c) f(x) = sin x . cos x

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

This function is the product of $g(x) = \sin x$ and $h(x) = \cos x$.

Discussion of Continuity:

Since $\sin x$ is continuous on $\mathbb{R}$ and $\cos x$ is continuous on $\mathbb{R}$, their product $f(x) = \sin x \cdot \cos x$ is also continuous on $\mathbb{R}$.

For any real number $c$, $\lim\limits_{x \to c} (\sin x \cos x) = (\lim\limits_{x \to c} \sin x) \cdot (\lim\limits_{x \to c} \cos x) = \sin c \cdot \cos c = f(c)$.

Conclusion:

The function $f(x) = \sin x \cdot \cos x$ is continuous for all real numbers ($\mathbb{R}$).

We discuss the continuity of each trigonometric function based on the properties of continuous functions and the known continuity of the sine and cosine functions.

We know that the sine function $g(x) = \sin x$ is continuous for all real numbers ($\mathbb{R}$).

We also know that the cosine function $h(x) = \cos x$ is continuous for all real numbers ($\mathbb{R}$).

A key property of continuous functions states that if two functions are continuous at a point, then their sum, difference, and product are also continuous at that point. Furthermore, their quotient is continuous at any point where both functions are continuous and the denominator is non-zero.

Continuity of the Cosine Function:

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

Its domain is $\mathbb{R}$.

To show it is continuous on its domain, consider an arbitrary real number $c$. We need to show $\lim\limits_{x \to c} \cos x = \cos c$.

Let $x = c+h$. As $x \to c$, $h \to 0$.

$\lim\limits_{x \to c} \cos x = \lim\limits_{h \to 0} \cos(c+h) = \lim\limits_{h \to 0} (\cos c \cos h - \sin c \sin h)$

$= \cos c \lim\limits_{h \to 0} \cos h - \sin c \lim\limits_{h \to 0} \sin h$

Using the standard limits $\lim\limits_{h \to 0} \cos h = 1$ and $\lim\limits_{h \to 0} \sin h = 0$:

$= \cos c \cdot 1 - \sin c \cdot 0 = \cos c$

Since $\lim\limits_{x \to c} \cos x = \cos c$ for any $c \in \mathbb{R}$, the cosine function is continuous at every real number.

Conclusion: The cosine function $f(x) = \cos x$ is continuous on its domain, which is all real numbers ($\mathbb{R}$).

Continuity of the Cosecant Function:

The cosecant function is $f(x) = \text{cosec } x = \frac{1}{\sin x}$.

Its domain is the set of all real numbers $x$ for which $\sin x \neq 0$. This excludes $x = n\pi$, where $n$ is any integer.

The numerator is the constant function $g(x) = 1$, which is continuous on $\mathbb{R}$.

The denominator is $h(x) = \sin x$, which is continuous on $\mathbb{R}$.

The cosecant function is the quotient of two continuous functions. It is continuous at all points where the denominator is non-zero.

The denominator $\sin x$ is non-zero exactly on the domain of $\text{cosec } x$.

Therefore, the cosecant function is continuous at every point in its domain.

Conclusion: The cosecant function $f(x) = \text{cosec } x$ is continuous on its domain, which is $\mathbb{R} \setminus \{n\pi \mid n \in \mathbb{Z}\}$.

Continuity of the Secant Function:

The secant function is $f(x) = \sec x = \frac{1}{\cos x}$.

Its domain is the set of all real numbers $x$ for which $\cos x \neq 0$. This excludes $x = \frac{\pi}{2} + n\pi$, where $n$ is any integer.

The numerator is the constant function $g(x) = 1$, which is continuous on $\mathbb{R}$.

The denominator is $h(x) = \cos x$, which is continuous on $\mathbb{R}$.

The secant function is the quotient of two continuous functions. It is continuous at all points where the denominator is non-zero.

The denominator $\cos x$ is non-zero exactly on the domain of $\sec x$.

Therefore, the secant function is continuous at every point in its domain.

Conclusion: The secant function $f(x) = \sec x$ is continuous on its domain, which is $\mathbb{R} \setminus \{\frac{\pi}{2} + n\pi \mid n \in \mathbb{Z}\}$.

Continuity of the Cotangent Function:

The cotangent function is $f(x) = \cot x = \frac{\cos x}{\sin x}$.

Its domain is the set of all real numbers $x$ for which $\sin x \neq 0$. This excludes $x = n\pi$, where $n$ is any integer.

The numerator is $g(x) = \cos x$, which is continuous on $\mathbb{R}$.

The denominator is $h(x) = \sin x$, which is continuous on $\mathbb{R}$.

The cotangent function is the quotient of two continuous functions. It is continuous at all points where the denominator is non-zero.

The denominator $\sin x$ is non-zero exactly on the domain of $\cot x$.

Therefore, the cotangent function is continuous at every point in its domain.

Conclusion: The cotangent function $f(x) = \cot x$ is continuous on its domain, which is $\mathbb{R} \setminus \{n\pi \mid n \in \mathbb{Z}\}$.

Given:

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

$f(x) = \begin{cases} \frac{\sin x}{x},& if\; x < 0\\x+1,& if\; x ≥ 0 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

To find the points of discontinuity, we examine the function's continuity in the open intervals defined by the piecewise function and at the point where the definition changes, which is $x = 0$.

We consider the continuity in the intervals $(-\infty, 0)$ and $(0, \infty)$, and at the boundary point $x = 0$.

Case 1: For $x < 0$

Let $c$ be any real number such that $c < 0$. In this interval, $f(x) = \frac{\sin x}{x}$. The sine function $g(x) = \sin x$ is continuous on $\mathbb{R}$, and the identity function $h(x) = x$ is continuous on $\mathbb{R}$. The quotient of two continuous functions is continuous wherever the denominator is non-zero. For $x < 0$, the denominator $x$ is never zero. Thus, $f(x) = \frac{\sin x}{x}$ is continuous for all $x < 0$.

Case 2: For $x > 0$

Let $c$ be any real number such that $c > 0$. In this interval, $f(x) = x+1$. This is a polynomial function, which is continuous everywhere. Thus, $f(x)$ is continuous for all $x > 0$.

Case 3: At $x = 0$

This is the critical point where the definition of the function changes. To check for continuity at $x=0$, we need to verify if $\lim\limits_{x \to 0} f(x) = f(0)$. We need to find $f(0)$, the left-hand limit (LHL) at $x=0$, and the right-hand limit (RHL) at $x=0$.

Function value at $x = 0$:

Using the definition for $x \geq 0$, $f(x) = x+1$. So, at $x=0$, $f(0) = 0+1 = 1$.

$f(0) = 1$

Left-hand limit at $x = 0$:

For $x < 0$, $f(x) = \frac{\sin x}{x}$.

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

We use the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$. The left-hand limit is equal to this standard limit.

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

Right-hand limit at $x = 0$:

For $x > 0$, $f(x) = x+1$.

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

Since $x+1$ is a polynomial expression, the limit can be found by direct substitution:

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

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

Comparing the LHL, RHL, and $f(0)$:

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

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

$f(0) = 1$

Since the left-hand limit is equal to the right-hand limit ($\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x)$), the limit $\lim\limits_{x \to 0} f(x)$ exists and is equal to 1.

Furthermore, the limit of the function at $x=0$ is equal to the function value at $x=0$ ($\lim\limits_{x \to 0} f(x) = f(0) = 1$).

Therefore, the function is continuous at $x = 0$.

Combining all cases, the function is continuous on $(-\infty, 0)$, $(0, \infty)$, and at $x=0$. Thus, the function is continuous for all real numbers.

Conclusion:

The function $f(x)$ is continuous at every real number. Therefore, there are no points of discontinuity for this function.

Given:

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

$f(x) = \begin{cases} x^2\sin \frac{1}{x},& if\; x \neq 0\\0,& if\; x = 0 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Determine:

Whether the function $f(x)$ is a continuous function (i.e., continuous at every point in its domain).

Solution:

To determine if $f(x)$ is continuous, we examine its continuity in the interval where it is defined by a single expression ($x \neq 0$) and at the point where the definition changes ($x = 0$).

Case 1: For $x \neq 0$

Let $c$ be any real number such that $c \neq 0$. For $x$ in an open interval around $c$ that does not contain $0$, $f(x) = x^2 \sin \frac{1}{x}$.

We know that the function $g(x) = x^2$ is a polynomial, hence continuous on $\mathbb{R}$.

The function $h(x) = \frac{1}{x}$ is a rational function, continuous on its domain $\mathbb{R} \setminus \{0\}$. For $c \neq 0$, $h(x)$ is continuous at $c$.

The function $k(u) = \sin u$ is the sine function, continuous on $\mathbb{R}$.

The function $\sin \frac{1}{x}$ is the composition of $h(x) = \frac{1}{x}$ and $k(u) = \sin u$. For $x \neq 0$, $h(x)$ is defined, and $k(h(x)) = \sin(\frac{1}{x})$. Since $h(x)$ is continuous for $x \neq 0$ and $k(u)$ is continuous for all real numbers, the composite function $\sin \frac{1}{x}$ is continuous for all $x \neq 0$.

Finally, $f(x) = x^2 \sin \frac{1}{x}$ is the product of $g(x) = x^2$ (continuous for $x \neq 0$) and $\sin \frac{1}{x}$ (continuous for $x \neq 0$). The product of continuous functions is continuous.

Therefore, $f(x)$ is continuous for all $x \neq 0$.

Case 2: At $x = 0$

This is the critical point where the definition of the function changes. To check for continuity at $x=0$, we need to verify if $\lim\limits_{x \to 0} f(x) = f(0)$.

Function value at $x = 0$:

From the definition, $f(0) = 0$.

$f(0) = 0$

Limit at $x = 0$:

We need to evaluate $\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} x^2 \sin \frac{1}{x}$.

We know that for any real number $u$, $-1 \leq \sin u \leq 1$. Therefore, for $x \neq 0$, we have:

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

Multiplying the inequality by $x^2$ (which is non-negative for real $x$), we get:

$-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$

Now, we consider the limits of the bounding functions as $x \to 0$:

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

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

Since $\lim\limits_{x \to 0} (-x^2) = 0$ and $\lim\limits_{x \to 0} x^2 = 0$, by the Squeeze Theorem (also known as the Sandwich Theorem), the limit of the function in between must also be $0$.

$\lim\limits_{x \to 0} x^2 \sin \frac{1}{x} = 0$

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

Now, we compare the limit of the function as $x$ approaches $0$ with the value of the function 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 function is continuous at $x = 0$.

Combining the cases, the function is continuous for all $x \neq 0$ and continuous at $x = 0$. Thus, the function is continuous for all real numbers.

Conclusion:

The function $f(x)$ is continuous for all real numbers. Therefore, $f(x)$ is a continuous function.

Given:

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

$f(x) = \begin{cases} \sin x − \cos x,& if\; x ≠ 0\\−1,& if\; x = 0 \end{cases}$

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Examine:

The continuity of the function $f(x)$ at all points in its domain.

Solution:

To examine the continuity of the function $f(x)$, we check its continuity for $x \neq 0$ and at the point where the definition changes, which is $x = 0$.

Case 1: For $x \neq 0$

Let $c$ be any real number such that $c \neq 0$. For $x$ in an open interval around $c$ that does not contain $0$, the function is defined as $f(x) = \sin x - \cos x$.

We know that the function $g(x) = \sin x$ is continuous for all real numbers.

We also know that the function $h(x) = \cos x$ is continuous for all real numbers.

The difference of two continuous functions is continuous on the intersection of their domains. Since both $\sin x$ and $\cos x$ are continuous on $\mathbb{R}$, their difference $f(x) = \sin x - \cos x$ is continuous on $\mathbb{R}$.

Therefore, $f(x)$ is continuous for all $x \neq 0$.

Case 2: At $x = 0$

This is the critical point where the definition of the function changes. To check for continuity at $x=0$, we need to verify if $\lim\limits_{x \to 0} f(x) = f(0)$.

Function value at $x = 0$:

From the definition, $f(0) = -1$.

$f(0) = -1$

Limit at $x = 0$:

We need to evaluate $\lim\limits_{x \to 0} f(x)$. For $x \neq 0$, $f(x) = \sin x - \cos x$.

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

Using the properties of limits (limit of a difference is the difference of limits):

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

We use the standard limits $\lim\limits_{x \to 0} \sin x = 0$ and $\lim\limits_{x \to 0} \cos x = 1$.

$= 0 - 1 = -1$

So, $\lim\limits_{x \to 0} f(x) = -1$.

Now, we compare the limit of the function as $x$ approaches $0$ with the value of the function at $x = 0$:

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

$f(0) = -1$

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the function is continuous at $x = 0$.

Combining the cases, the function is continuous for all $x \neq 0$ and continuous at $x = 0$. Thus, the function is continuous for all real numbers.

Conclusion:

The function $f(x)$ is continuous at every real number. Therefore, $f(x)$ is a continuous function.

Given:

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

$f(x) = \begin{cases} \frac{k\cos x}{π−2x},& if\; x ≠ \frac{π}{2} \\3,& if\; x = \frac{π}{2} \end{cases}$

The function is continuous at $x = \frac{\pi}{2}$.

To Find:

The value of $k$ for which the function $f(x)$ is continuous at $x = \frac{\pi}{2}$.

Solution:

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

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

In this problem, the point of interest is $x = \frac{\pi}{2}$. We are given that $f(x)$ is continuous at $x = \frac{\pi}{2}$. Therefore, the continuity condition must hold at $x = \frac{\pi}{2}$:

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

Let's evaluate each part:

1. Function value at $x = \frac{\pi}{2}$:

From the definition, when $x = \frac{\pi}{2}$, $f(x) = 3$.

$f(\frac{π}{2}) = 3$

2. Limit at $x = \frac{\pi}{2}$:

We need to evaluate $\lim\limits_{x \to \frac{π}{2}} f(x) = \lim\limits_{x \to \frac{π}{2}} \frac{k\cos x}{π−2x}$.

As $x \to \frac{\pi}{2}$, the numerator approaches $k\cos(\frac{\pi}{2}) = k \cdot 0 = 0$, and the denominator approaches $\pi - 2(\frac{\pi}{2}) = \pi - \pi = 0$. This is an indeterminate form $\frac{0}{0}$. We can evaluate this limit using substitution or L'Hopital's Rule.

Using substitution: Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$, $h \to 0$.

$\lim\limits_{x \to \frac{π}{2}} \frac{k\cos x}{π−2x} = \lim\limits_{h \to 0} \frac{k\cos(\frac{π}{2} + h)}{π−2(\frac{π}{2} + h)}$

Using the trigonometric identity $\cos(\frac{\pi}{2} + h) = -\sin h$ and simplifying the denominator:

$= \lim\limits_{h \to 0} \frac{k(-\sin h)}{π−π−2h} = \lim\limits_{h \to 0} \frac{-k\sin h}{-2h} = \lim\limits_{h \to 0} \frac{k\sin h}{2h}$

We can rewrite this as:

$= \frac{k}{2} \lim\limits_{h \to 0} \frac{\sin h}{h}$

Using the standard limit $\lim\limits_{h \to 0} \frac{\sin h}{h} = 1$:

$= \frac{k}{2} \cdot 1 = \frac{k}{2}$

So, $\lim\limits_{x \to \frac{π}{2}} f(x) = \frac{k}{2}$.

For the function to be continuous at $x = \frac{\pi}{2}$, the limit must equal the function value:

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

$\frac{k}{2} = 3$

Solving for $k$:

$k = 3 \times 2 = 6$

Conclusion:

For the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$, the value of $k$ must be 6.

Given:

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

$f(x) = \begin{cases} kx^2,& if\; x ≤ 2\\3,& if\; x > 2 \end{cases}$

The function is continuous at $x = 2$.

To Find:

The value of $k$ for which the function $f(x)$ is continuous at $x = 2$.

Solution:

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

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

This implies that the left-hand limit (LHL), the right-hand limit (RHL), and the function value at $x=c$ must all be equal:

$\lim\limits_{x \to c^-} f(x) = \lim\limits_{x \to c^+} f(x) = f(c)$

In this problem, the point of interest is $x = 2$. We are given that $f(x)$ is continuous at $x = 2$. Therefore, the continuity condition must hold at $x = 2$:

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^+} f(x) = f(2)$

Let's evaluate each part:

1. Function value at $x = 2$:

Using the definition for $x \leq 2$, $f(x) = kx^2$. So, $f(2) = k(2)^2 = 4k$.

$f(2) = 4k$

2. Left-hand limit at $x = 2$:

For $x < 2$, $f(x) = kx^2$.

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (kx^2)$

Since $kx^2$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to 2^-} (kx^2) = k(2)^2 = 4k$

LHL at $x=2$ is $4k$

3. Right-hand limit at $x = 2$:

For $x > 2$, $f(x) = 3$.

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

The limit of a constant function is the constant itself:

$\lim\limits_{x \to 2^+} 3 = 3$

RHL at $x=2$ is $3$

For $f(x)$ to be continuous at $x=2$, we must have LHL = RHL = $f(2)$.

This means:

$4k = 3 = 4k$

We need to find the value of $k$ that satisfies $4k = 3$.

$4k = 3$

Dividing by 4:

$k = \frac{3}{4}$

Conclusion:

For the function $f(x)$ to be continuous at $x=2$, the value of $k$ must be $\frac{3}{4}$.

Given:

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

$f(x) = \begin{cases} kx+1,& if\; x ≤ π\\\cos x,& if\; x > π \end{cases}$

The function is continuous at $x = \pi$.

To Find:

The value of $k$ for which the function $f(x)$ is continuous at $x = \pi$.

Solution:

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

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

This implies that the left-hand limit (LHL), the right-hand limit (RHL), and the function value at $x=c$ must all be equal:

$\lim\limits_{x \to c^-} f(x) = \lim\limits_{x \to c^+} f(x) = f(c)$

In this problem, the point of interest is $x = \pi$. We are given that $f(x)$ is continuous at $x = \pi$. Therefore, the continuity condition must hold at $x = \pi$:

$\lim\limits_{x \to π^-} f(x) = \lim\limits_{x \to π^+} f(x) = f(π)$

Let's evaluate each part:

1. Function value at $x = \pi$:

Using the definition for $x \leq \pi$, $f(x) = kx+1$. So, $f(\pi) = k(\pi)+1 = k\pi+1$.

$f(\pi) = k\pi+1$

2. Left-hand limit at $x = \pi$:

For $x < \pi$, $f(x) = kx+1$.

$\lim\limits_{x \to π^-} f(x) = \lim\limits_{x \to π^-} (kx+1)$

Since $kx+1$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to π^-} (kx+1) = k(\pi)+1 = k\pi+1$

LHL at $x=\pi$ is $k\pi+1$

3. Right-hand limit at $x = \pi$:

For $x > \pi$, $f(x) = \cos x$.

$\lim\limits_{x \to π^+} f(x) = \lim\limits_{x \to π^+} \cos x$

Since $\cos x$ is a continuous function, the limit can be found by direct substitution:

$\lim\limits_{x \to π^+} \cos x = \cos(\pi) = -1$

RHL at $x=\pi$ is $-1$

For $f(x)$ to be continuous at $x=\pi$, we must have LHL = RHL = $f(\pi)$.

This means:

$k\pi+1 = -1 = k\pi+1$

We need to find the value of $k$ that satisfies $k\pi+1 = -1$.

$k\pi+1 = -1$

Subtract 1 from both sides:

$k\pi = -1 - 1$

$k\pi = -2$

Divide by $\pi$ (since $\pi \neq 0$):

$k = -\frac{2}{\pi}$

Conclusion:

For the function $f(x)$ to be continuous at $x=\pi$, the value of $k$ must be $-\frac{2}{\pi}$.

Given:

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

$f(x) = \begin{cases} kx+1,& if\; x ≤ 5\\3x-5,& if\; x > 5 \end{cases}$

The function is continuous at $x = 5$.

To Find:

The value of $k$ for which the function $f(x)$ is continuous at $x = 5$.

Solution:

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

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

This implies that the left-hand limit (LHL), the right-hand limit (RHL), and the function value at $x=c$ must all be equal:

$\lim\limits_{x \to c^-} f(x) = \lim\limits_{x \to c^+} f(x) = f(c)$

In this problem, the point of interest is $x = 5$. We are given that $f(x)$ is continuous at $x = 5$. Therefore, the continuity condition must hold at $x = 5$:

$\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^+} f(x) = f(5)$

Let's evaluate each part:

1. Function value at $x = 5$:

Using the definition for $x \leq 5$, $f(x) = kx+1$. So, $f(5) = k(5)+1 = 5k+1$.

$f(5) = 5k+1$

2. Left-hand limit at $x = 5$:

For $x < 5$, $f(x) = kx+1$.

$\lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^-} (kx+1)$

Since $kx+1$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to 5^-} (kx+1) = k(5)+1 = 5k+1$

LHL at $x=5$ is $5k+1$

3. Right-hand limit at $x = 5$:

For $x > 5$, $f(x) = 3x-5$.

$\lim\limits_{x \to 5^+} f(x) = \lim\limits_{x \to 5^+} (3x-5)$

Since $3x-5$ is a polynomial expression, the limit can be found by direct substitution:

$\lim\limits_{x \to 5^+} (3x-5) = 3(5)-5 = 15-5 = 10$

RHL at $x=5$ is $10$

For $f(x)$ to be continuous at $x=5$, we must have LHL = RHL = $f(5)$.

This means:

$5k+1 = 10 = 5k+1$

We need to find the value of $k$ that satisfies $5k+1 = 10$.

$5k+1 = 10$

Subtract 1 from both sides:

$5k = 10 - 1$

$5k = 9$

Divide by 5:

$k = \frac{9}{5}$

Conclusion:

For the function $f(x)$ to be continuous at $x=5$, the value of $k$ must be $\frac{9}{5}$.

Given:

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

$f(x) = \begin{cases} 5,& if\; x ≤ 2\\ax+b,& if\; 2 < x < 10\\21,& if\; x ≥ 10 \end{cases}$

The function $f(x)$ is continuous for all real numbers.

To Find:

The values of $a$ and $b$ for which the function $f(x)$ is continuous.

Solution:

The function $f(x)$ is defined piecewise. The function is continuous on the open intervals $(-\infty, 2)$, $(2, 10)$, and $(10, \infty)$ because the expressions defining it ($5$, $ax+b$, and $21$) are polynomial (or constant) functions, which are continuous on these open intervals.

For $f(x)$ to be continuous over its entire domain ($\mathbb{R}$), it must also be continuous at the points where the definition changes, which are $x=2$ and $x=10$.

The condition for continuity at a point $x=c$ is $\lim\limits_{x \to c} f(x) = f(c)$. This requires that the left-hand limit, the right-hand limit, and the function value at $c$ are all equal: $\lim\limits_{x \to c^-} f(x) = \lim\limits_{x \to c^+} f(x) = f(c)$.

We apply this condition at $x=2$ and $x=10$.

Continuity at $x = 2$:

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

1. Function value at $x = 2$:

For $x \leq 2$, $f(x) = 5$. So, $f(2) = 5$.

$f(2) = 5$

2. Left-hand limit at $x = 2$:

For $x < 2$, $f(x) = 5$.

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} 5 = 5$

3. Right-hand limit at $x = 2$:

For $x > 2$, $f(x) = ax+b$.

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (ax+b) = a(2)+b = 2a+b$

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

$5 = 2a+b = 5$

This gives us the first equation:

Continuity at $x = 10$:

For continuity at $x=10$, we must have $\lim\limits_{x \to 10^-} f(x) = \lim\limits_{x \to 10^+} f(x) = f(10)$.

1. Function value at $x = 10$:

For $x \geq 10$, $f(x) = 21$. So, $f(10) = 21$.

$f(10) = 21$

2. Left-hand limit at $x = 10$:

For $x < 10$, $f(x) = ax+b$.

$\lim\limits_{x \to 10^-} f(x) = \lim\limits_{x \to 10^-} (ax+b) = a(10)+b = 10a+b$

3. Right-hand limit at $x = 10$:

For $x > 10$, $f(x) = 21$.

$\lim\limits_{x \to 10^+} f(x) = \lim\limits_{x \to 10^+} 21 = 21$

For continuity at $x=10$, we require $\lim\limits_{x \to 10^-} f(x) = \lim\limits_{x \to 10^+} f(x) = f(10)$.

$10a+b = 21 = 21$

This gives us the second equation:

Now we have a system of two linear equations with two variables, $a$ and $b$:

$2a + b = 5$ $\quad$ ...(1)

$10a + b = 21$ $\quad$ ...(2)

Subtract Equation (1) from Equation (2) to eliminate $b$:

$(10a + b) - (2a + b) = 21 - 5$

$10a + b - 2a - b = 16$

$8a = 16$

$a = \frac{16}{8}$

$a = 2$

Substitute the value of $a=2$ into Equation (1):

$2(2) + b = 5$

$4 + b = 5$

$b = 5 - 4$

$b = 1$

Thus, the values of $a$ and $b$ that make the function continuous are $a=2$ and $b=1$.

Conclusion:

For the function $f(x)$ to be continuous, the values of $a$ and $b$ must be $a = 2$ and $b = 1$.

Given:

The function $f(x) = \cos(x^2)$.

The domain of the function $f(x) = \cos(x^2)$ is the set of all real numbers ($\mathbb{R}$).

To Show:

The function $f(x) = \cos(x^2)$ is a continuous function on its domain.

Proof:

We can express the function $f(x) = \cos(x^2)$ as a composition of two simpler functions.

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

Then, $f(x) = h(g(x)) = h(x^2) = \cos(x^2)$.

To show that $f(x)$ is continuous, we can use the theorem on the continuity of composite functions.

The theorem states that if a function $g$ is continuous at a point $c$, and a function $h$ is continuous at $g(c)$, then the composite function $h \circ g$, defined by $(h \circ g)(x) = h(g(x))$, is continuous at $c$.

Let's examine the continuity of the individual functions $g(x)$ and $h(u)$.

1. Continuity of $g(x) = x^2$:

The function $g(x) = x^2$ is a polynomial function. We know that every polynomial function is continuous at every real number (as shown in Example 14). Therefore, $g(x)$ is continuous for all $x \in \mathbb{R}$.

2. Continuity of $h(u) = \cos(u)$:

The function $h(u) = \cos(u)$ is the cosine function. We know that the cosine function is continuous at every real number (as shown in Question 22). Therefore, $h(u)$ is continuous for all $u \in \mathbb{R}$.

Now, consider an arbitrary real number $c \in \mathbb{R}$.

Since $g(x) = x^2$ is continuous at $c$, the value $g(c) = c^2$ is a real number.

Since $h(u) = \cos(u)$ is continuous for all real numbers, it is continuous at the real number $g(c) = c^2$.

By the theorem on the continuity of composite functions, since $g(x)$ is continuous at $c$ and $h(u)$ is continuous at $g(c)$, the composite function $f(x) = h(g(x))$ is continuous at $c$.

Since $c$ was an arbitrary real number, this proves that $f(x) = \cos(x^2)$ is continuous at every point in its domain, which is $\mathbb{R}$.

Conclusion:

The function $f(x) = \cos(x^2)$ is a composition of the polynomial function $g(x) = x^2$ and the cosine function $h(u) = \cos(u)$, both of which are continuous on $\mathbb{R}$. By the theorem on continuity of composite functions, $f(x)$ is continuous at every real number.

Given:

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

The domain of the function $f(x) = |\cos x|$ is the set of all real numbers ($\mathbb{R}$).

To Show:

The function $f(x) = |\cos x|$ is a continuous function on its domain.

Proof:

We can express the function $f(x) = |\cos x|$ as a composition of two simpler functions.

Let $g(x) = \cos x$ and $h(u) = |u|$.

Then, $f(x) = h(g(x)) = h(\cos x) = |\cos x|$.

To show that $f(x)$ is continuous, we can use the theorem on the continuity of composite functions.

The theorem states that if a function $g$ is continuous at a point $c$, and a function $h$ is continuous at $g(c)$, then the composite function $h \circ g$, defined by $(h \circ g)(x) = h(g(x))$, is continuous at $c$.

Let's examine the continuity of the individual functions $g(x)$ and $h(u)$.

1. Continuity of $g(x) = \cos x$:

The function $g(x) = \cos x$ is the cosine function. We know that the cosine function is continuous at every real number (as shown in Question 22). Therefore, $g(x)$ is continuous for all $x \in \mathbb{R}$.

2. Continuity of $h(u) = |u|$:

The function $h(u) = |u|$ is the absolute value function. We know that the absolute value function is continuous at every real number (as discussed in Example 7). Therefore, $h(u)$ is continuous for all $u \in \mathbb{R}$.

Now, consider an arbitrary real number $c \in \mathbb{R}$.

Since $g(x) = \cos x$ is continuous at $c$, the value $g(c) = \cos c$ is a real number.

Since $h(u) = |u|$ is continuous for all real numbers, it is continuous at the real number $g(c) = \cos c$.

By the theorem on the continuity of composite functions, since $g(x)$ is continuous at $c$ and $h(u)$ is continuous at $g(c)$, the composite function $f(x) = h(g(x))$ is continuous at $c$.

Since $c$ was an arbitrary real number, this proves that $f(x) = |\cos x|$ is continuous at every point in its domain, which is $\mathbb{R}$.

Conclusion:

The function $f(x) = |\cos x|$ is a composition of the cosine function $g(x) = \cos x$ and the absolute value function $h(u) = |u|$, both of which are continuous on $\mathbb{R}$. By the theorem on continuity of composite functions, $f(x)$ is continuous at every real number.

Given:

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

The domain of the function $f(x) = \sin |x|$ is the set of all real numbers ($\mathbb{R}$).

To Examine:

Whether the function $f(x) = \sin |x|$ is a continuous function on its domain.

Solution:

We can express the function $f(x) = \sin |x|$ as a composition of two simpler functions.

Let $g(x) = |x|$ and $h(u) = \sin u$.

Then, $f(x) = h(g(x)) = h(|x|) = \sin |x|$.

To examine the continuity of $f(x)$, we can use the theorem on the continuity of composite functions.

The theorem states that if a function $g$ is continuous at a point $c$, and a function $h$ is continuous at $g(c)$, then the composite function $h \circ g$, defined by $(h \circ g)(x) = h(g(x))$, is continuous at $c$.

Let's examine the continuity of the individual functions $g(x)$ and $h(u)$.

1. Continuity of $g(x) = |x|$:

The function $g(x) = |x|$ is the absolute value function. We know that the absolute value function is continuous at every real number (as discussed in Example 7). Therefore, $g(x)$ is continuous for all $x \in \mathbb{R}$.

2. Continuity of $h(u) = \sin u$:

The function $h(u) = \sin u$ is the sine function. We know that the sine function is continuous at every real number (as shown in Example 17). Therefore, $h(u)$ is continuous for all $u \in \mathbb{R}$.

Now, consider an arbitrary real number $c \in \mathbb{R}$.

Since $g(x) = |x|$ is continuous at $c$, the value $g(c) = |c|$ is a real number.

Since $h(u) = \sin u$ is continuous for all real numbers, it is continuous at the real number $g(c) = |c|$.

By the theorem on the continuity of composite functions, since $g(x)$ is continuous at $c$ and $h(u)$ is continuous at $g(c)$, the composite function $f(x) = h(g(x))$ is continuous at $c$.

Since $c$ was an arbitrary real number, this proves that $f(x) = \sin |x|$ is continuous at every point in its domain, which is $\mathbb{R}$.

Alternatively, we can examine the function piecewise:

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

For $x > 0$, $f(x) = \sin x$. The sine function is continuous, so $f(x)$ is continuous for $x > 0$.

For $x < 0$, $f(x) = -\sin x$. Since $\sin x$ is continuous, $-\sin x$ is also continuous (by the constant multiple property of continuous functions), so $f(x)$ is continuous for $x < 0$.

Now, check continuity at $x=0$:

Function value at $x=0$: For $x \geq 0$, $f(x) = \sin x$. $f(0) = \sin(0) = 0$.

$f(0) = 0$

Left-hand limit at $x=0$:

For $x < 0$, $f(x) = -\sin x$.

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

Right-hand limit at $x=0$:

For $x > 0$, $f(x) = \sin x$.

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

Comparing the LHL, RHL, and $f(0)$:

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

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

$f(0) = 0$

Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0) = 0$, the function is continuous at $x = 0$.

Since the function is continuous for $x<0$, $x>0$, and at $x=0$, it is continuous for all real numbers.

Conclusion:

The function $f(x) = \sin |x|$ is a composition of the absolute value function $g(x) = |x|$ and the sine function $h(u) = \sin u$, both of which are continuous on $\mathbb{R}$. By the theorem on continuity of composite functions, or by piecewise analysis, $f(x)$ is continuous at every real number.

Given:

The function $f(x) = |x| - |x+1|$.

The domain of the function is the set of all real numbers ($\mathbb{R}$).

To Find:

All points of discontinuity of the function $f(x)$.

Solution:

The absolute value function $|y|$ is continuous for all real numbers $y$. The functions $g(x) = x$ and $h(x) = x+1$ are polynomial functions, hence continuous for all real numbers.

The function $|x|$ is a composition of the absolute value function and the identity function $x$, both of which are continuous. Thus, $|x|$ is continuous for all $x \in \mathbb{R}$.

The function $|x+1|$ is a composition of the absolute value function and the polynomial function $x+1$. Since both are continuous, $|x+1|$ is continuous for all $x \in \mathbb{R}$.

The function $f(x) = |x| - |x+1|$ is the difference of two continuous functions, $|x|$ and $|x+1|$. The difference of two continuous functions is continuous on the intersection of their domains. Since both $|x|$ and $|x+1|$ are continuous on $\mathbb{R}$, their difference $f(x)$ is continuous on $\mathbb{R}$.

Alternatively, we can analyze the function using its piecewise definition based on the points where the expressions inside the absolute values become zero, i.e., $x=0$ and $x+1=0 \implies x=-1$. These points divide the real line into three intervals: $(-\infty, -1)$, $[-1, 0)$, and $[0, \infty)$.

Let's define $f(x)$ in each interval:

For $x < -1$:

$|x| = -x$

$|x+1| = -(x+1)$

$f(x) = -x - (-(x+1)) = -x + x + 1 = 1$

For $-1 \leq x < 0$:

$|x| = -x$

$|x+1| = x+1$

$f(x) = -x - (x+1) = -x - x - 1 = -2x - 1$

For $x \geq 0$:

$|x| = x$

$|x+1| = x+1$

$f(x) = x - (x+1) = x - x - 1 = -1$

So, the piecewise definition of $f(x)$ is:

$f(x) = \begin{cases} 1 & \text{if } x < -1 \\ -2x - 1 & \text{if } -1 \leq x < 0 \\ -1 & \text{if } x \geq 0 \end{cases}$

We examine continuity in the open intervals and at the boundary points $x=-1$ and $x=0$.

Case 1: For $x < -1$

Let $c < -1$. In this interval, $f(x) = 1$, which is a constant function and continuous. So $f$ is continuous for $x < -1$.

Case 2: For $-1 < x < 0$

Let $-1 < c < 0$. In this interval, $f(x) = -2x - 1$, which is a polynomial function and continuous. So $f$ is continuous for $-1 < x < 0$.

Case 3: For $x > 0$

Let $c > 0$. In this interval, $f(x) = -1$, which is a constant function and continuous. So $f$ is continuous for $x > 0$.

Case 4: At $x = -1$

Function value: $f(-1) = -2(-1) - 1 = 2 - 1 = 1$ (using the definition for $-1 \leq x < 0$).

Left-hand limit: $\lim\limits_{x \to -1^-} f(x) = \lim\limits_{x \to -1^-} 1 = 1$ (using the definition for $x < -1$).

Right-hand limit: $\lim\limits_{x \to -1^+} f(x) = \lim\limits_{x \to -1^+} (-2x - 1) = -2(-1) - 1 = 2 - 1 = 1$ (using the definition for $-1 \leq x < 0$).

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

Case 5: At $x = 0$

Function value: $f(0) = -1$ (using the definition for $x \geq 0$).

Left-hand limit: $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} (-2x - 1) = -2(0) - 1 = -1$ (using the definition for $-1 \leq x < 0$).

Right-hand limit: $\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} (-1) = -1$ (using the definition for $x \geq 0$).

Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0) = -1$, the function is continuous at $x = 0$.

Combining all cases, the function is continuous on $(-\infty, -1)$, $(-1, 0)$, $(0, \infty)$, at $x=-1$, and at $x=0$. Thus, the function is continuous for all real numbers.

Conclusion:

The function $f(x) = |x| - |x+1|$ is continuous at every real number. Therefore, there are no points of discontinuity for this function.

Given:

The function $f(x) = \sin(x^2)$.

To Find:

The derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \sin(x^2)$ is a composite function. We can find its derivative using the Chain Rule.

The Chain Rule states that if $f(x) = h(g(x))$, then $f'(x) = h'(g(x)) \cdot g'(x)$.

Let the outer function be $h(u) = \sin u$ and the inner function be $g(x) = x^2$.

Then, $f(x) = h(g(x))$.

First, find the derivative of the outer function $h(u)$ with respect to $u$:

$h'(u) = \frac{d}{du}(\sin u) = \cos u$

Next, find the derivative of the inner function $g(x)$ with respect to $x$:

$g'(x) = \frac{d}{dx}(x^2)$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$g'(x) = 2x^{2-1} = 2x$

Now, apply the Chain Rule: $f'(x) = h'(g(x)) \cdot g'(x)$.

Substitute $g(x) = x^2$ into $h'(u) = \cos u$ to get $h'(g(x)) = \cos(x^2)$.

Substitute $g'(x) = 2x$ into the Chain Rule formula.

$f'(x) = \cos(x^2) \cdot (2x)$

Rearranging the terms, we get:

$f'(x) = 2x \cos(x^2)$

Conclusion:

The derivative of the function $f(x) = \sin(x^2)$ is $f'(x) = 2x \cos(x^2)$.

Given:

The function $f(x) = \tan(2x+3)$.

To Find:

The derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \tan(2x+3)$ is a composite function. We can find its derivative using the Chain Rule.

The Chain Rule states that if $f(x) = h(g(x))$, then $f'(x) = h'(g(x)) \cdot g'(x)$.

Let the outer function be $h(u) = \tan u$ and the inner function be $g(x) = 2x+3$.

Then, $f(x) = h(g(x))$.

First, find the derivative of the outer function $h(u)$ with respect to $u$:

$h'(u) = \frac{d}{du}(\tan u) = \sec^2 u$

Next, find the derivative of the inner function $g(x)$ with respect to $x$:

$g'(x) = \frac{d}{dx}(2x+3)$

Using the sum rule and constant multiple rule:

$g'(x) = \frac{d}{dx}(2x) + \frac{d}{dx}(3) = 2 \frac{d}{dx}(x) + 0 = 2(1) + 0 = 2$

$g'(x) = 2$

Now, apply the Chain Rule: $f'(x) = h'(g(x)) \cdot g'(x)$.

Substitute $g(x) = 2x+3$ into $h'(u) = \sec^2 u$ to get $h'(g(x)) = \sec^2(2x+3)$.

Substitute $g'(x) = 2$ into the Chain Rule formula.

$f'(x) = \sec^2(2x+3) \cdot (2)$

Rearranging the terms, we get:

$f'(x) = 2 \sec^2(2x+3)$

Conclusion:

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

Given:

The function $f(x) = \sin(\cos(x^2))$.

To Find:

The derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \sin(\cos(x^2))$ is a composite function involving multiple layers. We will use the Chain Rule repeatedly to find its derivative.

Let's break down the function into layers:

Outer function: $\sin(\text{expression})$

Middle function: $\cos(\text{inner expression})$

Inner function: $x^2$

Let $u = \cos(x^2)$. Then $f(x) = \sin u$.

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

First, find $\frac{df}{du}$:

$\frac{df}{du} = \frac{d}{du}(\sin u) = \cos u$

Now, we need to find $\frac{du}{dx}$, where $u = \cos(x^2)$. This is another composite function.

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

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

Find $\frac{du}{dv}$:

$\frac{du}{dv} = \frac{d}{dv}(\cos v) = -\sin v$

Find $\frac{dv}{dx}$:

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

Now, combine these to find $\frac{du}{dx}$:

$\frac{du}{dx} = (-\sin v) \cdot (2x)$

Substitute back $v = x^2$:

$\frac{du}{dx} = -\sin(x^2) \cdot (2x) = -2x \sin(x^2)$

Finally, combine $\frac{df}{du}$ and $\frac{du}{dx}$ to find $\frac{df}{dx}$:

$\frac{df}{dx} = \cos u \cdot (-2x \sin(x^2))$

Substitute back $u = \cos(x^2)$:

$\frac{df}{dx} = \cos(\cos(x^2)) \cdot (-2x \sin(x^2))$

Rearranging the terms, we get the derivative:

$f'(x) = -2x \sin(x^2) \cos(\cos(x^2))$

Conclusion:

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

Given:

The function $f(x) = \sin(x^2+5)$.

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \sin(x^2+5)$ is a composite function. We will use the Chain Rule to find its derivative.

The Chain Rule states that if $y = h(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Let the outer function be $h(u) = \sin u$ and the inner function be $g(x) = x^2+5$.

Then $f(x) = h(g(x))$.

Step 1: Find the derivative of the outer function with respect to its argument $u$:

$\frac{d}{du}(h(u)) = \frac{d}{du}(\sin u) = \cos u$

Step 2: Find the derivative of the inner function $g(x)$ with respect to $x$:

$\frac{d}{dx}(g(x)) = \frac{d}{dx}(x^2+5)$

Using the sum rule and power rule:

$= \frac{d}{dx}(x^2) + \frac{d}{dx}(5) = 2x + 0 = 2x$

Step 3: Apply the Chain Rule. Substitute $u = g(x) = x^2+5$ into the derivative of the outer function.

$f'(x) = \frac{d}{du}(h(u))|_{u=g(x)} \cdot \frac{d}{dx}(g(x))$

$f'(x) = \cos(u) \cdot (2x)$

Substitute back $u = x^2+5$:

$f'(x) = \cos(x^2+5) \cdot (2x)$

Rearranging the terms:

$f'(x) = 2x \cos(x^2+5)$

Conclusion:

The derivative of $\sin(x^2+5)$ is $2x \cos(x^2+5)$.

Given:

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

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \cos(\sin x)$ is a composite function. We will use the Chain Rule to find its derivative.

The Chain Rule states that if $y = h(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Let the outer function be $h(u) = \cos u$ and the inner function be $g(x) = \sin x$.

Then $f(x) = h(g(x))$.

Step 1: Find the derivative of the outer function with respect to its argument $u$:

$\frac{d}{du}(h(u)) = \frac{d}{du}(\cos u) = -\sin u$

Step 2: Find the derivative of the inner function $g(x)$ with respect to $x$:

$\frac{d}{dx}(g(x)) = \frac{d}{dx}(\sin x) = \cos x$

Step 3: Apply the Chain Rule. Substitute $u = g(x) = \sin x$ into the derivative of the outer function.

$f'(x) = \frac{d}{du}(h(u))|_{u=g(x)} \cdot \frac{d}{dx}(g(x))$

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

Substitute back $u = \sin x$:

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

Rearranging the terms:

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

Conclusion:

The derivative of $\cos(\sin x)$ is $-\cos x \sin(\sin x)$.

Given:

The function $f(x) = \sin(ax+b)$, where $a$ and $b$ are constants.

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \sin(ax+b)$ is a composite function. We will use the Chain Rule to find its derivative.

The Chain Rule states that if $y = h(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Let the outer function be $h(u) = \sin u$ and the inner function be $g(x) = ax+b$.

Then $f(x) = h(g(x))$.

Step 1: Find the derivative of the outer function with respect to its argument $u$:

$\frac{d}{du}(h(u)) = \frac{d}{du}(\sin u) = \cos u$

Step 2: Find the derivative of the inner function $g(x)$ with respect to $x$:

$\frac{d}{dx}(g(x)) = \frac{d}{dx}(ax+b)$

Using the sum rule, constant multiple rule, and power rule:

$= \frac{d}{dx}(ax) + \frac{d}{dx}(b) = a \frac{d}{dx}(x) + 0 = a(1) + 0 = a$

Step 3: Apply the Chain Rule. Substitute $u = g(x) = ax+b$ into the derivative of the outer function.

$f'(x) = \frac{d}{du}(h(u))|_{u=g(x)} \cdot \frac{d}{dx}(g(x))$

$f'(x) = (\cos u) \cdot (a)$

Substitute back $u = ax+b$:

$f'(x) = \cos(ax+b) \cdot a$

Rearranging the terms:

$f'(x) = a \cos(ax+b)$

Conclusion:

The derivative of $\sin(ax+b)$ is $a \cos(ax+b)$.

Given:

The function $f(x) = \sec(\tan(\sqrt{x}))$.

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \sec(\tan(\sqrt{x}))$ is a composite function with multiple layers. We will use the Chain Rule repeatedly.

Let's break down the function into layers:

Outer function: $\sec(\text{expression})$

Middle function: $\tan(\text{inner expression})$

Inner function: $\sqrt{x} = x^{1/2}$

Let $u = \tan(\sqrt{x})$. Then $f(x) = \sec u$.

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

Step 1: Find the derivative of the outermost function with respect to $u$:

$\frac{df}{du} = \frac{d}{du}(\sec u) = \sec u \tan u$

Step 2: Find $\frac{du}{dx}$, where $u = \tan(\sqrt{x})$. This is another composite function.

Let $v = \sqrt{x} = x^{1/2}$. Then $u = \tan v$.

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

Find $\frac{du}{dv}$:

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

Find $\frac{dv}{dx}$:

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

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

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

Now, combine these to find $\frac{du}{dx}$:

$\frac{du}{dx} = (\sec^2 v) \cdot (\frac{1}{2\sqrt{x}})$

Substitute back $v = \sqrt{x}$:

$\frac{du}{dx} = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\sec^2(\sqrt{x})}{2\sqrt{x}}$

Step 3: Apply the outermost Chain Rule: $f'(x) = \frac{df}{du} \cdot \frac{du}{dx}$.

Substitute $u = \tan(\sqrt{x})$ into $\frac{df}{du} = \sec u \tan u$ to get $\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x}))$.

Substitute $\frac{du}{dx} = \frac{\sec^2(\sqrt{x})}{2\sqrt{x}}$ into the formula.

$f'(x) = (\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x}))) \cdot \left(\frac{\sec^2(\sqrt{x})}{2\sqrt{x}}\right)$

Rearranging the terms:

$f'(x) = \frac{\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \sec^2(\sqrt{x})}{2\sqrt{x}}$

Conclusion:

The derivative of $\sec(\tan(\sqrt{x}))$ is $\frac{\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \sec^2(\sqrt{x})}{2\sqrt{x}}$.

Given:

The function $f(x) = \frac{\sin (ax+b)}{\cos (cx+d)}$, where $a, b, c, d$ are constants.

This function is defined wherever $\cos(cx+d) \neq 0$.

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function is in the form of a quotient of two functions. We will use the Quotient Rule along with the Chain Rule for the numerator and denominator.

The Quotient Rule 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 $u(x) = \sin(ax+b)$ and $v(x) = \cos(cx+d)$.

We need to find $u'(x)$ and $v'(x)$ using the Chain Rule.

Find $u'(x) = \frac{d}{dx}(\sin(ax+b))$:

Using the Chain Rule with outer function $\sin(u)$ and inner function $ax+b$:

$u'(x) = \cos(ax+b) \cdot \frac{d}{dx}(ax+b) = \cos(ax+b) \cdot a = a \cos(ax+b)$

Find $v'(x) = \frac{d}{dx}(\cos(cx+d))$:

Using the Chain Rule with outer function $\cos(u)$ and inner function $cx+d$:

$v'(x) = -\sin(cx+d) \cdot \frac{d}{dx}(cx+d) = -\sin(cx+d) \cdot c = -c \sin(cx+d)$

Now, apply the Quotient Rule:

$f'(x) = \frac{(a \cos(ax+b))(\cos(cx+d)) - (\sin(ax+b))(-c \sin(cx+d))}{(\cos(cx+d))^2}$

$f'(x) = \frac{a \cos(ax+b)\cos(cx+d) + c \sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$

We can also split the fraction:

$f'(x) = \frac{a \cos(ax+b)\cos(cx+d)}{\cos^2(cx+d)} + \frac{c \sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$

$f'(x) = a \cos(ax+b) \frac{1}{\cos(cx+d)} + c \sin(ax+b) \frac{\sin(cx+d)}{\cos(cx+d)} \frac{1}{\cos(cx+d)}$

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

$f'(x) = a \cos(ax+b) \sec(cx+d) + c \sin(ax+b) \tan(cx+d) \sec(cx+d)$

Alternatively, we can write:

$f'(x) = a \frac{\cos(ax+b)}{\cos(cx+d)} + c \frac{\sin(ax+b)}{\cos(cx+d)} \frac{\sin(cx+d)}{\cos(cx+d)} = a \frac{\cos(ax+b)}{\cos(cx+d)} + c \tan(ax+b) \tan(cx+d)$ (This simplification seems incorrect)

Let's stick to the correct simplified form:

$f'(x) = \frac{a \cos(ax+b)\cos(cx+d) + c \sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$

We can also write $\frac{1}{\cos^2(cx+d)} = \sec^2(cx+d)$.

$f'(x) = [a \cos(ax+b)\cos(cx+d) + c \sin(ax+b)\sin(cx+d)] \sec^2(cx+d)$

Conclusion:

The derivative of $\frac{\sin (ax+b)}{\cos (cx+d)}$ is $\frac{a \cos(ax+b)\cos(cx+d) + c \sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$.

Given:

The function $f(x) = \cos(x^3) \cdot \sin^2(x^5)$.

We can write $\sin^2(x^5)$ as $(\sin(x^5))^2$.

So, $f(x) = \cos(x^3) (\sin(x^5))^2$.

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x)$ is a product of two functions, $u(x) = \cos(x^3)$ and $v(x) = (\sin(x^5))^2$. We will use the Product Rule and the Chain Rule.

The Product Rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Step 1: Find the derivative of $u(x) = \cos(x^3)$ using the Chain Rule.

Let $g(x) = x^3$, so $u(x) = \cos(g(x))$.

$u'(x) = \frac{d}{dx}(\cos(x^3)) = -\sin(x^3) \cdot \frac{d}{dx}(x^3) = -\sin(x^3) \cdot (3x^2) = -3x^2 \sin(x^3)$

Step 2: Find the derivative of $v(x) = (\sin(x^5))^2$ using the Chain Rule twice.

Let $h(y) = y^2$ and $y = \sin(x^5)$. So $v(x) = h(y)$.

$\frac{dv}{dy} = \frac{d}{dy}(y^2) = 2y$

Now, find $\frac{dy}{dx} = \frac{d}{dx}(\sin(x^5))$. This is another Chain Rule application.

Let $k(z) = \sin z$ and $z = x^5$. So $y = k(z)$.

$\frac{dy}{dz} = \frac{d}{dz}(\sin z) = \cos z$

$\frac{dz}{dx} = \frac{d}{dx}(x^5) = 5x^4$

So, $\frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} = (\cos z) \cdot (5x^4)$. Substitute back $z = x^5$:

$\frac{dy}{dx} = 5x^4 \cos(x^5)$

Now, combine the derivatives for $v'(x)$: $v'(x) = \frac{dv}{dy} \cdot \frac{dy}{dx}$. Substitute back $y = \sin(x^5)$:

$v'(x) = (2y) \cdot (5x^4 \cos(x^5)) = 2(\sin(x^5)) (5x^4 \cos(x^5)) = 10x^4 \sin(x^5) \cos(x^5)$

Step 3: Apply the Product Rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$.

$f'(x) = (-3x^2 \sin(x^3)) \cdot (\sin(x^5))^2 + (\cos(x^3)) \cdot (10x^4 \sin(x^5) \cos(x^5))$

Rearranging the terms:

$f'(x) = -3x^2 \sin(x^3) \sin^2(x^5) + 10x^4 \cos(x^3) \sin(x^5) \cos(x^5)$

We can use the identity $2 \sin A \cos A = \sin (2A)$ to simplify the second term if desired, but the current form is also acceptable.

Second term: $10x^4 \sin(x^5) \cos(x^5) = 5x^4 (2 \sin(x^5) \cos(x^5)) = 5x^4 \sin(2x^5)$.

So, an alternative form of the derivative is:

$f'(x) = -3x^2 \sin(x^3) \sin^2(x^5) + 5x^4 \cos(x^3) \sin(2x^5)$

Conclusion:

The derivative of $\cos(x^3) \cdot \sin^2(x^5)$ is $-3x^2 \sin(x^3) \sin^2(x^5) + 10x^4 \cos(x^3) \sin(x^5) \cos(x^5)$ (or $-3x^2 \sin(x^3) \sin^2(x^5) + 5x^4 \cos(x^3) \sin(2x^5)$).

Given:

The function $f(x) = 2\sqrt{\cot (x^2)}$.

We can rewrite the function using exponent notation and the constant multiple rule for derivatives:

$f(x) = 2 (\cot (x^2))^{1/2}$

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

We will use the Chain Rule multiple times and the constant multiple rule.

The function $f(x)$ has the form $2 \cdot h(g(k(x)))$, where the layers are:

Constant multiple: 2

Outermost function: $(\text{expression})^{1/2}$ (square root)

Middle function: $\cot(\text{inner expression})$

Inner function: $x^2$

We use the formula $\frac{d}{dx} (c \cdot g(x)) = c \cdot \frac{d}{dx} (g(x))$ and the Chain Rule $\frac{d}{dx} h(u) = h'(u) \frac{du}{dx}$.

Let $y = 2(\cot(x^2))^{1/2}$.

$\frac{dy}{dx} = 2 \frac{d}{dx} ((\cot(x^2))^{1/2})$

Now, differentiate $(\cot(x^2))^{1/2}$ using the Chain Rule.

Let $u = \cot(x^2)$. The outer function is $u^{1/2}$.

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

Now, we need to find $\frac{du}{dx} = \frac{d}{dx}(\cot(x^2))$. This is another Chain Rule application.

Let $v = x^2$. The outer function is $\cot v$.

$\frac{d}{dv}(\cot v) = -\text{cosec}^2 v$

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

Combining these to find $\frac{du}{dx}$:

$\frac{du}{dx} = \frac{d}{dv}(\cot v) \cdot \frac{dv}{dx} = (-\text{cosec}^2 v) \cdot (2x)$

Substitute back $v = x^2$:

$\frac{du}{dx} = -2x \text{cosec}^2(x^2)$

Now, combine the derivatives for $\frac{d}{dx}((\cot(x^2))^{1/2})$: Use $\frac{d}{du}(u^{1/2}) \cdot \frac{du}{dx}$. Substitute back $u = \cot(x^2)$:

$\frac{d}{dx}((\cot(x^2))^{1/2}) = \left(\frac{1}{2\sqrt{u}}\right) \cdot (-2x \text{cosec}^2(x^2))$

$= \frac{1}{2\sqrt{\cot(x^2)}} \cdot (-2x \text{cosec}^2(x^2))$

$= \frac{-2x \text{cosec}^2(x^2)}{2\sqrt{\cot(x^2)}} = \frac{-x \text{cosec}^2(x^2)}{\sqrt{\cot(x^2)}}$

Finally, multiply by the constant 2:

$\frac{dy}{dx} = 2 \cdot \left(\frac{-x \text{cosec}^2(x^2)}{\sqrt{\cot(x^2)}}\right)$

$f'(x) = \frac{-2x \text{cosec}^2(x^2)}{\sqrt{\cot(x^2)}}$

Conclusion:

The derivative of $2\sqrt{\cot (x^2)}$ is $\frac{-2x \text{cosec}^2(x^2)}{\sqrt{\cot(x^2)}}$.

Given:

The function $f(x) = \cos(\sqrt{x})$.

We can rewrite $\sqrt{x}$ as $x^{1/2}$.

So, $f(x) = \cos(x^{1/2})$.

To Differentiate:

Find the derivative of $f(x)$ with respect to $x$, i.e., $f'(x)$ or $\frac{d}{dx} f(x)$.

Solution:

The function $f(x) = \cos(x^{1/2})$ is a composite function. We will use the Chain Rule to find its derivative.

The Chain Rule states that if $y = h(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Let the outer function be $h(u) = \cos u$ and the inner function be $g(x) = \sqrt{x} = x^{1/2}$.

Then $f(x) = h(g(x))$.

Step 1: Find the derivative of the outer function with respect to its argument $u$:

$\frac{d}{du}(h(u)) = \frac{d}{du}(\cos u) = -\sin u$

Step 2: Find the derivative of the inner function $g(x)$ with respect to $x$:

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

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

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

Step 3: Apply the Chain Rule. Substitute $u = g(x) = \sqrt{x}$ into the derivative of the outer function.

$f'(x) = \frac{d}{du}(h(u))|_{u=g(x)} \cdot \frac{d}{dx}(g(x))$

$f'(x) = (-\sin u) \cdot (\frac{1}{2\sqrt{x}})$

Substitute back $u = \sqrt{x}$:

$f'(x) = (-\sin (\sqrt{x})) \cdot (\frac{1}{2\sqrt{x}})$

Rearranging the terms:

$f'(x) = -\frac{\sin(\sqrt{x})}{2\sqrt{x}}$

Conclusion:

The derivative of $\cos(\sqrt{x})$ is $-\frac{\sin(\sqrt{x})}{2\sqrt{x}}$.

Given:

The function $f(x) = |x - 1|$, for $x \in \mathbb{R}$.

To Prove:

The function $f(x)$ is not differentiable at $x = 1$.

Proof:

A function $f(x)$ is differentiable at a point $x=c$ if and only if the limit of the difference quotient exists at that point, i.e., the following limit exists:

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

For the function $f(x) = |x - 1|$ at $x = 1$, we need to evaluate the limit of the difference quotient at $c=1$:

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

First, let's find $f(1)$:

$f(1) = |1 - 1| = |0| = 0$

Next, let's find $f(1+h)$:

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

Now, substitute these into the difference quotient:

$\lim\limits_{h \to 0} \frac{|h| - 0}{h} = \lim\limits_{h \to 0} \frac{|h|}{h}$

For the limit $\lim\limits_{h \to 0} \frac{|h|}{h}$ to exist, the left-hand limit (LHL) and the right-hand limit (RHL) must be equal.

Left-hand limit (as $h$ approaches 0 from the left, $h < 0$):

If $h < 0$, then $|h| = -h$.

$\lim\limits_{h \to 0^-} \frac{|h|}{h} = \lim\limits_{h \to 0^-} \frac{-h}{h} = \lim\limits_{h \to 0^-} (-1) = -1$

Right-hand limit (as $h$ approaches 0 from the right, $h > 0$):

If $h > 0$, then $|h| = h$.

$\lim\limits_{h \to 0^+} \frac{|h|}{h} = \lim\limits_{h \to 0^+} \frac{h}{h} = \lim\limits_{h \to 0^+} (1) = 1$

Comparing the LHL and RHL:

$\lim\limits_{h \to 0^-} \frac{|h|}{h} = -1$

$\lim\limits_{h \to 0^+} \frac{|h|}{h} = 1$

Since the left-hand limit is not equal to the right-hand limit ($-1 \neq 1$), the limit $\lim\limits_{h \to 0} \frac{|h|}{h}$ does not exist.

Therefore, the derivative of $f(x)$ at $x = 1$, $f'(1)$, does not exist.

This shows that the function $f(x) = |x-1|$ is not differentiable at $x = 1$. Graphically, this corresponds to a sharp corner or cusp at $x=1$.

Conclusion:

Since the limit of the difference quotient at $x=1$ does not exist (as the left-hand and right-hand limits are different), the function $f(x) = |x - 1|$ is not differentiable at x = 1.

Given:

The greatest integer function $f(x) = [x]$, defined for $0 < x < 3$.

The domain of interest is the open interval $(0, 3)$. The integers within this domain are $1$ and $2$.

To Prove:

The function $f(x) = [x]$ is not differentiable at $x = 1$ and $x = 2$.

Proof:

A function $f(x)$ is differentiable at a point $x=c$ if and only if the limit of the difference quotient exists at that point, i.e., the following limit exists:

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

For this limit to exist, the left-hand limit (LHL) and the right-hand limit (RHL) of the difference quotient must be equal.

Differentiability at $x = 1$:

We need to evaluate $\lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$.

First, find $f(1)$:

$f(1) = [1] = 1$

Now, evaluate the left-hand limit of the difference quotient as $h \to 0^-$ (meaning $h < 0$ and $h$ is close to 0). This means $1+h$ is slightly less than 1 (e.g., $0.9, 0.99$).

For $h < 0$ and $|h|$ small, $1+h < 1$. The greatest integer less than or equal to $1+h$ is 0.

$f(1+h) = [1+h] = 0$ for $h \in (-1, 0)$

LHL: $\lim\limits_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim\limits_{h \to 0^-} \frac{[1+h] - [1]}{h} = \lim\limits_{h \to 0^-} \frac{0 - 1}{h} = \lim\limits_{h \to 0^-} \frac{-1}{h}$

As $h \to 0^-$ (a small negative number), $\frac{-1}{h}$ approaches $\frac{-1}{\text{small negative}} \to \infty$. The limit does not exist.

Let's check the right-hand limit of the difference quotient as $h \to 0^+$ (meaning $h > 0$ and $h$ is close to 0). This means $1+h$ is slightly greater than 1 (e.g., $1.1, 1.01$).

For $h > 0$ and $|h|$ small, $1+h > 1$. The greatest integer less than or equal to $1+h$ is 1.

$f(1+h) = [1+h] = 1$ for $h \in (0, 2)$

RHL: $\lim\limits_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim\limits_{h \to 0^+} \frac{[1+h] - [1]}{h} = \lim\limits_{h \to 0^+} \frac{1 - 1}{h} = \lim\limits_{h \to 0^+} \frac{0}{h} = \lim\limits_{h \to 0^+} 0 = 0$

Comparing the LHL and RHL:

LHL = $\infty$ (or DNE)

RHL = 0

Since the left-hand limit and the right-hand limit of the difference quotient are not equal, the limit $\lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$ does not exist. Therefore, $f(x)$ is not differentiable at $x = 1$.

Differentiability at $x = 2$:

We need to evaluate $\lim\limits_{h \to 0} \frac{f(2+h) - f(2)}{h}$.

First, find $f(2)$:

$f(2) = [2] = 2$

Now, evaluate the left-hand limit of the difference quotient as $h \to 0^-$ (meaning $h < 0$ and $h$ is close to 0). This means $2+h$ is slightly less than 2 (e.g., $1.9, 1.99$).

For $h < 0$ and $|h|$ small, $2+h < 2$. The greatest integer less than or equal to $2+h$ is 1.

$f(2+h) = [2+h] = 1$ for $h \in (-1, 0)$

LHL: $\lim\limits_{h \to 0^-} \frac{f(2+h) - f(2)}{h} = \lim\limits_{h \to 0^-} \frac{[2+h] - [2]}{h} = \lim\limits_{h \to 0^-} \frac{1 - 2}{h} = \lim\limits_{h \to 0^-} \frac{-1}{h}$

As $h \to 0^-$ (a small negative number), $\frac{-1}{h}$ approaches $\frac{-1}{\text{small negative}} \to \infty$. The limit does not exist.

Let's check the right-hand limit of the difference quotient as $h \to 0^+$ (meaning $h > 0$ and $h$ is close to 0). This means $2+h$ is slightly greater than 2 (e.g., $2.1, 2.01$).

For $h > 0$ and $|h|$ small, $2+h > 2$. The greatest integer less than or equal to $2+h$ is 2.

$f(2+h) = [2+h] = 2$ for $h \in (0, 1)$ (since the domain is $0

RHL: $\lim\limits_{h \to 0^+} \frac{f(2+h) - f(2)}{h} = \lim\limits_{h \to 0^+} \frac{[2+h] - [2]}{h} = \lim\limits_{h \to 0^+} \frac{2 - 2}{h} = \lim\limits_{h \to 0^+} \frac{0}{h} = \lim\limits_{h \to 0^+} 0 = 0$

Comparing the LHL and RHL:

LHL = $\infty$ (or DNE)

RHL = 0

Since the left-hand limit and the right-hand limit of the difference quotient are not equal, the limit $\lim\limits_{h \to 0} \frac{f(2+h) - f(2)}{h}$ does not exist. Therefore, $f(x)$ is not differentiable at $x = 2$.

This proves that the greatest integer function $f(x) = [x]$ is not differentiable at $x = 1$ and $x = 2$ within the domain $0 < x < 3$. Note that the greatest integer function is discontinuous at all integers, and differentiability implies continuity. Therefore, discontinuity at integers also implies non-differentiability at integers.

Conclusion:

For both $x=1$ and $x=2$, the limit of the difference quotient does not exist (as the left-hand and right-hand limits are different). Therefore, the function $f(x) = [x]$ is not differentiable at x = 1 and x = 2.

Given:

The equation $x - y = \pi$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

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

Solution:

We can find $\frac{dy}{dx}$ using 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 $x - y = \pi$ with respect to $x$:

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

Using the difference rule on the left side and the constant rule on the right side:

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

The derivative of $x$ with respect to $x$ is 1:

$1 - \frac{d}{dx}(y) = 0$

The derivative of $y$ with respect to $x$ is denoted as $\frac{dy}{dx}$:

$1 - \frac{dy}{dx} = 0$

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

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

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

Alternatively, we could solve the given equation explicitly for $y$ first:

$x - y = \pi$

$-y = \pi - x$

$y = x - \pi$

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

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

Using the difference rule and constant rule:

$\frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(\pi) = 1 - 0 = 1$

Both methods yield the same result.

Conclusion:

If $x - y = \pi$, then $\frac{dy}{dx} = 1$.

Given:

The equation $y + \sin y = \cos x$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

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

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. 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 $y + \sin y = \cos x$ with respect to $x$:

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

Using the sum rule on the left side:

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

The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$.

For the term $\frac{d}{dx}(\sin y)$, we use the Chain Rule. Treat $\sin y$ as a function of $y$, where $y$ is a function of $x$. The derivative of $\sin u$ with respect to $u$ is $\cos u$. So, $\frac{d}{dx}(\sin y) = \frac{d}{dy}(\sin y) \cdot \frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$.

The derivative of $\cos x$ with respect to $x$ is $-\sin x$.

Substituting these derivatives back into the equation:

$\frac{dy}{dx} + \cos y \cdot \frac{dy}{dx} = -\sin x$

Now, we need to solve for $\frac{dy}{dx}$. Factor out $\frac{dy}{dx}$ from the terms on the left side:

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

Divide both sides by $(1 + \cos y)$, assuming $1 + \cos y \neq 0$ (which means $\cos y \neq -1$, or $y \neq \pi + 2n\pi$ for integer $n$):

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

Conclusion:

If $y + \sin y = \cos x$, then $\frac{dy}{dx} = \frac{-\sin x}{1 + \cos y}$, provided $1 + \cos y \neq 0$.

Given:

The function $f(x) = \sin^{-1} x$ (also written as $\arcsin x$).

The domain of $f(x)$ is $[-1, 1]$ and the standard range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We assume the derivative exists, which is for $x \in (-1, 1)$.

To Find:

The derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{d}{dx}(\sin^{-1} x)$.

Solution:

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

By the definition of the inverse sine function, this means:

$x = \sin y$

where $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $x \in [-1, 1]$. We are assuming the derivative exists, so we consider $x \in (-1, 1)$, which implies $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

We differentiate both sides of the equation $x = \sin y$ with respect to $x$. We use implicit differentiation on the right side, treating $y$ as a function of $x$ and applying the Chain Rule.

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

$1 = \frac{d}{dy}(\sin y) \cdot \frac{dy}{dx}$

We know that $\frac{d}{dy}(\sin y) = \cos y$.

$1 = \cos y \cdot \frac{dy}{dx}$

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

$\frac{dy}{dx} = \frac{1}{\cos y}$

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

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

Since $x = \sin y$, we have $\sin^2 y = x^2$.

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

Taking the square root of both sides:

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

Since the range of $y = \sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and for $y$ in the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\cos y$ is positive, we take the positive square root.

$\cos y = \sqrt{1 - x^2}$ for $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$ (which corresponds to $x \in (-1, 1)$)

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

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

This derivative exists for all $x$ in the interval $(-1, 1)$. At the endpoints $x = \pm 1$, the derivative is undefined, corresponding to a vertical tangent line on the graph of $y = \sin^{-1} x$.

Conclusion:

The derivative of the function $f(x) = \sin^{-1} x$, where it exists (for $x \in (-1, 1)$), is $f'(x) = \frac{1}{\sqrt{1 - x^2}}$.

Given:

The function $f(x) = \tan^{-1} x$ (also written as $\arctan x$).

The domain of $f(x)$ is the set of all real numbers ($\mathbb{R}$), and the standard range is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We assume the derivative exists.

To Find:

The derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$ or $\frac{d}{dx}(\tan^{-1} x)$.

Solution:

Let $y = \tan^{-1} x$.

By the definition of the inverse tangent function, this means:

$x = \tan y$

where $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Since $x$ can be any real number, the derivative exists for all $x \in \mathbb{R}$.

We differentiate both sides of the equation $x = \tan y$ with respect to $x$. We use implicit differentiation on the right side, treating $y$ as a function of $x$ and applying the Chain Rule.

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

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

We know that $\frac{d}{dy}(\tan y) = \sec^2 y$.

$1 = \sec^2 y \cdot \frac{dy}{dx}$

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

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

We need to express $\sec^2 y$ in terms of $x$. We use the trigonometric identity relating tangent and secant: $\sec^2 y = 1 + \tan^2 y$.

Since $x = \tan y$, we have $\tan^2 y = x^2$.

$\sec^2 y = 1 + x^2$

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

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

This derivative exists for all real numbers $x$.

Conclusion:

The derivative of the function $f(x) = \tan^{-1} x$ is $f'(x) = \frac{1}{1 + x^2}$.

Given:

The equation $2x + 3y = \sin x$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

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

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. 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 $2x + 3y = \sin x$ with respect to $x$:

$\frac{d}{dx}(2x + 3y) = \frac{d}{dx}(\sin x)$

Using the sum rule on the left side:

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x)$

Using the constant multiple rule:

$2 \frac{d}{dx}(x) + 3 \frac{d}{dx}(y) = \frac{d}{dx}(\sin x)$

We know that $\frac{d}{dx}(x) = 1$. The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$. The derivative of $\sin x$ with respect to $x$ is $\cos x$.

$2(1) + 3 \frac{dy}{dx} = \cos x$

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

Now, we need to solve for $\frac{dy}{dx}$. Subtract 2 from both sides:

$3 \frac{dy}{dx} = \cos x - 2$

Divide both sides by 3:

$\frac{dy}{dx} = \frac{\cos x - 2}{3}$

Conclusion:

If $2x + 3y = \sin x$, then $\frac{dy}{dx} = \frac{\cos x - 2}{3}$.

Given:

The equation $2x + 3y = \sin y$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

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

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. 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 $2x + 3y = \sin y$ with respect to $x$:

$\frac{d}{dx}(2x + 3y) = \frac{d}{dx}(\sin y)$

Using the sum rule on the left side and the Chain Rule on the right side:

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dy}(\sin y) \cdot \frac{dy}{dx}$

Using the constant multiple rule on the left side:

$2 \frac{d}{dx}(x) + 3 \frac{d}{dx}(y) = \cos y \cdot \frac{dy}{dx}$

We know that $\frac{d}{dx}(x) = 1$. The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$.

$2(1) + 3 \frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$

$2 + 3 \frac{dy}{dx} = \cos y \frac{dy}{dx}$

Now, we need to solve for $\frac{dy}{dx}$. Gather all terms containing $\frac{dy}{dx}$ on one side and other terms on the other side.

$2 = \cos y \frac{dy}{dx} - 3 \frac{dy}{dx}$

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

$2 = \frac{dy}{dx} (\cos y - 3)$

Divide both sides by $(\cos y - 3)$, assuming $\cos y - 3 \neq 0$ (which is always true since $-1 \leq \cos y \leq 1$, so $\cos y - 3$ is between $-4$ and $-2$).

$\frac{dy}{dx} = \frac{2}{\cos y - 3}$

Conclusion:

If $2x + 3y = \sin y$, then $\frac{dy}{dx} = \frac{2}{\cos y - 3}$.

Given:

The equation $ax + by^2 = \cos y$, where $a$ and $b$ are constants.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

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

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the Chain Rule where necessary.

Differentiate both sides of the equation $ax + by^2 = \cos y$ with respect to $x$:

$\frac{d}{dx}(ax + by^2) = \frac{d}{dx}(\cos y)$

Using the sum rule on the left side:

$\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y)$

Using the constant multiple rule:

$a \frac{d}{dx}(x) + b \frac{d}{dx}(y^2) = \frac{d}{dx}(\cos y)$

We know that $\frac{d}{dx}(x) = 1$.

For the term $\frac{d}{dx}(y^2)$, we use the Chain Rule. Treat $y^2$ as a function of $y$, where $y$ is a function of $x$. The derivative of $u^2$ with respect to $u$ is $2u$. So, $\frac{d}{dx}(y^2) = \frac{d}{dy}(y^2) \cdot \frac{dy}{dx} = 2y \cdot \frac{dy}{dx}$. Thus, $b \frac{d}{dx}(y^2) = b(2y \frac{dy}{dx}) = 2by \frac{dy}{dx}$.

For the term $\frac{d}{dx}(\cos y)$, we use the Chain Rule. Treat $\cos y$ as a function of $y$, where $y$ is a function of $x$. The derivative of $\cos u$ with respect to $u$ is $-\sin u$. So, $\frac{d}{dx}(\cos y) = \frac{d}{dy}(\cos y) \cdot \frac{dy}{dx} = -\sin y \cdot \frac{dy}{dx}$.

Substituting these derivatives back into the equation:

$a(1) + 2by \frac{dy}{dx} = -\sin y \frac{dy}{dx}$

$a + 2by \frac{dy}{dx} = -\sin y \frac{dy}{dx}$

Now, we need to solve for $\frac{dy}{dx}$. Gather all terms containing $\frac{dy}{dx}$ on one side and other terms on the other side. Move the $2by \frac{dy}{dx}$ term to the right side:

$a = -\sin y \frac{dy}{dx} - 2by \frac{dy}{dx}$

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

$a = \frac{dy}{dx} (-\sin y - 2by)$

Divide both sides by $(-\sin y - 2by)$, assuming this expression is not zero:

$\frac{dy}{dx} = \frac{a}{-\sin y - 2by}$

We can factor out $-1$ from the denominator:

$\frac{dy}{dx} = \frac{a}{-(2by + \sin y)}$

$\frac{dy}{dx} = -\frac{a}{2by + \sin y}$

Conclusion:

If $ax + by^2 = \cos y$, then $\frac{dy}{dx} = -\frac{a}{2by + \sin y}$, provided $2by + \sin y \neq 0$.

Given:

The equation $xy + y^2 = \tan x + y$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

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

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the Chain Rule where necessary.

Differentiate both sides of the equation $xy + y^2 = \tan x + y$ with respect to $x$:

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

Using the sum rule on both sides:

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

For the term $\frac{d}{dx}(xy)$, we use the Product Rule, treating $x$ as a function of $x$ and $y$ as a function of $x$. $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=x$ and $v=y$. $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \frac{dy}{dx}$.

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

For the term $\frac{d}{dx}(y^2)$, we use the Chain Rule. $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$.

For the term $\frac{d}{dx}(\tan x)$, the derivative is $\sec^2 x$.

For the term $\frac{d}{dx}(y)$, the derivative is $\frac{dy}{dx}$.

Substituting these derivatives back into the equation:

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

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

Now, we need to solve for $\frac{dy}{dx}$. Gather all terms containing $\frac{dy}{dx}$ on one side and other terms on the other side. Let's move $\frac{dy}{dx}$ terms to the left side and the $y$ term to the right side:

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

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

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

Divide both sides by $(x + 2y - 1)$, assuming this expression is not zero:

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

Conclusion:

If $xy + y^2 = \tan x + y$, then $\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}$, provided $x + 2y - 1 \neq 0$.

Given:

The equation $x^2 + xy + y^2 = 100$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

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

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the Chain Rule where necessary.

Differentiate both sides of the equation $x^2 + xy + y^2 = 100$ with respect to $x$:

$\frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(100)$

Using the sum rule on the left side and the constant rule on the right side:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0$

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

For the term $\frac{d}{dx}(xy)$, we use the Product Rule, treating $x$ as a function of $x$ and $y$ as a function of $x$. $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=x$ and $v=y$. $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \frac{dy}{dx}$.

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

For the term $\frac{d}{dx}(y^2)$, we use the Chain Rule. Treat $y^2$ as a function of $y$, where $y$ is a function of $x$. The derivative of $u^2$ with respect to $u$ is $2u$. So, $\frac{d}{dx}(y^2) = \frac{d}{dy}(y^2) \cdot \frac{dy}{dx} = 2y \cdot \frac{dy}{dx}$.

Substituting these derivatives back into the equation:

$2x + (y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0$

$2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$

Now, we need to solve for $\frac{dy}{dx}$. Gather all terms containing $\frac{dy}{dx}$ on one side and other terms on the other side. Move the terms without $\frac{dy}{dx}$ to the right side:

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

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

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

Divide both sides by $(x + 2y)$, assuming this expression is not zero:

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

We can factor out $-1$ from the numerator:

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

Conclusion:

If $x^2 + xy + y^2 = 100$, then $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$, provided $x + 2y \neq 0$.

Given:

The equation $x^3 + x^2y + xy^2 + y^3 = 81$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

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

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the given equation with respect to $x$, treating $y$ as a function of $x$ and applying the Chain Rule and Product Rule where necessary.

Differentiate both sides of the equation $x^3 + x^2y + xy^2 + y^3 = 81$ with respect to $x$:

$\frac{d}{dx}(x^3 + x^2y + xy^2 + y^3) = \frac{d}{dx}(81)$

Using the sum rule on the left side and the constant rule on the right side:

$\frac{d}{dx}(x^3) + \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) + \frac{d}{dx}(y^3) = 0$

Differentiate each term:

$\frac{d}{dx}(x^3) = 3x^2$ (using the power rule)

$\frac{d}{dx}(x^2y)$: Use the Product Rule $\frac{d}{dx}(uv) = u'v + uv'$ with $u=x^2$ and $v=y$.

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

$\frac{d}{dx}(xy^2)$: Use the Product Rule $\frac{d}{dx}(uv) = u'v + uv'$ with $u=x$ and $v=y^2$. Note that $\frac{d}{dx}(y^2)$ requires the Chain Rule.

$\frac{d}{dx}(y^2) = \frac{d}{dy}(y^2) \cdot \frac{dy}{dx} = 2y \cdot \frac{dy}{dx}$

So, $\frac{d}{dx}(xy^2) = \frac{d}{dx}(x) \cdot y^2 + x \cdot \frac{d}{dx}(y^2) = 1 \cdot y^2 + x \cdot (2y \frac{dy}{dx}) = y^2 + 2xy \frac{dy}{dx}$

$\frac{d}{dx}(y^3)$: Use the Chain Rule. $\frac{d}{dx}(y^3) = \frac{d}{dy}(y^3) \cdot \frac{dy}{dx} = 3y^2 \cdot \frac{dy}{dx}$

Substitute these derivatives back into the differentiated equation:

$3x^2 + (2xy + x^2 \frac{dy}{dx}) + (y^2 + 2xy \frac{dy}{dx}) + 3y^2 \frac{dy}{dx} = 0$

$3x^2 + 2xy + y^2 + x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0$

Now, group the terms containing $\frac{dy}{dx}$ and the terms that do not contain $\frac{dy}{dx}$. Move the terms without $\frac{dy}{dx}$ to the right side of the equation:

$x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = -3x^2 - 2xy - y^2$

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

$\frac{dy}{dx} (x^2 + 2xy + 3y^2) = -(3x^2 + 2xy + y^2)$

Finally, divide both sides by $(x^2 + 2xy + 3y^2)$ to solve for $\frac{dy}{dx}$, assuming that $x^2 + 2xy + 3y^2 \neq 0$:

$\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}$

Conclusion:

If $x^3 + x^2y + xy^2 + y^3 = 81$, then $\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}$, provided $x^2 + 2xy + 3y^2 \neq 0$.

Given:

The equation $\sin^2 y + \cos(xy) = \kappa$, where $\kappa$ is a constant.

We can write $\sin^2 y$ as $(\sin y)^2$.

So, the equation is $(\sin y)^2 + \cos(xy) = \kappa$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

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

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the given equation with respect to $x$, treating $y$ as a function of $x$ and applying the Chain Rule and Product Rule where necessary.

Differentiate both sides of the equation $(\sin y)^2 + \cos(xy) = \kappa$ with respect to $x$:

$\frac{d}{dx}((\sin y)^2 + \cos(xy)) = \frac{d}{dx}(\kappa)$

Using the sum rule on the left side and the constant rule on the right side:

$\frac{d}{dx}((\sin y)^2) + \frac{d}{dx}(\cos(xy)) = 0$

Differentiate each term:

$\frac{d}{dx}((\sin y)^2)$: Use the Chain Rule twice. Outer function is $u^2$, inner function is $\sin y$. $\frac{d}{du}(u^2) = 2u$. $\frac{d}{dx}(\sin y) = \cos y \frac{dy}{dx}$.

$\frac{d}{dx}((\sin y)^2) = 2(\sin y) \cdot \frac{d}{dx}(\sin y) = 2\sin y \cdot (\cos y \frac{dy}{dx}) = 2\sin y \cos y \frac{dy}{dx}$

This can also be written as $\sin(2y) \frac{dy}{dx}$.

$\frac{d}{dx}(\cos(xy))$: Use the Chain Rule with outer function $\cos u$ and inner function $u=xy$. Note that $\frac{d}{dx}(xy)$ requires the Product Rule.

$\frac{d}{du}(\cos u) = -\sin u$.

$\frac{d}{dx}(xy) = y + x \frac{dy}{dx}$ (from previous questions)

So, $\frac{d}{dx}(\cos(xy)) = \frac{d}{du}(\cos u)|_{u=xy} \cdot \frac{d}{dx}(xy) = (-\sin(xy)) \cdot (y + x \frac{dy}{dx})$

$= -y \sin(xy) - x \sin(xy) \frac{dy}{dx}$

Substitute these derivatives back into the differentiated equation:

$2\sin y \cos y \frac{dy}{dx} + (-y \sin(xy) - x \sin(xy) \frac{dy}{dx}) = 0$

$2\sin y \cos y \frac{dy}{dx} - y \sin(xy) - x \sin(xy) \frac{dy}{dx} = 0$

Now, group the terms containing $\frac{dy}{dx}$ and the terms that do not contain $\frac{dy}{dx}$. Move the term without $\frac{dy}{dx}$ to the right side:

$2\sin y \cos y \frac{dy}{dx} - x \sin(xy) \frac{dy}{dx} = y \sin(xy)$

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

$\frac{dy}{dx} (2\sin y \cos y - x \sin(xy)) = y \sin(xy)$

Using the identity $2\sin y \cos y = \sin(2y)$:

$\frac{dy}{dx} (\sin(2y) - x \sin(xy)) = y \sin(xy)$

Finally, divide both sides by $(\sin(2y) - x \sin(xy))$ to solve for $\frac{dy}{dx}$, assuming that $\sin(2y) - x \sin(xy) \neq 0$:

$\frac{dy}{dx} = \frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}$

Conclusion:

If $\sin^2 y + \cos xy = \kappa$, then $\frac{dy}{dx} = \frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}$, provided $\sin(2y) - x \sin(xy) \neq 0$.

Given:

The equation $\sin^2 x + \cos^2 y = 1$.

We can rewrite this as $(\sin x)^2 + (\cos y)^2 = 1$.

Here, $y$ is defined implicitly as a function of $x$.

To Find:

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

Solution:

We will use implicit differentiation to find $\frac{dy}{dx}$. We differentiate both sides of the given equation with respect to $x$, treating $y$ as a function of $x$ and applying the Chain Rule where necessary.

Differentiate both sides of the equation $(\sin x)^2 + (\cos y)^2 = 1$ with respect to $x$:

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

Using the sum rule on the left side and the constant rule on the right side:

$\frac{d}{dx}((\sin x)^2) + \frac{d}{dx}((\cos y)^2) = 0$

Differentiate each term using the Chain Rule:

For the term $\frac{d}{dx}((\sin x)^2)$: The outer function is $u^2$ and the inner function is $u = \sin x$.

$\frac{d}{dx}((\sin x)^2) = 2(\sin x) \cdot \frac{d}{dx}(\sin x) = 2 \sin x (\cos x)$

Using the double angle identity $2 \sin x \cos x = \sin(2x)$:

$\frac{d}{dx}((\sin x)^2) = \sin(2x)$

For the term $\frac{d}{dx}((\cos y)^2)$: The outer function is $v^2$ and the inner function is $v = \cos y$. Since $y$ is a function of $x$, $\cos y$ is a function of $x$. We apply the Chain Rule, and then the Chain Rule again for $\frac{d}{dx}(\cos y)$.

$\frac{d}{dx}((\cos y)^2) = 2(\cos y) \cdot \frac{d}{dx}(\cos y)$

$= 2 \cos y \cdot \left(\frac{d}{dy}(\cos y) \cdot \frac{dy}{dx}\right)$

$= 2 \cos y \cdot \left((-\sin y) \cdot \frac{dy}{dx}\right)$

$= -2 \sin y \cos y \frac{dy}{dx}$

Using the double angle identity $2 \sin y \cos y = \sin(2y)$:

$\frac{d}{dx}((\cos y)^2) = -\sin(2y) \frac{dy}{dx}$

Substitute these derivatives back into the differentiated equation:

$\sin(2x) + (-\sin(2y) \frac{dy}{dx}) = 0$

$\sin(2x) - \sin(2y) \frac{dy}{dx} = 0$

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

$\sin(2x) = \sin(2y) \frac{dy}{dx}$

Finally, divide both sides by $\sin(2y)$ to solve for $\frac{dy}{dx}$, assuming that $\sin(2y) \neq 0$:

$\frac{dy}{dx} = \frac{\sin(2x)}{\sin(2y)}$

The condition $\sin(2y) \neq 0$ means $2y \neq n\pi$, where $n$ is an integer. This implies $y \neq \frac{n\pi}{2}$ for any integer $n$.

Note: The original equation $\sin^2 x + \cos^2 y = 1$ is not an identity that holds for all $x, y$. For example, if $x=\pi/4$, $\sin^2(\pi/4) = (1/\sqrt{2})^2 = 1/2$. The equation becomes $1/2 + \cos^2 y = 1$, so $\cos^2 y = 1/2$. This means $\cos y = \pm 1/\sqrt{2}$, so $y = \frac{\pi}{4} + \frac{n\pi}{2}$ for integer $n$. The equation implicitly defines a relationship between $x$ and $y$.

Conclusion:

If $\sin^2 x + \cos^2 y = 1$, then $\frac{dy}{dx} = \frac{\sin(2x)}{\sin(2y)}$, provided $\sin(2y) \neq 0$.

Given:

The function $y = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$.

To Find:

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

Solution:

We can differentiate this function using the Chain Rule directly, or by using a trigonometric substitution to simplify the expression before differentiating.

Method 1: Using the Chain Rule Directly

The function is of the form $y = \sin^{-1}(u)$, where $u = \frac{2x}{1+x^2}$.

The Chain Rule states $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

We know that $\frac{d}{du}(\sin^{-1} u) = \frac{1}{\sqrt{1-u^2}}$. So, $\frac{dy}{du} = \frac{1}{\sqrt{1-u^2}}$.

Now, we need to find $\frac{du}{dx} = \frac{d}{dx}\left(\frac{2x}{1+x^2}\right)$. We use the Quotient Rule.

Let $p(x) = 2x$ and $q(x) = 1+x^2$. $p'(x) = 2$ and $q'(x) = 2x$.

$\frac{du}{dx} = \frac{p'(x)q(x) - p(x)q'(x)}{(q(x))^2} = \frac{(2)(1+x^2) - (2x)(2x)}{(1+x^2)^2}$

$= \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} = \frac{2(1 - x^2)}{(1+x^2)^2}$

Now, substitute $u = \frac{2x}{1+x^2}$ into $\frac{dy}{du} = \frac{1}{\sqrt{1-u^2}}$:

$\frac{dy}{du} = \frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}} = \frac{1}{\sqrt{1 - \frac{4x^2}{(1+x^2)^2}}} = \frac{1}{\sqrt{\frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2}}}$

$= \frac{1}{\sqrt{\frac{1 + 2x^2 + x^4 - 4x^2}{(1+x^2)^2}}} = \frac{1}{\sqrt{\frac{1 - 2x^2 + x^4}{(1+x^2)^2}}} = \frac{1}{\sqrt{\frac{(1 - x^2)^2}{(1+x^2)^2}}} = \frac{1}{\left|\frac{1 - x^2}{1+x^2}\right|}$

$= \frac{|1+x^2|}{|1-x^2|}$

Since $1+x^2 \ge 1$ for real $x$, $|1+x^2| = 1+x^2$. So, $\frac{dy}{du} = \frac{1+x^2}{|1-x^2|}$.

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

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

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

This derivative depends on the sign of $1-x^2$.

If $|x| < 1$, then $1-x^2 > 0$, so $|1-x^2| = 1-x^2$.

$\frac{dy}{dx} = \frac{2(1 - x^2)}{(1 - x^2)(1+x^2)} = \frac{2}{1+x^2}$ for $|x| < 1$.

If $|x| > 1$, then $1-x^2 < 0$, so $|1-x^2| = -(1-x^2) = x^2-1$.

$\frac{dy}{dx} = \frac{2(1 - x^2)}{-(1 - x^2)(1+x^2)} = \frac{2}{-(1+x^2)} = -\frac{2}{1+x^2}$ for $|x| > 1$.

At $|x|=1$ (i.e., $x=1$ or $x=-1$), $1-x^2 = 0$, and the derivative is undefined. These are points where the simplified function has a change in derivative.

The domain of $\sin^{-1}(u)$ is $[-1, 1]$. So we need $-1 \le \frac{2x}{1+x^2} \le 1$. This inequality holds for all real $x$.

However, the principal value range of $\sin^{-1}$ is $[-\pi/2, \pi/2]$. The substitution $x = \tan \theta$ where $\theta \in (-\pi/2, \pi/2)$ gives $\frac{2 \tan \theta}{1+\tan^2 \theta} = \frac{2 \tan \theta}{\sec^2 \theta} = 2 \tan \theta \cos^2 \theta = 2 \frac{\sin \theta}{\cos \theta} \cos^2 \theta = 2 \sin \theta \cos \theta = \sin(2\theta)$.

So $y = \sin^{-1}(\sin(2\theta))$. For $-\pi/4 \le \theta \le \pi/4$, $\sin^{-1}(\sin(2\theta)) = 2\theta$. This corresponds to $-1 \le x \le 1$.

Method 2: Using Trigonometric Substitution

Let $x = \tan \theta$, where $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Then $dx = \sec^2 \theta \, d\theta$.

$y = \sin^{-1} \left( \frac{2 \tan \theta}{1+\tan^2 \theta} \right) = \sin^{-1} \left( \frac{2 \tan \theta}{\sec^2 \theta} \right)$

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

$= \sin^{-1} (\sin(2\theta))$

Now, $y = \sin^{-1}(\sin(2\theta))$. The value of $\sin^{-1}(\sin u)$ is $u$ only when $u$ is in the principal value range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So, we need $2\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

$-\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2}$

$-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$

If $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$, then $y = 2\theta$. Since $x = \tan \theta$, $\theta = \tan^{-1} x$. For this range of $\theta$, $x = \tan \theta$ is between $\tan(-\pi/4) = -1$ and $\tan(\pi/4) = 1$. So, if $-1 \leq x \leq 1$, then $-\pi/4 \leq \theta \leq \pi/4$, and $y = 2\tan^{-1} x$.

Differentiating $y = 2\tan^{-1} x$ with respect to $x$ for $|x| < 1$:

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

This matches the result from Method 1 for $|x| < 1$.

If $\theta$ is outside $[-\frac{\pi}{4}, \frac{\pi}{4}]$, the derivative is different. For example, if $\frac{\pi}{4} < \theta < \frac{3\pi}{4}$, then $2\theta \in (\frac{\pi}{2}, \frac{3\pi}{2})$. In this range, $\sin^{-1}(\sin(2\theta)) = \pi - 2\theta$. This corresponds to $x = \tan \theta$ where $\theta \in (\pi/4, 3\pi/4)$, i.e., $x \in (-\infty, -1) \cup (1, \infty)$. For $\theta \in (\pi/4, \pi/2)$, $x \in (1, \infty)$ and $y = \pi - 2\theta = \pi - 2\tan^{-1} x$. $\frac{dy}{dx} = -2 \cdot \frac{1}{1+x^2}$. For $\theta \in (\pi/2, 3\pi/4)$, $x \in (-\infty, -1)$ and $y = \pi - 2\theta = \pi - 2\tan^{-1} x$. $\frac{dy}{dx} = -2 \cdot \frac{1}{1+x^2}$.

So, the derivative is $\frac{2}{1+x^2}$ for $|x| < 1$ and $-\frac{2}{1+x^2}$ for $|x| > 1$. At $x=\pm 1$, the derivative does not exist.

However, typically in such problems in this context, the derivative is asked for the principal branch where $|x| < 1$. If the question assumes the principal value and existence of the derivative, the simpler case is intended.

Conclusion:

Assuming the question refers to the principal value branch where $|x| < 1$, the derivative of $y = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$ is $\frac{dy}{dx} = \frac{2}{1+x^2}$.

Given:

The function $y = \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$, with the condition $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$.

To Find:

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

Solution:

The expression inside the inverse tangent function suggests using a trigonometric substitution. Recall the triple angle formula for tangent: $\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$.

Let $x = \tan \theta$. Since $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$, this implies:

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

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

Substitute $x = \tan \theta$ into the expression:

$\frac{3x - x^3}{1 - 3x^2} = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan(3\theta)$

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

The value of $\tan^{-1}(\tan u)$ is equal to $u$ if $u$ is in the principal value range of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Given the range of $\theta$: $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$.

Multiply by 3: $-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$.

Since $3\theta$ is within the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have:

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

Now, substitute back $\theta = \tan^{-1} x$:

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

Now we can differentiate $y$ with respect to $x$:

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

Using the constant multiple rule:

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

We know that $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$.

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

This derivative is valid for the given range of $x$, which ensures that the substitution $y = 3\theta$ is correct.

Conclusion:

The derivative of $y = \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$ for $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$ is $\frac{dy}{dx} = \frac{3}{1+x^2}$.

Given:

The function $y = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$, with the condition $0 < x < 1$.

To Find:

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

Solution:

The expression inside the inverse cosine function suggests using a trigonometric substitution related to $\tan \theta$. Recall the double angle formula for cosine in terms of tangent: $\cos(2\theta) = \frac{1 - \tan^2\theta}{1 + \tan^2\theta}$.

Let $x = \tan \theta$. Since $0 < x < 1$, this implies:

$\tan^{-1}(0) < \theta < \tan^{-1}(1)$

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

Substitute $x = \tan \theta$ into the expression:

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

So, $y = \cos^{-1}(\cos(2\theta))$.

The value of $\cos^{-1}(\cos u)$ is equal to $u$ if $u$ is in the principal value range of $\cos^{-1}$, which is $[0, \pi]$.

Given the range of $\theta$: $0 < \theta < \frac{\pi}{4}$.

Multiply by 2: $0 < 2\theta < \frac{\pi}{2}$.

Since $2\theta$ is within the interval $[0, \pi]$ (specifically, it is in $(0, \pi/2)$), we have:

$y = \cos^{-1}(\cos(2\theta)) = 2\theta$

Now, substitute back $\theta = \tan^{-1} x$. For $0 < x < 1$, the corresponding $\theta$ is in $(0, \pi/4)$, so $y = 2\tan^{-1} x$.

$y = 2\tan^{-1} x$

Now we can differentiate $y$ with respect to $x$:

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

Using the constant multiple rule:

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

We know that $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$.

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

This derivative is valid for the given range of $x$, which ensures that the substitution $y = 2\theta$ is correct.

Conclusion:

The derivative of $y = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$ for $0 < x < 1$ is $\frac{dy}{dx} = \frac{2}{1+x^2}$.

Given:

The function $y = \sin^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$, with the condition $0 < x < 1$.

To Find:

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

Solution:

The expression inside the inverse sine function, $\frac{1 - x^2}{1 + x^2}$, is the formula for $\cos(2\theta)$ when $x = \tan \theta$. Let's use this substitution.

Let $x = \tan \theta$. Since $0 < x < 1$, this implies:

$\tan^{-1}(0) < \theta < \tan^{-1}(1)$

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

Substitute $x = \tan \theta$ into the expression:

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

So, $y = \sin^{-1}(\cos(2\theta))$.

We can use the identity $\cos(u) = \sin(\frac{\pi}{2} - u)$ to express $\cos(2\theta)$ in terms of sine:

$y = \sin^{-1}(\sin(\frac{\pi}{2} - 2\theta))$

The value of $\sin^{-1}(\sin u)$ is equal to $u$ if $u$ is in the principal value range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Given the range of $\theta$: $0 < \theta < \frac{\pi}{4}$.

Multiply by $-2$: $-\frac{\pi}{2} < -2\theta < 0$.

Add $\frac{\pi}{2}$: $\frac{\pi}{2} - \frac{\pi}{2} < \frac{\pi}{2} - 2\theta < \frac{\pi}{2} + 0$.

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

Since $\frac{\pi}{2} - 2\theta$ is within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ (specifically, it is in $(0, \pi/2)$), we have:

$y = \frac{\pi}{2} - 2\theta$

Now, substitute back $\theta = \tan^{-1} x$. For $0 < x < 1$, the corresponding $\theta$ is in $(0, \pi/4)$.

$y = \frac{\pi}{2} - 2\tan^{-1} x$

Now we can differentiate $y$ with respect to $x$:

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

Using the difference rule and constant multiple rule:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) - \frac{d}{dx}(2\tan^{-1} x) = 0 - 2 \frac{d}{dx}(\tan^{-1} x)$

We know that $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$.

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

This derivative is valid for the given range of $x$, which ensures that the substitution $y = \frac{\pi}{2} - 2\theta$ is correct.

Conclusion:

The derivative of $y = \sin^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$ for $0 < x < 1$ is $\frac{dy}{dx} = -\frac{2}{1+x^2}$.

Given:

The function $y = \cos^{-1} \left( \frac{2x}{1 + x^2} \right)$, with the condition $-1 < x < 1$.

To Find:

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

Solution:

The expression inside the inverse cosine function, $\frac{2x}{1 + x^2}$, is related to trigonometric identities. Recall the double angle formula for sine in terms of tangent: $\sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta}$.

Let $x = \tan \theta$. Since $-1 < x < 1$, this implies:

$\tan^{-1}(-1) < \theta < \tan^{-1}(1)$

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

Substitute $x = \tan \theta$ into the expression:

$\frac{2x}{1 + x^2} = \frac{2\tan\theta}{1 + \tan^2\theta} = \sin(2\theta)$

So, $y = \cos^{-1}(\sin(2\theta))$.

We can use the identity $\sin(u) = \cos(\frac{\pi}{2} - u)$ to express $\sin(2\theta)$ in terms of cosine:

$y = \cos^{-1}(\cos(\frac{\pi}{2} - 2\theta))$

The value of $\cos^{-1}(\cos u)$ is equal to $u$ if $u$ is in the principal value range of $\cos^{-1}$, which is $[0, \pi]$.

Given the range of $\theta$: $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.

Multiply by $-2$: $-\frac{\pi}{2} < -2\theta < \frac{\pi}{2}$. (The order of inequality reverses when multiplying by a negative number)

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

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

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

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

Since $\frac{\pi}{2} - 2\theta$ is within the interval $[0, \pi]$ (specifically, it is in $(0, \pi)$), we have:

$y = \frac{\pi}{2} - 2\theta$

Now, substitute back $\theta = \tan^{-1} x$. For $-1 < x < 1$, the corresponding $\theta$ is in $(-\pi/4, \pi/4)$, so $y = \frac{\pi}{2} - 2\tan^{-1} x$.

$y = \frac{\pi}{2} - 2\tan^{-1} x$

Now we can differentiate $y$ with respect to $x$:

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

Using the difference rule and constant multiple rule:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) - \frac{d}{dx}(2\tan^{-1} x) = 0 - 2 \frac{d}{dx}(\tan^{-1} x)$

We know that $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$.

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

This derivative is valid for the given range of $x$, which ensures that the substitution $y = \frac{\pi}{2} - 2\theta$ is correct.

Conclusion:

The derivative of $y = \cos^{-1} \left( \frac{2x}{1 + x^2} \right)$ for $-1 < x < 1$ is $\frac{dy}{dx} = -\frac{2}{1+x^2}$.

Given:

The function $y = \sin^{-1} \left( 2x \sqrt{1 - x^2} \right)$, with the condition $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$.

To Find:

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

Solution:

The expression inside the inverse sine function, $2x \sqrt{1 - x^2}$, suggests using a trigonometric substitution involving $\sin \theta$ or $\cos \theta$. Recall the identity $\sin(2\theta) = 2\sin\theta \cos\theta$ and $\cos^2\theta = 1 - \sin^2\theta$, so $\cos\theta = \sqrt{1-\sin^2\theta}$ for $\theta \in [-\pi/2, \pi/2]$.

Let $x = \sin \theta$. Since $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$, this implies:

$\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) < \theta < \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$

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

For $\theta \in (-\pi/4, \pi/4)$, $\cos \theta = \sqrt{1-\sin^2\theta} > 0$. So $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} = |\cos\theta| = \cos\theta$.

Substitute $x = \sin \theta$ and $\sqrt{1-x^2} = \cos \theta$ into the expression:

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

So, $y = \sin^{-1}(\sin(2\theta))$.

The value of $\sin^{-1}(\sin u)$ is equal to $u$ if $u$ is in the principal value range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Given the range of $\theta$: $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.

Multiply by 2: $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$.

Since $2\theta$ is within the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have:

$y = \sin^{-1}(\sin(2\theta)) = 2\theta$

Now, substitute back $\theta = \sin^{-1} x$. For $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$, the corresponding $\theta$ is in $(-\pi/4, \pi/4)$.

$y = 2\sin^{-1} x$

Now we can differentiate $y$ with respect to $x$:

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

Using the constant multiple rule:

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

We know that $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$.

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

This derivative is valid for the given range of $x$, which ensures that the substitution $y = 2\theta$ is correct and $\sqrt{1-x^2} = \cos\theta > 0$.

Conclusion:

The derivative of $y = \sin^{-1} \left( 2x \sqrt{1 - x^2} \right)$ for $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$ is $\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$.

Given:

The function is $y = \sec^{-1}\left( \frac{1}{2x^2 - 1} \right)$.

The domain for $x$ is $0 < x < \frac{1}{\sqrt{2}}$.

Solution:

We are asked to find the derivative of the function $y = \sec^{-1}\left( \frac{1}{2x^2 - 1} \right)$ with respect to $x$, given the domain $0 < x < \frac{1}{\sqrt{2}}$.

Let's simplify the expression inside the inverse secant function using a substitution.

Consider the substitution $x = \cos\theta$.

Since $0 < x < \frac{1}{\sqrt{2}}$, we have $0 < \cos\theta < \frac{1}{\sqrt{2}}$.

For the interval $0 < x < \frac{1}{\sqrt{2}}$, where $x = \cos\theta$, the corresponding range for $\theta$ in $[0, \pi]$ is $\frac{\pi}{4} < \theta < \frac{\pi}{2}$, because $\cos(\frac{\pi}{2}) = 0$ and $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, and the cosine function is decreasing in $[0, \pi]$.

From $x = \cos\theta$, we can write $\theta = \cos^{-1}x$.

Now substitute $x = \cos\theta$ into the expression $\frac{1}{2x^2 - 1}$:

$\frac{1}{2x^2 - 1} = \frac{1}{2\cos^2\theta - 1}$

Using the double angle identity for cosine, $\cos(2\theta) = 2\cos^2\theta - 1$, we get:

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

Since $\frac{1}{\cos(2\theta)} = \sec(2\theta)$, the original function becomes:

$y = \sec^{-1}(\sec(2\theta))$

We need to evaluate $\sec^{-1}(\sec(2\theta))$. The principal value branch of $\sec^{-1}u$ is $[0, \pi] \setminus \{\frac{\pi}{2}\}$.

We found earlier that $\frac{\pi}{4} < \theta < \frac{\pi}{2}$. Multiplying by 2, we get $\frac{\pi}{2} < 2\theta < \pi$.

The range of $2\theta$ is $(\frac{\pi}{2}, \pi)$. This range lies within the principal value branch of $\sec^{-1}u$ and does not include $\frac{\pi}{2}$.

Therefore, for $\frac{\pi}{2} < 2\theta < \pi$, we have $\sec^{-1}(\sec(2\theta)) = 2\theta$.

So, the function simplifies to:

$y = 2\theta$

Substitute back $\theta = \cos^{-1}x$:

$y = 2\cos^{-1}x$

Now, we differentiate $y$ with respect to $x$:

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

Using the constant multiple rule and the derivative of $\cos^{-1}x$, which is $\frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}$, we have:

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

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

Final Answer:

The derivative of the given function is $\frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}}$.

Given:

The equation $x = e^{\log x}$.

The assertion that this equation is true for all real $x$.

Solution:

We are asked to determine if the equation $x = e^{\log x}$ is true for all real values of $x$.

Let's consider the definition and domain of the function $\log x$. In mathematics, $\log x$ usually refers to the natural logarithm, which is the logarithm with base $e$. It is also often written as $\ln x$.

The natural logarithm function, $y = \log x$ (or $y = \ln x$), is defined as the inverse function of the exponential function $y = e^x$.

The domain of the exponential function $e^x$ is all real numbers ($\mathbb{R}$), and its range is the set of positive real numbers ($(0, \infty)$).

Since $\log x$ is the inverse of $e^x$, the domain of $\log x$ is the range of $e^x$, which is the set of positive real numbers. That is, $\log x$ is defined only for $x > 0$.

The property of logarithms states that for any positive base $b$ ($b \neq 1$) and any positive number $x$, $b^{\log_b x} = x$. When the base is $e$, this property becomes $e^{\log x} = x$, where $\log x$ denotes $\log_e x$.

Now let's look at the given equation: $x = e^{\log x}$.

The left side of the equation, $x$, is defined for all real numbers.

The right side of the equation, $e^{\log x}$, is defined only when $\log x$ is defined. As discussed, $\log x$ is only defined for $x > 0$.

Therefore, the expression $e^{\log x}$ is only defined for $x > 0$.

For the equality $x = e^{\log x}$ to hold, both sides must be defined and equal. Since the right side is only defined for $x > 0$, the equality can only hold for $x > 0$.

For any $x \leq 0$, $\log x$ is undefined in the real number system, and thus $e^{\log x}$ is also undefined.

Hence, the equation $x = e^{\log x}$ is true only for $x > 0$. It is not true for all real $x$.

Conclusion:

It is false that $x = e^{\log x}$ for all real $x$. The equality is only true for $x > 0$.

(i) Differentiate $e^{-x}$ w.r.t. x

Given:

The function is $y = e^{-x}$.

Solution:

We need to find the derivative of $y = e^{-x}$ with respect to $x$.

Let $u = -x$. 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 $u$:

$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u$

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

$\frac{du}{dx} = \frac{d}{dx}(-x) = -1$

Now, multiply the results:

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

Final Answer:

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

(ii) Differentiate $\sin(\log x)$, x > 0 w.r.t. x

Given:

The function is $y = \sin(\log x)$, for $x > 0$.

Solution:

We need to find the derivative of $y = \sin(\log x)$ with respect to $x$.

Let $u = \log x$. Then $y = \sin u$.

Using the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

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

$\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u$

Next, find the derivative of $u$ with respect to $x$. Assuming $\log x$ is the natural logarithm (base $e$):

$\frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

Now, multiply the results:

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

Final Answer:

The derivative of $\sin(\log x)$ with respect to $x$ is $\frac{\cos(\log x)}{x}$.

(iii) Differentiate $\cos^{-1}(e^x)$ w.r.t. x

Given:

The function is $y = \cos^{-1}(e^x)$.

Solution:

We need to find the derivative of $y = \cos^{-1}(e^x)$ with respect to $x$.

Let $u = e^x$. Then $y = \cos^{-1} u$.

Using the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

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

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

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

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

Now, multiply the results:

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

Note: The domain of $\cos^{-1}(e^x)$ requires $-1 \leq e^x \leq 1$. Since $e^x > 0$, this simplifies to $0 < e^x \leq 1$, which implies $x \leq 0$. The derivative is valid for $x < 0$ (to avoid division by zero).

Final Answer:

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

(iv) Differentiate $e^{\cos x}$ w.r.t. x

Given:

The function is $y = e^{\cos x}$.

Solution:

We need to find the derivative of $y = e^{\cos x}$ with respect to $x$.

Let $u = \cos x$. 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 $u$:

$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u$

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

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

Now, multiply the results:

$\frac{dy}{dx} = e^u \cdot (-\sin x) = e^{\cos x} \cdot (-\sin x) = -\sin x \cdot e^{\cos x}$

Final Answer:

The derivative of $e^{\cos x}$ with respect to $x$ is $-\sin x \cdot e^{\cos x}$.

Given:

The function to differentiate is $y = \frac{e^x}{\sin x}$.

To Find:

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

Solution:

We need to find the derivative of $y = \frac{e^x}{\sin x}$ with respect to $x$. This function is in the form of a quotient $\frac{u}{v}$, where $u = e^x$ and $v = \sin x$.

We will use the quotient rule for differentiation, which states that if $y = \frac{u(x)}{v(x)}$, then $\frac{dy}{dx} = \frac{v(x) \frac{du}{dx} - u(x) \frac{dv}{dx}}{(v(x))^2}$.

Let $u = e^x$. The derivative of $u$ with respect to $x$ is:

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

Let $v = \sin x$. The derivative of $v$ with respect to $x$ is:

$\frac{dv}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the quotient rule:

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

Substitute the derivatives we found:

$\frac{dy}{dx} = \frac{(\sin x) \cdot (e^x) - (e^x) \cdot (\cos x)}{\sin^2 x}$

Factor out $e^x$ from the numerator:

$\frac{dy}{dx} = \frac{e^x (\sin x - \cos x)}{\sin^2 x}$

Final Answer:

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

Given:

The function to differentiate is $y = e^{\sin^{-1} x}$.

Note: The domain of $\sin^{-1} x$ is $[-1, 1]$. The function $e^u$ is defined for all real $u$. Thus, the domain of the given function is $[-1, 1]$. For the derivative to exist, we consider $x \in (-1, 1)$.

To Find:

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

Solution:

We need to find the derivative of $y = e^{\sin^{-1} x}$ with respect to $x$. We will use the chain rule.

Let $u = \sin^{-1} x$. Then the function becomes $y = e^u$.

According to the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

First, differentiate $y = e^u$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u$

Next, differentiate $u = \sin^{-1} x$ with respect to $x$:

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

Now, multiply these results to find $\frac{dy}{dx}$:

$\frac{dy}{dx} = e^u \cdot \frac{1}{\sqrt{1-x^2}}$

Substitute back $u = \sin^{-1} x$:

$\frac{dy}{dx} = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$

So, the derivative is:

$\frac{dy}{dx} = \frac{e^{\sin^{-1} x}}{\sqrt{1-x^2}}$

Final Answer:

The derivative of $e^{\sin^{-1} x}$ with respect to $x$ is $\frac{e^{\sin^{-1} x}}{\sqrt{1-x^2}}$.

Given:

The function to differentiate is $y = e^{x^3}$.

To Find:

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

Solution:

We need to find the derivative of $y = e^{x^3}$ with respect to $x$. We will use the chain rule.

The function is in the form $e^u$, where $u$ is a function of $x$. Let $u = x^3$.

Then the function is $y = e^u$.

According to the chain rule, the derivative $\frac{dy}{dx}$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

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

$\frac{dy}{du} = \frac{d}{du}(e^u)$

The derivative of $e^u$ with respect to $u$ is $e^u$ itself.

$\frac{dy}{du} = e^u$

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

$\frac{du}{dx} = \frac{d}{dx}(x^3)$

Using the power rule for differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$:

$\frac{du}{dx} = 3x^{3-1} = 3x^2$

Now, multiply $\frac{dy}{du}$ and $\frac{du}{dx}$ to get $\frac{dy}{dx}$:

$\frac{dy}{dx} = e^u \cdot 3x^2$

Substitute back $u = x^3$ into the expression:

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

We can write this as:

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

Final Answer:

The derivative of $e^{x^3}$ with respect to $x$ is $3x^2 e^{x^3}$.

Given:

The function to differentiate is $y = \sin(\tan^{-1} e^{-x})$.

To Find:

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

Solution:

We need to find the derivative of $y = \sin(\tan^{-1} e^{-x})$ with respect to $x$. We will use the chain rule.

Let's apply the chain rule step by step.

The outermost function is $\sin(u)$, where $u = \tan^{-1} e^{-x}$.

The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$.

So, $\frac{dy}{du} = \cos(\tan^{-1} e^{-x})$.

Now we need to differentiate $u = \tan^{-1} e^{-x}$ with respect to $x$. This is of the form $\tan^{-1}(v)$, where $v = e^{-x}$.

The derivative of $\tan^{-1}(v)$ with respect to $v$ is $\frac{1}{1+v^2}$.

So, $\frac{du}{dv} = \frac{1}{1+(e^{-x})^2} = \frac{1}{1+e^{-2x}}$.

Next, we need to differentiate $v = e^{-x}$ with respect to $x$. This is of the form $e^w$, where $w = -x$.

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

So, $\frac{dv}{dw} = e^{-x}$.

Finally, we differentiate $w = -x$ with respect to $x$.

$\frac{dw}{dx} = \frac{d}{dx}(-x) = -1$.

Now, combining these using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}$

$\frac{dy}{dx} = \cos(\tan^{-1} e^{-x}) \cdot \frac{1}{1+e^{-2x}} \cdot e^{-x} \cdot (-1)$

Multiplying these terms gives:

$\frac{dy}{dx} = -\frac{e^{-x} \cos(\tan^{-1} e^{-x})}{1+e^{-2x}}$

Final Answer:

The derivative of $\sin(\tan^{-1} e^{-x})$ with respect to $x$ is $-\frac{e^{-x} \cos(\tan^{-1} e^{-x})}{1+e^{-2x}}$.

Given:

The function to differentiate is $y = \log(\cos e^x)$.

Note: For the logarithm to be defined, its argument must be positive. Thus, $\cos e^x > 0$. Also, $e^x$ is defined for all real $x$. The domain of $\cos(u)$ is all real numbers. So, the domain of the function is the set of $x$ such that $\cos(e^x) > 0$.

To Find:

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

Solution:

We need to find the derivative of $y = \log(\cos e^x)$ with respect to $x$. We will use the chain rule.

The function is a composition of functions. Let's break it down:

Let $u = \cos e^x$. Then $y = \log u$.

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

The chain rule states that $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.

First, differentiate $y = \log u$ with respect to $u$. Assuming $\log u$ is the natural logarithm (base $e$):

$\frac{dy}{du} = \frac{d}{du}(\log u) = \frac{1}{u}$

Substitute $u = \cos e^x$ back:

$\frac{dy}{du} = \frac{1}{\cos e^x}$

Next, differentiate $u = \cos v$ with respect to $v$:

$\frac{du}{dv} = \frac{d}{dv}(\cos v) = -\sin v$

Substitute $v = e^x$ back:

$\frac{du}{dv} = -\sin e^x$

Finally, differentiate $v = e^x$ with respect to $x$:

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

Now, multiply the results together:

$\frac{dy}{dx} = \frac{1}{\cos e^x} \cdot (-\sin e^x) \cdot e^x$

Simplify the expression:

$\frac{dy}{dx} = -\frac{\sin e^x}{\cos e^x} \cdot e^x$

Recall that $\frac{\sin \theta}{\cos \theta} = \tan \theta$. So, $\frac{\sin e^x}{\cos e^x} = \tan e^x$.

$\frac{dy}{dx} = -\tan(e^x) \cdot e^x$

Rearranging the terms:

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

Final Answer:

The derivative of $\log(\cos e^x)$ with respect to $x$ is $-e^x \tan(e^x)$.

Given:

The function to differentiate is $y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}$.

To Find:

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

Solution:

We need to find the derivative of the function $y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}$ with respect to $x$.

We can differentiate each term of the sum separately:

$\frac{dy}{dx} = \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{x^2}) + \frac{d}{dx}(e^{x^3}) + \frac{d}{dx}(e^{x^4}) + \frac{d}{dx}(e^{x^5})$

We use the chain rule for differentiation, which states that $\frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x)$.

For the first term, $\frac{d}{dx}(e^x)$: Here $f(x) = x$, so $f'(x) = \frac{d}{dx}(x) = 1$.

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

For the second term, $\frac{d}{dx}(e^{x^2})$: Here $f(x) = x^2$, so $f'(x) = \frac{d}{dx}(x^2) = 2x$.

$\frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot 2x = 2xe^{x^2}$

For the third term, $\frac{d}{dx}(e^{x^3})$: Here $f(x) = x^3$, so $f'(x) = \frac{d}{dx}(x^3) = 3x^2$.

$\frac{d}{dx}(e^{x^3}) = e^{x^3} \cdot 3x^2 = 3x^2e^{x^3}$

For the fourth term, $\frac{d}{dx}(e^{x^4})$: Here $f(x) = x^4$, so $f'(x) = \frac{d}{dx}(x^4) = 4x^3$.

$\frac{d}{dx}(e^{x^4}) = e^{x^4} \cdot 4x^3 = 4x^3e^{x^4}$

For the fifth term, $\frac{d}{dx}(e^{x^5})$: Here $f(x) = x^5$, so $f'(x) = \frac{d}{dx}(x^5) = 5x^4$.

$\frac{d}{dx}(e^{x^5}) = e^{x^5} \cdot 5x^4 = 5x^4e^{x^5}$

Now, sum the derivatives of all terms:

$\frac{dy}{dx} = e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x^3e^{x^4} + 5x^4e^{x^5}$

Final Answer:

The derivative of $e^x + e^{x^2} + … + e^{x^5}$ with respect to $x$ is $e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x^3e^{x^4} + 5x^4e^{x^5}$.

Given:

The function to differentiate is $y = \sqrt{e^{\sqrt{x}}}$.

The domain for $x$ is $x > 0$.

To Find:

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

Solution:

We need to find the derivative of $y = \sqrt{e^{\sqrt{x}}}$ with respect to $x$. We can rewrite the function using exponent notation:

$y = (e^{\sqrt{x}})^{1/2}$

$y = e^{\frac{1}{2}\sqrt{x}}$

Now we use the chain rule. The function is of the form $e^{f(x)}$, where $f(x) = \frac{1}{2}\sqrt{x}$.

The derivative of $e^{f(x)}$ is $e^{f(x)} \cdot f'(x)$.

So, $\frac{dy}{dx} = \frac{d}{dx}\left(e^{\frac{1}{2}\sqrt{x}}\right) = e^{\frac{1}{2}\sqrt{x}} \cdot \frac{d}{dx}\left(\frac{1}{2}\sqrt{x}\right)$.

Now we need to find the derivative of $\frac{1}{2}\sqrt{x}$.

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

We know that $\sqrt{x} = x^{1/2}$. Using the power rule, $\frac{d}{dx}(x^n) = nx^{n-1}$:

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

So, $\frac{d}{dx}\left(\frac{1}{2}\sqrt{x}\right) = \frac{1}{2} \cdot \left(\frac{1}{2\sqrt{x}}\right) = \frac{1}{4\sqrt{x}}$.

Now substitute this back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = e^{\frac{1}{2}\sqrt{x}} \cdot \frac{1}{4\sqrt{x}}$.

We can rewrite $e^{\frac{1}{2}\sqrt{x}}$ back as $\sqrt{e^{\sqrt{x}}}$.

$\frac{dy}{dx} = \sqrt{e^{\sqrt{x}}} \cdot \frac{1}{4\sqrt{x}}$.

This can also be written as a single fraction:

$\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$.

Final Answer:

The derivative of $\sqrt{e^{\sqrt{x}}}$ with respect to $x$ is $\frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$.

Given:

The function to differentiate is $y = \log(\log x)$.

The domain for $x$ is $x > 1$. This ensures that $\log x > 0$, so $\log(\log x)$ is defined.

To Find:

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

Solution:

We need to find the derivative of $y = \log(\log x)$ with respect to $x$. We will use the chain rule.

Let the outer function be $f(u) = \log u$, and the inner function be $u = \log x$. Then $y = f(u(x))$.

According to the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

First, we find the derivative of the outer function $y = \log u$ with respect to $u$. Assuming $\log u$ is the natural logarithm (base $e$):

$\frac{dy}{du} = \frac{d}{du}(\log u) = \frac{1}{u}$

Substitute back $u = \log x$ into this expression:

$\frac{dy}{du} = \frac{1}{\log x}$

Next, we find the derivative of the inner function $u = \log x$ with respect to $x$. Assuming $\log x$ is the natural logarithm (base $e$):

$\frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

Now, multiply $\frac{dy}{du}$ and $\frac{du}{dx}$ to find $\frac{dy}{dx}$:

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

Simplify the expression:

$\frac{dy}{dx} = \frac{1}{x \log x}$

Final Answer:

The derivative of $\log(\log x)$ with respect to $x$ is $\frac{1}{x \log x}$.

Given:

The function to differentiate is $y = \frac{\cos x}{\log x}$.

The domain for $x$ is $x > 0$. Note that for the function to be defined and the derivative to exist, we also require $\log x \neq 0$, which means $x \neq 1$. Thus, the domain for differentiation is $x > 0, x \neq 1$.

To Find:

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

Solution:

We need to find the derivative of $y = \frac{\cos x}{\log x}$ with respect to $x$. This function is in the form of a quotient $\frac{u}{v}$, where $u = \cos x$ and $v = \log x$. We assume $\log x$ refers to the natural logarithm (base $e$).

We will use the quotient rule for differentiation, which states that if $y = \frac{u(x)}{v(x)}$, then:

$\frac{dy}{dx} = \frac{v(x) \frac{du}{dx} - u(x) \frac{dv}{dx}}{(v(x))^2}$

Let $u = \cos x$. The derivative of $u$ with respect to $x$ is:

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

Let $v = \log x$. The derivative of $v$ with respect to $x$ is:

$\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

Now, apply the quotient rule:

$\frac{dy}{dx} = \frac{(\log x) \cdot \frac{d}{dx}(\cos x) - (\cos x) \cdot \frac{d}{dx}(\log x)}{(\log x)^2}$

Substitute the derivatives we found:

$\frac{dy}{dx} = \frac{(\log x) \cdot (-\sin x) - (\cos x) \cdot \left(\frac{1}{x}\right)}{(\log x)^2}$

Simplify the numerator:

$\frac{dy}{dx} = \frac{-\sin x \log x - \frac{\cos x}{x}}{(\log x)^2}$

To express the numerator as a single fraction, find a common denominator:

$\frac{dy}{dx} = \frac{\frac{-x \sin x \log x - \cos x}{x}}{(\log x)^2}$

Multiply the numerator by the reciprocal of the denominator $(\log x)^2 = \frac{(\log x)^2}{1}$:

$\frac{dy}{dx} = \frac{-x \sin x \log x - \cos x}{x} \cdot \frac{1}{(\log x)^2}$

$\frac{dy}{dx} = \frac{-(x \sin x \log x + \cos x)}{x(\log x)^2}$

Final Answer:

The derivative of $\frac{\cos x}{\log x}$ with respect to $x$ is $\frac{-(x \sin x \log x + \cos x)}{x(\log x)^2}$.

Given:

The function to differentiate is $y = \cos(\log x + e^x)$.

The domain for $x$ is $x > 0$. This ensures that $\log x$ is defined.

To Find:

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

Solution:

We need to find the derivative of $y = \cos(\log x + e^x)$ with respect to $x$. We will use the chain rule.

Let the outer function be $\cos(u)$, where $u$ is the argument of the cosine function.

Let $u = \log x + e^x$.

Then the function is $y = \cos u$.

According to the chain rule, the derivative $\frac{dy}{dx}$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

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

$\frac{dy}{du} = \frac{d}{du}(\cos u) = -\sin u$

Substitute back $u = \log x + e^x$:

$\frac{dy}{du} = -\sin(\log x + e^x)$

Next, we find the derivative of $u$ with respect to $x$. We differentiate each term in the sum $\log x + e^x$ separately:

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

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

Assuming $\log x$ is the natural logarithm (base $e$):

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

The derivative of $e^x$ is:

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

So, the derivative of $u$ with respect to $x$ is:

$\frac{du}{dx} = \frac{1}{x} + e^x$

Now, multiply $\frac{dy}{du}$ and $\frac{du}{dx}$ to find $\frac{dy}{dx}$:

$\frac{dy}{dx} = (-\sin(\log x + e^x)) \cdot \left(\frac{1}{x} + e^x\right)$

We can write the term $\left(\frac{1}{x} + e^x\right)$ with a common denominator:

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

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

$\frac{dy}{dx} = -\sin(\log x + e^x) \cdot \left(\frac{1 + xe^x}{x}\right)$

Rearranging the terms gives:

$\frac{dy}{dx} = -\frac{1 + xe^x}{x} \sin(\log x + e^x)$

Final Answer:

The derivative of $\cos(\log x + e^x)$ with respect to $x$ is $-\frac{1 + xe^x}{x} \sin(\log x + e^x)$.

Given:

The function to differentiate is $y = \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}}$.

To Find:

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

Solution:

We need to find the derivative of $y = \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}}$. This differentiation can be simplified by using logarithmic differentiation.

First, take the natural logarithm of both sides of the equation:

$\log y = \log \left( \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}} \right)$

Rewrite the square root as a power of $\frac{1}{2}$:

$\log y = \log \left( \left(\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}\right)^{1/2} \right)$

Using the logarithm property $\log(a^b) = b \log a$:

$\log y = \frac{1}{2} \log \left( \frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5} \right)$

Using the logarithm property $\log(\frac{a}{b}) = \log a - \log b$:

$\log y = \frac{1}{2} \left[ \log((x − 3) (x^2 + 4)) - \log(3x^2 + 4x + 5) \right]$

Using the logarithm property $\log(ab) = \log a + \log b$:

$\log y = \frac{1}{2} \left[ \log(x − 3) + \log(x^2 + 4) - \log(3x^2 + 4x + 5) \right]$

Now, differentiate both sides with respect to $x$. On the left side, use the chain rule $\frac{d}{dx}(\log y) = \frac{1}{y}\frac{dy}{dx}$.

On the right side, differentiate each term within the brackets:

$\frac{d}{dx}(\log y) = \frac{d}{dx} \left( \frac{1}{2} \left[ \log(x − 3) + \log(x^2 + 4) - \log(3x^2 + 4x + 5) \right] \right)$

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx}(\log(x − 3)) + \frac{d}{dx}(\log(x^2 + 4)) - \frac{d}{dx}(\log(3x^2 + 4x + 5)) \right]$

Differentiate each term using the chain rule $\frac{d}{dx}(\log f(x)) = \frac{f'(x)}{f(x)}$:

$\frac{d}{dx}(\log(x − 3)) = \frac{\frac{d}{dx}(x-3)}{x-3} = \frac{1}{x-3}$

$\frac{d}{dx}(\log(x^2 + 4)) = \frac{\frac{d}{dx}(x^2+4)}{x^2+4} = \frac{2x}{x^2+4}$

$\frac{d}{dx}(\log(3x^2 + 4x + 5)) = \frac{\frac{d}{dx}(3x^2+4x+5)}{3x^2+4x+5} = \frac{6x+4}{3x^2+4x+5}$

Substitute these derivatives back into the equation:

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]$

Finally, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \cdot \frac{1}{2} \left[ \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]$

Substitute the original expression for $y$ back into the equation:

$\frac{dy}{dx} = \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}} \cdot \frac{1}{2} \left[ \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]$

Final Answer:

The derivative of $\sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}}$ with respect to $x$ is:

$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x − 3) (x^2 + 4)}{3x^2 + 4x + 5}} \left( \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right)$

Given:

The function to differentiate is $y = a^x$, where $a$ is a positive constant.

To Find:

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

Solution:

We need to find the derivative of $y = a^x$ with respect to $x$. We can use logarithmic differentiation to solve this.

Take the natural logarithm of both sides of the equation:

$\log y = \log (a^x)$

Using the logarithm property $\log(b^c) = c \log b$:

$\log y = x \log a$

Here, $\log a$ is a constant since $a$ is a constant.

Now, differentiate both sides with respect to $x$. On the left side, we use the chain rule:

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

For the left side, let $u = y$. Then $\log y = \log u$. $\frac{d}{dx}(\log u) = \frac{d}{du}(\log u) \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx}$.

For the right side, since $\log a$ is a constant, we use the constant multiple rule:

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

The derivative of $x$ with respect to $x$ is 1:

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

So, the derivative of the right side is:

$\frac{d}{dx}(x \log a) = \log a \cdot 1 = \log a$

Equating the derivatives of both sides:

$\frac{1}{y} \frac{dy}{dx} = \log a$

Now, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \log a$

Substitute the original expression for $y$, which is $a^x$:

$\frac{dy}{dx} = a^x \log a$

Note that $\log a$ represents the natural logarithm of $a$, which is $\log_e a$ or $\ln a$. The formula can also be written as $\frac{d}{dx}(a^x) = a^x \ln a$.

Final Answer:

The derivative of $a^x$ with respect to $x$ is $a^x \log a$.

Given:

The function to differentiate is $y = x^{\sin x}$.

The domain for $x$ is $x > 0$.

To Find:

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

Solution:

We need to find the derivative of $y = x^{\sin x}$ with respect to $x$. This is a function of the form $f(x)^{g(x)}$, which can be differentiated using logarithmic differentiation.

Take the natural logarithm of both sides of the equation:

$\log y = \log (x^{\sin x})$

Using the logarithm property $\log(a^b) = b \log a$:

$\log y = \sin x \log x$

Now, differentiate both sides with respect to $x$. On the left side, we use the chain rule:

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

On the right side, we have a product of two functions, $\sin x$ and $\log x$. We use the product rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

Let $u = \sin x$ and $v = \log x$.

The derivative of $u$ with respect to $x$ is:

$\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$

The derivative of $v$ with respect to $x$ is:

$\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$

Now, apply the product rule to the right side of the equation $\log y = \sin x \log x$:

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

$\frac{d}{dx}(\sin x \log x) = (\sin x) \cdot \left(\frac{1}{x}\right) + (\log x) \cdot (\cos x)$

$\frac{d}{dx}(\sin x \log x) = \frac{\sin x}{x} + \cos x \log x$

Equating the derivatives of both sides:

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

Now, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \left( \frac{\sin x}{x} + \cos x \log x \right)$

Substitute the original expression for $y$, which is $x^{\sin x}$:

$\frac{dy}{dx} = x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right)$

Final Answer:

The derivative of $x^{\sin x}$ with respect to $x$ is $x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right)$.

Given:

The equation is $y^x + x^y + x^x = a^b$, where $a$ and $b$ are constants.

To Find:

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

Solution:

We are given the implicit relation $y^x + x^y + x^x = a^b$. To find $\frac{dy}{dx}$, we will differentiate both sides of the equation with respect to $x$.

Let $u = y^x$, $v = x^y$, and $w = x^x$. The equation becomes $u + v + w = a^b$.

Differentiating both sides with respect to $x$:

$\frac{d}{dx}(u + v + w) = \frac{d}{dx}(a^b)$

$\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0$

(Since $a^b$ is a constant, its derivative is 0).

Now we need to find the derivative of each term separately using logarithmic differentiation.

For $u = y^x$:

Take the natural logarithm of both sides:

$\log u = \log(y^x)$

$\log u = x \log y$

Differentiate both sides with respect to $x$ using the chain rule on the left and the product rule on the right:

$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x \log y)$

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

$\frac{du}{dx} = u \left( \log y + \frac{x}{y} \frac{dy}{dx} \right)$

Substitute back $u = y^x$:

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

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

For $v = x^y$:

Take the natural logarithm of both sides:

$\log v = \log(x^y)$

$\log v = y \log x$

Differentiate both sides with respect to $x$ using the chain rule on the left and the product rule on the right:

$\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(y \log x)$

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

$\frac{dv}{dx} = v \left( \log x \frac{dy}{dx} + \frac{y}{x} \right)$

Substitute back $v = x^y$:

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

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

For $w = x^x$:

Take the natural logarithm of both sides:

$\log w = \log(x^x)$

$\log w = x \log x$

Differentiate both sides with respect to $x$ using the chain rule on the left and the product rule on the right:

$\frac{1}{w} \frac{dw}{dx} = \frac{d}{dx}(x \log x)$

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

$\frac{1}{w} \frac{dw}{dx} = \log x + 1$

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

Substitute back $w = x^x$:

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

Now substitute $\frac{du}{dx}$, $\frac{dv}{dx}$, and $\frac{dw}{dx}$ back into the equation $\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0$:

$(y^x \log y + x y^{x-1} \frac{dy}{dx}) + (x^y \log x \frac{dy}{dx} + y x^{y-1}) + x^x (\log x + 1) = 0$

Group the terms containing $\frac{dy}{dx}$ and the terms without $\frac{dy}{dx}$:

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

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

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

Solve for $\frac{dy}{dx}$ by dividing by the coefficient of $\frac{dy}{dx}$:

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

Final Answer:

The derivative $\frac{dy}{dx}$ is $- \frac{y^x \log y + y x^{y-1} + x^x (\log x + 1)}{x y^{x-1} + x^y \log x}$.

Given:

Let the given function be denoted by $y$.

$y = \cos x \cdot \cos 2x \cdot \cos 3x$

Solution:

To differentiate the product of multiple functions, we can use logarithmic differentiation.

Taking natural logarithm on both sides, we get:

$\ln y = \ln (\cos x \cdot \cos 2x \cdot \cos 3x)$

Using the property of logarithms $\ln(abc) = \ln a + \ln b + \ln c$, we can write:

$\ln y = \ln(\cos x) + \ln(\cos 2x) + \ln(\cos 3x)$

Now, differentiate both sides with respect to $x$:

$\frac{d}{dx}(\ln y) = \frac{d}{dx}(\ln(\cos x)) + \frac{d}{dx}(\ln(\cos 2x)) + \frac{d}{dx}(\ln(\cos 3x))$

Using the chain rule $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$ and $\frac{d}{dx}(\cos u) = -\sin u \frac{du}{dx}$, we have:

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{\cos x} \frac{d}{dx}(\cos x) + \frac{1}{\cos 2x} \frac{d}{dx}(\cos 2x) + \frac{1}{\cos 3x} \frac{d}{dx}(\cos 3x)$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{\cos x} (-\sin x \cdot 1) + \frac{1}{\cos 2x} (-\sin 2x \cdot 2) + \frac{1}{\cos 3x} (-\sin 3x \cdot 3)$

Simplifying the terms, remembering that $\frac{\sin \theta}{\cos \theta} = \tan \theta$:

$\frac{1}{y} \frac{dy}{dx} = -\tan x - 2\tan 2x - 3\tan 3x$

Now, multiply both sides by $y$ to find $\frac{dy}{dx}$:

$\frac{dy}{dx} = y (-\tan x - 2\tan 2x - 3\tan 3x)$

Substitute the original expression for $y$ back into the equation:

$\frac{dy}{dx} = (\cos x \cdot \cos 2x \cdot \cos 3x) (-\tan x - 2\tan 2x - 3\tan 3x)$

We can factor out the negative sign:

$\frac{dy}{dx} = - \cos x \cdot \cos 2x \cdot \cos 3x (\tan x + 2\tan 2x + 3\tan 3x)$

Final Answer:

The derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = - \cos x \cdot \cos 2x \cdot \cos 3x (\tan x + 2\tan 2x + 3\tan 3x)$

Given:

Let the given function be denoted by $y$.

$y = \sqrt{\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}}$

We can rewrite the square root as a power of $\frac{1}{2}$.

$y = \left(\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}\right)^{1/2}$

Solution:

To differentiate this function, logarithmic differentiation is the most suitable method due to the product and quotient structure raised to a power.

Take the natural logarithm of both sides:

$\ln y = \ln \left[\left(\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}\right)^{1/2}\right]$

Using the logarithm property $\ln(a^n) = n \ln a$:

$\ln y = \frac{1}{2} \ln \left(\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}\right)$

Using the logarithm property $\ln(a/b) = \ln a - \ln b$:

$\ln y = \frac{1}{2} [\ln((x − 1) (x − 2)) - \ln((x − 3) (x − 4) (x − 5))]$

Using the logarithm property $\ln(ab) = \ln a + \ln b$:

$\ln y = \frac{1}{2} [\ln(x − 1) + \ln(x − 2) - (\ln(x − 3) + \ln(x − 4) + \ln(x − 5))]$

$\ln y = \frac{1}{2} [\ln(x − 1) + \ln(x − 2) - \ln(x − 3) - \ln(x − 4) - \ln(x − 5)]$

Now, differentiate both sides with respect to $x$. Remember that $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$ and $\frac{d}{dx}(x-c) = 1$ for any constant $c$.

$\frac{d}{dx}(\ln y) = \frac{d}{dx} \left(\frac{1}{2} [\ln(x − 1) + \ln(x − 2) - \ln(x − 3) - \ln(x − 4) - \ln(x − 5)]\right)$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x − 1} \cdot \frac{d}{dx}(x-1) + \frac{1}{x − 2} \cdot \frac{d}{dx}(x-2) - \frac{1}{x − 3} \cdot \frac{d}{dx}(x-3) - \frac{1}{x − 4} \cdot \frac{d}{dx}(x-4) - \frac{1}{x − 5} \cdot \frac{d}{dx}(x-5) \right]$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x − 1} \cdot 1 + \frac{1}{x − 2} \cdot 1 - \frac{1}{x − 3} \cdot 1 - \frac{1}{x − 4} \cdot 1 - \frac{1}{x − 5} \cdot 1 \right]$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x − 1} + \frac{1}{x − 2} - \frac{1}{x − 3} - \frac{1}{x − 4} - \frac{1}{x − 5} \right]$

Now, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \cdot \frac{1}{2} \left[ \frac{1}{x − 1} + \frac{1}{x − 2} - \frac{1}{x − 3} - \frac{1}{x − 4} - \frac{1}{x − 5} \right]$

Substitute the original expression for $y$ back into the equation:

$\frac{dy}{dx} = \sqrt{\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}} \cdot \frac{1}{2} \left[ \frac{1}{x − 1} + \frac{1}{x − 2} - \frac{1}{x − 3} - \frac{1}{x − 4} - \frac{1}{x − 5} \right]$

This can also be written as:

$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}} \left( \frac{1}{x − 1} + \frac{1}{x − 2} - \frac{1}{x − 3} - \frac{1}{x − 4} - \frac{1}{x − 5} \right)$

Final Answer:

The derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x − 1) (x − 2)}{(x − 3) (x − 4) (x − 5)}} \left( \frac{1}{x − 1} + \frac{1}{x − 2} - \frac{1}{x − 3} - \frac{1}{x − 4} - \frac{1}{x − 5} \right)$

Given:

Let the given function be denoted by $y$.

$y = (\log x)^{\cos x}$

We assume $\log x$ refers to the natural logarithm, $\ln x$.

Solution:

The function is in the form of $f(x)^{g(x)}$, so we use logarithmic differentiation.

Taking the natural logarithm on both sides:

$\ln y = \ln((\log x)^{\cos x})$

Using the logarithm property $\ln(a^b) = b \ln a$:

$\ln y = \cos x \cdot \ln(\log x)$

Now, differentiate both sides with respect to $x$. Use the chain rule on the left side and the product rule on the right side.

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

Left side: $\frac{1}{y} \frac{dy}{dx}$

Right side (using product rule $\frac{d}{dx}(uv) = u'v + uv'$, where $u = \cos x$ and $v = \ln(\log x)$):

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

$\frac{dv}{dx} = \frac{d}{dx}(\ln(\log x))$

Using the chain rule for $\frac{dv}{dx}$, let $w = \log x$. Then $\frac{dv}{dx} = \frac{d}{dw}(\ln w) \cdot \frac{dw}{dx}$.

$\frac{d}{dw}(\ln w) = \frac{1}{w} = \frac{1}{\log x}$

$\frac{dw}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$ (assuming natural logarithm)

So, $\frac{dv}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$.

Applying the product rule on the right side:

$\frac{d}{dx}(\cos x \cdot \ln(\log x)) = (-\sin x) \cdot \ln(\log x) + (\cos x) \cdot \left(\frac{1}{x \log x}\right)$

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

Equating the derivatives of both sides:

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

Now, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \left(-\sin x \ln(\log x) + \frac{\cos x}{x \log x}\right)$

Substitute the original expression for $y$ back into the equation:

$\frac{dy}{dx} = (\log x)^{\cos x} \left(-\sin x \ln(\log x) + \frac{\cos x}{x \log x}\right)$

Final Answer:

The derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = (\log x)^{\cos x} \left(-\sin x \ln(\log x) + \frac{\cos x}{x \log x}\right)$

Given:

Let the given function be denoted by $y$.

$y = x^x - 2^{\sin x}$

We can write this as $y = u - v$, where $u = x^x$ and $v = 2^{\sin x}$.

Solution:

We need to find $\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$.

Step 1: Differentiate $u = x^x$

This is of the form $f(x)^{g(x)}$, so we use logarithmic differentiation.

Take natural logarithm on both sides:

$\ln u = \ln (x^x)$

Using the property $\ln(a^b) = b \ln a$:

$\ln u = x \ln x$

Differentiate both sides with respect to $x$. Use the chain rule on the left and the product rule on the right.

$\frac{d}{dx}(\ln u) = \frac{d}{dx}(x \ln x)$

$\frac{1}{u} \frac{du}{dx} = (1) \cdot \ln x + x \cdot \left(\frac{1}{x}\right)$

$\frac{1}{u} \frac{du}{dx} = \ln x + 1$

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

$\frac{du}{dx} = u (\ln x + 1)$

Substitute $u = x^x$ back:

$\frac{du}{dx} = x^x (\ln x + 1)$

Step 2: Differentiate $v = 2^{\sin x}$

This is of the form $a^{f(x)}$. The derivative is $a^{f(x)} \ln a \cdot f'(x)$.

Here, $a=2$ and $f(x) = \sin x$. So $f'(x) = \frac{d}{dx}(\sin x) = \cos x$.

Using the formula:

$\frac{dv}{dx} = 2^{\sin x} \ln 2 \cdot \cos x$

Alternatively, using logarithmic differentiation:

$\ln v = \ln (2^{\sin x})$

$\ln v = \sin x \ln 2$

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(\ln v) = \frac{d}{dx}(\sin x \ln 2)$

$\frac{1}{v} \frac{dv}{dx} = \ln 2 \cdot \frac{d}{dx}(\sin x)$

$\frac{1}{v} \frac{dv}{dx} = \ln 2 \cdot \cos x$

$\frac{dv}{dx} = v (\ln 2 \cdot \cos x)$

Substitute $v = 2^{\sin x}$ back:

$\frac{dv}{dx} = 2^{\sin x} \ln 2 \cos x$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$

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

Final Answer:

The derivative of the given function with respect to $x$ is:

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

Given:

Let the given function be denoted by $y$.

$y = (x + 3)^2 (x + 4)^3 (x + 5)^4$

Solution:

Since the function is a product of terms raised to powers, logarithmic differentiation is a suitable method.

Taking the natural logarithm on both sides:

$\ln y = \ln[(x + 3)^2 (x + 4)^3 (x + 5)^4]$

Using the logarithm property $\ln(abc) = \ln a + \ln b + \ln c$:

$\ln y = \ln(x + 3)^2 + \ln(x + 4)^3 + \ln(x + 5)^4$

Using the logarithm property $\ln(a^n) = n \ln a$:

$\ln y = 2 \ln(x + 3) + 3 \ln(x + 4) + 4 \ln(x + 5)$

Now, differentiate both sides with respect to $x$. Remember that $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$ and $\frac{d}{dx}(x+c) = 1$ for any constant $c$.

$\frac{d}{dx}(\ln y) = \frac{d}{dx}[2 \ln(x + 3) + 3 \ln(x + 4) + 4 \ln(x + 5)]$

$\frac{1}{y} \frac{dy}{dx} = 2 \frac{d}{dx}(\ln(x + 3)) + 3 \frac{d}{dx}(\ln(x + 4)) + 4 \frac{d}{dx}(\ln(x + 5))$

$\frac{1}{y} \frac{dy}{dx} = 2 \left(\frac{1}{x + 3} \cdot \frac{d}{dx}(x+3)\right) + 3 \left(\frac{1}{x + 4} \cdot \frac{d}{dx}(x+4)\right) + 4 \left(\frac{1}{x + 5} \cdot \frac{d}{dx}(x+5)\right)$

$\frac{1}{y} \frac{dy}{dx} = 2 \left(\frac{1}{x + 3} \cdot 1\right) + 3 \left(\frac{1}{x + 4} \cdot 1\right) + 4 \left(\frac{1}{x + 5} \cdot 1\right)$

$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}$

Now, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \left( \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5} \right)$

Substitute the original expression for $y$ back into the equation:

$\frac{dy}{dx} = (x + 3)^2 (x + 4)^3 (x + 5)^4 \left( \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5} \right)$

Final Answer:

The derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = (x + 3)^2 (x + 4)^3 (x + 5)^4 \left( \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5} \right)$

Given:

Let the given function be denoted by $y$.

$y = \left( x + \frac{1}{x} \right)^x + x^{\left(1 + \frac{1}{x} \right)}$

We can write this as $y = u + v$, where $u = \left( x + \frac{1}{x} \right)^x$ and $v = x^{\left(1 + \frac{1}{x} \right)}$.

Solution:

To find $\frac{dy}{dx}$, we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ separately and then add them.

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = \left( x + \frac{1}{x} \right)^x$

This is of the form $f(x)^{g(x)}$, so we use logarithmic differentiation.

Taking natural logarithm on both sides:

$\ln u = \ln \left( \left( x + \frac{1}{x} \right)^x \right)$

Using the property $\ln(a^b) = b \ln a$:

$\ln u = x \ln \left( x + \frac{1}{x} \right)$

Differentiate both sides with respect to $x$ using the product rule on the right side:

$\frac{d}{dx}(\ln u) = \frac{d}{dx}(x \ln \left( x + \frac{1}{x} \right))$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{d}{dx} x\right) \ln \left( x + \frac{1}{x} \right) + x \left(\frac{d}{dx} \ln \left( x + \frac{1}{x} \right)\right)$

We know $\frac{d}{dx} x = 1$. For the second term, use the chain rule. Let $w = x + \frac{1}{x} = x + x^{-1}$.

$\frac{dw}{dx} = \frac{d}{dx}(x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2}$

$\frac{d}{dx}(\ln w) = \frac{1}{w} \frac{dw}{dx} = \frac{1}{x + \frac{1}{x}} \left(1 - \frac{1}{x^2}\right) = \frac{1}{\frac{x^2+1}{x}} \left(\frac{x^2-1}{x^2}\right) = \frac{x}{x^2+1} \cdot \frac{x^2-1}{x^2} = \frac{x^2-1}{x(x^2+1)}$

Substituting these back:

$\frac{1}{u} \frac{du}{dx} = 1 \cdot \ln \left( x + \frac{1}{x} \right) + x \cdot \frac{x^2-1}{x(x^2+1)}$

$\frac{1}{u} \frac{du}{dx} = \ln \left( x + \frac{1}{x} \right) + \frac{x^2-1}{x^2+1}$

Solving for $\frac{du}{dx}$:

$\frac{du}{dx} = u \left( \ln \left( x + \frac{1}{x} \right) + \frac{x^2-1}{x^2+1} \right)$

Substitute $u = \left( x + \frac{1}{x} \right)^x$ back:

$\frac{du}{dx} = \left( x + \frac{1}{x} \right)^x \left( \ln \left( x + \frac{1}{x} \right) + \frac{x^2-1}{x^2+1} \right)$

Step 2: Differentiate $v = x^{\left(1 + \frac{1}{x} \right)}$

This is also of the form $f(x)^{g(x)}$. Use logarithmic differentiation.

Taking natural logarithm on both sides:

$\ln v = \ln \left( x^{\left(1 + \frac{1}{x} \right)} \right)$

Using the property $\ln(a^b) = b \ln a$:

$\ln v = \left(1 + \frac{1}{x} \right) \ln x$

Differentiate both sides with respect to $x$ using the product rule on the right side:

$\frac{d}{dx}(\ln v) = \frac{d}{dx}\left(\left(1 + \frac{1}{x}\right) \ln x\right)$

$\frac{1}{v} \frac{dv}{dx} = \left(\frac{d}{dx}\left(1 + \frac{1}{x}\right)\right) \ln x + \left(1 + \frac{1}{x}\right) \left(\frac{d}{dx} \ln x\right)$

We know $\frac{d}{dx}\left(1 + \frac{1}{x}\right) = \frac{d}{dx}(1 + x^{-1}) = 0 - x^{-2} = -\frac{1}{x^2}$ and $\frac{d}{dx} \ln x = \frac{1}{x}$.

Substituting these back:

$\frac{1}{v} \frac{dv}{dx} = \left(-\frac{1}{x^2}\right) \ln x + \left(1 + \frac{1}{x}\right) \left(\frac{1}{x}\right)$

$\frac{1}{v} \frac{dv}{dx} = -\frac{\ln x}{x^2} + \frac{1}{x} + \frac{1}{x^2}$

Combining terms on the right side:

$\frac{1}{v} \frac{dv}{dx} = \frac{- \ln x + x + 1}{x^2} = \frac{x + 1 - \ln x}{x^2}$

Solving for $\frac{dv}{dx}$:

$\frac{dv}{dx} = v \left( \frac{x + 1 - \ln x}{x^2} \right)$

Substitute $v = x^{\left(1 + \frac{1}{x} \right)}$ back:

$\frac{dv}{dx} = x^{\left(1 + \frac{1}{x} \right)} \left( \frac{x + 1 - \ln x}{x^2} \right)$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

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

Final Answer:

The derivative of the given function with respect to $x$ is:

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

Given:

Let the given function be denoted by $y$.

$y = (\log x)^x + x^{\log x}$

We interpret $\log x$ as the natural logarithm, $\ln x$.

Solution:

We can write the function as $y = u + v$, where $u = (\ln x)^x$ and $v = x^{\ln x}$.

To find the derivative $\frac{dy}{dx}$, we find the derivatives of $u$ and $v$ separately and then add them:

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = (\ln x)^x$

This is of the form $f(x)^{g(x)}$. Use logarithmic differentiation.

Take the natural logarithm of both sides:

$\ln u = \ln((\ln x)^x)$

Using the property $\ln(a^b) = b \ln a$:

$\ln u = x \ln(\ln x)$

Differentiate both sides with respect to $x$ using the product rule on the right side:

$\frac{d}{dx}(\ln u) = \frac{d}{dx}(x \ln(\ln x))$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{d}{dx} x\right) \ln(\ln x) + x \left(\frac{d}{dx} \ln(\ln x)\right)$

We know $\frac{d}{dx} x = 1$. For the second term, use the chain rule. Let $w = \ln x$. Then $\frac{d}{dx}(\ln w) = \frac{1}{w} \frac{dw}{dx} = \frac{1}{\ln x} \frac{d}{dx}(\ln x)$.

Since $\frac{d}{dx}(\ln x) = \frac{1}{x}$, we have:

$\frac{d}{dx}(\ln(\ln x)) = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}$

Substitute these back:

$\frac{1}{u} \frac{du}{dx} = 1 \cdot \ln(\ln x) + x \cdot \left(\frac{1}{x \ln x}\right)$

$\frac{1}{u} \frac{du}{dx} = \ln(\ln x) + \frac{1}{\ln x}$

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

$\frac{du}{dx} = u \left( \ln(\ln x) + \frac{1}{\ln x} \right)$

Substitute $u = (\ln x)^x$ back:

$\frac{du}{dx} = (\ln x)^x \left( \ln(\ln x) + \frac{1}{\ln x} \right)$

Step 2: Differentiate $v = x^{\ln x}$

This is also of the form $f(x)^{g(x)}$. Use logarithmic differentiation.

Take the natural logarithm of both sides:

$\ln v = \ln(x^{\ln x})$

Using the property $\ln(a^b) = b \ln a$:

$\ln v = (\ln x) (\ln x) = (\ln x)^2$

Differentiate both sides with respect to $x$ using the chain rule on the right side:

$\frac{d}{dx}(\ln v) = \frac{d}{dx}((\ln x)^2)$

$\frac{1}{v} \frac{dv}{dx} = 2 (\ln x) \cdot \frac{d}{dx}(\ln x)$

$\frac{1}{v} \frac{dv}{dx} = 2 (\ln x) \cdot \frac{1}{x} = \frac{2 \ln x}{x}$

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

$\frac{dv}{dx} = v \left( \frac{2 \ln x}{x} \right)$

Substitute $v = x^{\ln x}$ back:

$\frac{dv}{dx} = x^{\ln x} \left( \frac{2 \ln x}{x} \right)$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

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

Final Answer:

The derivative of the given function with respect to $x$ is:

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

Given:

Let the given function be denoted by $y$.

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

We can express this as the sum of two functions, $u$ and $v$, such that $y = u + v$.

Let $u = (\sin x)^x$ and $v = \sin^{-1} \sqrt{x}$.

Solution:

To find the derivative $\frac{dy}{dx}$, we find the derivatives of $u$ and $v$ with respect to $x$ separately and then add them.

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = (\sin x)^x$

This function is in the form of $f(x)^{g(x)}$, so we use logarithmic differentiation.

Take the natural logarithm of both sides:

$\ln u = \ln((\sin x)^x)$

Using the logarithm property $\ln(a^b) = b \ln a$:

$\ln u = x \ln(\sin x)$

Differentiate both sides with respect to $x$. Use the product rule on the right side, $\frac{d}{dx}(ab) = a'b + ab'$, where $a=x$ and $b=\ln(\sin x)$.

$\frac{d}{dx}(\ln u) = \frac{d}{dx}(x \ln(\sin x))$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{d}{dx} x\right) \ln(\sin x) + x \left(\frac{d}{dx} \ln(\sin x)\right)$

We know $\frac{d}{dx} x = 1$. For the second term, use the chain rule: $\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)$.

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

So, $\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x$.

Substitute these back into the derivative equation for $\ln u$:

$\frac{1}{u} \frac{du}{dx} = 1 \cdot \ln(\sin x) + x \cdot (\cot x)$

$\frac{1}{u} \frac{du}{dx} = \ln(\sin x) + x \cot x$

Solve for $\frac{du}{dx}$ by multiplying both sides by $u$:

$\frac{du}{dx} = u (\ln(\sin x) + x \cot x)$

Substitute the expression for $u$ back:

$\frac{du}{dx} = (\sin x)^x (\ln(\sin x) + x \cot x)$

Step 2: Differentiate $v = \sin^{-1} \sqrt{x}$

This requires the chain rule. The derivative of $\sin^{-1} w$ with respect to $w$ is $\frac{1}{\sqrt{1 - w^2}}$. Here $w = \sqrt{x} = x^{1/2}$.

The derivative of $\sin^{-1} \sqrt{x}$ with respect to $x$ is $\frac{d}{dw}(\sin^{-1} w) \cdot \frac{dw}{dx}$, where $w = \sqrt{x}$.

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

The derivative of $\sin^{-1} w$ with $w = \sqrt{x}$ is $\frac{1}{\sqrt{1 - (\sqrt{x})^2}} = \frac{1}{\sqrt{1 - x}}$.

Combine these using the chain rule:

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

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

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Substitute the expressions for $\frac{du}{dx}$ and $\frac{dv}{dx}$:

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

Final Answer:

The derivative of the given function with respect to $x$ is:

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

Given:

Let the given function be denoted by $y$.

$y = x^{\sin x} + (\sin x)^{\cos x}$

We can write this as the sum of two functions, $u$ and $v$, such that $y = u + v$.

Let $u = x^{\sin x}$ and $v = (\sin x)^{\cos x}$.

Solution:

To find the derivative of $y$ with respect to $x$, we need to find the derivatives of $u$ and $v$ separately and then add them.

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = x^{\sin x}$

This is of the form $f(x)^{g(x)}$, so we use logarithmic differentiation.

Taking natural logarithm on both sides:

$\ln u = \ln(x^{\sin x})$

Using the property $\ln(a^b) = b \ln a$:

$\ln u = \sin x \ln x$

Differentiate both sides with respect to $x$ using the product rule on the right side:

$\frac{d}{dx}(\ln u) = \frac{d}{dx}(\sin x \ln x)$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{d}{dx} \sin x\right) \ln x + \sin x \left(\frac{d}{dx} \ln x\right)$

We know $\frac{d}{dx} \sin x = \cos x$ and $\frac{d}{dx} \ln x = \frac{1}{x}$.

Substitute these back:

$\frac{1}{u} \frac{du}{dx} = (\cos x) \ln x + \sin x \left(\frac{1}{x}\right)$

$\frac{1}{u} \frac{du}{dx} = \cos x \ln x + \frac{\sin x}{x}$

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

$\frac{du}{dx} = u \left(\cos x \ln x + \frac{\sin x}{x}\right)$

Substitute $u = x^{\sin x}$ back:

$\frac{du}{dx} = x^{\sin x} \left(\cos x \ln x + \frac{\sin x}{x}\right)$

Step 2: Differentiate $v = (\sin x)^{\cos x}$

This is also of the form $f(x)^{g(x)}$. Use logarithmic differentiation.

Taking natural logarithm on both sides:

$\ln v = \ln((\sin x)^{\cos x})$

Using the property $\ln(a^b) = b \ln a$:

$\ln v = \cos x \ln(\sin x)$

Differentiate both sides with respect to $x$ using the product rule on the right side:

$\frac{d}{dx}(\ln v) = \frac{d}{dx}(\cos x \ln(\sin x))$

$\frac{1}{v} \frac{dv}{dx} = \left(\frac{d}{dx} \cos x\right) \ln(\sin x) + \cos x \left(\frac{d}{dx} \ln(\sin x)\right)$

We know $\frac{d}{dx} \cos x = -\sin x$. For the second term, use the chain rule: $\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.

Substitute these back:

$\frac{1}{v} \frac{dv}{dx} = (-\sin x) \ln(\sin x) + \cos x (\cot x)$

$\frac{1}{v} \frac{dv}{dx} = -\sin x \ln(\sin x) + \cos x \frac{\cos x}{\sin x}$

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

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

$\frac{dv}{dx} = v \left(-\sin x \ln(\sin x) + \frac{\cos^2 x}{\sin x}\right)$

Substitute $v = (\sin x)^{\cos x}$ back:

$\frac{dv}{dx} = (\sin x)^{\cos x} \left(-\sin x \ln(\sin x) + \frac{\cos^2 x}{\sin x}\right)$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Substitute the expressions for $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$\frac{dy}{dx} = x^{\sin x} \left(\cos x \ln x + \frac{\sin x}{x}\right) + (\sin x)^{\cos x} \left(-\sin x \ln(\sin x) + \frac{\cos^2 x}{\sin x}\right)$

Final Answer:

The derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = x^{\sin x} \left(\cos x \ln x + \frac{\sin x}{x}\right) + (\sin x)^{\cos x} \left(-\sin x \ln(\sin x) + \frac{\cos^2 x}{\sin x}\right)$

Given:

Let the given function be denoted by $y$.

$y = x^{x \cos x} + \frac{x^2 + 1}{x^2 − 1}$

We can write this as the sum of two functions, $u$ and $v$, such that $y = u + v$.

Let $u = x^{x \cos x}$ and $v = \frac{x^2 + 1}{x^2 − 1}$.

Solution:

To find the derivative of $y$ with respect to $x$, we need to find the derivatives of $u$ and $v$ separately and then add them.

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = x^{x \cos x}$

This function is of the form $f(x)^{g(x)}$, so we use logarithmic differentiation.

Take the natural logarithm of both sides:

$\ln u = \ln(x^{x \cos x})$

Using the logarithm property $\ln(a^b) = b \ln a$:

$\ln u = (x \cos x) \ln x$

Differentiate both sides with respect to $x$. Use the product rule on the right side, $\frac{d}{dx}(AB) = A'B + AB'$, where $A = x \cos x$ and $B = \ln x$.

$\frac{d}{dx}(\ln u) = \frac{d}{dx}((x \cos x) \ln x)$

First, find the derivative of $A = x \cos x$ using the product rule:

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

The derivative of $B = \ln x$ is:

$\frac{dB}{dx} = \frac{d}{dx}(\ln x) = \frac{1}{x}$

Now, apply the product rule to $\ln u$:

$\frac{1}{u} \frac{du}{dx} = (\cos x - x \sin x) (\ln x) + (x \cos x) \left(\frac{1}{x}\right)$

$\frac{1}{u} \frac{du}{dx} = (\cos x - x \sin x) \ln x + \cos x$

Solve for $\frac{du}{dx}$ by multiplying both sides by $u$:

$\frac{du}{dx} = u \left((\cos x - x \sin x) \ln x + \cos x\right)$

Substitute the expression for $u$ back:

$\frac{du}{dx} = x^{x \cos x} \left((\cos x - x \sin x) \ln x + \cos x\right)$

Step 2: Differentiate $v = \frac{x^2 + 1}{x^2 − 1}$

This function is a quotient. Use the quotient rule: $\frac{d}{dx}\left(\frac{p}{q}\right) = \frac{p'q - pq'}{q^2}$.

Here, $p = x^2 + 1$ and $q = x^2 - 1$.

Find the derivatives of $p$ and $q$:

$p' = \frac{d}{dx}(x^2 + 1) = 2x$

$q' = \frac{d}{dx}(x^2 - 1) = 2x$

Apply the quotient rule:

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

Simplify the numerator:

Numerator = $2x(x^2 - 1) - 2x(x^2 + 1) = (2x^3 - 2x) - (2x^3 + 2x) = 2x^3 - 2x - 2x^3 - 2x = -4x$

So, the derivative of $v$ is:

$\frac{dv}{dx} = \frac{-4x}{(x^2 - 1)^2}$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Substitute the expressions for $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$\frac{dy}{dx} = x^{x \cos x} \left((\cos x - x \sin x) \ln x + \cos x\right) + \frac{-4x}{(x^2 - 1)^2}$

$\frac{dy}{dx} = x^{x \cos x} \left((\cos x - x \sin x) \ln x + \cos x\right) - \frac{4x}{(x^2 - 1)^2}$

Final Answer:

The derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = x^{x \cos x} \left((\cos x - x \sin x) \ln x + \cos x\right) - \frac{4x}{(x^2 - 1)^2}$

Given:

Let the given function be denoted by $y$.

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

We can express this as the sum of two functions, $u$ and $v$, such that $y = u + v$.

Let $u = (x \cos x)^x$ and $v = (x \sin x)^{\frac{1}{x}}$.

Solution:

To find the derivative of $y$ with respect to $x$, we need to find the derivatives of $u$ and $v$ separately and then add them.

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Step 1: Differentiate $u = (x \cos x)^x$

This function is in the form of $f(x)^{g(x)}$, so we use logarithmic differentiation.

Take the natural logarithm of both sides:

$\ln u = \ln((x \cos x)^x)$

Using the logarithm property $\ln(a^b) = b \ln a$:

$\ln u = x \ln(x \cos x)$

Using the logarithm property $\ln(ab) = \ln a + \ln b$:

$\ln u = x (\ln x + \ln(\cos x))$

$\ln u = x \ln x + x \ln(\cos x)$

Differentiate both sides with respect to $x$. Use the product rule on both terms on the right side.

$\frac{d}{dx}(\ln u) = \frac{d}{dx}(x \ln x) + \frac{d}{dx}(x \ln(\cos x))$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{d}{dx} x\right) (\ln x) + x \left(\frac{d}{dx} \ln x\right) + \left(\frac{d}{dx} x\right) (\ln(\cos x)) + x \left(\frac{d}{dx} \ln(\cos x)\right)$

$\frac{1}{u} \frac{du}{dx} = (1)(\ln x) + x\left(\frac{1}{x}\right) + (1)(\ln(\cos x)) + x\left(\frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)\right)$

$\frac{1}{u} \frac{du}{dx} = \ln x + 1 + \ln(\cos x) + x\left(\frac{1}{\cos x} \cdot (-\sin x)\right)$

$\frac{1}{u} \frac{du}{dx} = \ln x + 1 + \ln(\cos x) - x \frac{\sin x}{\cos x}$

$\frac{1}{u} \frac{du}{dx} = \ln x + 1 + \ln(\cos x) - x \tan x$

Solve for $\frac{du}{dx}$ by multiplying both sides by $u$:

$\frac{du}{dx} = u (\ln x + 1 + \ln(\cos x) - x \tan x)$

Substitute the expression for $u$ back:

$\frac{du}{dx} = (x \cos x)^x (\ln x + 1 + \ln(\cos x) - x \tan x)$

Step 2: Differentiate $v = (x \sin x)^{\frac{1}{x}}$

This is also of the form $f(x)^{g(x)}$. Use logarithmic differentiation.

Take the natural logarithm of both sides:

$\ln v = \ln((x \sin x)^{\frac{1}{x}})$

Using the property $\ln(a^b) = b \ln a$:

$\ln v = \frac{1}{x} \ln(x \sin x)$

Using the property $\ln(ab) = \ln a + \ln b$:

$\ln v = \frac{1}{x} (\ln x + \ln(\sin x))$

$\ln v = \frac{\ln x}{x} + \frac{\ln(\sin x)}{x}$

Differentiate both sides with respect to $x$. Use the quotient rule on both terms on the right side.

$\frac{d}{dx}(\ln v) = \frac{d}{dx}\left(\frac{\ln x}{x}\right) + \frac{d}{dx}\left(\frac{\ln(\sin x)}{x}\right)$

For the first term, $\frac{d}{dx}\left(\frac{\ln x}{x}\right) = \frac{(\frac{1}{x})(x) - (\ln x)(1)}{x^2} = \frac{1 - \ln x}{x^2}$.

For the second term, $\frac{d}{dx}\left(\frac{\ln(\sin x)}{x}\right) = \frac{(\frac{d}{dx}\ln(\sin x))(x) - (\ln(\sin x))(\frac{d}{dx}x)}{x^2}$.

Using the chain rule, $\frac{d}{dx}\ln(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.

So the second term derivative is: $\frac{(\cot x)(x) - (\ln(\sin x))(1)}{x^2} = \frac{x \cot x - \ln(\sin x)}{x^2}$.

Combine the derivatives for $\frac{1}{v} \frac{dv}{dx}$:

$\frac{1}{v} \frac{dv}{dx} = \frac{1 - \ln x}{x^2} + \frac{x \cot x - \ln(\sin x)}{x^2}$

$\frac{1}{v} \frac{dv}{dx} = \frac{1 - \ln x + x \cot x - \ln(\sin x)}{x^2}$

Solve for $\frac{dv}{dx}$ by multiplying both sides by $v$:

$\frac{dv}{dx} = v \left(\frac{1 - \ln x + x \cot x - \ln(\sin x)}{x^2}\right)$

Substitute the expression for $v$ back:

$\frac{dv}{dx} = (x \sin x)^{\frac{1}{x}} \left(\frac{1 - \ln x + x \cot x - \ln(\sin x)}{x^2}\right)$

Step 3: Combine the derivatives

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Substitute the expressions for $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$\frac{dy}{dx} = (x \cos x)^x (\ln x + 1 + \ln(\cos x) - x \tan x) + (x \sin x)^{\frac{1}{x}} \left(\frac{1 - \ln x + x \cot x - \ln(\sin x)}{x^2}\right)$

Final Answer:

The derivative of the given function with respect to $x$ is:

$\frac{dy}{dx} = (x \cos x)^x (\ln x + 1 + \ln(\cos x) - x \tan x) + (x \sin x)^{\frac{1}{x}} \left(\frac{1 - \ln x + x \cot x - \ln(\sin x)}{x^2}\right)$

Given:

The implicit equation is $x^y + y^x = 1$.

Solution:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$.

Let the given equation be

Differentiating both sides of equation (1) with respect to $x$, we get:

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

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

Let's evaluate each derivative separately.

For the term $x^y$, let $u = x^y$. Taking natural logarithm on both sides:

$\ln u = \ln(x^y)$

$\ln u = y \ln x$

Differentiating with respect to $x$ using the chain rule and product rule:

$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(y \ln x)$

$\frac{1}{u} \frac{du}{dx} = \left(\frac{dy}{dx}\right) \ln x + y \left(\frac{1}{x}\right)$

$\frac{du}{dx} = u \left(\frac{dy}{dx} \ln x + \frac{y}{x}\right)$

Substitute $u = x^y$ back:

$\frac{du}{dx} = x^y \left(\frac{dy}{dx} \ln x + \frac{y}{x}\right)$

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

$\frac{du}{dx} = x^y \ln x \frac{dy}{dx} + y x^{y-1}$

For the term $y^x$, let $v = y^x$. Taking natural logarithm on both sides:

$\ln v = \ln(y^x)$

$\ln v = x \ln y$

Differentiating with respect to $x$ using the chain rule and product rule:

$\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(x \ln y)$

$\frac{1}{v} \frac{dv}{dx} = (1) \ln y + x \left(\frac{1}{y} \frac{dy}{dx}\right)$

$\frac{dv}{dx} = v \left(\ln y + \frac{x}{y} \frac{dy}{dx}\right)$

Substitute $v = y^x$ back:

$\frac{dv}{dx} = y^x \left(\ln y + \frac{x}{y} \frac{dy}{dx}\right)$

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

$\frac{dv}{dx} = y^x \ln y + x y^{x-1} \frac{dy}{dx}$

Now substitute these derivatives back into the differentiated equation:

$(x^y \ln x \frac{dy}{dx} + y x^{y-1}) + (y^x \ln y + x y^{x-1} \frac{dy}{dx}) = 0$

Group terms containing $\frac{dy}{dx}$ on one side and the remaining terms on the other side:

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

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

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

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

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

Final Answer:

The derivative $\frac{dy}{dx}$ of the given equation is:

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

Given:

The implicit equation is $y^x = x^y$.

Solution:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$.

Given equation:

Since we have variables in the exponents, we take the natural logarithm of both sides of equation (1):

$\ln(y^x) = \ln(x^y)$

Using the logarithm property $\ln(a^b) = b \ln a$:

Now, differentiate both sides of equation (2) with respect to $x$. We will use the product rule on both sides, remembering that $y$ is a function of $x$, so we apply the chain rule when differentiating terms involving $y$.

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

Apply the product rule $\frac{d}{dx}(uv) = u'v + uv'$:

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

Calculate the individual derivatives:

$\frac{d}{dx} x = 1$

$\frac{d}{dx} \ln y = \frac{1}{y} \frac{dy}{dx}$ (using chain rule)

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

$\frac{d}{dx} \ln x = \frac{1}{x}$

Substitute these derivatives back into the differentiated equation:

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

$\ln y + \frac{x}{y} \frac{dy}{dx} = \ln x \frac{dy}{dx} + \frac{y}{x}$

Now, collect all terms containing $\frac{dy}{dx}$ on one side and all other terms on the other side.

$\frac{x}{y} \frac{dy}{dx} - \ln x \frac{dy}{dx} = \frac{y}{x} - \ln y$

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

$\frac{dy}{dx} \left(\frac{x}{y} - \ln x\right) = \frac{y}{x} - \ln y$

Combine the terms within the parentheses and on the right side by finding a common denominator:

$\frac{dy}{dx} \left(\frac{x - y \ln x}{y}\right) = \frac{y - x \ln y}{x}$

Solve for $\frac{dy}{dx}$ by multiplying both sides by $\frac{y}{x - y \ln x}$:

$\frac{dy}{dx} = \frac{y - x \ln y}{x} \cdot \frac{y}{x - y \ln x}$

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

Final Answer:

The derivative $\frac{dy}{dx}$ of the given equation is:

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

Given:

The implicit equation is $(\cos x)^y = (\cos y)^x$.

Solution:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$.

Given equation:

Since we have variables in the exponents, we take the natural logarithm of both sides of equation (1):

$\ln((\cos x)^y) = \ln((\cos y)^x)$

Using the logarithm property $\ln(a^b) = b \ln a$:

Now, differentiate both sides of equation (2) with respect to $x$. We will use the product rule on both sides, remembering that $y$ is a function of $x$, so we apply the chain rule when differentiating terms involving $y$.

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

Apply the product rule $\frac{d}{dx}(uv) = u'v + uv'$:

$\left(\frac{d}{dx} y\right) (\ln(\cos x)) + y \left(\frac{d}{dx} \ln(\cos x)\right) = \left(\frac{d}{dx} x\right) (\ln(\cos y)) + x \left(\frac{d}{dx} \ln(\cos y)\right)$

Calculate the individual derivatives:

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

$\frac{d}{dx} \ln(\cos x)$: Use the chain rule. Let $u = \cos x$, then $\frac{d}{du}(\ln u) = \frac{1}{u}$ and $\frac{du}{dx} = -\sin x$. So, $\frac{d}{dx}\ln(\cos x) = \frac{1}{\cos x}(-\sin x) = -\tan x$.

$\frac{d}{dx} x = 1$

$\frac{d}{dx} \ln(\cos y)$: Use the chain rule. Let $v = \cos y$, then $\frac{d}{dv}(\ln v) = \frac{1}{v}$ and $\frac{dv}{dx} = \frac{d}{dx}(\cos y) = -\sin y \frac{dy}{dx}$ (since $y$ is a function of $x$). So, $\frac{d}{dx}\ln(\cos y) = \frac{1}{\cos y}(-\sin y \frac{dy}{dx}) = -\tan y \frac{dy}{dx}$.

Substitute these derivatives back into the differentiated equation:

$\left(\frac{dy}{dx}\right) (\ln(\cos x)) + y (-\tan x) = (1) (\ln(\cos y)) + x \left(-\tan y \frac{dy}{dx}\right)$

$\ln(\cos x) \frac{dy}{dx} - y \tan x = \ln(\cos y) - x \tan y \frac{dy}{dx}$

Collect all terms containing $\frac{dy}{dx}$ on one side and all other terms on the other side.

$\ln(\cos x) \frac{dy}{dx} + x \tan y \frac{dy}{dx} = \ln(\cos y) + y \tan x$

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

$\frac{dy}{dx} (\ln(\cos x) + x \tan y) = \ln(\cos y) + y \tan x$

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

$\frac{dy}{dx} = \frac{\ln(\cos y) + y \tan x}{\ln(\cos x) + x \tan y}$

Final Answer:

The derivative $\frac{dy}{dx}$ of the given equation is:

$\frac{dy}{dx} = \frac{\ln(\cos y) + y \tan x}{\ln(\cos x) + x \tan y}$

Given:

The implicit equation is $xy = e^{(x – y)}$.

Solution:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$.

Given equation:

Differentiate both sides of equation (1) with respect to $x$:

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

For the left side, use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ with $u=x$ and $v=y$:

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

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

For the right side, use the chain rule $\frac{d}{dx}(e^w) = e^w \frac{dw}{dx}$ with $w = x-y$.

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

Differentiate the exponent:

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

Substitute the exponent's derivative back:

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

Now, equate the derivatives of both sides:

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

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

Group terms containing $\frac{dy}{dx}$ on one side and terms without $\frac{dy}{dx}$ on the other side:

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

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

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

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

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

From the original equation (1), we know that $e^{x-y} = xy$. We can substitute this into the expression for $\frac{dy}{dx}$ to simplify it:

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

Factor out common terms in the numerator and denominator:

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

Final Answer:

The derivative $\frac{dy}{dx}$ of the given equation is:

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

or equivalently,

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

Given:

The function is $f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$.

Solution:

To find the derivative of $f(x)$, we can use logarithmic differentiation, which is convenient for functions involving products.

Let $y = f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$.

Taking the natural logarithm of both sides:

$\ln y = \ln[(1 + x) (1 + x^2) (1 + x^4) (1 + x^8)]$

Using the logarithm property $\ln(abcd) = \ln a + \ln b + \ln c + \ln d$:

$\ln y = \ln(1 + x) + \ln(1 + x^2) + \ln(1 + x^4) + \ln(1 + x^8)$

Now, differentiate both sides with respect to $x$. Recall that $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$ and $\frac{d}{dx}(1+x^n) = nx^{n-1}$.

$\frac{d}{dx}(\ln y) = \frac{d}{dx}(\ln(1 + x)) + \frac{d}{dx}(\ln(1 + x^2)) + \frac{d}{dx}(\ln(1 + x^4)) + \frac{d}{dx}(\ln(1 + x^8))$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} \cdot \frac{d}{dx}(1+x) + \frac{1}{1+x^2} \cdot \frac{d}{dx}(1+x^2) + \frac{1}{1+x^4} \cdot \frac{d}{dx}(1+x^4) + \frac{1}{1+x^8} \cdot \frac{d}{dx}(1+x^8)$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} \cdot (1) + \frac{1}{1+x^2} \cdot (2x) + \frac{1}{1+x^4} \cdot (4x^3) + \frac{1}{1+x^8} \cdot (8x^7)$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}$

Now, solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \left( \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right)$

Substitute the original expression for $y = f(x)$ back into the equation to get $f'(x)$:

$f'(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \left( \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right)$

Now, we need to find $f'(1)$. Substitute $x=1$ into the expression for $f'(x)$:

$f'(1) = (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8) \left( \frac{1}{1+1} + \frac{2(1)}{1+1^2} + \frac{4(1)^3}{1+1^4} + \frac{8(1)^7}{1+1^8} \right)$

$f'(1) = (2) (1 + 1) (1 + 1) (1 + 1) \left( \frac{1}{2} + \frac{2}{1+1} + \frac{4}{1+1} + \frac{8}{1+1} \right)$

$f'(1) = (2) (2) (2) (2) \left( \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} \right)$

$f'(1) = 16 \left( \frac{1}{2} + 1 + 2 + 4 \right)$

$f'(1) = 16 \left( \frac{1}{2} + 7 \right)$

$f'(1) = 16 \left( \frac{1}{2} + \frac{14}{2} \right)$

$f'(1) = 16 \left( \frac{1 + 14}{2} \right)$

$f'(1) = 16 \left( \frac{15}{2} \right)$

$f'(1) = \cancel{16}^{8} \cdot \frac{15}{\cancel{2}_{1}}$

$f'(1) = 8 \cdot 15$

$f'(1) = 120$

Final Answer:

The derivative of the function $f(x)$ is:

$f'(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \left( \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right)$

The value of the derivative at $x=1$ is:

$f'(1) = 120$

Given:

The function to differentiate is $f(x) = (x^2 – 5x + 8) (x^3 + 7x + 9)$.

Solution - Method (i): Using Product Rule

Let $u(x) = x^2 - 5x + 8$ and $v(x) = x^3 + 7x + 9$.

Then $f(x) = u(x)v(x)$.

The derivative of $u(x)$ is $u'(x) = \frac{d}{dx}(x^2 - 5x + 8) = 2x - 5$.

The derivative of $v(x)$ is $v'(x) = \frac{d}{dx}(x^3 + 7x + 9) = 3x^2 + 7$.

Using the product rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (2x - 5)(x^3 + 7x + 9) + (x^2 - 5x + 8)(3x^2 + 7)$

Expand the first term:

$(2x - 5)(x^3 + 7x + 9) = 2x(x^3 + 7x + 9) - 5(x^3 + 7x + 9)$

$= 2x^4 + 14x^2 + 18x - 5x^3 - 35x - 45$

$= 2x^4 - 5x^3 + 14x^2 - 17x - 45$

Expand the second term:

$(x^2 - 5x + 8)(3x^2 + 7) = x^2(3x^2 + 7) - 5x(3x^2 + 7) + 8(3x^2 + 7)$

$= 3x^4 + 7x^2 - 15x^3 - 35x + 24x^2 + 56$

$= 3x^4 - 15x^3 + 31x^2 - 35x + 56$

Combine the expanded terms:

$f'(x) = (2x^4 - 5x^3 + 14x^2 - 17x - 45) + (3x^4 - 15x^3 + 31x^2 - 35x + 56)$

$f'(x) = (2x^4 + 3x^4) + (-5x^3 - 15x^3) + (14x^2 + 31x^2) + (-17x - 35x) + (-45 + 56)$

$f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Solution - Method (ii): By Expanding the Product

First, expand the product $f(x) = (x^2 - 5x + 8)(x^3 + 7x + 9)$:

$f(x) = x^2(x^3 + 7x + 9) - 5x(x^3 + 7x + 9) + 8(x^3 + 7x + 9)$

$f(x) = (x^5 + 7x^3 + 9x^2) - (5x^4 + 35x^2 + 45x) + (8x^3 + 56x + 72)$

$f(x) = x^5 + 7x^3 + 9x^2 - 5x^4 - 35x^2 - 45x + 8x^3 + 56x + 72$

Combine like terms:

$f(x) = x^5 - 5x^4 + (7x^3 + 8x^3) + (9x^2 - 35x^2) + (-45x + 56x) + 72$

$f(x) = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72$

Now, differentiate this polynomial term by term:

$f'(x) = \frac{d}{dx}(x^5) - \frac{d}{dx}(5x^4) + \frac{d}{dx}(15x^3) - \frac{d}{dx}(26x^2) + \frac{d}{dx}(11x) + \frac{d}{dx}(72)$

$f'(x) = 5x^{5-1} - 5(4x^{4-1}) + 15(3x^{3-1}) - 26(2x^{2-1}) + 11(1) + 0$

$f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Solution - Method (iii): By Logarithmic Differentiation

Let $y = f(x) = (x^2 - 5x + 8)(x^3 + 7x + 9)$.

Take the natural logarithm of both sides:

$\ln y = \ln[(x^2 - 5x + 8)(x^3 + 7x + 9)]$

Using the logarithm property $\ln(ab) = \ln a + \ln b$:

$\ln y = \ln(x^2 - 5x + 8) + \ln(x^3 + 7x + 9)$

Differentiate both sides with respect to $x$. Use the chain rule $\frac{d}{dx}(\ln w) = \frac{1}{w} \frac{dw}{dx}$.

$\frac{d}{dx}(\ln y) = \frac{d}{dx}(\ln(x^2 - 5x + 8)) + \frac{d}{dx}(\ln(x^3 + 7x + 9))$

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x^2 - 5x + 8} \cdot \frac{d}{dx}(x^2 - 5x + 8) + \frac{1}{x^3 + 7x + 9} \cdot \frac{d}{dx}(x^3 + 7x + 9)$

$\frac{1}{y} \frac{dy}{dx} = \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9}$

Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = y \left( \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \right)$

Substitute the original expression for $y$ back:

$\frac{dy}{dx} = (x^2 - 5x + 8)(x^3 + 7x + 9) \left( \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \right)$

Distribute the term $(x^2 - 5x + 8)(x^3 + 7x + 9)$ into the parenthesis:

$\frac{dy}{dx} = (x^2 - 5x + 8)(x^3 + 7x + 9) \frac{2x - 5}{x^2 - 5x + 8} + (x^2 - 5x + 8)(x^3 + 7x + 9) \frac{3x^2 + 7}{x^3 + 7x + 9}$

Cancel the common factors:

$\frac{dy}{dx} = (x^3 + 7x + 9) (2x - 5) + (x^2 - 5x + 8) (3x^2 + 7)$

This is the same expression obtained in Method (i) before combining terms. Expanding this leads to the same polynomial:

$\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Comparison of Results:

From Method (i), we got $f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$.

From Method (ii), we got $f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$.

From Method (iii), we got $f'(x) = 5x^4 - 20x^3 + 45x^2 - 52x + 11$.

All three methods yield the same derivative for the function.

Final Answer:

Yes, all three methods give the same answer. The derivative is $5x^4 - 20x^3 + 45x^2 - 52x + 11$.

Given:

u, v, and w are functions of x.

We need to show that $\frac{d}{dx} (uvw) = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$.

Solution - Method 1: By Repeated Application of Product Rule

Let $y = uvw$. We can treat $uv$ as a single function first. Let $z = uv$. Then $y = zw$.

Using the product rule $\frac{d}{dx}(zw) = \frac{dz}{dx} w + z \frac{dw}{dx}$:

Now, we need to find $\frac{d}{dx}(uv)$ using the product rule for two functions:

Substitute the expression for $\frac{d}{dx}(uv)$ from equation (2) into equation (1):

$\frac{dy}{dx} = \left(\frac{du}{dx} v + u \frac{dv}{dx}\right) w + uv \frac{dw}{dx}$

Distribute the term $w$ into the parentheses:

$\frac{dy}{dx} = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$

Thus, by repeated application of the product rule:

$\frac{d}{dx} (uvw) = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$

Solution - Method 2: By Logarithmic Differentiation

Let $y = uvw$.

Take the natural logarithm on both sides:

Using the logarithm property $\ln(abc) = \ln a + \ln b + \ln c$ for the right side of equation (3):

Now, differentiate both sides of equation (4) with respect to $x$. Remember to use the chain rule for $\ln u$, $\ln v$, and $\ln w$ since $u$, $v$, and $w$ are functions of $x$.

$\frac{d}{dx}(\ln y) = \frac{d}{dx}(\ln u) + \frac{d}{dx}(\ln v) + \frac{d}{dx}(\ln w)$

Using $\frac{d}{dx}(\ln f(x)) = \frac{1}{f(x)} f'(x)$, we get:

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}$

To solve for $\frac{dy}{dx}$, multiply both sides by $y$:

$\frac{dy}{dx} = y \left(\frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}\right)$

Substitute $y = uvw$ back into the equation:

$\frac{dy}{dx} = (uvw) \left(\frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}\right)$

Distribute the term $(uvw)$ into the parentheses:

$\frac{dy}{dx} = (uvw) \frac{1}{u} \frac{du}{dx} + (uvw) \frac{1}{v} \frac{dv}{dx} + (uvw) \frac{1}{w} \frac{dw}{dx}$

Cancel the common terms in each part:

$\frac{dy}{dx} = \cancel{u}vw \frac{1}{\cancel{u}} \frac{du}{dx} + u\cancel{v}w \frac{1}{\cancel{v}} \frac{dv}{dx} + uv\cancel{w} \frac{1}{\cancel{w}} \frac{dw}{dx}$

$\frac{dy}{dx} = vw \frac{du}{dx} + uw \frac{dv}{dx} + uv \frac{dw}{dx}$

This can be written in the desired form:

$\frac{d}{dx} (uvw) = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$

Conclusion:

Both methods, repeated application of the product rule and logarithmic differentiation, yield the same formula for the derivative of the product of three functions:

$\frac{d}{dx} (uvw) = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$

Given:

The parametric equations are $x = a \cos \theta$ and $y = a \sin \theta$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find $\frac{dy}{dx}$ for parametric equations, we use the formula:

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

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

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

Now, divide equation (ii) by equation (i):

Simplify the expression:

Using the trigonometric identity $\cot \theta = \frac{\cos \theta}{\sin \theta}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is $-\cot \theta$.

Given:

The parametric equations are $x = at^2$ and $y = 2at$, where $t$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, we find the derivative of $x$ with respect to the parameter $t$:

Next, we find the derivative of $y$ with respect to the parameter $t$:

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$:

Simplify the expression by cancelling the common terms:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

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

Given:

The parametric equations are $x = a (\theta + \sin \theta)$ and $y = a (1 – \cos \theta)$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

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

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

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

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

Cancel the common factor $a$:

Use the trigonometric half-angle identities: $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 + \cos \theta = 2 \cos^2(\theta/2)$:

Simplify the expression:

Using the identity $\tan \phi = \frac{\sin \phi}{\cos \phi}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \tan(\theta/2)$

Given:

The implicit equation is $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$.

Given equation:

Differentiating both sides of equation (1) with respect to $x$:

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

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

For the term $\frac{d}{dx}(x^{\frac{2}{3}})$, use the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

For the term $\frac{d}{dx}(y^{\frac{2}{3}})$, use the power rule and the chain rule (since $y$ is a function of $x$):

For the term $\frac{d}{dx}(a^{\frac{2}{3}})$, since $a$ is a constant, $a^{\frac{2}{3}}$ is also a constant:

Substitute these derivatives back into the differentiated equation:

Now, we solve equation (2) for $\frac{dy}{dx}$. Subtract $\frac{2}{3} x^{-\frac{1}{3}}$ from both sides:

Divide both sides by $\frac{2}{3} y^{-\frac{1}{3}}$:

Cancel the common factor $\frac{2}{3}$:

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

This can also be written using radicals:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = - \left(\frac{y}{x}\right)^{\frac{1}{3}}$

or equivalently,

$\frac{dy}{dx} = - \sqrt[3]{\frac{y}{x}}$

Given:

The parametric equations are $x = 2at^2$ and $y = at^4$, where $t$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, find the derivative of $x$ with respect to the parameter $t$:

Next, find the derivative of $y$ with respect to the parameter $t$:

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (equation (ii) divided by equation (i)):

Simplify the expression by cancelling the common terms:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = t^2$

Given:

The parametric equations are $x = a \cos \theta$ and $y = b \cos \theta$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

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

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

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

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

Simplify the expression by cancelling the common terms:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{b}{a}$

Given:

The parametric equations are $x = \sin t$ and $y = \cos 2t$, where $t$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, find the derivative of $x$ with respect to the parameter $t$:

Next, find the derivative of $y$ with respect to the parameter $t$:

Using the chain rule, let $u = 2t$. Then $\frac{d}{dt}(\cos u) = -\sin u \frac{du}{dt}$.

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (equation (ii) divided by equation (i)):

Using the double angle identity $\sin(2t) = 2 \sin t \cos t$:

Cancel the common $\cos t$ term (assuming $\cos t \neq 0$):

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = -4 \sin t$

Given:

The parametric equations are $x = 4t$ and $y = \frac{4}{t} = 4t^{-1}$, where $t$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, find the derivative of $x$ with respect to the parameter $t$:

Next, find the derivative of $y$ with respect to the parameter $t$:

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (equation (ii) divided by equation (i)):

Simplify the expression:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

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

Given:

The parametric equations are $x = \cos \theta – \cos 2\theta$ and $y = \sin \theta – \sin 2\theta$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

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

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

Using the derivative rule $\frac{d}{d\theta}(\cos u) = -\sin u \frac{du}{d\theta}$:

$\frac{d}{d\theta}(\cos \theta) = -\sin \theta \cdot \frac{d}{d\theta}(\theta) = -\sin \theta \cdot 1 = -\sin \theta$

$\frac{d}{d\theta}(\cos 2\theta) = -\sin(2\theta) \cdot \frac{d}{d\theta}(2\theta) = -\sin(2\theta) \cdot 2 = -2\sin 2\theta$

Substitute these back into the expression for $\frac{dx}{d\theta}$:

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

Using the derivative rule $\frac{d}{d\theta}(\sin u) = \cos u \frac{du}{d\theta}$:

$\frac{d}{d\theta}(\sin \theta) = \cos \theta \cdot \frac{d}{d\theta}(\theta) = \cos \theta \cdot 1 = \cos \theta$

$\frac{d}{d\theta}(\sin 2\theta) = \cos(2\theta) \cdot \frac{d}{d\theta}(2\theta) = \cos(2\theta) \cdot 2 = 2\cos 2\theta$

Substitute these back into the expression for $\frac{dy}{d\theta}$:

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

The expression can be left in terms of $\theta$ as requested.

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{\cos \theta - 2\cos 2\theta}{2\sin 2\theta - \sin \theta}$

Given:

The parametric equations are $x = a (\theta – \sin \theta)$ and $y = a (1 + \cos \theta)$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

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

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

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

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

Cancel the common factor $a$:

Use the trigonometric half-angle identities: $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 - \cos \theta = 2 \sin^2(\theta/2)$:

Simplify the expression:

Using the identity $\cot \phi = \frac{\cos \phi}{\sin \phi}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = -\cot(\theta/2)$

Given:

The parametric equations are $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$, where $t$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, find the derivative of $x$ with respect to the parameter $t$:

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

Using the quotient rule $\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = \sin^3 t$ and $v = (\cos 2t)^{1/2}$.

$u' = \frac{d}{dt}(\sin^3 t) = 3\sin^2 t \cos t$

$v' = \frac{d}{dt}((\cos 2t)^{1/2}) = \frac{1}{2}(\cos 2t)^{-1/2} (-2\sin 2t) = -\frac{\sin 2t}{\sqrt{\cos 2t}}$

$\frac{dx}{dt} = \frac{(3\sin^2 t \cos t)(\cos 2t)^{1/2} - (\sin^3 t)(-\frac{\sin 2t}{\sqrt{\cos 2t}})}{(\sqrt{\cos 2t})^2}$

$\frac{dx}{dt} = \frac{3\sin^2 t \cos t \sqrt{\cos 2t} + \frac{\sin^3 t \sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}$

Multiply numerator and denominator by $\sqrt{\cos 2t}$ to simplify:

$\frac{dx}{dt} = \frac{3\sin^2 t \cos t (\cos 2t) + \sin^3 t \sin 2t}{(\cos 2t)^{3/2}}$

Simplify the numerator using $\sin 2t = 2\sin t \cos t$:

Numerator $= 3\sin^2 t \cos t \cos 2t + \sin^3 t (2\sin t \cos t)$

Numerator $= 3\sin^2 t \cos t \cos 2t + 2\sin^4 t \cos t$

Factor out $\sin^2 t \cos t$:

Numerator $= \sin^2 t \cos t (3\cos 2t + 2\sin^2 t)$

Using $\cos 2t = 1 - 2\sin^2 t$:

Numerator $= \sin^2 t \cos t (3(1 - 2\sin^2 t) + 2\sin^2 t)$

Numerator $= \sin^2 t \cos t (3 - 6\sin^2 t + 2\sin^2 t)$

Numerator $= \sin^2 t \cos t (3 - 4\sin^2 t)$

Next, find the derivative of $y$ with respect to the parameter $t$:

$y = \frac{\cos^3 t}{(\cos 2t)^{1/2}}$

Using the quotient rule, where $p = \cos^3 t$ and $q = (\cos 2t)^{1/2}$.

$p' = \frac{d}{dt}(\cos^3 t) = 3\cos^2 t (-\sin t) = -3\cos^2 t \sin t$

$q' = -\frac{\sin 2t}{\sqrt{\cos 2t}}$

$\frac{dy}{dt} = \frac{(-3\cos^2 t \sin t)(\cos 2t)^{1/2} - (\cos^3 t)(-\frac{\sin 2t}{\sqrt{\cos 2t}})}{(\sqrt{\cos 2t})^2}$

$\frac{dy}{dt} = \frac{-3\cos^2 t \sin t \sqrt{\cos 2t} + \frac{\cos^3 t \sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}$

Multiply numerator and denominator by $\sqrt{\cos 2t}$ to simplify:

$\frac{dy}{dt} = \frac{-3\cos^2 t \sin t (\cos 2t) + \cos^3 t \sin 2t}{(\cos 2t)^{3/2}}$

Simplify the numerator using $\sin 2t = 2\sin t \cos t$:

Numerator $= -3\cos^2 t \sin t \cos 2t + \cos^3 t (2\sin t \cos t)$

Numerator $= -3\cos^2 t \sin t \cos 2t + 2\cos^4 t \sin t$

Factor out $\cos^2 t \sin t$:

Numerator $= \cos^2 t \sin t (-3\cos 2t + 2\cos^2 t)$

Using $\cos 2t = 2\cos^2 t - 1$:

Numerator $= \cos^2 t \sin t (-3(2\cos^2 t - 1) + 2\cos^2 t)$

Numerator $= \cos^2 t \sin t (-6\cos^2 t + 3 + 2\cos^2 t)$

Numerator $= \cos^2 t \sin t (3 - 4\cos^2 t)$

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (equation (ii) divided by equation (i)):

Cancel the common denominator $(\cos 2t)^{3/2}$:

Cancel common factors $\sin t$ and $\cos t$ (assuming $\sin t \neq 0$ and $\cos t \neq 0$):

Using the triple angle identities $\cos 3t = \cos t (4\cos^2 t - 3)$ and $\sin 3t = \sin t (3 - 4\sin^2 t)$:

$3 - 4\cos^2 t = -(4\cos^2 t - 3) = -\frac{\cos 3t}{\cos t}$

$3 - 4\sin^2 t = \frac{\sin 3t}{\sin t}$

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

Cancel $\cos t$ from the numerator and $\sin t$ from the denominator:

Using the identity $\cot \phi = \frac{\cos \phi}{\sin \phi}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = -\cot 3t$

Given:

The parametric equations are $x = a \left( \cos t + \log \tan \frac{t}{2} \right)$ and $y = a \sin t$, where $t$ is the parameter. We assume $\log$ denotes the natural logarithm $\ln$.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

First, find the derivative of $x$ with respect to the parameter $t$:

We know $\frac{d}{dt}(\cos t) = -\sin t$.

For the term $\frac{d}{dt}\left(\ln \tan \frac{t}{2}\right)$, use the chain rule. Let $u = \tan \frac{t}{2}$.

$\frac{d}{dt}\left(\ln u\right) = \frac{1}{u} \frac{du}{dt} = \frac{1}{\tan \frac{t}{2}} \cdot \frac{d}{dt}\left(\tan \frac{t}{2}\right)$

Now, differentiate $\tan \frac{t}{2}$ using the chain rule. Let $v = \frac{t}{2}$.

$\frac{d}{dt}\left(\tan v\right) = \sec^2 v \frac{dv}{dt} = \sec^2 \frac{t}{2} \cdot \frac{d}{dt}\left(\frac{t}{2}\right)$

$\frac{d}{dt}\left(\frac{t}{2}\right) = \frac{1}{2}$

So, $\frac{d}{dt}\left(\tan \frac{t}{2}\right) = \sec^2 \frac{t}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{t}{2}$.

Substitute this back into the derivative of $\ln \tan \frac{t}{2}$:

$\frac{d}{dt}\left(\ln \tan \frac{t}{2}\right) = \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2} \sec^2 \frac{t}{2}$

Rewrite in terms of $\sin$ and $\cos$: $\tan \frac{t}{2} = \frac{\sin(t/2)}{\cos(t/2)}$ and $\sec^2 \frac{t}{2} = \frac{1}{\cos^2(t/2)}$.

$\frac{d}{dt}\left(\ln \tan \frac{t}{2}\right) = \frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{2} \cdot \frac{1}{\cos^2(t/2)} = \frac{1}{2 \sin(t/2) \cos(t/2)}$

Using the identity $\sin t = 2 \sin(t/2) \cos(t/2)$:

$\frac{d}{dt}\left(\ln \tan \frac{t}{2}\right) = \frac{1}{\sin t} = \text{cosec } t$

Now substitute the derivatives of $\cos t$ and $\ln \tan \frac{t}{2}$ back into the expression for $\frac{dx}{dt}$:

Rewrite $\text{cosec } t$ as $\frac{1}{\sin t}$:

Combine the terms inside the parenthesis:

Using the identity $1 - \sin^2 t = \cos^2 t$:

Next, find the derivative of $y$ with respect to the parameter $t$:

Now, divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (equation (ii) divided by equation (i)):

Simplify the expression:

Using the identity $\tan t = \frac{\sin t}{\cos t}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \tan t$

Given:

The parametric equations are $x = a \sec \theta$ and $y = b \tan \theta$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

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

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

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

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

Simplify the expression by cancelling common terms and using identities $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

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

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{b}{a} \text{ cosec } \theta$

Given:

The parametric equations are $x = a (\cos \theta + \theta \sin \theta)$ and $y = a (\sin \theta – \theta \cos \theta)$, where $\theta$ is the parameter.

To Find:

We need to find $\frac{dy}{dx}$.

Solution:

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the formula:

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

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

The derivative of $\cos \theta$ is $-\sin \theta$.

To differentiate $\theta \sin \theta$, we use the product rule $\frac{d}{d\theta}(uv) = u'v + uv'$ with $u = \theta$ and $v = \sin \theta$:

$\frac{d}{d\theta}(\theta \sin \theta) = (1)(\sin \theta) + (\theta)(\cos \theta) = \sin \theta + \theta \cos \theta$

Substitute these derivatives back into the expression for $\frac{dx}{d\theta}$:

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

The derivative of $\sin \theta$ is $\cos \theta$.

To differentiate $\theta \cos \theta$, use the product rule $\frac{d}{d\theta}(uv) = u'v + uv'$ with $u = \theta$ and $v = \cos \theta$:

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

Substitute these derivatives back into the expression for $\frac{dy}{d\theta}$:

Now, divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ (equation (ii) divided by equation (i)):

Simplify the expression by cancelling the common terms $a$ and $\theta$ (assuming $a \neq 0$ and $\theta \neq 0$):

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

Final Answer:

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \tan \theta$

Given:

The parametric equations are $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$, where $t$ is the parameter and $a$ is a constant (presumably $a > 0$ for the expressions to be defined). We need to find $\frac{dy}{dx}$ and show that $\frac{dy}{dx} = -\frac{y}{x}$.

To Show:

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

Solution - Method 1: Using Parametric Differentiation

We are given $x = \sqrt{a^{\sin^{-1} t}} = (a^{\sin^{-1} t})^{1/2} = a^{\frac{1}{2}\sin^{-1} t}$.

To find $\frac{dx}{dt}$, we use the derivative rule $\frac{d}{du}(a^u) = a^u \ln a \cdot \frac{du}{dt}$. Here $u = \frac{1}{2}\sin^{-1} t$.

Substituting back $x = a^{\frac{1}{2}\sin^{-1} t}$:

Similarly, for $y = \sqrt{a^{\cos^{-1} t}} = (a^{\cos^{-1} t})^{1/2} = a^{\frac{1}{2}\cos^{-1} t}$.

To find $\frac{dy}{dt}$, we use the derivative rule $\frac{d}{du}(a^u) = a^u \ln a \cdot \frac{du}{dt}$. Here $u = \frac{1}{2}\cos^{-1} t$.

Substituting back $y = a^{\frac{1}{2}\cos^{-1} t}$:

Now, find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

Cancel the common factor $\frac{\ln a}{2\sqrt{1-t^2}}$ (assuming $\ln a \neq 0$ and $\sqrt{1-t^2} \neq 0$):

Solution - Method 2: Eliminating the Parameter (Implicit Differentiation)

We are given $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$.

Square both equations:

Multiply these two equations:

Using the property of exponents $a^m \cdot a^n = a^{m+n}$:

Using the inverse trigonometric identity $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$ (for $t \in [-1, 1]$):

Since $a$ is a constant, $a^{\pi/2}$ is also a constant. Let $C = a^{\pi/2}$.

Now, differentiate this implicit equation with respect to $x$. Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ on the left side, where $u=x^2$ and $v=y^2$. Remember that $y$ is a function of $x$, so use the chain rule for the term involving $y^2$.

$\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(C)$

$\frac{d}{dx}(x^2) y^2 + x^2 \frac{d}{dx}(y^2) = 0$

$(2x) y^2 + x^2 (2y \frac{dy}{dx}) = 0$

$2xy^2 + 2x^2y \frac{dy}{dx} = 0$

Divide the entire equation by $2xy$ (assuming $x \neq 0$ and $y \neq 0$):

$\frac{2xy^2}{2xy} + \frac{2x^2y}{2xy} \frac{dy}{dx} = 0$

$y + x \frac{dy}{dx} = 0$

Subtract $y$ from both sides:

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

Divide both sides by $x$ (assuming $x \neq 0$):

Conclusion:

Both methods demonstrate that the derivative $\frac{dy}{dx}$ of the given parametric equations is equal to $-\frac{y}{x}$.

Given:

The function is $y = x^3 + \tan x$.

To Find:

We need to find the second derivative $\frac{d^2y}{dx^2}$.

Solution:

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

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

For the first term, $\frac{d}{dx}(3x^2) = 3 \cdot 2x = 6x$ (using the constant multiple rule and power rule).

For the second term, $\frac{d}{dx}(\sec^2 x)$, use the chain rule. Let $u = \sec x$. Then $\sec^2 x = u^2$.

$\frac{d}{dx}(\sec^2 x) = \frac{d}{du}(u^2) \cdot \frac{du}{dx}$

Combine the derivatives of the terms:

Final Answer:

The second derivative of $y$ with respect to $x$ is:

$\frac{d^2y}{dx^2} = 6x + 2 \sec^2 x \tan x$

Given:

$y = A \sin x + B \cos x$, where $A$ and $B$ are constants.

To Prove:

$\frac{d^2y}{dx^2} + y = 0$

Proof:

We are given the function $y = A \sin x + B \cos x$.

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

$\frac{dy}{dx} = \frac{d}{dx}(A \sin x + B \cos x)$

Using the properties of differentiation (linearity):

$\frac{dy}{dx} = A \frac{d}{dx}(\sin x) + B \frac{d}{dx}(\cos x)$

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

Substituting these derivatives:

$\frac{dy}{dx} = A (\cos x) + B (-\sin x)$

$\frac{dy}{dx} = A \cos x - B \sin x$

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

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(A \cos x - B \sin x)$

Using the properties of differentiation:

$\frac{d^2y}{dx^2} = A \frac{d}{dx}(\cos x) - B \frac{d}{dx}(\sin x)$

Substituting the derivatives of $\cos x$ and $\sin x$ again:

$\frac{d^2y}{dx^2} = A (-\sin x) - B (\cos x)$

$\frac{d^2y}{dx^2} = -A \sin x - B \cos x$

Now, consider the expression $\frac{d^2y}{dx^2} + y$. Substitute the expressions we found for $\frac{d^2y}{dx^2}$ and the original expression for $y$:

$\frac{d^2y}{dx^2} + y = (-A \sin x - B \cos x) + (A \sin x + B \cos x)$

Rearranging the terms:

$\frac{d^2y}{dx^2} + y = (-A \sin x + A \sin x) + (-B \cos x + B \cos x)$

$\frac{d^2y}{dx^2} + y = 0 + 0$

$\frac{d^2y}{dx^2} + y = 0$

This shows that the given relation holds true.

Conclusion:

We have shown that if $y = A \sin x + B \cos x$, then $\frac{d^2y}{dx^2} + y = 0$.

Given:

$y = 3e^{2x} + 2e^{3x}$

To Prove:

$\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$

Proof:

We are given the function:

First, find the first derivative $\frac{dy}{dx}$ by differentiating equation (1) with respect to $x$:

Using the chain rule $\frac{d}{dx}(e^{kx}) = ke^{kx}$:

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating equation (2) with respect to $x$:

Using the chain rule again:

Now, substitute the expressions for $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ from equations (1), (2), and (3) into the left side of the equation to be proven:

LHS = $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y$

Distribute the constants $-5$ and $6$:

Group the terms with $e^{2x}$ and $e^{3x}$:

Combine the coefficients for each exponential term:

Since the Left Hand Side equals 0, which is the Right Hand Side of the equation to be proven, the statement is true.

Conclusion:

We have shown that if $y = 3e^{2x} + 2e^{3x}$, then $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$.

Given:

The function is $y = \sin^{-1} x$.

To Prove:

$(1 – x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 0$

Proof:

We are given the function $y = \sin^{-1} x$.

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

We can rewrite this using a negative exponent:

Now, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating equation (1) with respect to $x$. Use the chain rule.

Let $u = 1 - x^2$. Then $\frac{d}{dx}(u^{-\frac{1}{2}}) = -\frac{1}{2} u^{-\frac{3}{2}} \frac{du}{dx}$.

$\frac{du}{dx} = \frac{d}{dx}(1 - x^2) = 0 - 2x = -2x$.

Substitute $u$ and $\frac{du}{dx}$ back:

Simplify the expression:

Rewrite using positive exponents:

Now, substitute the expressions for $\frac{dy}{dx}$ from equation (1) and $\frac{d^2y}{dx^2}$ from equation (2) into the left-hand side of the equation to be proven: $(1 – x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx}$.

In the first term, the $(1 - x^2)$ in the numerator and denominator cancel out (assuming $x^2 \neq 1$).

Since the Left Hand Side equals 0, which is the Right Hand Side of the equation to be proven, the statement is true.

Conclusion:

We have shown that if $y = \sin^{-1} x$, then $(1 – x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 0$.

Given:

Let the given function be $y = x^2 + 3x + 2$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

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

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

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = 2$

Given:

Let the given function be $y = x^{20}$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$. Use the power rule again, along with the constant multiple rule $\frac{d}{dx}(cf(x)) = c \frac{d}{dx}(f(x))$.

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = 380 x^{18}$

Given:

Let the given function be $y = x \cos x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=x$ and $v=\cos x$.

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

The derivative of $\cos x$ is $-\sin x$.

To differentiate $x \sin x$, use the product rule again. Let $u=x$ and $v=\sin x$.

$\frac{d}{dx}(x \sin x) = \left(\frac{d}{dx} x\right)(\sin x) + (x)\left(\frac{d}{dx} \sin x\right) = (1)(\sin x) + (x)(\cos x) = \sin x + x \cos x$.

Substitute these derivatives back:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = -2\sin x - x \cos x$

Given:

Let the given function be $y = \log x$. We assume $\log x$ represents the natural logarithm, i.e., $y = \ln x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y = \ln x$ with respect to $x$.

We can write $\frac{1}{x}$ as $x^{-1}$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx} = x^{-1}$ with respect to $x$. Use the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$.

Rewrite using a positive exponent:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = -\frac{1}{x^2}$

Given:

Let the given function be $y = x^3 \log x$. We assume $\log x$ represents the natural logarithm, i.e., $y = x^3 \ln x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=x^3$ and $v=\ln x$.

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx} = 3x^2 \ln x + x^2$ with respect to $x$. Use the sum rule and the product rule again for the term $3x^2 \ln x$.

For the first term, $\frac{d}{dx}(3x^2 \ln x)$, use the product rule with $u=3x^2$ and $v=\ln x$.

$\frac{d}{dx}(3x^2 \ln x) = \left(\frac{d}{dx} 3x^2\right) (\ln x) + (3x^2) \left(\frac{d}{dx} \ln x\right)$

$= (6x)(\ln x) + (3x^2)\left(\frac{1}{x}\right)$

$= 6x \ln x + 3x$

For the second term, $\frac{d}{dx}(x^2) = 2x$.

Substitute these derivatives back into the expression for $\frac{d^2y}{dx^2}$:

Factor out $x$:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = x(6 \ln x + 5)$

Given:

Let the given function be $y = e^x \sin 5x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=e^x$ and $v=\sin 5x$.

The derivative of $u = e^x$ is $u' = \frac{d}{dx}(e^x) = e^x$.

The derivative of $v = \sin 5x$ requires the chain rule. Let $w=5x$, then $\frac{d}{dx}(\sin w) = \cos w \frac{dw}{dx} = \cos(5x) \cdot 5 = 5\cos 5x$. So $v' = 5\cos 5x$.

Factor out $e^x$:

$\frac{dy}{dx} = e^x (\sin 5x + 5 \cos 5x)$

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$. Use the product rule again for the expression $e^x (\sin 5x + 5 \cos 5x)$. Let $A=e^x$ and $B=(\sin 5x + 5 \cos 5x)$.

The derivative of $A = e^x$ is $A' = e^x$.

The derivative of $B = \sin 5x + 5 \cos 5x$ is:

$B' = \frac{d}{dx}(\sin 5x) + \frac{d}{dx}(5 \cos 5x)$

$B' = (5 \cos 5x) + 5(-\sin 5x \cdot 5)$

$B' = 5 \cos 5x - 25 \sin 5x$

Apply the product rule for $\frac{d^2y}{dx^2} = \frac{d}{dx}(AB) = A'B + AB'$:

Factor out $e^x$:

Combine like terms inside the bracket:

Rearrange the terms:

Final Answer:

The second order derivative of the given function is:

$\frac{d^2y}{dx^2} = e^x (10 \cos 5x - 24 \sin 5x)$

Given:

Let the given function be $y = e^{6x} \cos 3x$.

To Find:

We need to find the second order derivative $\frac{d^2y}{dx^2}$.

Solution:

First, find the first derivative $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$. Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$. Let $u=e^{6x}$ and $v=\cos 3x$.

The derivative of $u = e^{6x}$ is $u' = \frac{d}{dx}(e^{6x})$. Using the chain rule, let $w=6x$. Then $\frac{d}{dx}(e^w) = e^w \frac{dw}{dx} = e^{6x} \cdot 6 = 6e^{6x}$. So $u' = 6e^{6x}$.

The derivative of $v = \cos 3x$ is $v' = \frac{d}{dx}(\cos 3x)$. Using the chain rule, let $z=3x$. Then $\frac{d}{dx}(\cos z) = -\sin z \frac{dz}{dx} = -\sin(3x) \cdot 3 = -3\sin 3x$. So $v' = -3\sin 3x$.

Using the product rule for $\frac{dy}{dx} = u'v + uv'$:

Next, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$. Differentiate each term in equation (1) using the product rule and chain rule.

Differentiate the first term $6e^{6x} \cos 3x$. Use product rule with $A=6e^{6x}$ and $B=\cos 3x$. $A'=36e^{6x}$ and $B'=-3\sin 3x$.