Calculus Examples

$f (x) = \ln (x^{2} + 1)$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x^{2} + 1$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $x^{2} + 1$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{2} + 1]$

Step 1.1.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$\frac{1}{u} \frac{d}{d x} [x^{2} + 1]$

Step 1.1.3

Replace all occurrences of $u$ with $x^{2} + 1$ .

$\frac{1}{x^{2} + 1} \frac{d}{d x} [x^{2} + 1]$

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$\frac{1}{x^{2} + 1} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1])$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$\frac{1}{x^{2} + 1} (2 x + \frac{d}{d x} [1])$

Step 1.2.3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{1}{x^{2} + 1} (2 x + 0)$

Step 1.2.4

Combine fractions.

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Step 1.2.4.1

Add $2 x$ and $0$ .

$\frac{1}{x^{2} + 1} (2 x)$

Step 1.2.4.2

Combine $2$ and $\frac{1}{x^{2} + 1}$ .

$\frac{2}{x^{2} + 1} x$

Step 1.2.4.3

Combine $\frac{2}{x^{2} + 1}$ and $x$ .

$\frac{2 x}{x^{2} + 1}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $2$ is constant with respect to $x$ , the derivative of $\frac{2 x}{x^{2} + 1}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{x}{x^{2} + 1}]$ .

$f'' (x) = 2 \frac{d}{d x} (\frac{x}{x^{2} + 1})$

Step 2.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = x^{2} + 1$ .

$f'' (x) = 2 (\frac{(x^{2} + 1) \frac{d}{d x} (x) - x \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}})$

Step 2.3

Differentiate.

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Step 2.3.1

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 2 (\frac{(x^{2} + 1) \cdot 1 - x \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}})$

Step 2.3.2

Multiply $x^{2} + 1$ by $1$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - x \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}})$

Step 2.3.3

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - x (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (1))}{{(x^{2} + 1)}^{2}})$

Step 2.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - x (2 x + \frac{d}{d x} (1))}{{(x^{2} + 1)}^{2}})$

Step 2.3.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - x (2 x + 0)}{{(x^{2} + 1)}^{2}})$

Step 2.3.6

Simplify the expression.

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Step 2.3.6.1

Add $2 x$ and $0$ .

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

Step 2.3.6.2

Multiply $2$ by $- 1$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - 2 x \cdot x}{{(x^{2} + 1)}^{2}})$

Step 2.4

Raise $x$ to the power of $1$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - 2 (x \cdot x)}{{(x^{2} + 1)}^{2}})$

Step 2.5

Raise $x$ to the power of $1$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - 2 (x \cdot x)}{{(x^{2} + 1)}^{2}})$

Step 2.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.7

Add $1$ and $1$ .

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

Step 2.8

Subtract $2 x^{2}$ from $x^{2}$ .

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

Step 2.9

Combine $2$ and $\frac{- x^{2} + 1}{{(x^{2} + 1)}^{2}}$ .

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

Step 2.10

Simplify.

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Step 2.10.1

Apply the distributive property.

$f'' (x) = \frac{2 (- x^{2}) + 2 \cdot 1}{{(x^{2} + 1)}^{2}}$

Step 2.10.2

Simplify each term.

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Step 2.10.2.1

Multiply $- 1$ by $2$ .

$f'' (x) = \frac{- 2 x^{2} + 2 \cdot 1}{{(x^{2} + 1)}^{2}}$

Step 2.10.2.2

Multiply $2$ by $1$ .

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

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$\frac{2 x}{x^{2} + 1} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x^{2} + 1$ .

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Step 4.1.1.1

To apply the Chain Rule, set $u$ as $x^{2} + 1$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{2} + 1]$

Step 4.1.1.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$\frac{1}{u} \frac{d}{d x} [x^{2} + 1]$

Step 4.1.1.3

Replace all occurrences of $u$ with $x^{2} + 1$ .

$\frac{1}{x^{2} + 1} \frac{d}{d x} [x^{2} + 1]$

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$\frac{1}{x^{2} + 1} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1])$

Step 4.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$\frac{1}{x^{2} + 1} (2 x + \frac{d}{d x} [1])$

Step 4.1.2.3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{1}{x^{2} + 1} (2 x + 0)$

Step 4.1.2.4

Combine fractions.

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Step 4.1.2.4.1

Add $2 x$ and $0$ .

$\frac{1}{x^{2} + 1} (2 x)$

Step 4.1.2.4.2

Combine $2$ and $\frac{1}{x^{2} + 1}$ .

$\frac{2}{x^{2} + 1} x$

Step 4.1.2.4.3

Combine $\frac{2}{x^{2} + 1}$ and $x$ .

$f' (x) = \frac{2 x}{x^{2} + 1}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{2 x}{x^{2} + 1}$ .

$\frac{2 x}{x^{2} + 1}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{2 x}{x^{2} + 1} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{2 x}{x^{2} + 1} = 0$

Step 5.2

Set the numerator equal to zero.

$2 x = 0$

Step 5.3

Divide each term in $2 x = 0$ by $2$ and simplify.

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Step 5.3.1

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $2$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Step 5.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Divide $0$ by $2$ .

$x = 0$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = 0$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{- 2 {(0)}^{2} + 2}{{({(0)}^{2} + 1)}^{2}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Raising $0$ to any positive power yields $0$ .

$\frac{- 2 \cdot 0 + 2}{{(0^{2} + 1)}^{2}}$

Step 9.1.2

Multiply $- 2$ by $0$ .

$\frac{0 + 2}{{(0^{2} + 1)}^{2}}$

Step 9.1.3

Add $0$ and $2$ .

$\frac{2}{{(0^{2} + 1)}^{2}}$

Step 9.2

Simplify the denominator.

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Step 9.2.1

Raising $0$ to any positive power yields $0$ .

$\frac{2}{{(0 + 1)}^{2}}$

Step 9.2.2

Add $0$ and $1$ .

$\frac{2}{1^{2}}$

Step 9.2.3

One to any power is one.

$\frac{2}{1}$

Step 9.3

Divide $2$ by $1$ .

$2$

Step 10

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Step 11

Find the y-value when $x = 0$ .

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Step 11.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = \ln ({(0)}^{2} + 1)$

Step 11.2

Simplify the result.

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Step 11.2.1

Raising $0$ to any positive power yields $0$ .

$f (0) = \ln (0 + 1)$

Step 11.2.2

Add $0$ and $1$ .

$f (0) = \ln (1)$

Step 11.2.3

The natural logarithm of $1$ is $0$ .

$f (0) = 0$

Step 11.2.4

The final answer is $0$ .

$y = 0$

Step 12

These are the local extrema for $f (x) = \ln (x^{2} + 1)$ .

$(0, 0)$ is a local minima

Step 13