Calculus Examples

$\ln (x^{4} + 1)$

Step 1

Write $\ln (x^{4} + 1)$ as a function.

$f (x) = \ln (x^{4} + 1)$

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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 $g (x) = x^{4} + 1$ .

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Step 2.1.1.1

To apply the Chain Rule, set $u$ as $x^{4} + 1$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{4} + 1]$

Step 2.1.1.2

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

$\frac{1}{u} \frac{d}{d x} [x^{4} + 1]$

Step 2.1.1.3

Replace all occurrences of $u$ with $x^{4} + 1$ .

$\frac{1}{x^{4} + 1} \frac{d}{d x} [x^{4} + 1]$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

By the Sum Rule, the derivative of $x^{4} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [1]$ .

$\frac{1}{x^{4} + 1} (\frac{d}{d x} [x^{4}] + \frac{d}{d x} [1])$

Step 2.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$\frac{1}{x^{4} + 1} (4 x^{3} + \frac{d}{d x} [1])$

Step 2.1.2.3

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

$\frac{1}{x^{4} + 1} (4 x^{3} + 0)$

Step 2.1.2.4

Combine fractions.

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Step 2.1.2.4.1

Add $4 x^{3}$ and $0$ .

$\frac{1}{x^{4} + 1} (4 x^{3})$

Step 2.1.2.4.2

Combine $4$ and $\frac{1}{x^{4} + 1}$ .

$\frac{4}{x^{4} + 1} x^{3}$

Step 2.1.2.4.3

Combine $\frac{4}{x^{4} + 1}$ and $x^{3}$ .

$f' (x) = \frac{4 x^{3}}{x^{4} + 1}$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{4 x^{3}}{x^{4} + 1}$ .

$\frac{4 x^{3}}{x^{4} + 1}$

Step 3

Set the first derivative equal to $0$ then solve the equation $\frac{4 x^{3}}{x^{4} + 1} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$\frac{4 x^{3}}{x^{4} + 1} = 0$

Step 3.2

Set the numerator equal to zero.

$4 x^{3} = 0$

Step 3.3

Solve the equation for $x$ .

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Step 3.3.1

Divide each term in $4 x^{3} = 0$ by $4$ and simplify.

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Step 3.3.1.1

Divide each term in $4 x^{3} = 0$ by $4$ .

$\frac{4 x^{3}}{4} = \frac{0}{4}$

Step 3.3.1.2

Simplify the left side.

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Step 3.3.1.2.1

Cancel the common factor of $4$ .

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Step 3.3.1.2.1.1

Cancel the common factor.

$\frac{4 x^{3}}{4} = \frac{0}{4}$

Step 3.3.1.2.1.2

Divide $x^{3}$ by $1$ .

$x^{3} = \frac{0}{4}$

Step 3.3.1.3

Simplify the right side.

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Step 3.3.1.3.1

Divide $0$ by $4$ .

$x^{3} = 0$

Step 3.3.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Step 3.3.3

Simplify $\sqrt[3]{0}$ .

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Step 3.3.3.1

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Step 3.3.3.2

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Step 4

The values which make the derivative equal to $0$ are $0$ .

$0$

Step 5

After finding the point that makes the derivative $f' (x) = \frac{4 x^{3}}{x^{4} + 1}$ equal to $0$ or undefined, the interval to check where $f (x) = \ln (x^{4} + 1)$ is increasing and where it is decreasing is $(- \infty, 0) \cup (0, \infty)$ .

$(- \infty, 0) \cup (0, \infty)$

Step 6

Substitute a value from the interval $(- \infty, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 1$ in the expression.

$f' (- 1) = \frac{4 {(- 1)}^{3}}{{(- 1)}^{4} + 1}$

Step 6.2

Simplify the result.

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Step 6.2.1

Raise $- 1$ to the power of $3$ .

$f' (- 1) = \frac{4 \cdot - 1}{{(- 1)}^{4} + 1}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Raise $- 1$ to the power of $4$ .

$f' (- 1) = \frac{4 \cdot - 1}{1 + 1}$

Step 6.2.2.2

Add $1$ and $1$ .

$f' (- 1) = \frac{4 \cdot - 1}{2}$

Step 6.2.3

Simplify the expression.

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Step 6.2.3.1

Multiply $4$ by $- 1$ .

$f' (- 1) = \frac{- 4}{2}$

Step 6.2.3.2

Divide $- 4$ by $2$ .

$f' (- 1) = - 2$

Step 6.2.4

The final answer is $- 2$ .

$- 2$

Step 6.3

At $x = - 1$ the derivative is $- 2$ . Since this is negative, the function is decreasing on $(- \infty, 0)$ .

Decreasing on $(- \infty, 0)$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(0, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = \frac{4 {(1)}^{3}}{{(1)}^{4} + 1}$

Step 7.2

Simplify the result.

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Step 7.2.1

One to any power is one.

$f' (1) = \frac{4 \cdot 1}{1^{4} + 1}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

One to any power is one.

$f' (1) = \frac{4 \cdot 1}{1 + 1}$

Step 7.2.2.2

Add $1$ and $1$ .

$f' (1) = \frac{4 \cdot 1}{2}$

Step 7.2.3

Simplify the expression.

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Step 7.2.3.1

Multiply $4$ by $1$ .

$f' (1) = \frac{4}{2}$

Step 7.2.3.2

Divide $4$ by $2$ .

$f' (1) = 2$

Step 7.2.4

The final answer is $2$ .

$2$

Step 7.3

At $x = 1$ the derivative is $2$ . Since this is positive, the function is increasing on $(0, \infty)$ .

Increasing on $(0, \infty)$ since $f' (x) > 0$

Step 8

List the intervals on which the function is increasing and decreasing.

Increasing on: $(0, \infty)$

Decreasing on: $(- \infty, 0)$

Step 9