Calculus Examples

$f (x) = \ln (\frac{e^{x} + 1}{e^{x} - 1})$

Step 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 $g (x) = \frac{e^{x} + 1}{e^{x} - 1}$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $\frac{e^{x} + 1}{e^{x} - 1}$ .

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

Step 1.2

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

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

Step 1.3

Replace all occurrences of $u$ with $\frac{e^{x} + 1}{e^{x} - 1}$ .

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

Step 2

Multiply by the reciprocal of the fraction to divide by $\frac{e^{x} + 1}{e^{x} - 1}$ .

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

Step 3

Multiply $\frac{e^{x} - 1}{e^{x} + 1}$ by $1$ .

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

Step 4

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 $f (x) = e^{x} + 1$ and $g (x) = e^{x} - 1$ .

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

Step 5

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

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

Step 6

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

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

Step 7

Differentiate.

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Step 7.1

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

$\frac{e^{x} - 1}{e^{x} + 1} \cdot \frac{(e^{x} - 1) (e^{x} + 0) - (e^{x} + 1) \frac{d}{d x} [e^{x} - 1]}{{(e^{x} - 1)}^{2}}$

Step 7.2

Add $e^{x}$ and $0$ .

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

Step 7.3

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

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

Step 8

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

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

Step 9

Differentiate using the Constant Rule.

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Step 9.1

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

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

Step 9.2

Combine fractions.

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Step 9.2.1

Add $e^{x}$ and $0$ .

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

Step 9.2.2

Multiply $\frac{e^{x} - 1}{e^{x} + 1}$ by $\frac{(e^{x} - 1) e^{x} - (e^{x} + 1) e^{x}}{{(e^{x} - 1)}^{2}}$ .

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

Step 10

Cancel the common factors.

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Step 10.1

Factor $e^{x} - 1$ out of $(e^{x} + 1) {(e^{x} - 1)}^{2}$ .

$\frac{(e^{x} - 1) ((e^{x} - 1) e^{x} - (e^{x} + 1) e^{x})}{(e^{x} - 1) ((e^{x} + 1) (e^{x} - 1))}$

Step 10.2

Cancel the common factor.

$\frac{(e^{x} - 1) ((e^{x} - 1) e^{x} - (e^{x} + 1) e^{x})}{(e^{x} - 1) ((e^{x} + 1) (e^{x} - 1))}$

Step 10.3

Rewrite the expression.

$\frac{(e^{x} - 1) e^{x} - (e^{x} + 1) e^{x}}{(e^{x} + 1) (e^{x} - 1)}$

Step 11

Simplify.

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Step 11.1

Apply the distributive property.

$\frac{e^{x} e^{x} - 1 e^{x} - (e^{x} + 1) e^{x}}{(e^{x} + 1) (e^{x} - 1)}$

Step 11.2

Apply the distributive property.

$\frac{e^{x} e^{x} - 1 e^{x} + (- e^{x} - 1 \cdot 1) e^{x}}{(e^{x} + 1) (e^{x} - 1)}$

Step 11.3

Apply the distributive property.

$\frac{e^{x} e^{x} - 1 e^{x} - e^{x} e^{x} - 1 \cdot 1 e^{x}}{(e^{x} + 1) (e^{x} - 1)}$

Step 11.4

Simplify the numerator.

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Step 11.4.1

Combine the opposite terms in $e^{x} e^{x} - 1 e^{x} - e^{x} e^{x} - 1 \cdot 1 e^{x}$ .

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Step 11.4.1.1

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

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

Step 11.4.1.2

Subtract $1 e^{x}$ from $0$ .

$\frac{- 1 e^{x} - 1 \cdot 1 e^{x}}{(e^{x} + 1) (e^{x} - 1)}$

Step 11.4.2

Simplify each term.

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Step 11.4.2.1

Rewrite $- 1 e^{x}$ as $- e^{x}$ .

$\frac{- e^{x} - 1 \cdot 1 e^{x}}{(e^{x} + 1) (e^{x} - 1)}$

Step 11.4.2.2

Multiply $- 1$ by $1$ .

$\frac{- e^{x} - 1 e^{x}}{(e^{x} + 1) (e^{x} - 1)}$

Step 11.4.2.3

Rewrite $- 1 e^{x}$ as $- e^{x}$ .

$\frac{- e^{x} - e^{x}}{(e^{x} + 1) (e^{x} - 1)}$

Step 11.4.3

Subtract $e^{x}$ from $- e^{x}$ .

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

Step 11.5

Move the negative in front of the fraction.

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