Calculus Examples

$y = \ln (\frac{9 + e^{x}}{9 - e^{x}})$

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 $g (x) = \frac{9 + e^{x}}{9 - e^{x}}$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $\frac{9 + e^{x}}{9 - e^{x}}$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [\frac{9 + e^{x}}{9 - e^{x}}]$

Step 1.2

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

$\frac{1}{u} \frac{d}{d x} [\frac{9 + e^{x}}{9 - e^{x}}]$

Step 1.3

Replace all occurrences of $u$ with $\frac{9 + e^{x}}{9 - e^{x}}$ .

$\frac{1}{\frac{9 + e^{x}}{9 - e^{x}}} \frac{d}{d x} [\frac{9 + e^{x}}{9 - e^{x}}]$

Step 2

Multiply by the reciprocal of the fraction to divide by $\frac{9 + e^{x}}{9 - e^{x}}$ .

$1 \frac{9 - e^{x}}{9 + e^{x}} \frac{d}{d x} [\frac{9 + e^{x}}{9 - e^{x}}]$

Step 3

Multiply $\frac{9 - e^{x}}{9 + e^{x}}$ by $1$ .

$\frac{9 - e^{x}}{9 + e^{x}} \frac{d}{d x} [\frac{9 + e^{x}}{9 - e^{x}}]$

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 $f (x) = 9 + e^{x}$ and $g (x) = 9 - e^{x}$ .

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) \frac{d}{d x} [9 + e^{x}] - (9 + e^{x}) \frac{d}{d x} [9 - e^{x}]}{{(9 - e^{x})}^{2}}$

Step 5

Differentiate.

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Step 5.1

By the Sum Rule, the derivative of $9 + e^{x}$ with respect to $x$ is $\frac{d}{d x} [9] + \frac{d}{d x} [e^{x}]$ .

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) (\frac{d}{d x} [9] + \frac{d}{d x} [e^{x}]) - (9 + e^{x}) \frac{d}{d x} [9 - e^{x}]}{{(9 - e^{x})}^{2}}$

Step 5.2

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) (0 + \frac{d}{d x} [e^{x}]) - (9 + e^{x}) \frac{d}{d x} [9 - e^{x}]}{{(9 - e^{x})}^{2}}$

Step 5.3

Add $0$ and $\frac{d}{d x} [e^{x}]$ .

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) \frac{d}{d x} [e^{x}] - (9 + e^{x}) \frac{d}{d x} [9 - e^{x}]}{{(9 - e^{x})}^{2}}$

Step 6

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

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) e^{x} - (9 + e^{x}) \frac{d}{d x} [9 - e^{x}]}{{(9 - e^{x})}^{2}}$

Step 7

Differentiate.

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Step 7.1

By the Sum Rule, the derivative of $9 - e^{x}$ with respect to $x$ is $\frac{d}{d x} [9] + \frac{d}{d x} [- e^{x}]$ .

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) e^{x} - (9 + e^{x}) (\frac{d}{d x} [9] + \frac{d}{d x} [- e^{x}])}{{(9 - e^{x})}^{2}}$

Step 7.2

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) e^{x} - (9 + e^{x}) (0 + \frac{d}{d x} [- e^{x}])}{{(9 - e^{x})}^{2}}$

Step 7.3

Add $0$ and $\frac{d}{d x} [- e^{x}]$ .

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) e^{x} - (9 + e^{x}) \frac{d}{d x} [- e^{x}]}{{(9 - e^{x})}^{2}}$

Step 7.4

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

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) e^{x} - (9 + e^{x}) (- \frac{d}{d x} [e^{x}])}{{(9 - e^{x})}^{2}}$

Step 7.5

Multiply.

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Step 7.5.1

Multiply $- 1$ by $- 1$ .

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) e^{x} + 1 (9 + e^{x}) \frac{d}{d x} [e^{x}]}{{(9 - e^{x})}^{2}}$

Step 7.5.2

Multiply $9 + e^{x}$ by $1$ .

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) e^{x} + (9 + e^{x}) \frac{d}{d x} [e^{x}]}{{(9 - e^{x})}^{2}}$

Step 8

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

$\frac{9 - e^{x}}{9 + e^{x}} \cdot \frac{(9 - e^{x}) e^{x} + (9 + e^{x}) e^{x}}{{(9 - e^{x})}^{2}}$

Step 9

Multiply $\frac{9 - e^{x}}{9 + e^{x}}$ by $\frac{(9 - e^{x}) e^{x} + (9 + e^{x}) e^{x}}{{(9 - e^{x})}^{2}}$ .

$\frac{(9 - e^{x}) ((9 - e^{x}) e^{x} + (9 + e^{x}) e^{x})}{(9 + e^{x}) {(9 - e^{x})}^{2}}$

Step 10

Cancel the common factors.

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Step 10.1

Factor $9 - e^{x}$ out of $(9 + e^{x}) {(9 - e^{x})}^{2}$ .

$\frac{(9 - e^{x}) ((9 - e^{x}) e^{x} + (9 + e^{x}) e^{x})}{(9 - e^{x}) ((9 + e^{x}) (9 - e^{x}))}$

Step 10.2

Cancel the common factor.

$\frac{(9 - e^{x}) ((9 - e^{x}) e^{x} + (9 + e^{x}) e^{x})}{(9 - e^{x}) ((9 + e^{x}) (9 - e^{x}))}$

Step 10.3

Rewrite the expression.

$\frac{(9 - e^{x}) e^{x} + (9 + e^{x}) e^{x}}{(9 + e^{x}) (9 - e^{x})}$

Step 11

Simplify.

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Step 11.1

Apply the distributive property.

$\frac{9 e^{x} - e^{x} e^{x} + (9 + e^{x}) e^{x}}{(9 + e^{x}) (9 - e^{x})}$

Step 11.2

Apply the distributive property.

$\frac{9 e^{x} - e^{x} e^{x} + 9 e^{x} + e^{x} e^{x}}{(9 + e^{x}) (9 - e^{x})}$

Step 11.3

Simplify the numerator.

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Step 11.3.1

Combine the opposite terms in $9 e^{x} - e^{x} e^{x} + 9 e^{x} + e^{x} e^{x}$ .

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Step 11.3.1.1

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

$\frac{9 e^{x} + 9 e^{x} + 0}{(9 + e^{x}) (9 - e^{x})}$

Step 11.3.1.2

Add $9 e^{x} + 9 e^{x}$ and $0$ .

$\frac{9 e^{x} + 9 e^{x}}{(9 + e^{x}) (9 - e^{x})}$

Step 11.3.2

Add $9 e^{x}$ and $9 e^{x}$ .

$\frac{18 e^{x}}{(9 + e^{x}) (9 - e^{x})}$