Calculus Examples

$y = \ln (\frac{5 + e^{x}}{5 - e^{x}})$

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

$g (x) = \frac{5 + e^{x}}{5 - e^{x}}$ .

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Step 1.1

To apply the Chain Rule, set

$u$ as

$\frac{5 + e^{x}}{5 - e^{x}}$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [\frac{5 + e^{x}}{5 - e^{x}}]$

Step 1.2

The derivative of

$\ln (u)$ with respect to

$u$ is

$\frac{1}{u}$ .

$\frac{1}{u} \frac{d}{d x} [\frac{5 + e^{x}}{5 - e^{x}}]$

Step 1.3

Replace all occurrences of

$u$ with

$\frac{5 + e^{x}}{5 - e^{x}}$ .

$\frac{1}{\frac{5 + e^{x}}{5 - e^{x}}} \frac{d}{d x} [\frac{5 + e^{x}}{5 - e^{x}}]$

Step 2

Multiply by the reciprocal of the fraction to divide by

$\frac{5 + e^{x}}{5 - e^{x}}$ .

$1 \frac{5 - e^{x}}{5 + e^{x}} \frac{d}{d x} [\frac{5 + e^{x}}{5 - e^{x}}]$

Step 3

Multiply

$\frac{5 - e^{x}}{5 + e^{x}}$ by

$1$ .

$\frac{5 - e^{x}}{5 + e^{x}} \frac{d}{d x} [\frac{5 + e^{x}}{5 - e^{x}}]$

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

$f (x) = 5 + e^{x}$ and

$g (x) = 5 - e^{x}$ .

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) \frac{d}{d x} [5 + e^{x}] - (5 + e^{x}) \frac{d}{d x} [5 - e^{x}]}{{(5 - e^{x})}^{2}}$

Step 5

Differentiate.

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Step 5.1

By the Sum Rule, the derivative of

$5 + e^{x}$ with respect to

$x$ is

$\frac{d}{d x} [5] + \frac{d}{d x} [e^{x}]$ .

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) (\frac{d}{d x} [5] + \frac{d}{d x} [e^{x}]) - (5 + e^{x}) \frac{d}{d x} [5 - e^{x}]}{{(5 - e^{x})}^{2}}$

Step 5.2

Since

$5$ is constant with respect to

$x$ , the derivative of

$5$ with respect to

$x$ is

$0$ .

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) (0 + \frac{d}{d x} [e^{x}]) - (5 + e^{x}) \frac{d}{d x} [5 - e^{x}]}{{(5 - e^{x})}^{2}}$

Step 5.3

Add

$0$ and

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

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) \frac{d}{d x} [e^{x}] - (5 + e^{x}) \frac{d}{d x} [5 - e^{x}]}{{(5 - e^{x})}^{2}}$

Step 6

Differentiate using the Exponential Rule which states that

$\frac{d}{d x} [a^{x}]$ is

$a^{x} \ln (a)$ where

$a$ =

$e$ .

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) e^{x} - (5 + e^{x}) \frac{d}{d x} [5 - e^{x}]}{{(5 - e^{x})}^{2}}$

Step 7

Differentiate.

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Step 7.1

By the Sum Rule, the derivative of

$5 - e^{x}$ with respect to

$x$ is

$\frac{d}{d x} [5] + \frac{d}{d x} [- e^{x}]$ .

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) e^{x} - (5 + e^{x}) (\frac{d}{d x} [5] + \frac{d}{d x} [- e^{x}])}{{(5 - e^{x})}^{2}}$

Step 7.2

Since

$5$ is constant with respect to

$x$ , the derivative of

$5$ with respect to

$x$ is

$0$ .

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) e^{x} - (5 + e^{x}) (0 + \frac{d}{d x} [- e^{x}])}{{(5 - e^{x})}^{2}}$

Step 7.3

Add

$0$ and

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

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) e^{x} - (5 + e^{x}) \frac{d}{d x} [- e^{x}]}{{(5 - e^{x})}^{2}}$

Step 7.4

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- e^{x}$ with respect to

$x$ is

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

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) e^{x} - (5 + e^{x}) (- \frac{d}{d x} [e^{x}])}{{(5 - e^{x})}^{2}}$

Step 7.5

Multiply.

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Step 7.5.1

Multiply

$- 1$ by

$- 1$ .

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) e^{x} + 1 (5 + e^{x}) \frac{d}{d x} [e^{x}]}{{(5 - e^{x})}^{2}}$

Step 7.5.2

Multiply

$5 + e^{x}$ by

$1$ .

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) e^{x} + (5 + e^{x}) \frac{d}{d x} [e^{x}]}{{(5 - e^{x})}^{2}}$

Step 8

Differentiate using the Exponential Rule which states that

$\frac{d}{d x} [a^{x}]$ is

$a^{x} \ln (a)$ where

$a$ =

$e$ .

$\frac{5 - e^{x}}{5 + e^{x}} \cdot \frac{(5 - e^{x}) e^{x} + (5 + e^{x}) e^{x}}{{(5 - e^{x})}^{2}}$

Step 9

Multiply

$\frac{5 - e^{x}}{5 + e^{x}}$ by

$\frac{(5 - e^{x}) e^{x} + (5 + e^{x}) e^{x}}{{(5 - e^{x})}^{2}}$ .

$\frac{(5 - e^{x}) ((5 - e^{x}) e^{x} + (5 + e^{x}) e^{x})}{(5 + e^{x}) {(5 - e^{x})}^{2}}$

Step 10

Cancel the common factors.

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Step 10.1

Factor

$5 - e^{x}$ out of

$(5 + e^{x}) {(5 - e^{x})}^{2}$ .

$\frac{(5 - e^{x}) ((5 - e^{x}) e^{x} + (5 + e^{x}) e^{x})}{(5 - e^{x}) ((5 + e^{x}) (5 - e^{x}))}$

Step 10.2

Cancel the common factor.

$\frac{(5 - e^{x}) ((5 - e^{x}) e^{x} + (5 + e^{x}) e^{x})}{(5 - e^{x}) ((5 + e^{x}) (5 - e^{x}))}$

Step 10.3

Rewrite the expression.

$\frac{(5 - e^{x}) e^{x} + (5 + e^{x}) e^{x}}{(5 + e^{x}) (5 - e^{x})}$

Step 11

Simplify.

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Step 11.1

Apply the distributive property.

$\frac{5 e^{x} - e^{x} e^{x} + (5 + e^{x}) e^{x}}{(5 + e^{x}) (5 - e^{x})}$

Step 11.2

Apply the distributive property.

$\frac{5 e^{x} - e^{x} e^{x} + 5 e^{x} + e^{x} e^{x}}{(5 + e^{x}) (5 - e^{x})}$

Step 11.3

Simplify the numerator.

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Step 11.3.1

Combine the opposite terms in

$5 e^{x} - e^{x} e^{x} + 5 e^{x} + e^{x} e^{x}$ .

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Step 11.3.1.1

Add

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

$e^{x} e^{x}$ .

$\frac{5 e^{x} + 5 e^{x} + 0}{(5 + e^{x}) (5 - e^{x})}$

Step 11.3.1.2

Add

$5 e^{x} + 5 e^{x}$ and

$0$ .

$\frac{5 e^{x} + 5 e^{x}}{(5 + e^{x}) (5 - e^{x})}$

Step 11.3.2

Add

$5 e^{x}$ and

$5 e^{x}$ .

$\frac{10 e^{x}}{(5 + e^{x}) (5 - e^{x})}$