Calculus Examples

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

Step 1

Write

$\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$ as a function.

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

Step 2

The function

$F (x)$ can be found by finding the indefinite integral of the derivative

$f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x$

Step 4

Let

$u_{2} = e^{x} + e^{- x}$ . Then

$d u_{2} = (e^{x} - e^{- x}) d x$ , so

$\frac{1}{e^{x} - e^{- x}} d u_{2} = d x$ . Rewrite using

$u_{2}$ and

$d$

$u_{2}$ .

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Step 4.1

Let

$u_{2} = e^{x} + e^{- x}$ . Find

$\frac{d u_{2}}{d x}$ .

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Step 4.1.1

Differentiate

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

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

Step 4.1.2

By the Sum Rule, the derivative of

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

$x$ is

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

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

Step 4.1.3

Differentiate using the Exponential Rule which states that

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

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

$a$ =

$e$ .

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

Step 4.1.4

Evaluate

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

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Step 4.1.4.1

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

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

$g (x) = - x$ .

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Step 4.1.4.1.1

To apply the Chain Rule, set

$u_{1}$ as

$- x$ .

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

Step 4.1.4.1.2

Differentiate using the Exponential Rule which states that

$\frac{d}{d u_{1}} [a^{u_{1}}]$ is

$a^{u_{1}} \ln (a)$ where

$a$ =

$e$ .

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

Step 4.1.4.1.3

Replace all occurrences of

$u_{1}$ with

$- x$ .

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

Step 4.1.4.2

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- x$ with respect to

$x$ is

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

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

Step 4.1.4.3

Differentiate using the Power Rule which states that

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

$n x^{n - 1}$ where

$n = 1$ .

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

Step 4.1.4.4

Multiply

$- 1$ by

$1$ .

$e^{x} + e^{- x} \cdot - 1$

Step 4.1.4.5

Move

$- 1$ to the left of

$e^{- x}$ .

$e^{x} - 1 \cdot e^{- x}$

Step 4.1.4.6

Rewrite

$- 1 e^{- x}$ as

$- e^{- x}$ .

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

Step 4.2

Rewrite the problem using

$u_{2}$ and

$d u_{2}$ .

$\int \frac{1}{u_{2}} d u_{2}$

Step 5

The integral of

$\frac{1}{u_{2}}$ with respect to

$u_{2}$ is

$\ln (| u_{2} |)$ .

$\ln (| u_{2} |) + C$

Step 6

Replace all occurrences of

$u_{2}$ with

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

$\ln (| e^{x} + e^{- x} |) + C$

Step 7

The answer is the antiderivative of the function

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

$F (x) =$

$\ln (| e^{x} + e^{- x} |) + C$