Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

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

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

Step 4

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

To apply the Chain Rule, set $u_{1}$ as $- x$ .

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

Step 4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 4.3

Replace all occurrences of $u_{1}$ with $- x$ .

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

Step 5

Differentiate.

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Simplify the expression.

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Step 5.3.1

Multiply $- 1$ by $1$ .

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

Step 5.3.2

Move $- 1$ to the left of $e^{- x}$ .

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

Step 5.3.3

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

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

Step 6

Raise $e^{x} - e^{- x}$ to the power of $1$ .

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

Step 7

Raise $e^{x} - e^{- x}$ to the power of $1$ .

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

Step 8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 9

Differentiate using the Sum Rule.

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Step 9.1

Add $1$ and $1$ .

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

Step 9.2

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

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

Step 10

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

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

Step 11

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

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

Step 12

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 12.1

To apply the Chain Rule, set $u_{2}$ as $- x$ .

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

Step 12.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

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

Step 12.3

Replace all occurrences of $u_{2}$ with $- x$ .

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

Step 13

Differentiate.

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Step 13.1

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

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

Step 13.2

Multiply.

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Step 13.2.1

Multiply $- 1$ by $- 1$ .

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

Step 13.2.2

Multiply $e^{- x}$ by $1$ .

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

Step 13.3

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

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

Step 13.4

Multiply $e^{- x}$ by $1$ .

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

Step 14

Raise $e^{x} + e^{- x}$ to the power of $1$ .

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

Step 15

Raise $e^{x} + e^{- x}$ to the power of $1$ .

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

Step 16

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 17

Add $1$ and $1$ .

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

Step 18

Simplify.

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Step 18.1

Simplify the numerator.

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Step 18.1.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = e^{x} - e^{- x}$ and $b = e^{x} + e^{- x}$ .

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

Step 18.1.2

Simplify.

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Step 18.1.2.1

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

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

Step 18.1.2.2

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

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

Step 18.1.2.3

Add $2 e^{x}$ and $0$ .

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

Step 18.1.2.4

Apply the distributive property.

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

Step 18.1.2.5

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

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

Step 18.1.2.6

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

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

Step 18.1.2.7

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

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

Step 18.1.2.8

Combine exponents.

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Step 18.1.2.8.1

Multiply $- 2$ by $2$ .

$\frac{- 4 e^{x} e^{- x}}{{(e^{x} - e^{- x})}^{2}}$

Step 18.1.2.8.2

Multiply $e^{x}$ by $e^{- x}$ by adding the exponents.

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Step 18.1.2.8.2.1

Move $e^{- x}$ .

$\frac{- 4 (e^{- x} e^{x})}{{(e^{x} - e^{- x})}^{2}}$

Step 18.1.2.8.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{- 4 e^{- x + x}}{{(e^{x} - e^{- x})}^{2}}$

Step 18.1.2.8.2.3

Add $- x$ and $x$ .

$\frac{- 4 e^{0}}{{(e^{x} - e^{- x})}^{2}}$

Step 18.1.2.8.3

Simplify $- 4 e^{0}$ .

$\frac{- 4}{{(e^{x} - e^{- x})}^{2}}$

Step 18.2

Move the negative in front of the fraction.

$- \frac{4}{{(e^{x} - e^{- x})}^{2}}$