Calculus Examples

$y = \frac{2}{e^{x} + e^{- x}}$

Step 1

Differentiate using the Constant Multiple Rule.

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Step 1.1

Since $2$ is constant with respect to $x$ , the derivative of $\frac{2}{e^{x} + e^{- x}}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{1}{e^{x} + e^{- x}}]$ .

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

Step 1.2

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

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = e^{x} + e^{- x}$ .

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

Differentiate using the Sum Rule.

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Step 3.1

Multiply $- 1$ by $2$ .

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

Step 3.2

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

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

Step 4

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

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

Step 5

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 5.1

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

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

Step 5.2

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

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

Step 5.3

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

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

Step 6

Differentiate.

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Simplify the expression.

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Step 6.3.1

Multiply $- 1$ by $1$ .

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

Step 6.3.2

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

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

Step 6.3.3

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

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

Step 7

Simplify.

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Step 7.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 7.2

Combine terms.

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Step 7.2.1

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

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

Step 7.2.2

Move the negative in front of the fraction.

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

Step 7.3

Reorder the factors of $- \frac{2}{{(e^{x} + e^{- x})}^{2}} (e^{x} - e^{- x})$ .

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

Step 7.4

Apply the distributive property.

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

Step 7.5

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

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Step 7.5.1

Multiply $- 1$ by $- 1$ .

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

Step 7.5.2

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

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

Step 7.6

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

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

Step 7.7

Move $2$ to the left of $- e^{x} + e^{- x}$ .

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

Step 7.8

Factor $- 1$ out of $- e^{x}$ .

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

Step 7.9

Factor $- 1$ out of $e^{- x}$ .

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

Step 7.10

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

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

Step 7.11

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

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

Step 7.12

Move the negative in front of the fraction.

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