Calculus Examples

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

Step 1

Differentiate.

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Step 1.1

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

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

Step 1.2

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

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

Step 2

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

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

Step 3

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

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

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 $u$ as $- x$ .

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

Step 4.2

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

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

Step 4.3

Replace all occurrences of $u$ with $- x$ .

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

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

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

Step 5.2

Multiply.

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Step 5.2.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2.2

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

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

Step 5.3

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

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

Step 5.4

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

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

Step 6

Simplify.

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Step 6.1

Apply the distributive property.

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

Step 6.2

Simplify each term.

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Step 6.2.1

Combine $\frac{1}{2}$ and $e^{x}$ .

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

Step 6.2.2

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

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