Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

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 1.2.1.1

To apply the Chain Rule, set $u$ as $- x$ .

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

Step 1.2.1.2

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

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

Step 1.2.1.3

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

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

Step 1.2.2

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

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

Step 1.2.3

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

$f' (x) = e^{- x} (- 1 \cdot 1) + \frac{d}{d x} (e^{x})$

Step 1.2.4

Multiply $- 1$ by $1$ .

$f' (x) = e^{- x} \cdot - 1 + \frac{d}{d x} (e^{x})$

Step 1.2.5

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

$f' (x) = - 1 \cdot e^{- x} + \frac{d}{d x} (e^{x})$

Step 1.2.6

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

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

Step 1.3

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

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

Step 1.4

Reorder terms.

$f' (x) = e^{x} - e^{- x}$

Step 2

Find the second derivative.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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Step 2.3.2.1

To apply the Chain Rule, set $u$ as $- x$ .

$f'' (x) = e^{x} - (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- x))$

Step 2.3.2.2

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

$f'' (x) = e^{x} - (e^{u} \frac{d}{d x} (- x))$

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

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

$f'' (x) = e^{x} - (e^{- x} (- 1 \cdot 1))$

Step 2.3.5

Multiply $- 1$ by $1$ .

$f'' (x) = e^{x} - (e^{- x} \cdot - 1)$

Step 2.3.6

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

$f'' (x) = e^{x} - (- 1 \cdot e^{- x})$

Step 2.3.7

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

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

Step 2.3.8

Multiply $- 1$ by $- 1$ .

$f'' (x) = e^{x} + 1 e^{- x}$

Step 2.3.9

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

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

Step 3

Find the third derivative.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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Step 3.3.1

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 3.3.1.1

To apply the Chain Rule, set $u$ as $- x$ .

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

Step 3.3.1.2

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

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

Step 3.3.1.3

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

Multiply $- 1$ by $1$ .

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

Step 3.3.5

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

$f''' (x) = e^{x} - 1 \cdot e^{- x}$

Step 3.3.6

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

$f''' (x) = e^{x} - e^{- x}$

Step 4

Find the fourth derivative.

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Step 4.1

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

$f^{4} (x) = \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- e^{- x})$

Step 4.2

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

$f^{4} (x) = e^{x} + \frac{d}{d x} (- e^{- x})$

Step 4.3

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

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Step 4.3.1

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

$f^{4} (x) = e^{x} - \frac{d}{d x} e^{- x}$

Step 4.3.2

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

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Step 4.3.2.1

To apply the Chain Rule, set $u$ as $- x$ .

$f^{4} (x) = e^{x} - (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- x))$

Step 4.3.2.2

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

$f^{4} (x) = e^{x} - (e^{u} \frac{d}{d x} (- x))$

Step 4.3.2.3

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

$f^{4} (x) = e^{x} - (e^{- x} \frac{d}{d x} (- x))$

Step 4.3.3

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

$f^{4} (x) = e^{x} - (e^{- x} (- \frac{d}{d x} x))$

Step 4.3.4

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

$f^{4} (x) = e^{x} - (e^{- x} (- 1 \cdot 1))$

Step 4.3.5

Multiply $- 1$ by $1$ .

$f^{4} (x) = e^{x} - (e^{- x} \cdot - 1)$

Step 4.3.6

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

$f^{4} (x) = e^{x} - (- 1 \cdot e^{- x})$

Step 4.3.7

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

$f^{4} (x) = e^{x} + e^{- x}$

Step 4.3.8

Multiply $- 1$ by $- 1$ .

$f^{4} (x) = e^{x} + 1 e^{- x}$

Step 4.3.9

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

$f^{4} (x) = e^{x} + e^{- x}$