Calculus Examples

e^{\frac{1}{x}}

$e^{\frac{1}{x}}$

Step 1

Differentiate using the chain rule, which states that

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

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

f' (g (x)) g' (x)

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

f (x) = e^{x}

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

g (x) = \frac{1}{x}

$g (x) = \frac{1}{x}$ .

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Step 1.1

To apply the Chain Rule, set

u

$u$ as

\frac{1}{x}

$\frac{1}{x}$ .

\frac{d}{d u} [e^{u}] \frac{d}{d x} [\frac{1}{x}]

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

Step 1.2

Differentiate using the Exponential Rule which states that

\frac{d}{d u} [a^{u}]

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

a^{u} \ln (a)

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

a

$a$ =

e

$e$ .

e^{u} \frac{d}{d x} [\frac{1}{x}]

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

Step 1.3

Replace all occurrences of

u

$u$ with

\frac{1}{x}

$\frac{1}{x}$ .

e^{\frac{1}{x}} \frac{d}{d x} [\frac{1}{x}]

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

e^{\frac{1}{x}} \frac{d}{d x} [\frac{1}{x}]

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

Step 2

Differentiate using the Power Rule.

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Step 2.1

Rewrite

\frac{1}{x}

$\frac{1}{x}$ as

x^{- 1}

$x^{- 1}$ .

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

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

Step 2.2

Differentiate using the Power Rule which states that

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

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

n x^{n - 1}

$n x^{n - 1}$ where

n = - 1

$n = - 1$ .

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

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

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

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

Step 3

Simplify.

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Step 3.1

Rewrite the expression using the negative exponent rule

b^{- n} = \frac{1}{b^{n}}

$b^{- n} = \frac{1}{b^{n}}$ .

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

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

Step 3.2

Combine

e^{\frac{1}{x}}

$e^{\frac{1}{x}}$ and

\frac{1}{x^{2}}

$\frac{1}{x^{2}}$ .

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

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

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

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