Calculus Examples

$f (x) = e^{\frac{1}{x}}$

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

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Step 1.1

To apply the Chain Rule, set $u$ as $\frac{1}{x}$ .

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

Step 1.2

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

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

Step 1.3

Replace all occurrences of $u$ with $\frac{1}{x}$ .

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

Step 2

Differentiate using the Power Rule.

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Step 2.1

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 2.2

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

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

Step 3

Simplify.

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Step 3.1

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

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

Step 3.2

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

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