Calculus Examples

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

Step 1

Move the negative in front of the fraction.

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

Step 2

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

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Step 2.1

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 2.2

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 2.3

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

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

Step 3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{1}{x}$ .

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

Step 4

Differentiate.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Multiply.

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Step 4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.2

Multiply $x^{- 2}$ by $1$ .

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

Step 4.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$e^{- \frac{1}{x}} (x^{- 2} + \frac{1}{x} \cdot 0)$

Step 4.5

Simplify the expression.

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Step 4.5.1

Multiply $\frac{1}{x}$ by $0$ .

$e^{- \frac{1}{x}} (x^{- 2} + 0)$

Step 4.5.2

Add $x^{- 2}$ and $0$ .

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

Step 5

Simplify.

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Step 5.1

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

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

Step 5.2

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

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