Calculus Examples

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

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 $g (x) = - \frac{1}{x^{2}}$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $- \frac{1}{x^{2}}$ .

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

Step 1.2

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

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

Step 1.3

Replace all occurrences of $u$ with $- \frac{1}{x^{2}}$ .

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

Step 2

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 $g (x) = \frac{1}{x^{2}}$ .

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

Step 3

Differentiate.

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Step 3.1

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

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

Step 3.2

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 3.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.2.2

Multiply $2$ by $- 1$ .

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

Step 3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 2$ .

$e^{- \frac{1}{x^{2}}} (- (- 2 x^{- 3}) + \frac{1}{x^{2}} \frac{d}{d x} [- 1])$

Step 3.4

Multiply $- 2$ by $- 1$ .

$e^{- \frac{1}{x^{2}}} (2 x^{- 3} + \frac{1}{x^{2}} \frac{d}{d x} [- 1])$

Step 3.5

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

$e^{- \frac{1}{x^{2}}} (2 x^{- 3} + \frac{1}{x^{2}} \cdot 0)$

Step 3.6

Simplify the expression.

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Step 3.6.1

Multiply $\frac{1}{x^{2}}$ by $0$ .

$e^{- \frac{1}{x^{2}}} (2 x^{- 3} + 0)$

Step 3.6.2

Add $2 x^{- 3}$ and $0$ .

$e^{- \frac{1}{x^{2}}} (2 x^{- 3})$

Step 4

Simplify.

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Step 4.1

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

$e^{- \frac{1}{x^{2}}} (2 \frac{1}{x^{3}})$

Step 4.2

Combine terms.

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Step 4.2.1

Combine $2$ and $\frac{1}{x^{3}}$ .

$e^{- \frac{1}{x^{2}}} \frac{2}{x^{3}}$

Step 4.2.2

Combine $e^{- \frac{1}{x^{2}}}$ and $\frac{2}{x^{3}}$ .

$\frac{e^{- \frac{1}{x^{2}}} \cdot 2}{x^{3}}$

Step 4.2.3

Move $2$ to the left of $e^{- \frac{1}{x^{2}}}$ .

$\frac{2 e^{- \frac{1}{x^{2}}}}{x^{3}}$