Find the Derivative – d/dx e^(1/(x^2))

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

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

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

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

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

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

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

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

Differentiate using the Power Rule.

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

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

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

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

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

Multiply $2$ by $- 1$ .

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

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

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

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

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

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

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

Combine terms.

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

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

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

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

Move the negative in front of the fraction.

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

Find the Derivative – d/dx e^(1/(x^2))

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