Find the Derivative minus d First, divide dx e to the power of(minus x to the power of two)

$e^{- x^{2}}$

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

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

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

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

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

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

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

Differentiate.

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Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Clarify the expression.

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

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

Should be rewritten $- 1 e^{- x^{2}}$ as $- e^{- x^{2}}$ .

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

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

$- e^{- x^{2}} (2 x)$

First, First, multiply $2$ by $- 1$ .

$- 2 e^{- x^{2}} x$

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