What is the derivative of e^((x^2)/2)?

Feb 22, 2016

Use the chain rule to find that

$\frac{d}{\mathrm{dx}} {e}^{{x}^{2} / 2} = x {e}^{{x}^{2} / 2}$

Explanation:

The chain rule states that given two differentiable functions $f$ and $g$

$\frac{d}{\mathrm{dx}} f \circ g \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

In this case, let $f \left(x\right) = {e}^{x}$ and $g \left(x\right) = \frac{1}{2} {x}^{2}$

Then $f ' \left(x\right) = {e}^{x}$ and $g ' \left(x\right) = \frac{1}{2} \left(2 x\right) = x$

So, by the chain rule, we have

$\frac{d}{\mathrm{dx}} {e}^{{x}^{2} / 2} = \frac{d}{\mathrm{dx}} f \circ g \left(x\right)$

$= f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

$= {e}^{g \left(x\right)} \cdot x$

$= x {e}^{{x}^{2} / 2}$