# How do you find the derivative of f(x)=xe^(-(x^2)/2)

Feb 22, 2015

You can use the Product Rule where:

if $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$

$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

in your case $g \left(x\right) = x$ but $h \left(x\right) = {e}^{- {x}^{2} / 2}$ which needs the Chain Rule to be derived (here you derive the $e$ as it is and multiply times the derivative of the exponent).

So you get:
$g \left(x\right) = x$
$g ' \left(x\right) = 1$
$h \left(x\right) = {e}^{- {x}^{2} / 2}$
$h ' \left(x\right) = {e}^{- {x}^{2} / 2} \left(- 2 \frac{x}{2}\right)$

$f ' \left(x\right) = 1 {e}^{- {x}^{2} / 2} + x {e}^{- {x}^{2} / 2} \left(- 2 \frac{x}{2}\right) =$
$= {e}^{- {x}^{2} / 2} \left(1 - {x}^{2}\right)$