# How do you find the derivative of x^2*e^-x?

May 31, 2015

$\left(u v\right) ' = u ' v + u v '$

$u = {x}^{2}$
$u ' = 2 x$

$v = {e}^{- x}$
We know that $\left({e}^{a x}\right) ' = \left(a x\right) ' \cdot {e}^{a x}$.
$a x = - x$
$\left(- x\right) ' = - 1$
$v ' = - {e}^{- x}$

Therefore :

$\left(u v\right) ' = 2 x \left({e}^{- x}\right) + {x}^{2} \left(- {e}^{- x}\right)$

$\left(u v\right) ' = \left(2 x - {x}^{2}\right) \left({e}^{- x}\right)$

$\left(u v\right) ' = \left(2 - x\right) x {e}^{- x}$.