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

Jun 15, 2016

$\frac{d}{\mathrm{dx}} \left({x}^{2} \cdot {e}^{-} x\right) = 2 x {e}^{-} x - {x}^{2} {e}^{-} x$

#### Explanation:

This problem requires use of the product rule, which states:
$\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$
Where $u$ and $v$ are functions of $x$.

In our case, $u = {x}^{2} \to u ' = 2 x$ and $v = {e}^{- x} \to v ' = - {e}^{- x}$. Thus
$\frac{d}{\mathrm{dx}} \left({x}^{2} \cdot {e}^{-} x\right) = \left(2 x\right) \left({e}^{-} x\right) + \left({x}^{2}\right) \left(- {e}^{-} x\right)$
$= 2 x {e}^{-} x - {x}^{2} {e}^{-} x$

We could simplify this a little further by, say, pulling out an $x {e}^{-} x$:
$\frac{d}{\mathrm{dx}} \left({x}^{2} \cdot {e}^{-} x\right) = x {e}^{-} x \left(2 - x\right)$

Either form is correct.