# How do you differentiate y=x^2e^-x?

Dec 14, 2016

Use the product rule and the chain rule.

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$

$\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x\right) \left({e}^{-} x\right) + \left({x}^{2}\right) \left({e}^{-} x \left(- 1\right)\right)$

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

If you like, remove the common factor.

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