# How do you find the derivative of y = x^2 e^(-x)?

Jul 6, 2016

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

#### Explanation:

$y = {x}^{2} {e}^{-} x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left({e}^{-} x\right) + {e}^{-} x \cdot \frac{d}{\mathrm{dx}} {x}^{2}$.... [Product Rule for Diffn.]
$= {x}^{2} \cdot {e}^{-} x \cdot \frac{d}{\mathrm{dx}} \left(- x\right) + {e}^{-} x \cdot 2 x$...................[Chain rule]
$= - {x}^{2} {e}^{-} x + 2 x \cdot {e}^{-} x = x {e}^{-} x \left(2 - x\right) .$

Jul 6, 2016

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

#### Explanation:

Differentiate using the $\textcolor{b l u e}{\text{product rule}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f \left(x\right) = g \left(x\right) h \left(x\right) \Rightarrow f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here $g \left(x\right) = {x}^{2} \Rightarrow g ' \left(x\right) = 2 x$

and $h \left(x\right) = {e}^{- x} \Rightarrow h ' \left(x\right) = {e}^{- x} \left(- 1\right) = - {e}^{- x}$
$\text{-------------------------------------------------------------------}$
Substitute these values into f'(x)

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

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