How do you differentiate g(x) = e^(-x) ( 2x-2)x^2 using the product rule?

Apr 17, 2018

$g ' \left(x\right) = 2 x {e}^{- x} \left[4 x - {x}^{2} - 2\right]$

Explanation:

let us first simplify the function by multiplying the ${x}^{2}$ into the bracket

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

$g \left(x\right) = {e}^{- x} \left(2 {x}^{3} - 2 {x}^{2}\right)$

the product rule

$g \left(x\right) = \textcolor{red}{u} v$

$g ' \left(x\right) = \textcolor{red}{u '} v + \textcolor{red}{u} v '$

$\textcolor{red}{u = 2 {x}^{3} - 2 {x}^{2} \implies u ' = 6 {x}^{2} - 4 x}$

$v = {e}^{-} x \implies v ' = - {e}^{- x}$

$g ' \left(x\right) = \textcolor{red}{\left(6 {x}^{2} - 4 x\right)} {e}^{- x} + \textcolor{red}{\left(2 {x}^{3} - 2 {x}^{2}\right)} \left(- {e}^{- x}\right)$

$g ' \left(x\right) = {e}^{- x} \left[\left(6 {x}^{2} - 4 x\right) - \left(2 {x}^{3} - 2 {x}^{2}\right)\right]$

$g ' \left(x\right) = {e}^{- x} \left[8 {x}^{2} - 2 {x}^{3} - 4 x\right]$

$g ' \left(x\right) = 2 x {e}^{- x} \left[4 x - {x}^{2} - 2\right]$