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

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Answer 1 · 2018-04-17T15:25:26

$g'(x)=2xe^(-x)[4x-x^2-2]$

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

$g(x)=e^(-x)(2x-2)x^2$

$g(x)=e^(-x)(2x^3-2x^2)$

$g(x)=color(red)(u)v$

$g'(x)=color(red)(u')v+color(red)(u)v'$

$color(red)(u=2x^3-2x^2=>u'=6x^2-4x)$

$v=e^-x=>v'=-e^(-x)$

$g'(x)=color(red)((6x^2-4x))e^(-x)+color(red)((2x^3-2x^2))(-e^(-x))$

$g'(x)=e^(-x)[(6x^2-4x)-(2x^3-2x^2)]$

$g'(x)=e^(-x)[8x^2-2x^3-4x]$

$g'(x)=2xe^(-x)[4x-x^2-2]$

How do you differentiate g(x) = e^(-x) ( 2x-2)x^2g(x)=e−x(2x−2)x2g(x) = e^(-x) ( 2x-2)x^2 using the product rule?