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

Question

1141 views around the world

You can reuse this answer
Creative Commons License

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]#