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

1 Answer
Apr 16, 2017

Answer:

#:. g'(x)=e^x{(e^x-1)(cosx-x^2)+e^x(cosx-x^2)-(e^x-1)(sinx+2x)}.#

Explanation:

#g(x)=(e^(2x) - e^x)(cosx - x^2)=e^x(e^x-1)(cosx-x^2).#

We will use the following Product Rule for Diffn. :

#(uvw)'=u'vw+uv'w+uvw'.#

#:. g'(x)#

#={e^x}'(e^x-1)(cosx-x^2)+e^x{(e^x-1)}'(cosx-x^2)+e^x(e^x-1){(cosx-x^2)}'.#

#={e^x}(e^x-1)(cosx-x^2)+e^x{e^x-0}(cosx-x^2)+e^x(e^x-1){-sinx-2x},#

#=e^x(e^x-1)(cosx-x^2)+e^(2x)(cosx-x^2)-e^x(e^x-1)(sinx+2x).#

#:. g'(x)=e^x{(e^x-1)(cosx-x^2)+e^x(cosx-x^2)-(e^x-1)(sinx+2x)}.#

Enjoy Maths.!