How do you differentiate f(x)= (2x+1)(2x-3)^3 using the product rule?

1 Answer
Jan 25, 2016

f'(x)=16x(2x-3)^2

Explanation:

Given:

f(x)=h(x)*g(x)

the Product Rule tells:

f'(x)=h'(x)*g(x)+g'(x)*h(x)

h(x)=(2x+1)=>h'(x)=2+0=2
g(x)=(2x-3)^3

To find g'(x) you can use the Chain Rule:

g(x)=[i(x)]^n=>g'(x)=n[i(x)]^(n-1)*i'(x)

:. g'(x)=3(2x-3)^2*(2+0)=2*3(2x-3)^2=
=6(2x-3)^2

Thus:

f'(x)=2*(2x-3)^3+6(2x-3)^2*(2x+1)=2(2x-3)^2(2x-3+3(2x+1))=
=2(2x-3)^2(2xcolor(green)cancel(-3)+6xcolor(green)cancel(+3))=2(2x-3)^2(8x)=
=16x(2x-3)^2