Question

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

Answer 1

`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`