Question #16171

1 Answer
May 12, 2017

f'(x) = 2(2x-1)^2(x^2+3) (7x^2-2x+9)

Explanation:

We can use a combination of the chain rule and the product rule:

We have:

f(x) = (2x-1)^3(x^2+3)^2

We can use the chain rule on each individual function. Let

{ (u=(2x-1)^3), (v=(x^2+3)^2) :} => { ((du)/dx=3(2x-1)^2(2),=6(2x-1)^2), ((dv)/dx=2(x^2+3)(2x),=4x(x^2+3)) :}

And then:

f(x) = uv

And by the product rule we have:

f'(x) = (u)((dv)/dx) + ((du)/dx)(v)
" " = (2x-1)^3 4x(x^2+3) + 6(2x-1)^2(x^2+3)^2)
" " = (2x-1)^2(x^2+3) {4x(2x-1)+6(x^2+3)}
" " = (2x-1)^2(x^2+3) (8x^2-4x+6x^2+18)
" " = (2x-1)^2(x^2+3) (14x^2-4x+18)
" " = 2(2x-1)^2(x^2+3) (7x^2-2x+9)