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)