Find f'(x) for f(x) = sin3(3x2).

Answer 1

Sorry it is sin^3(3x^2)

Answer 2

look at it as:
f(x) = (sin(3x^2))^3
f ' (x) = 3(sin(3x^2))^2 cos(3x^2) (6x)
= 18x sin^2 (3x^2) cos(3x^2)
or
= 9x sin(6x^2) cos(3x^2), using property sin 2A = 2sinAcosA

Answer 3

Hmm. I see an extra sin, not cos:

9x sin(6x^2) sin(3x^2)