f(x) = sin^3 2x, then f '(x) =?

Question

possible solutions....

-6(sin 2x)(cos^2 2x)

-4(sin 2x)(cos 2x)

4(sin 2x)(cos 2x)

6(sin^2 2x)(cos 2x)

Answer 1

f(x) = sin^3(2x)

You have three functions to deal with; the "cubed", the sin and the 2x.

Let's rewrite this slightly, to make the differentiation more obvious.

f(x) = [sin(2x)]^3
f'(x) = 3[sin(2x)]^2 (cos(2x)) (2)

Simplifying this,

f'(x) = 6[sin^2(2x)] (cos(2x))

Answer 2

f '(x)
= 3(sin^2 2x)(cos2x) (2) , use chain rule
= 6(sin^2 2x)(cos 2x)

Answer 3

f'(x)=

3(sin(2x))^(2)(cos(2x)(2)

= 6sin^2(2x)(cos2x)

Chain Rule and power rule combination,

remember sin^3(2x)= (sin(2x))^(3)