How do you find the derivative of sin^3(2x)?

chain rule

3 sin^2 (2X) * 2 cos (2X) = 6 sin^2 (2X) cos (2X)

Put sin[2x] = p
So dp/dx = 2*cos[2x]

now d/dx { sin^3[2x] } = d/dp [p^3] * dp/dx --- chain rule.
= 3*p^2 * 2* cos[2x]
= 6*p^2*cos[2x] = 6*sin^2[2x]cos[2x]

using chain rule and substitution method we can differentiate this

let y =sin^3(2x) and let u =2x
then y= sin^3u and u =2x

now dy/dx = dy/du * du/dx substitution differentiation]

dy/du = 3sin^2u cosu (chain rule)

du/dx =2
so now dy/dx = 3sin^2u cosu (2) = 2 (3) sin^2(2x) cos(2x) = 6 sin^2x cos2x

f (x) = [ sin (2x) ] ³

f ` (x) = 3 [ sin (2x) ] ² [ 2 sin x cos x ]

f ` (x) = 3 [ sin (2x) ] ² [ sin (2x) ]

f ` (x) = 3 sin ³ (2x)