How to differentiate y=sin^3 (3x)?

Question

answer is -9sin (3x) cos^2 (3x) but can u show me the working please?

Answer 1

u = sin 3x

gives du/dx = 3 cos 3x

y  = u^3

gives dy/du = 3 u^2 = 3 sin ^2 3x

so dy/dx = (3 sin ^2 3x)(3 cos 3x) = 9 sin ^2 x cos 3x

But i think u mean y = cos ^3 3x

then u = cos 3x => du/dx = - 3 sin 3x

to get it let t = 3x then du/dt = - sin t and dt/dx = 3 so du/dx = - 3 sin 3x

dy/du = 3 u^2 = 3 cos^2 3x

hence dy/dx = ( 3 cos^2 3x)(-3 sin 3x) = -9 cos^2 3x sin 3x