Question

what is derivative of sin^2 (3x-1)^2 ?

dy/dx (sin^2 (x))= 2 sin x cos x
dy/dx (3x-1)^2= 6(3x-1)

chain rule?
= 2 sin ((3x-1)^2) cos ((3x-1)^2) * 6(3x-1)

Please help. Thank you

Answer 1

I'm not sure if this is correct. Can someone clarify thanks.

Answer 2

you are correct
Or, you could go on and say

sin (2(3x-1)^2) * 6(3x-1)

Answer 3

I got 12 sin (3x-1)^2 cos (3x-1)^2 * (3x-1) as my final answer. Is this in the correct simplified form ?

Answer 4

it is ok, but you can use

2sinu cosu = sin(2u)

to make it what I had:

6(3x-1) sin(2(3x-1)^2)

If you want to retain your original argument of sin and cos, then leave it as you had it.

Answer 5

I have taken that into consideration; thank you!