Question

How do you differentiate `f(x)=ln(1/sin(3x))` using the chain rule?

Answer 1

`f'(x)= -3cot3x`

Explanation:

`f'(x)=(ln(1/(sin3x)))'=1/((1/(sin3x))) * (1/(sin3x))'`

`f'(x)=sin3x * ((sin3x)^-1)' = sin3x * (-1)(sin3x)^-2 * (sin3x)'`

`f'(x)=-sin3x * 1/(sin3x)^2 * cos3x * (3x)'`

`f'(x)=-1/(sin3x) * cos3x * 3`

`f'(x)= -3cot3x`