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

1 Answer
Nov 15, 2015

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