How do you differentiate #f(x)=ln(1/sin(3x))# using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer Sasha P. 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# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 2548 views around the world You can reuse this answer Creative Commons License