How do you differentiate #f(x)= ln(1-x)^4#? Calculus Basic Differentiation Rules Chain Rule 1 Answer sjc Dec 27, 2016 #(dy)/(dx)=(-4(1-x)^3)/((1-x)^4)# Explanation: #f(x)=ln(1-x)^4# Using the chain rule. #(dy)/(dx)=(dy)/(du)xx(du)/(dx)# let#" "u=(1-x)^4=>(du)/(dx)=-4(1-x)^3# #y=lnu=>(du)/(dx)=1/u# #(dy)/(dx)=(dy)/(du)xx(du)/(dx)# #(dy)/(dx)=1/uxx-4(1-x)^3# #(dy)/(dx)=(-4(1-x)^3)/((1-x)^4)# 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 1990 views around the world You can reuse this answer Creative Commons License