How do you differentiate #f(x)=(x-sin3x)^3/9# using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer Dr. S Jan 9, 2018 #1/3*(1-3cos3x)*(x-sin3x)^2# Explanation: #d/dy(((x-sin3x)^3)/9)# = #d/dy(1/9(x-sin3x)^3)# = #1/9*3*(1-cos3x*3)*(x-sin3x)^2# = #1/3*(1-3cos3x)*(x-sin3x)^2# 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 1541 views around the world You can reuse this answer Creative Commons License