How do you find the derivative of #y = cos^3(3x+1)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Henry W. Oct 25, 2016 #(dy)/(dx)=-9sin(3x+1)cos^2(3x+1)# Explanation: #y=cos^3(3x+1)# Applying chain rule: #(dy)/(dx)=(dy)/(du)*(du)/(dx)# Let #u=cos(3x+1),# then #(du)/(dx)=-3sin(3x+1)# #(dy)/(du)=d/(du)u^3=3u^2=3cos^2(3x+1)# #:.(dy)/(dx)=3cos^2(3x+1)*-3sin(3x+1)# #=-9sin(3x+1)cos^2(3x+1)# 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 7463 views around the world You can reuse this answer Creative Commons License