How do you differentiate # y =(3x-2)^10 # using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer mason m Dec 24, 2015 #y'=30(3x-2)^9# Explanation: According to the chain rule, #d/dx[u^10]=10u^9*(du)/dx# Thus, #y'=10(3x-2)^9*d/dx[3x-2]# #=>10(3x-2)^9*3# #=>30(3x-2)^9# 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 2640 views around the world You can reuse this answer Creative Commons License