How do you find the derivative of #(1+4x)^5(3+x-x^2)^8# using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer Bdub Nov 1, 2016 #y'=8(1-2x)(1+4x)^5(3+x-x^2)^7 +20(1+4x)^4(3+x-x^2)^8# Explanation: #y=(1+4x)^5 (3+x-x^2)^8# Use product rule #(fg)'=fg'+gf'# and chain rule #(f(g(x))'=f'(g(x))*g'(x)# #f=(1+4x)^5,# #g=(3+x-x^2)^8# #f'=5(1+4x)^4 * 4,# #g'=8(3+x-x^2)^7 * (1-2x)# #y'=fg'+gf'# #y'=8(1-2x)(1+4x)^5(3+x-x^2)^7 +20(1+4x)^4(3+x-x^2)^8# 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 2999 views around the world You can reuse this answer Creative Commons License