How do you differentiate #g(x)=(1+4x)^5(3+x-x^2)^8#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Anjali G Nov 21, 2016 #g'(x)=8(4x+1)^5(-x^2+x+3)^7(-2x+1)+20(4x+1)^4(-x^2+x+3)^8# Explanation: #g(x)=(4x+1)^5(-x^2+x+3)^8# #g'(x)=(4x+1)^5(8)(-x^2+x+3)^7(-2x+1)+5(4x+1)^4(4)(-x^2+x+3)^8# #g'(x)=8(4x+1)^5(-x^2+x+3)^7(-2x+1)+20(4x+1)^4(-x^2+x+3)^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 2059 views around the world You can reuse this answer Creative Commons License