How do you differentiate #f(x)=(4x-x^2)^100#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Steve M Oct 24, 2016 #f'(x) = 100(4x-x^2)^99 (4-2x) # Explanation: Let # u=4x-x^2 # and then # f(x)=u^100 # # (du)/dx = 4-2x # Then, # f'(x)=d/dx(f(x)) # # :. f'(x) = d/dx u^100 # # :. f'(x) = d/(du) u^100 (du)/dx # (the chain rule) # :. f'(x) = 100u^99 (4-2x) # # :. f'(x) = 100(4x-x^2)^99 (4-2x) # 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 3937 views around the world You can reuse this answer Creative Commons License