How do you differentiate #(t-2)^3 (t-6)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Bdub Oct 31, 2016 see below Explanation: #f(t)=(t-2)^3 (t-6)# Use product rule #(fg)'=fg'+gf'# #f = (t-2)^3, g=t-6# #f'=3(t-2)^2*1 , g' = 1# #f'(t)=(t-2)^3+3(t-2)^2(t-6)# #f'(t)=(t-2)^2[t-2+3(t-6)]# #f'(t)=(t-2)^2[t-2+3t-18]# #f'(t)=(t-2)^2(4t-20)# 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 1490 views around the world You can reuse this answer Creative Commons License