How do you find the derivative of #(x^2-1)^3#? Calculus Basic Differentiation Rules Chain Rule 1 Answer GiĆ³ Jul 6, 2015 I found: #6x(x^2-1)^2# Explanation: I would use the Chain Rule to derive #()^3# first and then multiply times the derivative of the argument and get: #y'=3(x^2-1)^2*2x=6x(x^2-1)^2# 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 4578 views around the world You can reuse this answer Creative Commons License