How do you differentiate #g(x) = (x-1)(x-2)(x^2)# using the product rule?

1 Answer
Mar 5, 2018

Answer:

#g'(x) = x(4x^2 -9x +4)#

Explanation:

The chain rule: #(dg(h(x)))/(dx)=(dg(h(x)))/(dh(x))*(dh(x))/(dx)#

Applying it to this problem: take the derivative of one term at a time, treating the other two terms as constant.

#g(x) = (x-1)(x-2)(x^2)#

#g'(x)= (d(x-1))/(dx)(x-2)(x^2) + (x-1)(d(x-2))/(dx)(x^2) + (x-1)(x-2)(d(x^2))/(dx)#

#g'(x) = (x-2)(x^2) + (x-1)(x^2) + (x-1)(x-2)(2x)#

#g'(x) = x((x-2)x + (x-1)x + 2(x-1)(x-2))#

#g'(x)=x(x^2-2x + x^2 - x + 2(x^2-3x+2))#

#g'(x) = x(4x^2 -9x +4)#