How do you differentiate #f(x)=(x^2-2x)^2# using the chain rule?

1 Answer
Nov 30, 2015

Answer:

You can do it like this:

Explanation:

Treat the expression inside the brackets as a single function. Differentiate that, then multiply by the derivative of what you have inside the brackets:

#f(x)=(x^2-2x)^2#

So you let #(x^2-2x)=u#

So:

#dy/dx=(dy)/(cancel(du))xx(cancel(du))/(dx)#

#:.f'(x)=2(x^2-2x)^(1)xx(2x-2)#

This simplifies to:

#f'(x)=4x(x-1)(x-2)#