Need help with calculus (derivatives)

Hi there, I need some help to answer a question and aas you answer can you please explain. THANK YOU

If y= (x^2 -x)^3 find y'(3)
*note y'=y prime

THANKS AGAIN....FULL POINTS TO BEST ANSWER

y = (x^2 - x)^3

y' = [3(x^2 - x)^2]*(2x - 1)

Now plug in 3:

y'(3) = 3(9 - 3)^2 * (6 - 1)
y'(3) = 3(36)(5)
y'(3) = 540

This is a chain rule problem. What we do is look at what is our outer function (^3) and our inner function (x^2 - x).

The best way I can explain this is with the notation: y' = f'(g(x)) * g'(x)

f is our outer function and g is our inner function. So if we look at this, f(x) = x^3 where x is just some variable. But, for x, we have a function, g(x). So, we take the derivative of x, but with g(x) subbed in:

f(x) = x^3
f'(x) = 3x^3
f'(g(x)) = 3(x^2 - x)^2

So, g(x) = x^2 - x, and then we have to multiply by g'(x) to complete the chain rule:

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

Now plug it in and that is all I did:

f'(g(x)) * g'(x)

= 3(x^2 - x)^2 * (2x - 1)

It's hard to explain, but I hope it is clear enough! Just practice a bunch of these and you will get it.