Calculus problem? Studying for test

Given f(x) = (x^2-3)^6, find f'(2)??

Could you help me solve and explain please?

So we're going to apply a bit of the chain rule to obtain the first derivative. For now treat the stuff inside the ( ) as a u.

So the derivative of (u)^6 = 6 (u)^5 times du/dx
What that means is that you're going to have to find the derivative of the inside. So if u = x^2 -3
du/dx = 2x

So we have gotten:
6([x^2]-3)^5 times 2x
= 12x ([x^2] - 3) ^ 5

Now plug in 2 and you should get....

24(4-3)^5 = 24

Check with other answers i am not a mathlete.
(x^2-3)^6 I think its the derivative of the outside * the derivative of the inside.
6(x^2-3)^5(2x)
6(2^2-3)^5*(2(2))
6(1)*(4)
24

f '(2) means evaluate the first derivative when x =2.

The first deriv is 6 [(x² -3)^5 ] (2x) which simplifies to (12x)[(x² -3)^5 ]

That means f '(2) = 24 [(1)^5] =24