(2x^3-4x^2+3)/x^2

Find f'(x)

I do the following:

Numerator:

2x^3-4x^2+3

6x^2-8x+3

Denominator

x^2

2x

so, (6x^2-8x+3)/2x is the f'(x) according to what I just did

but the book says the answer is (2x^3-6)/(x^3)

????

I'll post more of this nature that I screwed up if however you help me solve this won't suffice for the other, similar problems

And,

Let g(x) = 9f(x) and let f'(-6) = -6. Find g'(-6).

I don't even know where to begin here.