Derivative of 8/x^2??????????????

I just want to know if I am doing it right.. and I'm getting somewhat stuck

1) 8/(x+h)^2 - 8/x^2 all over h
2 )common denominators 8x^2/x^2(x^2+2hx+h^2) - 8x^2+16hx+8h^2/x^2(x^2+2hx+h^2)
3) come out with -16hx-8h^2/x^4+2hx^3+h^2x^2 all over h

this is where I get stuck because I suck at simplifying expressions... am I supposed to get all these exponents??

You are on the right track, but don't expand the denominator. Just leave it as (h * x^2 * (x+h)^2)

So the top should look like:
8x^2 - 8x^2 - 16xh - 8h^2

The 8x^2 's cancel and you are left with -16xh - 8h^2

Factor out an h:
h*(-16x - 8h)

The h's from the top and bottom are h/h = 1

You should have (-16x - 8h) / (x^2 * (x+h)^2)

Now take the limit letting h = 0

Answer:
-16 / x^3

you can do this problem in your head using the quotient rule, obviously you don't know how.

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Use a substitute variable:

Let u = 1/x
then du/dx = -1/x^2 (we will need this alter)

then you are looking for the derivative of

8u^2

Derivative = Limit, as h goes to zero, of

[8(u+h)^2 - 8u^2] / h
[ 8u^2 + 16hu + 8h^2 - 8u^2 ] / h
[ 16hu + 8h^2] / h
[ 16u + 8h]

The limit, as h goes to zero, of 16u + 8h is.... 16u (the 8h disappears)

As a derivative, this is written as
D(8u^2) = 16u du (because we are trained to always write du/dx on the left, we keep forgetting that until we are finished, there is always a du (or a dx or a "d-something") somewhere in the picture.

In our case, we have

du/dx = -1/x^2, so we need to replace both:
u = 1/x
and
du = (-1/x^2)dx

So that our answer becomes:

D(8/x^2) = 16u du = 16(1/x) (-1/x^2)dx
= -16/x^3 dx

If we had been given, at the start:

y = 8/x^2

then we would derive both sides to get:
dy = -16/x^3 dx
we would then move the dx to the left (where it goes to the denominator), giving us:

dy/dx = -16/x^3

the derivative is -16/x^3