# Re: really quick help

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Posted by Joel on September 23, 2002 at 17:56:06:

In Reply to: really quick help posted by sean on September 23, 2002 at 16:08:58:

: what is the derivative of 1/x² ?
: i keep getting lost in all the x and h garbage. i get to (x²-x²-2hx-h²)/[x²(x²+2hx+h²)] and then i get a bit confused. is that much even right?

Almost...
but you lost an h.

Should be:
(x²-x²-2hx-h²)/[hx²(x²+2hx+h²)] remember, in the definition of a derivative you started out with h as the denominator.

Now, the x²-x² becomes 0 and you factor -2hx-h² into -h(h+2x). The h in the numerator cancels the h I just put back into the denominator, so you're left with:
-2x/[x²(x²+2hx+h²)]

and I'm sure you can take it from there.

The answer, as you will see in VM's post, is -2/x^3.

Your question should have stated: "Using the definition of a derivative, what is the derivative of 1/x^2?

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