# Sorry, I jumped slightly ahead there...

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Posted by Joel on September 23, 2002 at 18:59:16:

In Reply to: Re: really quick help posted by Joel on September 23, 2002 at 17:56:06:

: : 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)] should be -(h+2x)/[x²(x²+2hx+h²)] (I already started substituting 0 for h)

: 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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