Hi.

How to find the derivative of 1/(x^2) using definition of the derivative;I know how to find it by using the power rule but not the definition(Happy)(Speechless).

Thanks.

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- Oct 17th 2010, 02:47 PMwowoDerivative
Hi.

How to find the derivative of 1/(x^2) using definition of the derivative;I know how to find it by using the power rule but not the definition(Happy)(Speechless).

Thanks. - Oct 17th 2010, 02:51 PMlvleph
The definition of the derivative is: If $\displaystyle f(x) = \frac{1}{x^2}$ then

$\displaystyle f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$

So try doing that and come back and tell us how you did. We can help you if you still need help. - Oct 17th 2010, 03:06 PMwowo
I need a hint, 'cause I'm still stuck.

- Oct 17th 2010, 03:07 PMlvleph
Tell me what f(x+h) and f(x) are, then tell post what $\displaystyle \frac{f(x+h) - f(x)}{h}$. Once you have all that you should be able to simplify things and take the derivative.

- Oct 17th 2010, 03:09 PMskeeter
$\displaystyle \displaystyle f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$

start by getting a common denominator and get those two fractions in the numerator together. - Oct 17th 2010, 03:35 PMwowo
I keep getting lost during simplifying process.(Angry)(what's wrong with me(Headbang))

- Oct 17th 2010, 03:43 PMskeeter
- Oct 17th 2010, 03:53 PMwowo
$\displaystyle x^2-1(x^2+2xh+h^2)/(x^2+2xh+h^2)(x^2)/h$

Am I right???? - Oct 17th 2010, 03:55 PMlvleph
Yes, continue simplification. Eventually an h will cancel.

- Oct 17th 2010, 03:55 PMskeeter
- Oct 17th 2010, 04:26 PMwowo
Okay, I'm completely lost with simplifying

- Oct 17th 2010, 04:32 PMskeeter
- Oct 17th 2010, 04:58 PMwowo
So, It's $\displaystyle 2x+h/(x^2(x+h)^2$

$\displaystyle 2x/(x^2)(x^2)$

$\displaystyle 2/x^3$ - Oct 17th 2010, 05:09 PMmr fantastic
- Oct 17th 2010, 05:22 PMwowo
Is it -2x+h/(x^2(x+h)^2 ??