Derivative using definition

**Latszer** · September 27th 2009, 11:48 PM

If f(x)=1/x^2 find f'(x) using the definition of the derivative.
(note: If you instead use the power rule, f'(x)=nx^(n-1), you will get zero credit for this problem. It is useful check of your answer though.)

The answer is -2/x^3 and the best I can get is -1/x^3.
Could someone get me on the right track? If you can't help by tomorrow, don't bother because it is due tomorrow.

Thanks!

**Prove It** · September 27th 2009, 11:56 PM

Originally Posted by Latszer

If f(x)=1/x^2 find f'(x) using the definition of the derivative.
(note: If you instead use the power rule, f'(x)=nx^(n-1), you will get zero credit for this problem. It is useful check of your answer though.)

The answer is -2/x^3 and the best I can get is -1/x^3.
Could someone get me on the right track? If you can't help by tomorrow, don't bother because it is due tomorrow.

Thanks!

$f(x) = \frac{1}{x^2}$

$f(x + h) = \frac{1}{(x + h)^2}$

$f(x + h) - f(x) = \frac{1}{(x + h)^2} - \frac{1}{x^2}$

$= \frac{x^2}{x^2(x + h)^2} - \frac{(x + h)^2}{x^2(x + h)^2}$

$= \frac{x^2 - (x + h)^2}{x^2(x + h)^2}$

$= \frac{x^2 - x^2 - 2xh - h^2}{x^2(x + h)^2}$

$= \frac{h(-2x - h)}{x^2(x + h)^2}$ .

$\frac{f(x + h) - f(x)}{h} = \frac{h(-2x - h)}{hx^2(x + h)^2}$

$= \frac{-2x - h}{x^2(x + h)^2}$ .

$\lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0}\frac{-2x - h}{x^2(x + h)^2}$

$= \frac{-2x}{x^2(x^2)}$

$= -\frac{2x}{x^4}$

$= -\frac{2}{x^3}$ .

Thus $f'(x) = -\frac{2}{x^3}$ .

**Latszer** · September 28th 2009, 12:06 AM

Thanks man, I factored wrong the first time I did it, I understand now.

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