Need help with derivative

**RobMcD** · January 19th 2011, 03:14 PM

Hello, this is my first post. I need help finding the derivative of $1/x^2$ at x=2. The answer is supposed to be $-2/x^3$ and i need to find the derivative by definition. I have worked this out a million times and i keep making mistakes in my algebra i suppose because no matter what i try, i cannot get $-2/x^3$ . any help would be greatly appreciated.

also, this is the first time ive ever heard of latex so hopefully i got it right, if i didnt i apologize

**pickslides** · January 19th 2011, 03:31 PM

$\displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{1}{(x+h)^2}-\frac{1}{x^2}\right)$

$\displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{1}{x^2+2xh+h^2}-\frac{1}{x^2}\right)$

$\displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{x^2-(x^2+2xh+h^2)}{x^2(x^2+2xh+h^2)}\right)$

$\displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{-2xh-h^2}{x^2(x^2+2xh+h^2)}\right)$

$\displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{h(-2x-h)}{x^2(x^2+2xh+h^2)}\right)$

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

$\displaystyle \frac{-2x+0}{x^2(x^2+2x(0)+(0)^2)} = \dots$

**TheCoffeeMachine** · January 19th 2011, 04:31 PM

I always use substitution when it comes to finding derivatives using the limit definition:

We have $\displaystyle f'(x) = \lim_{h \to 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}.$ Let $x+h = a$ and $x = b$ , then $h = b-x = a-b$ , thus:

$\begin{aligned} f'(x) & = \lim_{a \to b}\frac{\frac{1}{a^2}-\frac{1}{b^2}}{a-b} = \lim_{a \to b}\frac{b^2-a^2}{a^2b^2(a-b)} = \lim_{a\to{b}}\frac{-(a-b)(a+b)}{a^2b^2(a-b)} \\& = -\lim_{a\to{b}} \frac{a+b}{a^2b^2} = -\frac{2b}{b^4} = -\frac{2}{b^3} = -\frac{2}{x^3} ~ \diamond \end{aligned}<br />$

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