# Definition of Derivative

1. Aug 15, 2007

### antinerd

1. The problem statement, all variables and given/known data
Find the derivative using the Definition of the Derivative:

f(x) = 1 / x^2

2. Relevant equations

The Definition:

f(a) = lim h->0 [f(a+h) - f(a)] / h

3. The attempt at a solution

This is what I did:

f(a) = lim h->0 $$(1/(x+h)^{2}) - 1/x^{2}) / (h)$$

f(a) = lim h->0 $$[((x^{2}) - 1 (x^{2} + 2xh + h^{2})) / x^{2}(x^{2} + 2xh + h^{2})] / h$$

f(a) = lim h->0 $$(2x + h) / (x^{4} + 2x^{3}h + x^{2}h^{2})$$

and finally

f`(a) = lim h->0 $$2 / x^{3}$$

So I got that as the derivative, and if I did it correctly it should be right. Did I use the definition properly?

2. Aug 15, 2007

### antinerd

lol nevermind im a retard i figured it out... sigh so much typing for nothing

3. Aug 15, 2007

### Feldoh

Should be

$$f'(a) = \lim_{h \to 0} \frac{-(2x + h)}{(x^{4} + 2x^{3}h + x^{2}h^{2})}$$

Also when you get to the final answer $\frac{-2}{x^3}$ you already took the limit so the answer is just:

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

Because:
$$f'(a) = \lim_{h \to 0} \frac{-(2x + 0)}{(x^{4} + 2x^{3}*0 + x^{2}*0^{2})}$$
$$f'(a) = \frac{-2x}{x^{4}}$$

Last edited: Aug 15, 2007