# Derivatives

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• Jul 10th 2007, 06:00 PM
SexyJenNeedsHelp
Derivatives
I need help solivng a few problems -- thanks in advance:

I'm trying to find the derivative using the limit definition?!?!

f(x) = X^2-4 :):):)
• Jul 10th 2007, 06:12 PM
galactus
$\lim_{h\rightarrow{0}}\frac{[(x+h)^{2}-4]-[x^{2}-4]}{h}$

= $\lim_{h\rightarrow{0}}\frac{2hx+h^{2}}{h}$

= $\lim_{h\rightarrow{0}}2x+h$

Now, if h approaches 0, what do you have?.
• Jul 10th 2007, 06:18 PM
SexyJenNeedsHelp
I have no idea :( I am doing some homework and I don't really understand this to well -- where di the H come from?
• Jul 10th 2007, 09:36 PM
Jhevon
Quote:

Originally Posted by SexyJenNeedsHelp
I have no idea :( I am doing some homework and I don't really understand this to well -- where di the H come from?

there are two mainstream definitions for the derivative in terms of limits.

Given a function $f(x)$ we can find it derivative, $f'(x)$ via one of the following definitions:

Definition 1:

For a function $f(x)$, it's derivative, $f'(x)$, at an arbitrary point $x$ is given by:

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

Defintion 2:

For a function $f(x)$, it's derivative, $f'(x)$, at an arbitrary point $a$ is given by:

$f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$

Obviously, galactus employed the first definition. Now the question stands as to whether you can actually evaluate the limit. In the form galactus left it, there was no problem in plugging in h = 0, so that is all that is required to evaluate the limit as h goes to zero. As the definition was at the beginning we could not plug in h = 0 since that would result in the fraction being undefined.