# Problem on finding the second derivative.

1. Feb 26, 2005

### jzq

I have a problem on finding the second derivative for this function:

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

For the first derivative, I got:

$$\frac {-x^2-4}{(x^2-4)^2}$$

Now here is where I am stuck! So far for the second derivative, I got this (Please check!):

$$\frac {-2x(x^2-4)^2-4x(-x^2-4)(x^2-4)}{(x^2-4)^4}$$

I need this simplified! I know, it's an easy question. I may have lost my mind! :rofl:
Also please show me the steps. Thanks!

BTW. I am new to this forum and just learned the latex system. It is very complicated. Took me a while just to write out the problems above. I guess I gotta get used to it.

2. Feb 26, 2005

### cronxeh

First derivative is:

$$\frac{1}{x^2 -4} - \frac{2x^2}{(x^2-4)^2}$$

Second derivative is:
$$\frac{-6x}{(x^2-4)^2} + \frac{8x^3}{(x^2-4)^3}$$

3. Feb 28, 2005

### Aki

it would be easier if you rewrite the original equation like this:
x(x^2 -4)^-1. So you don't need to deal with fractions

4. Feb 28, 2005

### jzq

Not necessarily. It is actually more complicated using the chain rule for this particular function. But either way is fine. I solved the problem already. Thanks for the advice though.

Here's a formula for finding quotient derivatives: (I'm sure you know it already)

Function: $$\frac {f(x)}{g(x)}$$

Formula: $$\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$$

Last edited: Feb 28, 2005