Problem on finding the second derivative.

[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!
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 got to get used to it.

 

[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}

 

[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

 

[jzq]

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

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}