Differentiate the following function:

$\displaystyle \frac{3x}{x^2+4}$

My attempt:

First I get rid of the fraction:

$\displaystyle f'(x) = (3x)(x^2+4)^-1$

Then I apply product rule F(x) = f(x)g(x) then F'(x) = f(x)g'(x)+f'(x)g(x):

$\displaystyle f'(x) = (3x)(x^2+4)'^-1 + (3x)'(x^2+4)^-1$

I differentiate the derivatives:

$\displaystyle (3x)(2x)^-1 + (3)(x^2+4)^-1$

I simplify:

$\displaystyle \frac{1}{6x^2} + \frac{1}{3x^2+12}$

This is the farthest I can get. What am I doing wrong? Thanks for all the help in advance.