Second Derivative of basic fraction using quotient rule I know this is a very basic question but I need some help.
I have to find the second derivative of: 
$$\frac{1}{3x^2 + 4}$$
I start by using the Quotient Rule and get the first derivative to be:
$$\frac{-6x}{(3x^2 + 4)^2}$$
This I believe to be correct.
Following that I proceed to find the second derivative in the same manner but I get this as my answer:
$$\frac{(54x^4 + 144x^2 +96) - (-36x^3 + 48x)}{(9x^4 +24x^2 +16)^2}$$ 
This I believe to be correct just not simplified.  However the answer I need to get is:
$$- \frac{6(4 - 9x^2)}{(3x^2 + 4)^3}$$
I do not know what the best way to approach this would be, should I multiply out the denominator and try to cancel?  Could someone point me in the right direction, I want to solve it myself but I need some guidance.
Thanks

The first derivative is correct. Now we want to differentiate $\frac{-6x}{(3x^2+4)^2}$. The main thing to remember is do not "simplify" unless there is good reason to do so. 
The derivative of $\frac{-6x}{(3x^2+4)^2}$ is
$$\frac{(3x^2+4)^2 (-6)-(-6x)(6x)(2)(3x^2+4)}{(3x^2+4)^4}.$$
Cancel a $3x^2+4$, and simplify the top. 
Remark: I probably would want to take out that ugly $-6$ from the top, which is an invitation to minus sign errors and other errors, and differentiate $\frac{x}{(3x^2+4)^2}$. 
