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

When you perform the quotient rule, it's often easier to not multiply everything out until the end, because there are a lot of cases where you can factor things out and cancel, and if you multiply out first, it will be much harder to see that. Your first derivative is indeed correct, but here's what I'd recommend to get the suggested answer for the second derivative:
\begin{align*}
\frac{d}{dx}\left(\frac{-6x}{\left(3x^2 + 4\right)^2}\right) &= \frac{-6\left(3x^2 + 4\right)^2 - (-6x)(2)\left(3x^2 + 4\right)(6x)}{\left(3x^2 + 4\right)^4}\\
&= \frac{6\left(3x^2 + 4\right)\left(-\left(3x^2 + 4\right) + 12x^2\right)}{\left(3x^2 + 4\right)^4}\\
&= \frac{6\left(9x^2 - 4\right)}{\left(3x^2 + 4\right)^3},
\end{align*}
which is what you wanted.
