How to derive this second derivative using the quotient rule? If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $
What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $
My Steps:
\begin{align*}
{d^2y \over dx^2}
&= {-48(x^2 +12)^2 - 2(x^2+12)(2x)(-48x)\over (x^2+12)^4} \\
&=-48{(x^2 +12)^2 - 4x^2(x^2+12)\over (x^2+12)^4} \\
&= -48{(x^2 +12)( - 4x^2 +(x^2+12))\over (x^2+12)^4} \\
&=  -48{(x^2 +12)( - 4x^2)\over (x^2+12)^3} \\
&= {???}
\end{align*}

We can factor out the constant term to make life easier (and just multiply by that $-48$ at the end of our calculation) as:
$$-48 \dfrac{x}{(x^2+12)^2}$$
This makes it easier to use the quotient and chain rule (I will assume you know these).
The derivative of $\ {dy \over dx} = {-48x \over (x^2+12)^2} $, using the quotient and chain rule is:
$\dfrac{d^2y}{dx^2} = -48 \dfrac{(1)(x^2+12)^2 - 2(x^2+12)(2x)x}{(x^2+12)^4} = -48 \dfrac{(x^2+12) - 2(2x)x}{(x^2+12)^3} = \dfrac{-144(4-x^2)}{(x^2+12)^3}$
