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*}

Recall the Quotient Rule:
$$(\frac{f}{g})' = \frac{f'\cdot g - g' \cdot f}{g^2}$$
So you have the first derivative, $\frac{dy}{dx}= \frac{-48x}{(x^2+12)^2}$. Just apply the quotient rule:
Let $f(x) = -48x$. So $f'(x) = -48$.
Let $g(x) = (x^2 + 12)^2$. So, by the chain rule, $g'(x) = 2(x^2 + 12) \cdot 2x = 4x(x^2+12)$.
$$\frac{d^2y}{dx^2} = \frac{(-48 \cdot (x^2+12)^2) - (4x(x^2+12) \cdot -48x)}{((x^2+12)^2)^2}$$
Notice that both elements of the numerator contain $-48$. We can factor that out.
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot ((x^2+12) - (4x^2(x^2+12)))}{(x^2+12)^4}$$
We also can factor out the $(x^2+12)^2$ term:
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot (x^2+12) \cdot ((x^2+12) - 4x^2)}{(x^2+12)^4}$$
And now we can cancel:
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot (12 - 3x^2)}{(x^2+12)^3}$$
Observe that we can factor the numerator some more:
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot 3 \cdot (4 - x^2)}{(x^2+12)^3}$$
And there we are:
$$\frac{d^2y}{dx^2} = \frac{-144 \cdot (4 - x^2)}{(x^2+12)^3}$$
