Finding the second derivative of $f(x) = \frac{4x}{x^2-4}$. What is the second derivative of 
$$f(x) = \frac{4x}{x^2-4}?$$ I have tried to use the quotient rule but I can't seem to get the answer.
 A: To simplify the differentiation, we can first rewrite the function as a sum of partial fractions. To do this, we assert that $\frac{4x}{x^2 - 4}$ can be written as $\frac{A}{x + 2} + \frac{B}{x - 2},$ with $A$ and $B$ constants. If this is the case, then clearly, $(A + B)x + 2(B - A) = 4x,$ from which we conclude that $A = B = 2.$ The decomposition is thus $\frac{2}{x + 2} + \frac{2}{x - 2}.$
Let us differentiate this expression twice. Notice that the first derivative can be computed as
$$\frac{d}{dx}\left[\frac{2}{x + 2} + \frac{2}{x - 2}\right]$$
$$= -2(x + 2)^{-2} - 2(x - 2)^{-2}.$$
The second derivative, likewise, is easy to compute:
$$f''(x) = \boxed{4(x + 2)^{-3} + 4(x - 2)^{-3}}.$$
And we are done! Hope this helped!
A: Avoiding the quotient rule, just for an option:
$$\begin{align}
\ln(f(x))
&=\ln(4)+\ln(x)-\ln(x+2)-\ln(x-2)\\
\frac{f'(x)}{f(x)}
&=x^{-1}-(x+2)^{-1}-(x-2)^{-1}\\
f'(x)
&=f(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)\\
f''(x)
&=f(x)\left(-x^{-2}+(x+2)^{-2}+(x-2)^{-2}\right)+f'(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)\\
&=f(x)\left(-x^{-2}+(x+2)^{-2}+(x-2)^{-2}\right)+f(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)^2\\
&=f(x)\left[-x^{-2}+(x+2)^{-2}+(x-2)^{-2}+\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)^2\right]\\
&=\frac{4x}{x^2-4}\left[-x^{-2}+(x+2)^{-2}+(x-2)^{-2}+\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)^2\right]\\
\end{align}$$
A: As Jonh said, $f(x)=\frac{2}{x+2}+\frac{2}{x-2}$, and so, 
$f'(x)=\frac{-2}{(x+2)^2}+\frac{-2}{(x-2)^2}$, and
$f''(x)=\frac{4}{(x+2)^3}+\frac{4}{(x-2)^3}$.
