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.

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!
