Finding the $N$-th derivative of $f(x)=\frac {x} {x^2-1}$ I'm practicing some problems from past exams and found this one:

Find the n-th derivative of this function:
$$f(x)=\frac {x} {x^2-1}$$


I have no idea how to start solving this problems. Is there any theorem for finding nth derivative?

Maybe we can add some more help -just in case you didn't succeed to find the answer yourself yet. Let 
$$
\frac{x}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1}
$$
be the splitting into partial fractions. (I'm too lazy to compute the coeffitients $A$ and $B$.) Then
$$
\frac{d}{dx} \frac{x}{x^2 - 1} = -\frac{A}{(x-1)^2} - \frac{B}{(x+1)^2}  \ .
$$
Differentiating again,
$$
\frac{d^2}{dx^2}\frac{x}{x^2 - 1}  = \frac{2A}{(x-1)^3} + \frac{2B}{(x+1)^3} \ .
$$
One more time:
$$
\frac{d^3}{dx^3} \frac{x}{x^2 - 1} = - \frac{3\cdot 2 A}{(x-1)^4} - \frac{3\cdot 2 B}{(x+1)^4} \ .
$$
And sure enough you can find the general pattern now, can't you? Then, use induction to prove your guess.
