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?

According to the Binomial Theorem (or using the usual formula for the sum of a geometric series with initial term 1 and common ratio $x^2$),
$f(x)=\frac {x} {x^2-1} = -x \frac {1} {1 - x^2} = -x \left( 1 + x^2 + x^4 + \cdots + x^{2n} + \cdots \right)$
$= -x - x^3 - x^5 - \cdots - x^{2n+1} - \cdots$.
Because the right side converges absolutely for $|x^2| < 1$ you can differentiate it term by term, introducing a coefficient $(2n+1)(2n) \cdots (2n+1-k+1)$ for $x^{2n+1-k}$; in other words, the coefficient of $x^j$ is $(j+1)(j+2) \cdots (j+k)$.  Dividing the entire thing through by $k!$ gives a series you can easily relate to the binomial expansion of $( 1 - x^2 ) $ to a negative integral value, yielding a closed form solution.
