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?

related problem: (I). I am referring you to this book where you will find the complete answer to the problem of finding the nth derivative of rational polynomials and other classes of functions.
Added Some people suggested the post should be self-contained. Here is the example I am referring to
\begin{equation}
f(x) = \frac{ x^2 + x - 1 } { x^4 + x^3 - x - 1 }\,.
\end{equation}
The roots of the denominator are
\begin{equation}\nonumber
x = -1, 1, -\frac{1}{2} + \frac{ i \sqrt{3}}{2}, -\frac{1}{2} -
\frac{ i \sqrt{3}}{2}\,.
\end{equation}
Expressing the function in the partial fraction form
\begin{equation}\nonumber
f(x) = \frac{ x^2 + x - 1 } { x^4 + x^3 - x - 1 } =
\frac{a_1}{x-1} + \frac{a_2}{x+1} + \frac{a_3}{ x + \frac{1 -
i\sqrt{3}}{2}} + \frac{a_4}{ x + \frac{1 + i\sqrt{3}}{2}}\,.
\end{equation} 
To find $a_i$, one can use the formula

$$ a_{ i_{\ell}} = \mathrm{Res}( ( x - \alpha_{\ell} )^{{i_\ell} - 1
} f(x)\,, x = \alpha_\ell) \,,\quad i_\ell = 1,2,...,m_\ell, $$

where $ \alpha_{\ell} $ are the roots of the denominator and the formula takes care of the multiplicities $m_{\ell}$ of the roots in case there is any. For instance,
\begin{align}\nonumber
a_1 & = \mathrm{Res}\left( \frac{ x^2 + x - 1 } { x^4 + x^3 - x -
1 }, x = 1 \right)&
\\ \nonumber
\\ \nonumber
& = \lim_{x = 1} (x-1)\frac{ x^2 + x - 1 } { x^4 + x^3 - x - 1 }&
\\ \nonumber
\\ \nonumber
& = \frac{1}{6}\,,&
\end{align} 
Once we have the function in the partial fraction form, the $nth$ derivative
can be found directly by using the formula
\begin{equation}
D^n ( x - \alpha )^{(-k)} =
{\frac { \left( -1 \right) ^{n}\Gamma  \left( n+k \right) }{\Gamma 
\left( k \right) }}(x-\alpha)^{-k-n}.
\end{equation}
Following the above techniques, the final answer is 
$$ \begin{align}\nonumber
f^{(n)}(x) = (-1)^n \Gamma(n+1) \, &\left(
\frac{1}{2\,(x+1)^{n+1}} + \frac{1}{6\,(x-1)^{n+1}}  \right. &
\\ \nonumber
\\ \nonumber
& \left. + \frac{\frac{ - 1 - i\sqrt{3}}{3}}{ \left(x + \frac{1 -
i\sqrt{3}}{2}\right)^{n+1} } + \frac{\frac{ - 1 + i\sqrt{3}}{3}}{
\left(x + \frac{1 + i\sqrt{3}}{2}\right)^{n+1}} \right) \,. &
\end{align}.$$ 
