Compute 100th derivative A friend suggested me a rather tricky problem, namely find the $100^{th}$ derivative of 
$$
f(x)=\frac{x^2+1}{x^3-x}.
$$
I have computed the zeroth derivative
$$
\frac{x^2+1}{x^3-x}
$$
and the first derivative
$$
\frac{2x(x^3-x)-(3x^2-1)(x^2+1)}{(x^3-x)^2}=\frac{1-x^4-4x^2}{(x^3-x)^2}
$$
but I don't see any obvious structure.

$$\frac{x^2+1}{x^3-x}=\frac{x^2-1+2}{x^3-x}=\frac{1}{x}+\frac{2}{x(x-1)(x+1)}=$$
$$=\frac{1}{x}+\frac{2}{x+1}\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x+1}+2\left(\frac{1}{x+1}-\frac{1}{x}\right)=$$
$$=-\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-1}.$$
Can you end it now?
