Compute 100th derivative

Question

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.

Answer 1

Write your function as $$ \frac{x^2 + 1}{x^3 - x} = \frac{1}{x-1} + \frac{1}{x+1} - \frac{1}{x} $$

Answer 2

Hint: since $x^3-x=x(x-1)(x+1)$ you can easily split $f(x)$ into partial fractions.

Answer 3

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