I am reading a collection of problems by the Russian mathematician Vladimir Arnol'd, titled A Mathematical Trivium. I am taking a stab at this one:
Calculate the $100$th derivative of the function $$\frac{x^2 + 1}{x^3 - x}.$$
The derivative is non-trivial (in the sense that I computed it for a few rounds, and it only became more assertive). My first thought was to let
$$f(x) = x^2 + 1, \text{ } g(x) = \frac{1}{x^3 - x}$$
and apply the Leibnitz rule for products,
$$fg^{(n)}(x) = \sum_{k=0}^n {n\choose k} f^{(n-k)}(x)g^{(k)}(x) .$$
Since $f$ is vanishing after the third differentiation, we get
$$fg^{(100)}(x) = {100 \choose 2}f^{(2)}g^{(98)} + {100 \choose 1}f^{(1)}g^{(99)} {100 \choose 0}f^{(0)}g^{(100)} \\= 9900g^{(98)} + 200xg^{(99)} + (x^2 + 1)g^{(100)}$$
This would be great if we could compute the last few derivatives of $g$. Indeed, we can boil this down: notice that
$$g(x) = h(x)i(x)j(x), \hspace{4mm} h(x) = \frac{1}{x-1}, \text{ } i(x) = \frac{1}{x}, \text{ } j(x) = \frac{1}{x+1};$$
further, $h, i,$ and $j$ have friendly behavior under repeated differentation, e.g. $h^{(n)}(x) = \frac{(-1)^n n!}{(x-1)^{n + 1}}$.
So overall, it is possible to use Leibnitz again to beat a lengthy derivative out of this function, (namely,
$$g^{(n)}(x) = \sum_{k=0}^n {n \choose k} h^{(n-k)}(x) \Bigl(\sum_{l=0}^k {k \choose l} i^{(k-l)}(x) j^{(l)}(x)\Bigr)$$
with the details filled in).
However, this is really pretty far from computing the derivative.
So, my question: does anyone know how to either improve the above argument, or generate a new one, which can resolve the problem?
