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 100th derivative of the function x2+1x3−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)=x2+1, g(x)=1x3−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

canboil this down: notice thatg(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

computingthe derivative.

So, my question: does anyone know how to either improve the above argument, or generate a new one, which can resolve the problem?

**Answer**

We have a partial fraction decomposition

\frac{x^2+1}{x^3-x}=\frac{-1}{x}+\frac{1}{x+1}+\frac{1}{x-1}

It follows that

\left(\frac{d}{dx}\right)^{100}\frac{x^2+1}{x^3-x}=\frac{-100!}{x^{101}}+\frac{100!}{(x+1)^{101}}+\frac{100!}{(x-1)^{101}}

**Attribution***Source : Link , Question Author : Chris , Answer Author : Julian Rosen*