Evaluating $\frac{d^{100}}{dx^{100}}\left(\frac{p(x)}{x^3-x}\right)$ I am given that $\dfrac{d^{100}}{dx^{100}}\left(\dfrac{p(x)}{x^3-x}\right) = \dfrac{f(x)}{g(x)}$ for some polynomials $f(x)$ and $g(x).$ $p(x)$ doesn't have the factor $x^3-x$ and I need to find the least possible degree of $f(x)$.
My Attempt: I am describing in short what I did. Used partial fraction to break up $\dfrac{1}{x^3-x}=\dfrac{A}{x+1}+\dfrac{B}{x}+\dfrac{C}{x-1}$
Now differentiating this $100$ times gave after simplification the denominator $[x(x+1)(x-1)]^{303}$ whle the numerator of Pairwise product of the factors of denominator.
That is, $\dfrac{d^{100}}{dx^{100}}\left(\dfrac{p(x)}{x+1}\right) =-A(100)!\left( \dfrac{a_0+a_1 x+\cdots +a_m x^m}{(x+1)^{101}}\right)$ . where degree of $p(x)$ is $m$ . The other two factors also produced a similar result and adding them the final expression had in numerator degree of $101+101+m=202+m$ .
Now the least possible degree is achieved if $m=0$ that is $p$ is a constant polynomial. So the answer is $202$.
I felt this solution was ok but I do need advices to make sure how much I would get out of 15. Thanks i Advance!

Using the chain rule we write:
\begin{eqnarray}
&&\frac{d^n}{d x^n} \left(\frac{p(x)}{x^3-x}\right) = 
\sum\limits_{p=0}^n \binom{n}{p} d^{n-p} p(x) \cdot d_x^p 
\underbrace{\left[
\frac{1}{2} \frac{1}{x-1} + \frac{1}{2} \frac{1}{x+1} - \frac{1}{x}
\right]}_{(-1)^p\left(\frac{1}{2} \frac{1}{(x-1)^{p+1}} + \frac{1}{2} \frac{1}{(x+1)^{p+1}} - \frac{1}{x^{p+1}}\right)}\\
&&\frac{1}{2} \frac{1}{\left(x^3-x\right)^{n+1}}\cdot 
\sum\limits_{p=0}^n \binom{n}{p} d^{n-p} p(x) (-1)^p \cdot 
\underbrace{\left[
\begin{array}{lll}
(x-1)^{n-p} (x+1)^{n+1} x^{n+1} +\\
(x-1)^{n+1} (x+1)^{n-p} x^{n+1} +\\
-2 (x-1)^{n+1} (x+1)^{n+1} x^{n-p}\\
\end{array}
\right]}_{{\mathfrak W}_{n,p}(x)}
\end{eqnarray}
where
\begin{eqnarray}
{\mathfrak W}_{n,p}(x) = 0 \cdot x^{3+n+2-p} + 0 \cdot x^{3n+1-p} + \left[(n-p)(n-p-1)+(n+1)(2p+2-n)\right]\cdot x^{3n-p} + \cdots
\end{eqnarray}
In other words the polynomial above is two orders lower than one would expect, i.e. the two highest order coefficients vanish.
Therefore the answer to your question is $2n + \mbox{deg$(p(x))$} $
