Find n'th derivative of polynomial and smallest possible degree Let $p(x)$ be a polynomial of degree strictly less than 100 and such that it does not have $x^3−x$ as a factor.If
$\frac{d^{100}}{d x^{100}} \frac{p(x)}{x^3-x} =\frac{f(x)}{g(x)}$
for some polynomials $f(x)$ and $g(x)$ then find the smallest possible degree of $f(x)$

Using the division algorithm we have
$$
\frac{p(x)}{x^3 − x}= q(x) +\frac{r(x)}{x^3 − x}$$
As the degree of $q(x)$ is strictly less than that of $p(x)$ its $100-$th derivative is certainly zero. As $x^3 − x$ is not a factor of $p(x)$ one may assume (without loss of generality) that $x^2 − 1$ divides $r(x)$. In that case we have
$$
\frac{\mathrm d^{100}}{\mathrm dx^{100}}\frac{p(x)}{x^3 − x}=
\frac{\mathrm d^{100}}{\mathrm dx^{100}}\frac{k}{x}=\frac{100!k}{x^{100}}
$$
Hence the least possible degree of $f(x)$ is $0$.
If one assumes that $x^3 − x$ doesn't divide $p(x)$ then we have
$$\frac{r(x)}{x^3 − x}=\frac{A'}{x}+\frac{B'}{x-1}+\frac{C'}{x+1}$$
Consequently,
$$
\begin{align}
\frac{f(x)}{g(x)}&=\frac{A}{x^{101}}+\frac{B}{(x-1)^{101}}+\frac{C}{(x+1)^{101}}
\end{align}
$$
and then
$$
\begin{align}
f(x)&=A(x^2-1)^{101}+{B}{(x^2+x)^{101}}+{C}{(x^2-x)^{101}}\\
&=(A+B+C)x^{202}+101(B-C)x^{201}+\left[\binom{101}{2}B+\binom{101}{2}C-101A\right]x^{200}+\cdots
\end{align}
$$
Choosing $B = C$ and $A+B+C = 0$ we see that the coefficient of $x^{200}$ is $101\cdot 102 \ne0$. Hence the least possible degree of $f(x)$ in this case is $200$.
