$\frac{d^{100}}{dx^{100}}\left[\frac{f(x)}{g(x)}\right]=\frac{p(x)}{q(x)}$,then find the degrees of the polynomials $p(x)$ and $q(x)$ Let $g(x)=x^3-x$,and $f(x)$ be a polynomial of degree $\leq100$.If $f(x)$ and $g(x)$ have no common factor and $\frac{d^{100}}{dx^{100}}\left[\frac{f(x)}{g(x)}\right]=\frac{p(x)}{q(x)}$,then find the degrees of the polynomials $p(x)$ and $q(x)$.
I tried this problem.SInce the degree of $f(x)$ is atmost 100,and degree of $g(x)$ is 3,so the atmost degree of $\frac{f(x)}{g(x)}$ is 97 and after differentiation we get degree of $\frac{p(x)}{q(x)}$ as $-3$ but i cannot exactly pinpoint the degree of $p(x)$ and $q(x)$.
Answer in my book says degree of $p(x)$ is $6\times 2^{99}-101$ and the degree of $q(x)$ is $6\times 2^{99}$.How can i get this answer.Please help me.
 A: Each time you derive $\dfrac{f(x)}{g(x)}$ the degree of the denominator will double, since $$\frac{d}{dx}\frac{N(x)}{D(x)}=\frac{N'(x)D(x)-D'(x)N(x)}{(D(x))^2}$$ So after a hundred iterations, the degree of  the denominator in your case is $3\times 2 ^{100}=6\times 2^{99}$
You can easily see that the degree of the numerator satisfies the following recurrence relation: $$\text{dg} (n)=\underbrace{\text{dg}(n-1)-1}_{\text{degree of }N'(x)}+\underbrace{3\times 2^n}_{\text{degree of } D(x)}$$
Or: 
$$\text{dg} (n)=\underbrace{\text{dg}(n-1)}_{\text{degree of }N(x)}+\underbrace{3\times 2^n -1}_{\text{degree of } D'(x)}$$
at the $n^{th}$ derivative, where $\text{dg}(0)=k$ ($k$ is the degree of $f(x)$).
Solving the recurrence relation yields:$$\text{dg}(n)=k-n+3\times 2^n-3$$
But if the degree of $f(x)$ is $100$ then the degree of $p(x)$ is $6\times 2^{99}-3$ which could've been obtained from your finding, namely that the degree of the final quotient is $-3$. To get the textbook's answer the degree of $f(x)$ should be $2$.
A: I think that I have a method, but it does not gives the answer in your book;
a) Write $f(x)=E(x)g(x)+R(x)$, with the degree of the polynomial $R$ $\leq 2$. Then the degree of $E$ is $\leq 97$. We have $f/g=E+R/g$, hence the $100$-th derivative of $f/g$ is equal to the $100$-th derivative of $R/g$.
b) Write $R/g=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$. We compute easily that $A=-R(0)=-f(0)$, $B=R(1)/2=f(1)/2$, $C=-R(-1)/2=-f(-1)/2$. As $f$ and $g$ have no commun factor, we have $ABC\not =0$.
c) We have now that the $100$-th derivative of $f/g$ is
$$(-1)^{100}(100!)(\frac{A}{x^{101}}+\frac{B}{(x-1)^{101}}+\frac{C}{(x+1)^{101}})$$
d) Hence, up to a multiplicative factor, we get $q(x)=(x(x-1)(x+1))^{101}$, and $p(x)=A((x-1)(x+1))^{101}+B(x(x+1))^{101}+C(x(x-1))^{101}$ (One verify easily as $ABC\not=0$ that they have no commun factor).
e) Problem: the degree of $q$ is $303$, and the degree of $p$ is $202$ if $A+B+C$ is not zero. But if $A+B+C$ is zero, the degree of $p$ is $\leq 201$....
