How to find nth derivative of $1/(1+x+x^2+x^3)$ I was trying to solve a differentiation question but unable to understand .
My question is :
find the $n^{th}$ derivative of  $1/(1+x+x^2+x^3)$
I know that if we divide the numerator by denominator then the expression would be :
$1- x(1+ x^2 + x )/(1+x+x^2+x^3)$
But now how to find the nth derivative?
Please somebody explain this..
Thanx :-)

Hint: Note that $(1-x)(1+x+x^2+x^3)=1-x^4$, so $$\frac1{1+x+x^2+x^3}=\frac{1-x}{1-x^4}=\frac{1-x}{(1-x^2)(1+x^2)}=\frac{1}{(1+x)(1+x^2)}.$$
(After the fact, this factorization is easy to see directly.)
