How to find nth derivative of $1/(1+x+x^2+x^3)$

Question

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 :-)

Answer 1

HINT:

Use Partial Fraction Decomposition

$$\frac1{1+x+x^2+x^3}=\frac1{(x+1)(x^2+1)}=\frac A{x+1}+\frac B{x+i}+\frac C{x-i}$$

Answer 2

Find the first derrivative and then the second and then the third, you might see a pattern emerge, then write down the general equation that follows that pattern the nth derrivative

Answer 3

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.)