Finding the $n^{th}$ derivative of $\frac{x^n}{(1+x)}$

Question

Find the $n^{th}$ derivative of $\frac{x^n}{(1+x)}$ . I think we have to use Leibnitz's Formula to evaluate this, but I haven't succeeded in it as well. I have already received an answer of $\frac {x^n}{(1+x)}$, that was a bit simpler maybe, but I could not get this one...hope some one can help.

Answer 1

One may recall that $$ 1-x+x^2-x^3+\dots+(-1)^{n-1}x^{n-1}=\frac1{x+1}-(-1)^n\frac{x^n}{x+1},\qquad x \neq-1, $$ then differentiating $n$ times gives easily $$ 0=\left(\frac1{x+1} \right)^{(n)}-(-1)^n\left(\frac{x^n}{x+1}\right)^{(n)} $$ or

$$ \left(\frac{x^n}{x+1}\right)^{(n)}=(-1)^n\left(\frac1{x+1} \right)^{(n)}=\color{blue}{\frac{n!}{(x+1)^{n+1}}},\quad \quad x \neq-1. $$