Find nth derivative of $\frac{x^{n}}{(1-x)^{2}}$, please? I need to find the nth derivative of $\frac{x^{n}}{(1-x)^{2}}$ for $0<x<1$
So far, I tried the same method used for $\frac{x^{n}}{1-x}$ and here's what I got:
\begin{equation}
\frac{x^{n}}{(1-x)^{2}}=x^{n}(1+2x+3x^{2}+4x^{3}+....)=(x^{n}+2x^{n+1}+3x^{n+2}+...)
\end{equation}
Take nth derivative to get:
\begin{align}
\frac{\partial^n }{\partial x^n} \frac{x^{n}}{(1-x)^{2}} \notag\\ = 
& (n!+2(n+1)!x+3\frac{(n+2)!}{2!}x^{2}+...) \notag\\ =
& n!(1+2(n+1)x+3\frac{(n+2)(n+1)}{2!}x^{2}+4\frac{(n+3)(n+2)(n+1)}{3!}x^{3}+..) 
\end{align}
Next would be finding a a function whose taylor expansion is the series in the parenthesis but I couldn't think of one. Any ideas for a function or for another method to do this? Thanks!
 A: Write $x = 1-t$, so you're looking at $$(-1)^n \dfrac{d^n}{dt^n} \dfrac{(1-t)^n}{t^2} = \dfrac{d^n}{dt^n}  \sum_{j=0}^n {n \choose j} (-1)^{n+j}   t^{j-2}$$
The terms with $j \ge 2$ all vanish, and you're left with
$$ \dfrac{d^n}{dt^n} \left((-1)^n (t^{-2} - n t^{-1})\right)$$
From there it's easy.  You can write the answer as 
$$\dfrac{n!(nx+1)}{(1-x)^{n+2}}$$
A: Consider the function
$$g(u):={(1-u)^n\over u^2}={1\over u^2}\sum_{k=0}^n{n\choose k}(-u)^k ={1\over u^2}-{n\over u}+\sum_{k=2}^n{n\choose k}(-u)^{k-2}\ .$$
It follows that
$$g^{(n)}(u)={(-1)^n (n+1)!\over u^{n+2}}-{n\>(-1)^n\> n!\over u^{n+1}}=(-1)^n n!{n(1-u)+1\over u^{n+2}}\ .$$
Now $f(x)=g(1-x)$, and therefore
$$f^{(n)}(x)=(-1)^n g^{(n)}(1-x)=n!{n x+1\over (1-x)^{n+2}}\ .$$
A: A related problem. Here is a closed form

$$ \left(\frac{x^n}{(1-x^2)}\right)^{(n)} = (-1)^{n} \frac{n!}{2} \left( \left( 1+x \right) ^{-1-n}+ \left( 1-x \right) ^{-1-n} \right).$$

Added: Here is the $n$th derivative for the wanted function

$$ \left(\frac{x^n}{(1-x)^2}\right)^{(n)}=\frac{n!(nx+1)}{(1-x)^{n+2}}. $$

You can apply the technique I used to solve this problem.
A: You could write
$$\frac{\partial^n}{\partial x^n}x^n(1-x)^{-2}=\sum_{k=0}^n\binom{n}{k}\left(\frac{\partial^k}{\partial x^k}x^n\right)\left(\frac{\partial^{n-k}}{\partial x^{n-k}}(1-x)^{-2}\right)$$ and proceed from there. You'd have it as a finite sum.
