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!
Best Answer
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 $$\dfrac{n!(nx+1)}{(1-x)^{n+2}}$$