Find the $n$-th derivative of $f(x)=\frac{x}{\sqrt{1-x}}$ Find the $n$-th derivative of
$$f(x)=\frac{x}{\sqrt{1-x}}$$
First I just calculated the first, second and 3-th, 4-th derivatives and now I want to summarize the general formula. But it seems too complicated. Then I want to use binomial theorem or Taylor expansion... Also got no more clues.

We have $f(x)=g(x)h(x)$ where $g(x)=x$ and $h(x)=(1-x)^{-1/2}$. Hence using Leibniz's rule we have
$$f^{(n)}(x)=\sum_{k=0}^n{n\choose k}h^{(k)}(x)g^{(n-k)}(x)={n\choose n}h^{(n)}(x)g(x)+{n\choose n-1}h^{(n-1)}(x)g'(x) \quad(*)$$
It remains to compute
$$h^{(n)}(x)=\frac{1}{2}\frac{3}{2}\cdots \frac{2n-1}{2}(1-x)^{(2n-3)/2} =\frac{(2n)!}{2^{2n}n!}(1-x)^{(2n-3)/2}$$
and then substitute it in $(*)$ and simplify.
