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.

$x(1-x)^{-\frac 12}$
Generalized binomial theorem.
$x(1-x)^{-\frac 12} = x(1^{-\frac {1}{2}} + \frac {-1}{ 2}1^{-\frac {3}{2}}(-x) + \frac {-1}{ 2}\frac {-3}{2}\frac {1}{2}1^{-\frac {5}{2}}(-x)^2+\frac {-1}{ 2}\frac {-3}{2}\frac {1}{2}\frac {-5}{2}\frac {1}{3}1^{-\frac {5}{2}}(-x)^3\cdots$
$x(1-x)^{-\frac 12} = x + \frac {1}{ 2}x^2 + \frac {3}{8}x^3+\frac {5}{16}x^4\cdots$
$a_1 = 1\\
a_{n+1} = a_n{\frac {2n-1}{2n}}\\
a_n= \frac {(2n-1)!}{2^{2n}((n-1)!)^2}$
