<s>[INST] 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.
 [/INST] We can prove that
$$f^{(n)}(x) = - \frac {(2n - 3)!!\, (x - 2n)} {2^n (1 - x)^{(2n + 1)/2}}$$
for $n \ge 2$ by induction on $n$.
The base case is easy. For the inductive step,
$$\begin{align*}
\frac d {dx} f^{(n)} (x) & = - \frac {(2n - 3)!!\, 2^n (1 - x)^{(2n + 1)/2} + (2n - 3)!!\, (x - 2n) 2^n \frac {2n+1} 2 (1 - x)^{(2n - 1)/2}} {2^{2n} (1 - x)^{2n + 1}} \\
& = - \frac {(2n - 3)!!\, 2^{n-1} (1 - x)^{(2n - 1)/2} \, [2 (1 - x) + (x - 2n) (2n + 1)]} {2^{2n} (1 - x)^{2n + 1}} \\
& = - \frac {(2n - 3)!!\, (2n - 1)(x - 2n - 2)} {2^{n+1} (1 - x)^{(2n + 3)/2}} \\
& = - \frac {[2(n + 1) - 1]!!\, [x - 2(n + 1)]} {2^{n+1} (1 - x)^{[2(n + 1) + 1]/2}}.
\end{align*}$$
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