I came across this question in my homework. I've tried to solve it with the regular induction of check for n=1 and then n=n+1 but without success. I would like to know how should I approach that kind of question.
Let $f(x) = (1+x)^{-1/2}$. Prove that $\forall n\in\mathbb N$, $$ f^n(x)= (-1)^n\frac{(2n)!(1+x)^{-\left(n+\frac12\right)}}{4^nn!}.$$
Thank you for your help!
You get $$f'(x)=-\frac{(1+x)^{-(1+1/2)}}{2}$$ Which satisfies the base case. If we assume that $f^n(x)$ is as stated, then \begin{align} f^{(n+1)}(x)=f^{(n)'}(x)&=(-1)^n\frac{(2n)!(1+x)^{-(n+1+1/2)}}{4^n n!}(-(n+1/2))\\ &=(-1)^{n+1}\frac{(2n)!(1+x)^{-(n+1+1/2)}}{4^n n!}\left(\frac{4n+2}{4}\right)\\ &=(-1)^{n+1}\frac{(2n)!(1+x)^{-(n+1+1/2)}}{4^{n+1} n!}(2(2n+1))\\ &=(-1)^{n+1}\frac{(2n)!(1+x)^{-(n+1+1/2)}}{4^{n+1} n!}\frac{2(2n+1)(n+1)}{n+1}\\ &=(-1)^{n+1}\frac{(2n)!(1+x)^{-(n+1+1/2)}}{4^{n+1} n!}\frac{(2n+1)(2n+2)}{n+1}\\ &=(-1)^{n+1}\frac{(2(n+1))!(1+x)^{-(n+1+1/2)}}{4^{n+1} (n+1)!}\\ \end{align} Completing the induction step
The derivative of $u_n(x)=(1+x)^{-\left(n+\frac12\right)}$ is $$-(n+\frac12)(1+x)^{-\left(n+1+\frac12\right)}=(-1)\frac{(2n+1)(2n+2)}{4(n+1)}u_{n+1}(x).$$
Therefore the $n$th derivative of $u_0(x)$ is $$(-1)^n\frac{(2n)!}{4^nn!}u_{n}(x).$$