What is the n-th derivative of $\sinh^{-1} x$ ?
I have been able to prove that it is $p_n(x)(x^2+1)^{-n-1/2}$, where $p_n$ is a polynomial of degree $n-1$, and I have calculated it leading and constant coefficient, but I can't figure out the rest.
Another similar topic is this .
Use the function $f(x)=\frac1{\sqrt{x+1}}$ and $g(x)=x^2$. The derivatives of order $n\geq1$ of $f$ and $g$ are $$f^{(n)}(x)=\frac{(-1)^n(2n)!}{2^{2n}n!(x+1)^{n+1/2}},\quad g^{(n)}(x)=\left\{\begin{array}{cc}2x&\text{if $n=1$}\\2&\text{if $n=2$}\\0&\text{otherwise.}\end{array}\right.$$
You can now use the formula of Arbogast (better known under the name of Faà di Bruno) which gives the $n^\text{th}$ derivative of the composed function $f\circ g$: $$ (f\circ g)^{(n)}(x)=\sum_{\substack{m_1,\dots,\,m_n\\m_1+2m_2+\cdots+nm_n=n}}\frac{n!}{m_1!\,\dots\,m_n!}f^{(m_1+\cdots+m_n)}(g(x))\;\cdot\prod_{j=1}^n\left(\frac{g^{(j)}(x)}{j!}\right)^{m_j}.$$ In this case, the right-hand product is simply $(2x)^{m_1}$ if $m_k=0$ for $k>2$ and $0$ otherwise. As a conclusion you get $$\frac{\mathrm d^n}{\mathrm dx^n}\frac1{\sqrt{x^2+1}}= \sum_{m_1,m_2\,| m_1+2m_2=n}\frac{n!}{m_1!m_2!}\frac{(-1)^{m_1+m_2}(2m_1+2m_2)!}{4^{m_1+m2}(m_1+m_2)!}\frac{(2x)^{m_1}}{(x^2+1)^{m_1+m_2+1/2}}.$$
The sum rewrites as a simple sum, in a way depending on the parity of $n$. I hope this will be helpful.
The Maple code $$diff(arcsinh(x), x$ n) $$ produces $${\frac {{2}^{n-1} G^{1, 3}_{3, 3}\left({x}^{2}\, \Big\vert\,^{0, 1/2, 1/2}_{0, -1/2+n/2, n/2}\right) {x}^{1-n}}{\sqrt {\pi }}} .$$ See MeijerG for info.
PS. Because $\sinh^{-1}(x)=\ln(x+\sqrt{x^2}+1)$ and $$( \sinh^{-1}(x))'=\frac {1}{\sqrt {{x}^{2}+1}},$$ use the formulas $$ \left(\frac f g \right)'= \frac {f'g-fg'}{g^2} $$ and $${\frac {d }{dx}} \left( \left( {x}^{2}+1 \right) ^{n/2 } \right)=\left( {x}^{2}+1 \right) ^{n/2-1}nx $$ and induction to this end.
