What is the formula for $n^{th}$ derivative of $ \sin^{-1} x, \quad \tan^{-1} x,\quad \sec x \quad \text{and}\quad \tan x$? Are there formulae for the nth derivatives of the following functions?
$1)\quad$ $sin^{-1} x$
$2)\quad$ $tan^{-1} x$
$3)\quad$ $sec x$
$4)\quad$ $tan x$
Thanks.
 A: A related problem. See Chapter 6 in this book for formulas for the nth derivative, $n$ is a non-negative integer, of $\tan(x)$ and $\sec(x)$ in terms of the $\psi$ function
\begin{equation}
{\tan}^{(n)} (z) = \frac{1}{{\pi}^{n+1}} \left({\psi}^{(n)} \left(
\frac{1}{2}+\frac{z}{\pi}\right) +  (-1)^{n+1} {\psi}^{(n)} \left(
\frac{1}{2}-\frac{z}{\pi}\right) \right)\,,
\end{equation}
\begin{align}\nonumber
{\sec}^{(n)}(z) = \frac{1}{ (2 \pi)^{ n + 1 } } &\left(
(-1)^{n}{\psi}^{(n)}\left( -\frac{2z-3\pi}{4\pi}\right) -
(-1)^{n}{\psi}^{(n)}\left( -\frac{2z-\pi}{4\pi}\right) \right.&
\\ \nonumber
& \left. - {\psi}^{(n)}\left( \frac{2z+\pi}{4\pi}\right) +
{\psi}^{(n)}\left( \frac{2z+3\pi}{4\pi}\right) \right) \,. &
\end{align}
A: For $\arcsin(x)$, the $n$-th derivative is a hypergeometric function, see here. In terms of Legendre polynomial:
$$
\frac{\partial ^n\sin ^{-1}(z)}{\partial z^n}=\frac{(-i)^{n-1} (n-1)! }{\left(1-z^2\right)^{n/2}} P_{n-1}\left(\frac{i z}{\sqrt{1-z^2}}\right)
$$
For $\arctan(x)$, see here. Here is a nice representative:
$$
\frac{\partial ^n\tan ^{-1}(z)}{\partial z^n}=\frac{1}{2} \left(i (-1)^n (n-1)!\right) \left((z-i)^{-n}-(z+i)^{-n}\right)
$$
For the tangent, the answer involved Stirling numbers of the second kind:
$$
\frac{\partial ^n\tan (z)}{\partial z^n}=-i^{n+1} 2^n (i \tan (z)-1+\delta _n) \sum _{k=0}^n \frac{(-1)^k k! }{2^k} \, \mathcal{S}_n^{(k)} \, \left(i \tan (z)+1\right)^k
$$
Results for $\sec(x)$ can be found here.
