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$?

Question

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.

Answer 1

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.

Answer 2

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}

Answer 3

One can find or derive all answers from results reviewed and surveyed in the paper

1. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.

and closely-related referecens therein. For expample, for $n\ge1$,

\begin{gather*} \frac{\operatorname{d}^n(\arcsin x)}{\operatorname{d} x^n}=-\frac{\operatorname{d}^n(\arccos x)}{\operatorname{d} x^n} =\frac{\operatorname{d}^{n-1}}{\operatorname{d} x^{n-1}}\biggl(\frac1{\sqrt{1-x^2}\,}\biggr)\\ =\frac1{(2x)^n}\sum_{k=0}^{n-1}2^{k+1}(2k-1)!! (n-k-1)!\binom{n-1}{k}\binom{k}{n-k-1}\biggl(\frac{x^2}{1-x^2}\biggr)^{k+1/2},\\ (\arctan z)^{(n)} =\frac{(n-1)!}{(2z)^{n-1}}\sum_{k=0}^{n-1}(-1)^k\binom{k}{n-k-1}\frac{(2z)^{2k}}{(1+z^2)^{k+1}},\\ (\tan x)^{(n)}=-(\ln\cos x)^{(n+1)} =\sum_{k=1}^{n+1} \frac{1}{k}\sum_{\ell=0}^k\binom{k}{\ell}\frac{(-1)^{\ell}}{(2\cos x)^\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^{n+1} \cos\biggl[(2q-\ell)x+\frac{(n+1)\pi}2\biggr], \end{gather*} where $\lfloor{t}\rfloor$ denotes the floor function whose value equals the largest integer less than or equal to $t$.

More references

2. https://math.stackexchange.com/a/4332741/
3. https://math.stackexchange.com/a/4672327/
4. https://math.stackexchange.com/a/4661266/
5. https://math.stackexchange.com/a/4249446/