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Are there formulae for the nth derivatives of the following functions? 1) arcsin x 2) arctan x 3) sec x, and 4) tan x

Thanks.

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If you want to determine the Taylor series at zero for these functions, you do not need formulae for the nth derivative. –  GEdgar Oct 23 '11 at 16:47

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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}

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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.

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...and a tiny piece of advice for the OP: the DLMF and the Wolfram Functions site really should be your first stops for answering questions like these. –  J. M. Oct 23 '11 at 13:34

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