$f(x) = \frac1{\cos x}$ then $f^{(n)}(x)$ is ...? $f(x) = \frac{1}{\cos x}$
$f'(x) = \frac{\sin(x)}{\cos^2(x)}$
$f''(x) = \frac{2\sin^2(x)+\cos^2(x)}{\cos^3(x)}$
$f^{(3)}(x) = \frac{6\sin^3(x)+5\cos^2(x)\sin(x)}{cos^4(x)}$
$\vdots$
$f^{(n)}(x) = \frac{ ?}{cos^{n+1}(x)}$
Some of these are easy: http://darkwing.uoregon.edu/~jcomes/251exn.pdf
Others are not. Why?

For completeness: the Wolfram Functions site gives a series representation for the $n$-th derivative of the secant:
$$\frac{\mathrm d^n}{\mathrm dx^n}\sec\,x=\sum_{j=0}^\infty \frac{(-1)^j}{(2j-n)!} E_{2j} x^{2j-n}$$
where the $E_{2j}$ are the Euler numbers mentioned in Robin's answer.
There is also a finite double series representation:
$$\frac{\mathrm d^n}{\mathrm dx^n}\sec\,x=(n+1)!\sec\,x \sum _{k=0}^n \sum _{j=0}^{\left\lfloor\frac{k-1}{2}\right\rfloor} \frac{\left((-1)^k 2^{1-k} (k-2 j)^n \sec ^k x\right) \cos\left(\frac{n\pi}{2}+(k-2j)x\right)}{(k+1) j! (k-j)! (n-k)!}$$
As Qiaochu says, there's no good reason to expect that computing higher derivatives of some function is an easy task...
