$f(x) = \frac1{\cos x}$ 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?
 A: 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...
A: This is asking for the $n$-th derivative of the secant function.
As the derivative of $\sec$ is $\sec\tan$ and that of $\tan$ is
$\sec^2=1+\tan^2$ then the $n$-th derivative of $\sec$ is
$\sec f_n(\tan)$ where $f_0(t)=1$ and $f_{n+1}(t)=tf_n(t)+(t^2+1)f_n'(t)$.
It's probably too much to hope to find a nice formula for the
coefficients of the $f_n$. The constant coefficients of $f_{2m}$
are essentially the Euler numbers.
A: That depends on what you mean by "easy."  As far as the examples in the .pdf you link to, I claim that the following are true for any reasonable definition of "easy":


*

*It is easy to compute the iterated derivatives of powers and logarithms.  

*It is easy to compute the iterated derivatives of solutions to homogeneous linear ODEs.

*It is easy to compute the iterated derivatives of $x^n f(x)$ if it is easy to compute the iterated derivatives of $f(x)$.

*It is easy to compute the iterated derivatives $f(x) + g(x)$ if it is easy to compute the iterated derivatives of $f(x)$ and of $g(x)$.


I think that covers all of them.  Otherwise, there is no general reason to expect that it should be easy to compute iterated derivatives.  Even at $x = 0$ they may define complicated and highly nontrivial sequences, for example the Bernoulli numbers (which are related to this problem).
A: You might want to check out Faa di Bruno's formula: http://mathworld.wolfram.com/FaadiBrunosFormula.html 
In your case you have $f(x) = g(h(x))$ where $g(x) = 1/x$ and $h(x) = \cos(x)$.  The answer is not simple unfortunately, and involves some combinatorics.  
Alternatively, notice $f(x) \cos(x) = 1$, and thus by taking the $m$th derivative of both sides we get
$$ \sum_{k=0}^m {m \choose k} f^{(k)}(x) \left( \frac{d^{m-k}}{dx^{m-k}} \cos(x) \right) = 0 $$
if $m \geq 1$.  Thus, you can view the vector $(f(x),f'(x),\dots,f^{(n)}(x))$ as the solution to the system of linear equations above with $m=0,1,\dots,n$.  You can invert this matrix numerically, if that if your interest.  A closed form might be possible too. 
