Is the $n$th derivative of a continuous function also continuous? Consider a differentiable (and hence continuous) function of order $n-1$. Is the $n$th derivative of such a function always continuous?
As an example, is the $n-th$ derivative of the function $f(s) = s \exp (n-1) (s-1)$ evaluated at $s=1$ continuous ?
 A: A derivative can be discontinuous but it always has the intermediate value property. Thus for instance $f(x)=\operatorname{sign}(x)$ is not the derivative of any function which is differentiable on $[-1,1]$; the "antiderivative" of $f$ is $F(x)=|x|$ which is simply not differentiable at all at $0$. On the other hand, the function 
$$g(x)=\begin{cases} 0 & x=0 \\
\int_0^x \sin(1/y) dy & x \neq 0 \end{cases}$$
is differentiable everywhere on $[-1,1]$, with a derivative which is not continuous at zero. (It is an interesting exercise to show that $g'(0)=0$.)
Reference: Wikipedia article on Darboux's theorem
A: You may verify yourself that the function $f$ given by
$$f(x) = \begin{cases}  x^2 \sin\tfrac1x &\text{if } x \neq 0 \\
0 & \text{if } x=0.\end{cases}$$
is continuous, is differentiable everywhere, but the derivative is not continuous at $0$.
If you integrate this function $n-1$ times, you obtain (using the main theorem of integration) a function that is $n$ times differentiable and the $n$th derivative is not continuous at $0$.
A: To go with the counterexamples given by Hagen von Eitzen and Ian, it is good to remember that there are explicitly differentiability classes for functions $f:\mathcal{I}\to\mathbb{R}$ on an interval $\mathcal{I}$ (with the case $\mathcal{I}=\mathbb{R}$ included). The class $C^n(\mathcal{I})$ comprises all functions that are $n$-fold differentiable and have continuous $n^{th}$ derivatives. Naturally $C^k(\mathcal{I})\subset C^j(\mathcal{I})$ whenever $k>j$ (a function $k$ times differentiable is also $j$ times differentiable when $j< k$) and the inclusion is strict. That is, there are always examples of functions whose $n^{th}$ derivatives are continuous, but whose $(n+1)^{th}$ derivatives are not. Differentiating a function always lowers its differentiability class by one. 
You also see the notations $C^\infty(\mathcal{I})$ for infinitely differentiable functions ("Smooth functions") and $C^\omega(\mathcal{I})$ for analytic functions (those whose values are given by convergent Taylor series in the interval). Note that $C^\infty\subset C^\omega$: there are smooth functions without convergent Taylor series:
$$f:\mathbb{R}\to[0,\,1);\,f(x) = \left\{\begin{array}{ll}\exp\left(-\frac{1}{x}\right);&x>0\\0;&x\leq0\end{array}\right.$$
being a standard example. See here
