Can the graph of a bounded function ever have an unbounded derivative? 
Can the graph of a bounded function ever have an unbounded derivative? 

I want to know if $f$ has bounded variation then its derivative is bounded. The converse is obvious. I think the answer is "yes". If the graph were to have an unbounded derivative, it would coincide with a vertical line. 

Oh, sure. I'm sure there are lots of examples, but because of the work I do, some modification of the entropy function comes to mind. Consider the following function:
$$
f:\mathbb{R}^n\rightarrow\mathbb{R}, \quad f(x) \triangleq \begin{cases} x \log |x| & x \neq 0 \\ 0 & x = 0 \end{cases}
$$
It is not difficult to verify that this function is continuous and has a derivative of 
$$f'(x) = \log|x| + 1$$ for nonzero $x$. So $f'(x)$ is unbounded at the origin; but $f(x)$ is unbounded as well, so we're not quite there. We can create the function we seek by multiplying $f$ by a well-chosen envelope that drives the function to $0$ at the extremes. For instance:
$$
g:\mathbb{R}^n\rightarrow\mathbb{R}, \quad g(x) \triangleq e^{-x^2} f(x) = \begin{cases} x e^{-x^2} \log |x| & x \neq 0 \\ 0 & x = 0 \end{cases}
$$
The first derivative for nonzero $x$ is
$$
g'(x) = e^{-x^2} \cdot \left( ( 1 - 2 x^2 ) \log|x| + 1 \right)
$$
which remains unbounded. Attached is a plot of $g$ and $g'$.
EDITED to add: I notice that a number of other answers have chosen a bounded domain. From my perspective that is a bit incomplete. After all, we often consider such functions using the extended real number line, and in that context they are not bounded. There are certainly many functions that satisfy the original poster's conditions without resorting to a bounded domain.
