Lots of examples have already been given. Perhaps a more qualitative explanation would also be useful: the value of $f'$ at $x$ is the slope of the graph of $f$ at $x$. So in order to make a function with unbounded derivative, we just have to choose some points $x_n$ such that at each $n$ the slope of the graph is $\geq n$. So: take a sheet of graph paper, and draw a very short line segment through each of the points $x = 1, 2, 3, 4,\ldots$ along the $x$-axis of slopes $+1, -2, +3, -4,\ldots$. Now draw a smooth curve passing through these points tangent to these line segmentss, but make sure that you "turn" before you cross the line $y = + 1$ or $y = -1$; you will have drawn the graph of an oscillating function, bounded between $\pm 1$, whose derivative is unbounded.
Thinking this through, and thinking about how much freedom you have in drawing such a graph, will show that there are many such examples!
Added: As Jonas Meyer pointed out in a comment, in the context of the OP's question (and the allusion to functions of BV), it might be more natural to fix attention on a bounded interval. If we work on an open interval, then we can make the same kind of construction as above, just letting the points $x_n$ tend towards the endpoint of the interval. If we work on a closed interval, and assume that $f'$ exists (in a one-sided sense at the endpoints) and is continuous, then it will be bounded (just because a continuous function on a closed interval is bounded), and this seems to be the intuition underlying the OP's comment about obtaining a vertical line from the assumption of an unbounded derivative. If we allow $f'$ to be discontinuous (even at one point), then we can again make many examples, just by crushing the function down to zero at the point of discontinuity of $f'$ (as in Jonas Meyer's $x^2 \sin(1/x^2)$ example).