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. 

$f(x)=\sin\left(x\sin x\right)$ is bounded and differentiable everywhere on $\mathbb{R}$, but it has not a bounded derivative.
Edit: To prove that $f'(x)=\cos(x\sin x)[\sin x+x\cos x]$ is unbounded, observe that $x\sin x=0$ when $x=2n\pi$ and $x\sin x=(2n+\frac12)\pi$ when $x=(2n+\frac12)\pi$. Hence there exists a sequence of positive numbers $x_n\to\infty$ such that $x_n\sin x_n=1$. Consequently,
$$
\frac{f'(x_n)}{\cos 1}
=\frac1{x_n}+x_n\sqrt{1-\frac1{x_n^2}}\to\infty \ \text{ as }\ n\to\infty.
$$
