I know that there exists bounded functions with unbounded derivatives. For example, $\sin(e^x)$ is bounded and differentiable everywhere on $\mathbb{R}$, but its derivative is unbounded.
Is it possible for an unbounded function to have a bounded derivative?
Best Answer
Yes, $f(x)=x$ whereas $f'(x)=1$.
Having seen the competition in here answering with the simplest possible function, why not define the function $$ f(x)=\int\sin(\ln(x))dx=-\frac{1}{2}x\cdot\left(\cos(\ln(x))-\sin(\ln(x))\right) $$ with its derivative $$ f'(x)=\sin(\ln(x))\in[-1,1] $$ Since $x$ is not bounded and $\cos(t)-\sin(t)=1$ whenever $t$ is a multiple of $2\pi$ and the equation $\ln(x)=t$ has solutions for all $t$ we see that $f(x)=-\frac{1}{2}x$ for an unbounded set of $x$-values thus being unbounded.