A differentiable function that is $\mathcal{O}(1)$ which has a derivative that is not Let $E$ be a compact subset of a bounded open set $\mathcal{D},$ both of which are subsets of the real numbers. Let a differentiable function $f(x)=\mathcal{O}(1), \forall x\in\mathcal{D}.$ Is there an $f(x)$ such that $f'(x)\neq\mathcal{O}(1)$, for some$ x\in E?$ I've attempted to construct functions that oscillate wildly at a point, but couldn't find one that didn't have a derivative that was $\mathcal{O}(1)$.

Take $E = [-1,1]$, and $\mathcal D = (-2, 2)$. Let $$ f(x) = \begin{cases} 0 &\text{if $x = 0$},\\\ x^2\sin(x^{-3}) &\text{else.}\end{cases}$$ Then $f$ is trivially bounded on $\mathcal D$. Also, the derivative of $f$ exists on all of $\mathcal D$: we have $$ f'(x) = \begin{cases} 0 &\text{if $x = 0$},\\\ 2x\sin(x^{-3}) - 3x^{-2}\cos(x^{-3}) &\text{else,}\end{cases}$$ which clearly is not bounded near $x = 0$.