Let $\Omega\subset\mathbb{R}^d$ be an open and bounded set. Is it possible that a function $\phi \in C_b^{\infty}(\Omega;\mathbb{R})$ (so $\phi$ is smooth, bounded and defined on a bounded open set) can have an unbounded derivative? I tried to find some examples but I could never have all the conditions in the same time.
example where the domain is closed, bounded, and the function smooth : $[-1,1]\to\mathbb{R}:x\mapsto\sqrt{(1-x^2)}$
example where the domain is open, bounded, function non-smooth : $(-1,1)\to \mathbb{R} : x\mapsto \sqrt{|x|}$
example where the domain is open, unbounded, function is smooth : $\mathbb{R}\to\mathbb{R} : x\mapsto\sin(x^2)$
For the record, $x\mapsto \sqrt{1-x^2}$ is not smooth on $[-1,1]$ (at least, not in the sense of a function being smooth on a manifold with boundary) because, since $\lim_{t\to 1^-}\frac{\sqrt{1-t^2}-1}{t}=-\infty$, the function cannot be extended to a differentiable function on any open neighbourhood of $[-1,1]$.
Anyways, you have the answer to the question right there: consider $\phi(x)=\sqrt{1-x^2}$ and $\Omega=(-1,1)$.