# Uniformly continuous with unbounded derivative

A bounded derivative is a sufficient condition for uniform continuity, but not necessary.

I know the counterexample $f(x) = \sqrt{x}$ on the interval $[0,\infty)$ where the derivative is unbounded at $0$, but the function is uniformly continuous.

Is there an example where $f$ is uniformly continuous and $f'(x)$ is unbounded as $x \to \infty$ but bounded on any compact interval?

Taking $$f(x)=\frac{\cos(x^3)}{x}$$ on $[1,\infty)$ suffices since it is uniformly continuous and has derivative
$$f’(x)=-3x\sin x-\frac{\cos(x^3)}{x^2}.$$
• Suppose if we put more one condition that function should be unbounded as $x$ goes to infinity. Can such uniformly continuous function exist? ($f^\prime(x)$ should be unbounded as $x$ goes to infinity.) – ramanujan Nov 28 '18 at 23:51
$$x^a logx$$ on $$[0, \infty)$$ for $$\alpha\in (0,1)$$.