Example of a uniformly continuous function on R but with derivative not uniformly continuous.

Question

Let $f$ be a smooth, bounded function such that the limits $\lim_{x\to\pm \infty}f(x)$ exist. Then is the derivative $f'$ uniformly continuous?

I am looking for a counterexample as I think such a situation not necessarily implies that $f'$ is uniformly continuous. I know that any such function $f$ is always uniformly continuous. But don't know about the derivative.

One example I am thinking is $$f(x)=\frac1{1+x^2}.$$ One can show that $f$ is uniformly continuous. Its derivative $$f'(x)=-\frac{2x}{(1+x^2)^2}.$$ Is the derivative uniformly continuous? Please help. If my example is wrong, please provide any other example.

Answer 1

$$f(x) = \frac{\sin(e^x)}{1+x^2}$$ is bounded with $\lim_{x\to\pm \infty}f(x) = 0$. The derivative $$ f'(x) = \frac{e^x \cos(e^x)}{1+x^2}- \frac{2x \sin(e^x)}{(1+x^2)^2} $$ is not uniformly continuous: For $x_k = \ln (k \pi)$ is $$ f'(x_k) = \frac{(-1)^k k \pi}{1 + (\ln(k \pi))^2} \, , $$ so that $x_{2k} - x_{2k-1} \to 0$, but $$ f'(x_{2k}) - f'(x_{2k-1}) = \frac{ 2k \pi}{1 + (\ln(2k \pi))^2} + \frac{ (2k-1) \pi}{1 + (\ln((2k-1) \pi))^2} \to \infty \, . $$