Is $f(x)= \cos(e^x)$ uniformly continuous? As in the topic, my quest is to check (and prove) whether the given function $$f(x)= \cos(e^x)$$is uniformly continuous  on $\left\{\begin{matrix}x \in (-\infty;0]
\\ 
x \in [0; +\infty)
\end{matrix}\right.$ . 
My problem is that I have absolutely no idea how to do it. Any hints will be appreciated and do not feel offended if I ask you a question which you consider stupid, but such are the beginnings. Thank you in advance.

It is uniformly continuous on the interval $(-\infty,0]$. To see that, use the mean value theorem twice.
Added: Exploiting the identies $ |\cos(t) - \cos(z)|\leq | t-z | $
and  $ |e^x - e^y| \leq | {x}-{y} | $ on $(-\infty,0]$, which they can be proved by the mean value theorem, we have 
$$ | \cos(e^{x}) - \cos(e^{y})|\leq | e^{x}-e^{y} |\leq| x-y |< \epsilon=\delta. $$
