Let $(f_n)$ be a sequence of functions, such that $f_n = \sin (x/n)$. Then $(f_n)$ does not converge uniformly to the zero function, because $$M_n =\sup_{x\in\Bbb{R}}\vert \sin(x/n)-0\vert=1$$
and now $M_n \to 1 \neq 0 $ when $n\to \infty$. However, in this case $(f'_n)=\frac{1}{n}\cos(x/n)$. Now for this derivative sequence we have $M'_n=1/n$, which does go to zero when $n$ tends to infinity. What I want to confirm is the following ; IF our interval was a closed interval of the form $[a,b]$, then the uniform convergence of $(f'_n)$ would also imply ufc for $(f_n)$ ( some other facts are also needed, but these are satisfied here, too ). However, $\Bbb{R}$ is not a closed interval like $[a,b]$ - the mean value theorem does not apply ( MVT is used to prove the statement I made about the derivative sequence ) . Hence, there is no contradiction -. Is this right?