Convergence of sequence of differentiable functions Provide an example or explain why the request is impossible.

a) A sequence of $(f_{n})$ of differentiable functions such that $(f'_n)$ converges uniformly but the original sequence $(f_n)$ does not converge for any $x \in \mathbb{R}$
b) A sequence of $(f_n)$ of differentiable functions such that both $(f_n)$ and $(f'_n)$ converge uniformly but $f = \lim f_n$ is not differentiable at some point.

I think both are not possible but I'm not quite sure how to prove it. Any help would be appreciated.
 A: For (a) try to find a sequence of constant functions that do not converge.  In this case the sequence of derivatives is trivially uniformly convergent to the zero function.
For (b) if $f_n$ is differentiable, $f_n \rightrightarrows f$ and $f_n' \rightrightarrows g,$ then we must have $f' = g$.  Hence,$f$ is everywhere differentiable.
Note that
$$\left|\frac{f(x) - f(c)}{x-c} - g(c)\right| \\ \leqslant \left|\frac{f(x) - f(c)}{x-c} - \frac{f_n(x) - f_n(c)}{x-c} \right| + \left|\frac{f_n(x) - f_n(c)}{x-c} - f_n'(c)\right| + \left|f_n'(c) - g(c)\right|. $$
The third term on the RHS can be made smaller than any $\epsilon/3$ for any $c$ by choosing $n$ sufficiently large.  With $n$ fixed, the second term can be made smaller than any $\epsilon/3$ by choosing $|x - c| < \delta$ for some delta that does not depend on $x$.  Finally, an application of the mean value theorem and the uniform convergence of $f_n'$ can be used to show that the first term can also be made smaller than any $\epsilon/3$ for $n$ sufficiently large.
