If $\{f_n\}$ is a sequence of continuously differentiable functions on $[a,b]$, $f_n \to f$ uniformly on $[a,b[$, and there is a function $g : [a,b] \to \Bbb{R}$ such that $f'_n \to g$ uniformly on $[a,b]$, the $f$ is continuous differentiable and $f'=g$.
So, when I first approached this problem, I misread the statement as asking me to prove such a $g$ exists. From this misunderstanding, I did the following:
Since $f_n$ is a continuously differentiable function on a compact interval, $f_n$ must be Riemann integrable and so $F_n(x) = \int_{a}^x f_n$ uniformly continuously and continuously differentiable. Note that $\{f_n\}$ is uniformly Cauchy, because it is uniformly convergent, so
$$|F_n(x)-F_m(x)| = |\int_{a}^{x}f_n - \int_{a}^{x} f_m| \le \int_{a}^{x}|f_n-f_m| < \epsilon(x-a) \le \epsilon(b-a)$$
which shows that $\{F_n\}$ is uniformly Cauchy. Hence there is some function $g : [a,b] \to \Bbb{R}$ such that $F_n \to g$ uniformly. Since $F'_n = f_n \to f$, then $F_n = \int F_n' \to \int f$, so by uniqueness of limits, it follows that $g = \int f$ and therefore $g'= f$ (hmm...this doesn't quite match the conclusion of the problem)
My question is, is the hypothesis that $g$ exists unnecessary? Unless I made a mistake, which is very probably, the above work suggests to my mind that proving $g$ exists is possible. I'm not very confident about the last sentence in my proof...I think I'm being careless about the distinction between definite and indefinite integrals.