I know the following relation between uniform convergence and derivatives:
If $(f_n)$ is a sequence of differentiable functions on $[a,b]$ such that $\lim_{n\to\infty} f_n(x_0)$ exists (and is finite) for some $x_0\in[a,b]$ and the sequence $(f'_n)$ converges uniformly on $[a,b]$, then $f_n$ converges uniformly to a function $f$ on $[a,b]$, and $f'(x) = \lim_{n\to \infty} f'_n(x)$ for $x \in [a, b]$.
Is it possible to replace $[a,b]$ with $(a,b)$ in the above result?
That is can we conclude that
If $(f_n)$ is a sequence of differentiable functions on $(a,b)$ such that $\lim_{n\to\infty} f_n(x_0)$ exists (and is finite) for some $x_0\in(a,b)$ and the sequence $(f'_n)$ converges uniformly on $(a,b)$, then $f_n$ converges uniformly to a function $f$ on $(a,b)$, and $f'(x) = \lim_{n\to \infty} f'_n(x)$ for $x \in (a, b)$.
The only problem is to prove that $f_n $ converge uniformly to their (pointwise) limit $f $ - everything else already follows from your first statement applied to the intervals $[a+\epsilon,b-\epsilon] $ for small enough $\epsilon $.
Here is my attempt: note $$f (x)-f_n (x)=(f (x_0)-f_n (x_0)) + (x-x_0)(f'(\mu)-f'_n (\mu))$$ for some $\mu \in (x_0, x) $. We can make the first addend $f (x_0)-f_n (x_0)$ arbitrarily small with big enough $n $. Also, due to supposed uniform convergence of $f'_n$, the factor $f'(\mu)-f'_n (\mu)$ can be made arbitrarily small with big enough $n $. Finally, the factor $x-x_0$ is bounded (by $b-a $). This means that with big enough $n $ we can get $f (x)-f_n (x) $ arbitrarily small, independently on $x $, which concludes the proof. The proof cannot extend to the case where $a=-\infty $ or $b=\infty $.