When does convergence of function imply convergence of its derivative? Let $F_n$ be a sequence of differentiable real valued functions.
Suppose that $$\lim_{n \to \infty} F_n(x) = F(x)$$
and that $F(x)$ is differentiable.
Under which conditions does that imply 
$$\lim_{n \to \infty} F'_n(x) = F'(x)$$?
Do I need some regularity, or maybe that the $F_n$ converges uniformly?
 A: You need to add the assumption that $F_n'$ converges uniformly on a closed interval $[a,b]$. In fact:
Theorem: Suppose $\{f_n\}$ is a sequence of functions, differentiable on $[a,b]$ and such that $\{f_n(x_0)\}$ converges for some point $x_0$ on $[a,b]$. If $\{f_n'\}$ converges uniformly on $[a,b]$, then $\{f_n\}$ converges uniformly on $[a,b]$, to a function $f$, and
$$f'(x)=\lim_{n\to\infty}f_n'(x),\quad(a\leq x\leq b).$$
Source: Rudin, Principles of Mathematical Analysis, Theorem 7.17.
A: This is a partial answer, there is a gap at the end.
Assume that $F_n$, $F$ are cumulative distribution functions (or can be scaled to be such), then convergence implies convergence in distribution, which in turn implies convergence of characteristic functions.
$$
\varphi_{f_n} \to \varphi_{f}
$$
Now the characteristic function of a distribution with a density is simply its inverse Fourier transform (up to constants). So slapping the Fourier transform over the characteristic functions $\varphi_{f_n}$ results in
$$
f_n = \hat{\varphi}_{f_n} = \int e^{-itx} \varphi_{f_n}(t) dt \overset{?}{\to} \int e^{-itx} \varphi_f(t) dt = f
$$
Unfortunately we only have pointwise convergence of the characteristic functions, so the convergence of the integrals is not a given. But if you can move the limits into the integral you would be done. Maybe there is a nice criterion for that which translates back to a regularity condition on $F$
