There are differentiable functions \(f_n\colon\RR\to\RR\) that converge uniformly on \(\RR\) to a function that is not differentiable.
There are differentiable functions \(f_n\colon\RR\to\RR\) that converge uniformly on \(\RR\) to a differentiable function \(f\) while \(f_n'(0)\not\to f'(0)\).
Let \(f_n(x)\coloneqq\sqrt{x^2+1/n}\). Each \(f_n\) is differentiable, and \[ 0\le f_n(x)-\abs{x}=\frac{1/n}{\sqrt{x^2+1/n}+\abs{x}}\le\frac{1}{\sqrt n}, \] so \(f_n\to\abs{x}\) uniformly on \(\RR\), and \(\abs{x}\) is not differentiable at \(0\).
Let \(f_n(x)\coloneqq n^{-1/2}\sin(nx)\). Then \(\sup_x\abs{f_n(x)}\le n^{-1/2}\), so \(f_n\to f=0\) uniformly, but \(f_n'(0)=n^{1/2}\to\infty\neq0=f'(0)\).
If \(f_n\colon[a,b]\to\RR\) are differentiable, \((f_n(x_0))\) converges for some \(x_0\in[a,b]\), and \((f_n')\) converges uniformly on \([a,b]\), then \((f_n)\) converges uniformly on \([a,b]\) to a differentiable function \(f\) with \(f'=\lim_n f_n'\). A uniform limit of differentiable functions can be differentiable at no point: by the Weierstrass approximation theorem, every continuous function on \([a,b]\), including a nowhere differentiable one, is a uniform limit of polynomials.