If $(f_n^\prime)$ converges uniformly and $(f_n(x_0))$ converges, then $f_n\to f$ uniformly, $f$ is differentiable, and $f^\prime=\lim f_n^\prime$.Suggest a slogan
Let \(a<b\), and let \(f_n\colon[a,b]\to\RR\) for \(n\geq1\) be differentiable on \([a,b]\). Suppose that \(\{f_n(x_0)\}\) converges for some \(x_0\in[a,b]\) and that \(\{f_n'\}\) converges uniformly on \([a,b]\). Then \(\{f_n\}\) converges uniformly on \([a,b]\) to a function \(f\colon[a,b]\to\RR\), \(f\) is differentiable on \([a,b]\), and \[ f'(x)=\lim_{n\to\infty}f_n'(x)\qquad\text{for all }x\in[a,b]. \] [@Rud76].