After defining
$BC^1(\mathbb R,\mathbb R):=\{f: \mathbb R \to \mathbb R \mid f \in C^1, \ \ \lVert f \rVert_\infty + \lVert f' \rVert_\infty < \infty \}$
and proving that it's complete, our lecturer made the following comment:
From the completeness of $BC^1$ immediately follows that if both $f_n$ and $f'_n$ converge uniformly then $(\lim f_n)'=\lim f'_n$.
I would highly appreciate an explanation on why should it follow.
If you have a sequence $f_n \in \mathscr{C}^1(\mathbb{R})$ such that both $f_n$ and $f_n'$ converge uniformly, there exists an $n_0 \in \mathbb{N}$ such that
$$\bigl(\forall n, m \geqslant n_0\bigr) \bigl(\lVert f_n - f_m\rVert_\infty \leqslant 1 \land \lVert f_n' - f_m'\rVert_\infty \leqslant 1\bigr).$$
Then $(g_k)_{k \in \mathbb{N}}$ defined by $g_k = f_{n_0 + k} - f_{n_0}$ is a Cauchy sequence in $BC^1$.
By the completeness of $BC^1$, there exists a limit $g \in BC^1$ of $(g_k)$, that means $g_k \to g$ uniformly, and $g_k' \to g'$ uniformly.
But of course $g_k$ converges uniformly to $\lim f_n - f_{n_0}$, and $g_k'$ converges uniformly to $\lim f_n' - f_{n_0}'$, hence $\lim f_n' = (g' + f_{n_0}') = (g + f_{n_0})' = \bigl(\lim f_n\bigr)'$.
(Note: Neither the $f_n$ nor the $f_n'$ need to be bounded, we subtracted $f_{n_0}$ to obtain a sequence in $BC^1$.)