Theorem: Let $f_n:[a,b]\to\mathbb{R}$, $n=1,2,\ldots$ be differentiable on $[a,b]$.
Suppose:
(i) $(f_n(x))$ converges for some $c \in [a,b]$
(ii) $(f'_n)$ converges uniformly on $[a,b]$.
Then $(f_n)$ converges uniformly on $[a,b]$, and if $\displaystyle f=\lim_{n\to\infty} f_n$ then $f$ is differentiable with $\displaystyle f'(x)=\lim_{n\to\infty}f'_n(x)$.
The proof in my lecture note starts like this:
Proof: Fix $x\in[a,b]$. Apply the Mean Value Theorem to $(f_n - f_m)$ to get $(f_n - f_m)(x) - (f_n-f_m)(c) = (x-c)((f'_n - f'_m)(t))$ for some $t$ between $x$ and $c$.
Then: $\displaystyle\sup_{x\in[a,b]} |f_n(x)-f_m(x)| \le |f_n(c) - f_m(c)| + (b-a)\sup_{t\in[a,b]} |f'_n(t)-f'_m(t)|$.
I do not understand how this last line follows though, in particular how we can have $t\in[a,b]$ when it must be between $x$ and $c$. Btw there could be a mistake there as my lecturer doesn't have great handwriting so please say if you think that is the case.
The line you don't understand follows from rearranging the previous line and taking the supremum. So: $$(f_n-f_m)(x)-(f_n-f_m)(c) = (x-c)(f'_n - f'_m)(t) $$ Rearranging gives us: $$(f_n-f_m)(x) = (f_n-f_m)(c) + (x-c)(f'_n - f'_m)(t) $$ Taking the supremum of both sides: $$\sup_{x\in[a,b]} \left| f_n(x)-f_m(x)\right| = \sup_{x\in[a,b]} \left|f_n(c)-f_m(c)\right| + \sup_{x\in[a,b]}\left|(x-c)(f'_n(t)-f'_m(t))\right| $$ Now, on the right hand side, the first term is independent of $x$ so we can drop the supremum from it. For the second term the supremum is going to affect $(x-c)$. The largest this can be is $b-a$ since that's the measure of the interval. It's also going to affect that $t$ since $t\in[x,c]$, so we can allow the supremum to refer to that explicitly.
So, replacing $(x-c)$ by ($b-a)$ and writing the suprema appropriately we see that the right hand side will never be smaller than the left hand side, but because we've extended $(x-c)$ out to $b-a$ it might be bigger. So equality becomes inequality and you get $$\sup_{x\in[a,b]} \left| f_n(x)-f_m(x)\right| \leq \left|f_n(c)-f_m(c)\right| + (b-a)\sup_{t\in[a,b]}\left|(f'_n(t)-f'_m(t))\right| $$