Showing the derivative of a differentiable function has a point of continuity The question goes like this - 
Let $f:[0,1]\rightarrow \mathbb{R}$ be a differentiable function. Show that $f'(x)$ has a continuity point.
Thanks for the help! 
 A: Here are a few thoughts. 
1) a theorem of Baire (according to one of my old textbooks): Let $(X,\mathcal{T})$ be a topological space and $f_n$ be a sequence of continuous functions $f_n : X \to \Bbb{R}$ with the property that there exists $f:X \to \Bbb{R},$ 
$$ f(x)=\lim_{n \to \infty} f_n(x),\ \forall x \in X.$$
[edit] A good reference for this (assuming $X$ is metrizable) is Ch. 24.B of Kechris's Classical descriptive set theory (pp. 192ff), given by Theo Buehler in the comments. Many thanks. My source textbook is very old and not known outside my university.
Then the set $D(f)$ of the discontinuity points of the function $f$ is of first Baire category type (it is a countable union of sets $E$ with $\text{int}(\text{cl}(E))=\emptyset$; int is interior, cl is closure).
2) $[0,1]$ with the standard topology is not a space of the first category, because every complete metric space is a Baire space.
3) Let
$$ f_n: [0,1] \to \Bbb{R},\ f_n(x)=\begin{cases}
\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}},\ \ x \in \Bigl[0,1-\frac{1}{n}\Bigr]\\
\frac{f(1)-f(1-\frac{1}{n})}{\frac{1}{n}},\ \ x \in \Bigl[1-\frac{1}{n},1\Bigr]
\end{cases}$$
The sequence of continuous functions $f_n$ converge pointwise to $f'$ on $[0,1]$. By the first two points, the set of discontinuity points of $f'$ cannot be the whole $[0,1]$.
