Discontinuity of differential functions

There is a corollary in Rudin analysis. But I am not able to understand it. Can someone help to understand it?

The Corollary is:

Let $f$ be a real differential function on $[a,b]$, then $f'$ cannot have any simple discontinuity.

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Did you understand the definition of simple discontinuity? Did you understand the statement to which this is a corollary? – Siminore Nov 24 '12 at 13:27
YES. It means both left and right hand limit exist, for simple discontinuity. Theorem proves the intermediate value theorem for derivative. – user38764 Nov 24 '12 at 13:32
Does "differential" mean "differentiable"? – Chris Eagle Nov 24 '12 at 13:46
yes............ – user38764 Nov 24 '12 at 13:49

I'll sketch the argument. If the left and right hand limits $f'(c-)$ and $f'(c+)$ both exist and are not equal, then we're in a situation similar to $f'(c-) < f'(c) < f'(c+)$. So working on the lefthand side, we can find an $\epsilon > 0$ $f'(x) < f'(c) - \epsilon$ for all $x \in (c-\delta, c)$. Applying the theorem, we have a contradiction.
If $f$ is continuous at $a$, differentiable for $x>a$, and $\lim_{x\to a+} f'(x)=p$, then one has $$\lim_{x\to a+}{f(x)-f(a)\over x-a}=p\ .$$
Proof. Given an $\epsilon>0$ there is a $\delta>0$ such that $$|f'(x)-p|<\epsilon\qquad\bigl(x\in\ ]a,a+\delta[\ \bigr)\ .$$ Let $x\in\ ]a,a+\delta[\$. Then by the mean value theorem there is a $\xi\in\ ]a,x[\ \subset \ ]a,a+\delta[\$ such that $$\left|{f(x)-f(a)\over x-a}- p\right|=\bigl|f'(\xi)-p\bigr|<\epsilon\ .\qquad\qquad\square$$ It follows that the limits $\lim_{x\to a+} f'(x)$ and $\lim_{x\to a-} f'(x)$ cannot both exist and be different, if $f$ is differentiable at $a$.