Proof: If $ f $ is differentiable on $ [a,b] $ then $ f' $ cannot have any simple discontinuities on $[a,b] $ This is a corollary following Rudin's theorem 5.12, which states: Suppose $ f $ is a real differentiable function on $ [a,b]$ and suppose that $f'(a)<\lambda<f'(b).$ Then there is a point $x \in (a,b) $ such that $ f'(x)=\lambda$. 
The corollary says if $ f $ is differentiable on $ [a,b] $ then $ f' $ cannot have any simple discontinuities on $[a,b] $. 
Intuitively this makes sense but I can't come up with a neat proof. 
I was thinking to do it by contradiction; to suppose $ f' $ has a simple discontinuity. This means the lefthand and righthand limits at $ x\in (a,b) $ exist but there are two cases;either $ f(x+)$ does not equal $ f(x)$ or $f(x-) $ does not equal $f(x) $.
Then I'm not sure where to go from here/how to apply the theorem above.
 A: Suppose $f'$ has a simple discontinuity at $x_0$. Assume without loss of generality that $f'(x^-)<f'(x^+)$. Then choose real numbers $\lambda_1,\lambda_2$ such that $f'(x^-)<\lambda_1<\lambda_2<f'(x^+)$. Since $f'$ has left and right limits at $x_0$, we may choose $c<x_0<d$ such that, we have $f'(x)\le \lambda_1$ and $\lambda_2\le f'(y)$ for all $x\in[c,x_0)$, $y\in(x_0,d]$. Now choose $\lambda$ so $\lambda_1<\lambda<\lambda_2$. By the given theorem, for some $x\in(c,d)$ we have $f'(x)=\lambda$. But this is impossible - if $x\in(c,d)$, either


*

*$c<x<x_0$, so $f'(x)\le\lambda_1<\lambda$;

*$x_0<x<d$, so $f'(x)\ge\lambda_2>\lambda$; or

*$x=x_0$, in which chase $f'(x)$ is either $f'(x^+)$ or $f'(x^-)$, neither of which are equal to $\lambda$.

A: Is this not just an application of the intermediate value theorem? Define $h(x) = f'(x) - \lambda$ then $h(a)<0$ and $h(b)>0$ so there exists $c  \in (a,b)$ s.t $h(c) = 0 \Rightarrow f'(c) = \lambda$.
A: I am trying to supplement two humble details to @Jason 's answer.
The first one is we need to prove only one side. Suppose $f'$ has a simple discontinuity at $x_0$, then either $f'(x_0) \neq f'(x_0+)$ or $f'(x_0) \neq f'(x_0-)$. Without loss of generality, we can assume $f'(x_0) < f'(x_0+)$.
The second one is since $f'(x_0) < f'(x_0+)=\lim_{t \to x_0} f'(t)$, $ t>x_0$. Let $\epsilon = \frac{f'(x_0+)-f'(x_0)}{2}>0$, then $\exists \delta>0$, such that for any $t-x_0<\delta$, we have $f'(x_0+)-f'(x_0)>\epsilon$.
The second one is trivial but it still took me some time to figure it out.
