Discontinuities of the derivative of a differentiable function on closed interval I have a question about the corollary to theorem 5.12 in Rudin's Principles of Mathematical Analysis (page 108):

Suppose $f$ is a real differentiable function on $[a,b]$ and suppose $f'(a)< \lambda < f'(b)$ then there is a point $x \in (a,b)$ such that $f'(x) = \lambda$
Corollary: If $f$ is differentiable on $[a,b]$ then $f'$ cannot have any simple discontinuities on $[a,b]$.

Can someone help me to show how he uses the result in the "main theorem" in the corollary?

(There are two cases of simple discontinuities  $f(x+) = f(x-) \neq f(x)$ and $f(x +) \neq f(x-)$
 A: The theorem in question (Darboux's theorem) basically states that the conclusion of the intermediate value theorem holds for the derivative of an everywhere differentiable function, even if the derivative is discontinuous.
For a simple discontinuity of either kind, it is true that either $f'(x-) \ne f'(x)$ or $f'(x+) \ne f'(x)$.
To be specific, let us treat the case where $f'(x)<f'(x+)$.
Let $\lambda\in(f'(x),f'(x+))$, and pick $y>x$ with $f'(z)>\lambda$ whenever $z\in(x,y]$. Thus $f'(x)<\lambda<f'(x+)$, and $f'(z)\ne\lambda$ for all $z\in[x,y]$, contradicting Darboux's theorem.
The case $f'(x)>f'(x+)$ is treated similarly (or replace $f$ by $-f$ and use the case already treated). The case $f'(x-) \ne f'(x)$ is also treated the same way (or replace $f(x)$ by $f(-x)$).
Edited to fix a flaw pointed out in the comments, no less than seven years later!
A: Assume, on the contrary, that $f^{\prime}$ has a simple discontinuity of type (i), i.e., jump discontinuity such that $f^{\prime}(x-)\ne f^{\prime}(x+)$  in which case $f^{\prime}(x)$ is immaterial.
Then it follows that $f$ is not differentiable at that point, contrary to the fact that $f$ is differentiable on the interval $[a,b]$.
So $f^{\prime}$ cannot have simple discontinuity of type (i).
Since $f$ is continuous, the derivatives of $f$ need not have a simple discontinuity of type (ii).
In addition, suppose $x∈[a,b]$ is a point of discontinuity of $f^{\prime}$ such that $f^{\prime}(x-)\ne f^{\prime}(x+)$, since $f$ is differentiable on $[a,b]$, we see that $f$ is differentiable on $[x-δ,x+δ]$ for every $δ>0$, suppose w.l.o.g $f^{\prime} (x-δ)<f^{\prime} (x+δ)$, then Theorem 5.12 shows that $f^{\prime} (x-δ)<f^{\prime} (x)<f^{\prime} (x+δ)$. Since $δ$ was arbitrary, it follows that $f^{\prime} (x-)<f^{\prime} (x)<f^{\prime} (x+)$. To this case, $f^{\prime} (x)$ is material.
So, we conclude that the function $f^{\prime}$ cannot have any simple discontinuity.
