Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function. Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.
I know that for $f$ is differentiable in $(a,b)$ and that it has one-sided derivative $f_+' (a)≠f_-' (b)$ at the endpoints. If $C$ is a real number between $f_+' (a)$  and $f_-' (b)$, then there exists $c∈(a,b)$ such that $f' (c)=C $. How can I use this to prove the above?
 A: Just trying to add some clarity to the tricky part of the proofs presented above.
Let $f:[a,b]\to\mathbb{R}$ be differentiable in $(a,b)$ and $c$ and interior point of the interval, $c\in(a,b)$. We'll prove that
$$\left.\begin{array}{rr}
\displaystyle\lim_{x\to c^+}f'(x)=A\\
\displaystyle\lim_{x\to c^-}f'(x)=B\\
\end{array}\right\}\,\Rightarrow A=B=f'(c)$$
For each $x\in(c,b)$, $f$ is continuous in $[c,x]$ and differentiable in $(c,x)$ so we can apply the mean value theorem to $f$ on the interval $[c,x]$. Thus, for each interval $[c,x]$, there exists a $d(x)\in (c,x)$ such that $$f'\big(d(x)\big)=\frac{f(x)-f(c)}{x-c}$$ 
Since $f$ is differentiable at $c$, making $x\to c^+$ on the RHS of the expression gives:
$$\displaystyle\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}=f'(c)$$
The LHS of the equality is a bit trickier, we would like to say that
$$\displaystyle\lim_{x\to c^+}f'\big(d(x)\big)=\displaystyle\lim_{x\to c^+}f'(x)=A$$
but that is not obvious, since we can't assume $f'$ to be continuous (on the right) at $c$. 
However we know that $\displaystyle\lim_{x\to c^+}f'(x)=A$, we'll use $\epsilon-\delta$ definition of limit. Let $\epsilon>0$, 
$$\exists\delta>0 \,:\, \text{ if } 0<x-c<\delta\, \text{ then }\, |f'(x)-A|<\epsilon$$
(notice that we write $0<x-c<\delta$ instead of $0<|x-c|<\delta$ because we're taking the right hand limit, that is $x>c$).
The MVT guarantees that $d(x)\in (c,x)$, that is
$$c<d(x)<x\,\Rightarrow\, 0<d(x)-c<x-c$$
But this means
$$0<x-c<\delta\,\Rightarrow\,0<d(x)-c<\delta\,\Rightarrow\,|f'\big(d(x)\big)-A|<\epsilon$$
thus $\displaystyle\lim_{x\to c^+}f'\big(d(x)\big)=A$ as desired.
We've seen 
$$f'\big(d(x)\big)=\frac{f(x)-f(c)}{x-c}\xrightarrow{x\rightarrow c^+} A=f'(c)$$
Similarly, we can prove $B=f'(c)$ and we're done.
The answer to your question follows directly from this result.
Let $f$ be differentiable in $(a,c)\cup(c,b)$. Assume $f'$ has a jump discontinuity at $c$: $$\displaystyle\lim_{x\to c^+}f'(x)=A\not= B=
\displaystyle\lim_{x\to c^-}f'(x)$$
(to say we have a jump we require the limits to \textbf{exist} and to be different!).
By the result, if $f$ was also differentiable at $c$, we'd have $A=B$, that contradicts the hypothesis of having a jump: $A\neq B$. Therefore $f$ can't be differentiable at $c$. 
A: This a standard result that derivatives don't have jump discontinuity.
Let $c \in (a, b)$ then $f'(c) = \lim_{x \to c}\dfrac{f(x) - f(c)}{x - c}$ exists. Let us assume that limits $\lim_{x \to c^{+}}f'(x) = A, \lim_{x \to c^{-}}f'(x) = B$ exist. Now let's handle the case for $x \to c^{+}$ first. Clearly then $x > c$ and we have $\dfrac{f(x) - f(c)}{x - c} = f'(d)$ for some $d \in (c, x)$. As $x \to c^{+}$, $d \to c^{+}$ and we get $$f'(c) = \lim_{x \to c^{+}}\dfrac{f(x) - f(c)}{x - c} = \lim_{d \to c^{+}}f'(d) = A$$
Similarly by considering $x \to c^{-}$ we can show that $B = f'(c)$ so that $A = B$ and $f'(x)$ is continuous at $c$ and therefore does not have jump discontinuity. It may happen however that one or both of the limits $A, B$ don't exist or are $\pm\infty$.

Update: I am bit surprised to see that in the comments to the question people have linked this result with IVT (intermediate value theorem) for derivatives. These two properties of derivatives (IVT and no jump discontinuity) are not derivable from each other. Rather they are both derived from Mean Value Theorem in completely different ways.
Further Update: I had a look at the Wikipedia article dealing with Darboux theorem (IVT for derivatives). Even the wikipedia makes a mistake that any function satisfying IVT can't have jump discontinuity. This is totally unexpected from wikipedia and I don't know whom to complain for this.
Here is a very simple example to prove my point. Let $f(0) = 0, f(1) = 1$ and $ f(x) = 1 - x$ for $x\in (0, 1)$. This function satisfies IVT on $[0, 1]$ and is yet having jumps at the end-points.
What is true is the following:
A function $f$ which is monotone and satisfies IVT on $[a,b]$ does not have jump discontinuity and is therefore continuous in $[a, b]$

Even more update: Due to the paraphrasing of the comment by copper.hat in the question I misinterpreted the Wikipedia article. According to copper.hat comment if $g(x)$ takes all values in interval $[g(a), g(b)]$ as $x$ varies in $[a, b]$ then $g(x)$ can't have jumps in $[a, b]$. This statement is wrong.
Wikipedia however has a different definition. It says that a function is Darboux function if it satisfied intermediate value property. The intermediate value property is defined as follows: let $f$ be defined on interval $I$. If for any $[a, b] \subseteq I$ the function $f$ takes all values between $f(a)$ and $f(b)$ for some value of $x \in (a, b)$ then it is said to have intermediate value property on $I$.
I missed the part of any $[a, b] \subseteq I$ and thought that intermediate value property of $f$ on an interval $[a, b]$ is supposed to mean that $f$ must take all values between $f(a)$ and $f(b)$ for some $x \in (a, b)$. Note the subtle difference in Wikipedia version and my interpretation. Wikipedia prescribes a very strong condition where we have to check every subinterval $[a, b]$ of the domain of defintion $I$ of function $f$ whereas in my interpretation we only need to check this for $I$ and not any subintervals of $I$.
To put formally let Wikipedia version of IVT be called WIVT and my version be called PIVT. Then a function $f$ satisfies WIVT if it satisfies PIVT on every sub-interval of $I$. A function satisfying WIVT does not have jumps whereas a function satisfying PIVT may have jumps.
