Problem
Assume that $f(x)$ is derivable over $(a,b)$. Prove that there exsits no discontinuity point of the first kind for $f'(x)$ over $(a,b)$.
Proof
Assume that $f'(x)$ has a discontinuity point of the first kind at $x=x_0$ where $x_0 \in (a,b)$. Then, at least, $\lim\limits_{x \to x_0+}f'(x)$ and $\lim\limits_{x \to x_0-}f'(x)$ both exist. Since $f(x)$ is derivable over $(a,b)$, $f(x)$ is continuous over $[x_0,x_0+h]\subset [x_0,b)$. Thus, by Largrange's Mean Value Theorem, we obtain $$\lim_{h \to 0+}\frac{f(x_0+h)-f(x_0)}{h}=\lim_{h \to 0+}\frac{f'(\xi)h}{h}=\lim_{h \to 0+}f'(\xi)=\lim_{\xi \to x_0+}f'(\xi),$$ where $\xi \in (x_0,x_0+h)$. This shows that, at $x=x_0$, the right derivative of $f(x)$ equals the right limit of $f'(x)$. Likewise, we may obatin, at the same point, the left derivative of $f(x)$ equals the left limit of $f'(x)$ too. But, the fact that $f(x)$ is derivable at $x=x_0$ requires that at $x=x_0$ the left and the right derivatives are equal, therefore, the left and the right limits of $f'(x)$ are also equal, whcih shows that $f'(x)$ is contionuous at $x=x_0$. Now, the contradiction comes out.