If $\lim_{x\to x_0+}\frac{df}{dx}=K$ exists then $f$ has right derivative and is equal to $K$ Let $f$ be continuous at $[x_0;x_0+h]$ and have finite derivative at $(x_0;x_0+h)$.
If the limit
$$
\lim_{x\to x_0^+}\frac{df}{dx}=K
$$
exists then $f$ has right derivative and is equal to $K$ ($K$ can be finite or infinite).

Can you help how from this we get that derivative of $f$ is continuous or has only second order discontinuous
points?


If $K$ finite, for all $\epsilon>0$, there is a constant $\delta>0$ such that
$$|f'(x)-K|<\epsilon$$
for all $x_0<x<x_0 +\delta$
Thus for all $x,y \in (x_0,x_0+\delta)$
$$\left| \frac{f(x)-f(y)}{x-y} -K\right|<\epsilon$$
By continuity of $f$, this implies
$$\left| \frac{f(x_0)-f(y)}{x_0-y} -K\right|<\epsilon$$
for all $y \in (x_0,x_0+\delta)$
Hence forth the conclusion.
