If derivative of $f$ is continuous, then $f$ is continuous. I have searched a lot, but i haven't found any proof about that statement. I have checked the proof of

If $f$ is differentiable, then $f$ is continuous

but it's not the same argument I think. Also, I want to know what's your opinion about the statement

If derivative of $f$ is not continuous, then $f$ is not continuous

 A: If $f$ is differentiable, then $f$ is continuous. The continuity of $f'$ is irrelevant here.
In particular, even if $f'$ is discontinuous, $f$ is continuous.
A: $f'$ need not be continuous.
Suppose that $f'(x)$ exists in the interval $(a,b)$. If $\xi \in (a,b)$, then $f'(\xi)$ exists. Hence $f$ is continuous at $\xi$. Since this is true for all $\xi$ in $(a,b)$, then $f$ is continuous on $(a,b)$.
A: Your problem seems to be the logical relationships between the statements


*

*If f is differentiable, then it is continuous 

*If the derivative of $f$ is continuous, then $f$ is continuous

*If the derivative of $f$ is not continuous, then $f$ is not continous.


The first statement trivially implies the second, since saying "the derivative of $f$ is continuous" is the same as saying "$f$ is differentiable and $f^{\prime}$ is continuous". 
The contrapositive of the third statement is "If $f$ is continuous, then the derivative of $f$ is continuous." This is false. For example, the function 
$$f(x)=x^2\sin\left(\frac{1}{x}\right)$$
is differentiable everywhere, with derivative
$$f^{\prime}(x)=\left\{\begin{array}{ll}
2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right)& x\neq 0
\\
0 & x=0
\end{array}\right.$$
But $\lim_{x\to 0}f^{\prime}(x)$ does not exist, hence $f^{\prime}$ is not continuous.
A: Answering only part of the question: "If derivative of f is not continuous, then f is not continuous": As perhaps the simplest counter-example, the absolute value function is continuous but not continuously differentiable. 
This does not disprove the opposite statement, of course.
