Suppose we want to prove that the derivative of a function across an interval exists at $C$, but the derivative at $C$ cannot be found. We know the function must be continuous. Can we take the limit of derivative from the negative and positive direction of $C$ and show that if they are equal, the derivative at $C$ exists and is equal to the limit obtained? Is this a necessary and sufficient condition?
EDIT:
Sufficiency – If a function is a derivative along some interval, it does not have a removable singularity at $C$.
Necessity – There is no interval of a derivative of some function in which a jump or essential discontinuity occurs.
There are two cases in which the condition is met if this is a necessary and sufficient condition. One is where the derivative is continuous, the other is where there is a removable discontinuity in the derivative. Is the latter possible?
Best Answer
If the $\lim_{x\to c^+}f^\prime(x)=\ell$ then using the mean value theorem you have that $$\frac{f(x)-f(c)}{x-c}=f'(d_x),$$ where $c<d_x<x$ and so $d_x\to c$ as $x\to c^+$. Hence, there is $$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c^+}f'(d_x)=\ell.$$ Same for the other side. So if the limits of the derivatives exist and are equal then $f$ is differentiable. The other implication is false because $f'$ could exist everywhere but not be continuous at $c$. The function $f(x)=x^2\sin{1/x}$ if $x\ne 0$ and $f(0)=0$ is differentiable everywhere but the derivative is discontinuous. Even worse you could take $f(x)=x^2$ if $x$ is rational and $f(x)=0$ otherwise. then $f$ is differentiable at $x=0$ and discontinuous at every other point.
EDIT: The original question has been modified and so my answer makes no sense anymore. Anyway the derivative of a function is Darboux continuous, so if it cannot have certain types of discontinuities. Darboux