Here is the definition of Caratheodory differentiability:
Let $I\subseteq \mathbb{R}$ be an open interval, $a\in I$, $f:I\to\mathbb{R}$. We say $f$ is
Caratheodory differentiable at $a$ if there is a continuous function $\varphi:I\to\mathbb{R}$ such that
$f(x)-f(a) = \varphi(x)(x-a)$ for all $x\in I$. If it exists, this $\varphi$ is unique; further, $f'(a)=\varphi(a)$.
It is a somewhat standard fact that CD is equivalent to differentiability.
Now if we consider $f:\mathbb{R}\to\mathbb{R}$, $f(x)=x^2 \sin(1/x),$ $x\neq 0$ and $f(0)=0$, the definition makes short work of showing $f$ is CD (take $\varphi(x) = x\sin(1/x)$, with $\varphi(0)=0$). However, as is well-known, $f$ is not continuously differentiable at $x=0$, since $\lim_{x\to 0}f'(x)$ doesn't exist. But doesn't the Caratheodory version imply $f'$ is continuous at $x=0$?
Best Answer
Just doing this so it can become answered; Daniel's answer was very clear.