While presenting Cauchy's theorem in their well-known textbook, the authors chose to state a stronger condition on the analytical function $f(z)$ that require it to be not only analytic within the simple domain $D$ and on its contour $C$, but also that the derivative $f'(z)$ is continuous on the same complex domain. They admitted that this condition is stronger than needed (only analyticity of $f(z)$ is actually needed) for the theorem, and that they gave such version for simpler proof within their book's scope.
However, this is a bit confusing to me, because the authors are here clearly distinguishing between being "analytic" versus being "of continuous derivative $f'(z)$" on all points of the domain under question.
But I had thought that being analytic would automatically mean that the derivative function $f'(z)$ [as a limit by definition, $\lim\limits_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}$] exists and that it exists uniquely (regardless of the direction in which $h$ approaches zero) at each point. And if I recall the basic requirement for a function to be continuous at any point, it is that the limit $\lim\limits_{x\rightarrow x_{0}}f(x)$ exists (same value from all directions) and is equal to the function's value there [$\lim\limits_{x\rightarrow x_{0}}f(x)=f(x_{0})$]. Similarly, a function $f(x)$ is differentiable if it has its derivative $f'(x)$ continuous using the same definiton just mentioned (again as existent limit from all directions)…
Furthermore, more modern definitions of the term "analytic" (e.g. here and here) refers to being representable as Taylor series, which again implies having existent higher derivatives everywhere in the domain, which therefore must mean that these derivatives are themselves continuous functions.
So drawing a distinction between being "analytic" and being "of continuous derivative" is somewhat confusing here. Any help to clarify this would be helpful.
Best Answer
A function that is differentiable (in the complex sense) in an open set does have a continuous derivative there, and is representable by Taylor series. These facts are not at all obvious, and require proof. More "theoretically" oriented texts on complex analysis contain such proofs.
In the case of functions on the real line, differentiability does not imply continuity of the derivative. A standard example is $$ f(x) = x^2 \sin(1/x)$$ If $x \ne 0$, $f'(x) = 2 x \sin(1/x) - \cos(1/x)$. This does not have a limit as $x \to 0$. On the other hand, $f'(0)$ does exist and is $0$, as can be seen from the limit definition of derivative and the fact that $|f(x)| \le x^2$.
I think your confusion arises from the fact that there are two quite different limits here: $$ \lim_{h \to 0} \frac{f(z+h) - f(z)}{h} = f'(z)\ \text{the definition of $f'(z)$}$$ and $$ \lim_{h \to 0} f'(z+h) = f'(z) \ \text{for its continuity}$$ In the case of functions on the reals, you can have one without the other.