I have had to think for a while about this and noticed some subtleties which I still cannot explain.
Firstly, I am wondering about a special case of the non differentiability of a discontinuous function along some interval. If the function is simply undefined at some point it is not a function and thus not differentiable at that point. But suppose it is defined and there is a jump discontinuity by a constant $C$. Then the derivative at that point and everywhere around that point is continuous and defined. So is this a discontinuous function with a continuous derivative?
Secondly, I have found some instances where a function is differentiable, continuous but its derivative is not continuous. For example, the node of a cusp on the y axis will have a unique derivative but the derivative function is not continuous. Or another example, $y = x, a < x < c$ $y=2x, c \leq x < b$. The derivative is jump discontinuous because it shifts up at c, but it is not undefined at c. Are these valid examples? I'm guessing it is possible to integrate jump discontinuous functions.
Answer
If a function is 'jumps up' at a point $x$, its derivative can't be defined there. Suppose we have a function which is equal to $a$ at $x$, and $b$ just before $x$, with $a\neq b$.
Now, use the definition of derivative:
$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$
If we work out
$$\lim_{h\to 0^-}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^-}\frac{b-a}{h}=\infty$$
we see that the derivative is not defined there.
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