What is the relationship, if any, between maximal domains of $f$ and $f'(x)$?

Question

(Edited question after some Calculus review)

Let $D$ be the maximal domain of function $f$. Let $D^d$ be the largest subset of $D$ where $f'(x)$ is defined.

1. Maximal domain of $f(x) = \ln(x)$ is $D_1 = (0, \infty)$

Its derivative $f'(x) = \frac{1}{x}$ is defined on $D_1$

2. Maximal domain of $f(x) = \sqrt{x}$ is $D_2 = [0, \infty)$

While $f' = \frac{1}{2\sqrt{x}}$ is not defined on $D_2$, it is defined on $D_2^d := (0, \infty)$

$D_2$ and $D_2^d$ differ only by one value.

3. Maximal domain of $f(x) = |x|$ is $D_3 = \mathbb R$

While $f' = \frac{|x|}{x}$ is not defined on $D_3$, it is defined on $D_3^d := \mathbb R \setminus \{0\}$

$D_3$ and $D_3^d$ differ only by one value.

4. Maximal domain of $f(x) = \lfloor x \rfloor$ is $D_4 = \mathbb R$

While $f'$ is not defined on $D_4$, it is defined on $D_4^d := \mathbb R \setminus \mathbb Z$

$D_4$ and $D_4^d$ differ only by countably many values.

It seems that $D$ and $D^d$ differ by only a set of Lebesgue measure zero.

However, the Weierstrass function disproves that conjecture.

So what kind of conditions of $f$ are sufficient or necessary to have such a relationship between $D$ and $D^d$?

Answer 1

Not without more hypotheses, anyway: The Weierstrass function(s) are the standard example of a function that is continuous everywhere but differentiable nowhere: Weierstrass' original family of examples were given by $$f(x) := \sum_{n = 1}^{\infty} a^n \cos (b^n \pi x),$$ where $0 < a < 1$ and $b$ a positive odd integer such that $ab > 1 + \frac{3 \pi}{2}$.