Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ a differentiable function such that $f'(x)=0$ for all $x\in\mathbb{Q}.$ $f$ is a constant function?

**Answer**

No, such a function is not necessarily constant.

At the bottom of page 351 of *Everywhere Differentiable, Nowhere Monotone Functions*, Katznelson and Stromberg give the following theorem:

Let $A$ and $B$ be disjoint countable subsets of $\mathbb{R}$. Then there exists an everywhere differentiable function $F: \mathbb{R} \to \mathbb{R}$ satisfying

- $F'(a) = 1$ for all $a \in A$,
- $F'(b) < 1$ for all $b \in B$,
- $0 < F'(x) \leq 1$ for all $x \in \mathbb{R}$.

Choosing $A = \mathbb{Q}$ and an arbitrary (nonempty*) countable set $B \subset \mathbb{R} \setminus \mathbb{Q}$, we get an everywhere differentiable function $F$ with $F'(q) = 1$ when $q \in \mathbb{Q}$. So if we define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = F(x) – x$, this then is the desired function satisfying $f'(q) = F'(q) – 1 = 1 – 1 = 0$ for $q \in \mathbb{Q}$, and $f$ is not constant (or else its derivative would be zero everywhere by the mean value theorem).

*(in case you take “countable” to mean either “countably infinite” or “finite”)

**Attribution***Source : Link , Question Author : felipeuni , Answer Author : Vandermonde*