Function $\Bbb Q\rightarrow\Bbb Q$ with everywhere irrational derivative

Yes there are many of them.

Let $\alpha$ be any irrational number and let's build a function whose derivative is $\alpha$.

We pick an enumeration of the rationals $\{r_1,r_2,r_3,\ldots\}$ and we will choose each $f(r_n)$ in order. At the same time in order to make $f'(r_n) = \alpha$ we will decide how to squeeze the graph of $f$ near $r_n$

Suppose we have chosen $n$ points and that we have restricted the remaining graph of $f$ to some open set $U_n \subset \Bbb R^2$ where $\pi(U_n) = \Bbb R \setminus \{r_1,\ldots,r_n\}$ ($\pi : \Bbb R^2 \to \Bbb R$ is the projection on the $x$-axis) .

First, we pick a rational value $y_{n+1}$ for $f(r_{n+1})$ such that $(r_{n+1},y_{n+1}) \in U_n$ ($U_n \cap \pi^{-1}(r_{n+1})$ is nonempty by the induction hypothesis, and $U_n$ is open, so we can find a rational value in there).

Next, we choose two parabolas tangent at $(r_{n+1},y_{n+1})$ with slope $\alpha$ (one of them upside down) and in particular we choose their leading coefficient large enough (in absolute value) so that the upper parabola doesn't meet the lower border of $U_n$ and the lower parabola doesn't meet the upper border of $U_n$ (those borders are a finite number of parabola pieces so this is possible).

Then we choose $U_{n+1}$ to be the interection of $U_n$ and the region between the two parabolas. Then $\pi(U_{n+1}) = \pi(U_n) \setminus \{r_{n+1}\}$, and any function whose graph stays in $U_{n+1}$ will have a derivative $\alpha$ at $r_{n+1}$.

Once we have done this for every $n$, we have a function $\Bbb Q \to \Bbb Q$ "differentiable" everywhere with derivative $\alpha$.

Though, it might not look good and it may not have a continuous extension to $\Bbb R$. Heck, you can even choose any function $g : \Bbb Q \to \Bbb R$ and force $f' = g$