Let $k$ be a nonarchimedean local field and $G$ a reductive $k$-group, which we assume to be semisimple and simply-connected. Recall that an abstract group $H$ is perfect if it is generated by commutators, that is, equals its derived subgroup.
Question: Is $G(k)$ perfect?
When $G$ is isotropic, $G(k)$ is known to be perfect. This is a consequence of the solution to the Kneser–Tits problem for $k$: the group $G(k)^+$ generated by (the rational points of) additive subgroups of $G$ is the same as the group generated by root subgroups of $G$ for a fixed maximal split torus, and the latter group is generated by commutators. It is known that $G(k)=G(k)^+$ when $G$ is simply-connected.
So the main thrust of my question is when $G$ is anistropic, or equivalently (hence the title), when $G(k)$ is compact in the analytic topology. In type $A$, Platonov and Jančevskiĭ have proved (On a conjecture of Harder) by computations with division rings that $G(k)$ is perfect. (Clarification: This result is for $D^\times$, not $\text{SL}_1(D)$.)
Related question: Are there any anisotropic simple $k$-groups not of type $A$?
I realize that if the answer to the related question is negative then this work of Platonov and Jančevskiĭ answers my first question. I once looked through the tables in Tits's Boulder notes Classification of semisimple algebraic groups and it seemed like there were compact simple reductive groups only in type $A$. But this result was so surprising, and the tables so hard for me to follow, that I was not confident in the correctness of my understanding.
Bonus points for questions that work with a more general $k$, say, a complete discretely valued field with residue field of dimension $\leq1$.
Clarification: I was too blithe about what happens in type $A$. As krl pointed out in the comments, $\operatorname{SL}_1(D)$ is not perfect, and in fact, its commutator subgroup is its pro-unipotent radical. Riehm's paper "The norm 1 group of a $\mathfrak P$-adic division algebra" carefully analyzes the situation. So in the end, the answer to my question is "no" overall, but "yes" in many cases.