Let $\Omega \subseteq \mathbb{R}^d$ (or for simplicity take $\Omega = \mathbb{R}^d$). Let $u\in L^1(\mathbb{R}^d)$ such that $u\in \text{BV}(\Omega)$ i.e its distributional derivative denoted $Du$ is representable as a finite Radon measure. How can I relate $D(u^m)$ to $Du$, for some power $m\in \mathbb{N}$. I guess I'm asking for a chain rule for distributional derivatives.
My main confusion with these things is that at this moment I cannot show $u \in W^{1,1}(\mathbb{R}^d)$, hence don't have access to the usual chain rule.
Setting $f(u) := u^m$ and $v := f\circ u = u^m$, the chain rule in BV says that $$ \begin{cases} \widetilde{D} v = f'(u) \nabla u \, \mathcal{L}^d + f'(\widetilde{u}) D^c u,\\ D^j v = [f(u^+) - f(u^-)]\, \mathcal{H}^{d-1}\lfloor J_u. \end{cases} $$ Since $f$ is not (globally) Lipschitz continuous, you should know beforehand that $v$ belongs to $BV$.
(For notations and the proof of the result, see Ambrosio-Fusco-Pallara, "Functions of Bounded Variations and Free Discontinuity Problems", Thm. 3.96.)