Chain rule for weak derivatives for composing a Sobolev function with a differentiable function Suppose that $ g: \mathbb R^n \rightarrow \mathbb R^n$ is a diffeomorphism and that $f \in L^{1}(\mathbb R^n)$ has a weak (distributional) gradient $\nabla f \in L^1(\mathbb R^n)^n$.  I am wondering whether the chain rule
$$
\nabla ( f \circ g ) = \nabla F(g(x)) \cdot D g(x)
$$
is valid in that case.
That is equivalent to showing that for all test functions $\psi \in C^\infty_c(\mathbb R^n)$ we have
$$
\int f(g(x)) \nabla \psi(x) \ dx
=
- \int \nabla F(g(x)) \cdot Dg(x) \psi(x) \ dx
$$
I struggle figuring out how to prove that identity.
Is there an argument building on local smooth approximations?
Can this proven via a chain of variables, similar to the technique discussed in another question?
To clarify, I am not asking for the case that $f$ is differentiable and $g$ is weakly differentiable, which seems to have received more attention.
 A: Consider the two linear maps
$$ A\colon W^{1,1}(\mathbb R^n)\mapsto \mathscr D’ (\mathbb R^n) $$
$$ Af=\nabla [f\circ g] $$
and
$$ B\colon W^{1,1} (\mathbb R^n)\mapsto L^1(\mathbb R^n)\subset \mathscr D’ (\mathbb R^n) $$
$$ Bf=[\nabla f]\circ g\cdot Dg. $$
These maps are continuous. Moreover, $Af=Bf$ whenever $f\in\mathscr D(\mathbb R^n)$ simply by the usual chain rule. By density, the two maps agree on the whole domain.

Edit.
Note, I wrote that $Bf\in L^1$. This is true from the change of variables formula, but it is not necessary to use it. We can simply say that $[\nabla f]\circ g\in L^1_{loc}$.
The latter is true because $g$ is locally bi-lipschitz, which in turn implies that for any compact set $K\subset \mathbb R^n$ there exists a constant $C=C(K)>0$ such that
$$ \frac{1}{C}|U|\leq|g(U)|\leq C|U|, $$
where $U\subseteq K$ is any measurable set and $|\cdot|$ is the Lebesgue measure (the same holds for $g^{-1}$). In fact, for any measurable function, one has
$$ \int f(x)dx=\int_0^\infty |\{x\in\mathbb R^n|\;|f(x)|>\lambda\}| d\lambda $$
and for any compact set $K$, one has
$$ \int_K f\circ g(x)dx= \int_0^\infty |K\cap\{x\in\mathbb R^n|\;|f\circ g(x)|>\lambda\}| d\lambda= $$
$$ = \int_0^\infty |g^{-1}(g(K)\cap\{x\in\mathbb R^n|\;|f(x)|>\lambda\})| d\lambda\lesssim_K \int_0^\infty |g(K)\cap\{x\in\mathbb R^n|\;|f(x)|>\lambda\}| d\lambda =$$
$$= \int_{g(K)} f(x)dx<\infty. $$
We then use the fact that $Dg$ is continuous (locally bounded is enough) to prove that $Bf\in L^1_{loc}$.

Finally, I don’t think that you need to use the question in the link you provided, since in your case $g$ is smooth enough to do everything you need to do. Moreover, unlike the question you linked, the question is not hopeless because both objects in your equality make sense a priori.
