Let $g:(0,1) \rightarrow \mathbb R^n$ be absolutely continuous, $F \in W^{1,2} (\mathbb R^n).$ Is it true that a.e. it holds $$ \dfrac{dF(g(t))}{dt} = \nabla F(g(t)) \cdot g'(t) \quad ? $$
What I want to prove is that $$ \int_0^1 F(g(t)) \phi'(t) \ dt = - \int_0^1 \nabla F(g(t)) \cdot g'(t) \phi(t) \ dt $$ for every $\phi \in C_c^{\infty}((0,1))$. My idea would be to use some sort of change of variable formula to obtain the thesis as in \begin{align} \int_0^1 F(g(t)) \phi'(t) \ dt \stackrel{*}{=} &~ \int_{\mathbb R^n} F(x) \phi'(g^{-1}(x)) (g^{-1})'(x) \cdot dx \\ = &~ \int_{\mathbb R^n} F(x) \nabla(\phi(g^{-1}(x))) \cdot dx \\ = &~ - \int_{\mathbb R^n} \nabla F(x) \phi(g^{-1}(x)) \cdot dx \\ = &~ - \int_{\mathbb R^n} \nabla F(g(t)) \cdot g'(t) \phi(t) \ dt, \end{align} but equality in $*$ would require $g$ to be a $C^1$-diffeomorphism or similar hypothesis. How could I prove the result for an absolutely continuous $g$?
There are some issues which must be considered. For instance, the image of $g$ is always a set of zero measure in $\mathbb{R}^n$, thus, you can redefine $F$ and $\nabla F$ in this set to be anything. How do you approach this?
Another issue is the following: Let $n=3$ and $x=(x_1,x_2,x_3)$. Define $F(x)=|x|^{-1/4}-1$ for $|x|\le 1$ and $F(x)=0$ for $|x|>1$. Note that $F\in W^{1,2}(\mathbb{R}^3)$. Let's take for $\nabla F$ the usual gradient of $F$.
Let $g:(0,1)\to\mathbb{R}^3$ be defined by $g(t)=(t-1/2,0,0)$. We have that $F(g(t))=|t-1/2|^{-1/4}$. Moreover, $$\nabla F(g(t)) g'(t)=\frac{-1}{4}|t-1/2|^{-5/4}.$$
In this case $\nabla F(g(t)) g'(t)$ does not belong to $L^1_{loc}((0,1))$.