Let $v \in L^1_\text{loc}(\mathbb{R})$ weakly differentiable, $b\in \mathbb{R} \setminus \{0\} $ constant and $u:\mathbb{R}^2 \to \mathbb{R}$ defined by $u(x,t):= v(x-bt)$.
Show that $u$ has weak derivatives $ \partial_x u$ and $ \partial_t u$ which satisfy the transport equation $\partial_t u + b \partial_x u = 0$ almost everywhere.
I know the definition of the weak derivative, but since I don´t have any explicit function given I don´t know how to test it. I also wonder if there is a chain rule for weak derivatives, since $u$ is a concatenation of $v$ with $x-bt$.
Could anyone help me out with this problem? Thank you very much in advance!