Im approaching PDE theory as a newbie, and there is a super simple question I cannot wrap my head around yet:
Let the PDE linear transport equation be defined on $\mathbb{R}^n \times (0,\infty)$ as $$ u_t + b \cdot \nabla u$$ with $b=(b_1,\ldots, b_n) \in \mathbb{R}^n$ and $t>0$.
If we define $z(s) := u(x+sb,t+s)$ with $s \in \mathbb{R}$, then why is
$$\frac{\partial}{\partial s}z(s) = \nabla u (x+sb,t+s) \cdot b + u_t(x+sb,t+s) \quad ?$$
Why is the derivation of $z$ that way? To be honest, I thought naively that
$$ \frac{\partial}{\partial s}z(s) = \frac{\partial}{\partial s} u (x+sb,t+s) = \begin{pmatrix} \dot{u}(x+sb,t+s) \cdot b \\ \dot{u}(x+sb,t+s) \end{pmatrix} $$ or something like that.
First notice that $z(s)$ has values in $\mathbb{R}$, so $\frac{\partial z}{\partial s}$ must be one-dimensional.
The function $u(x+sb, t+s)$ is a composition of $u$ and the mapping $s\mapsto (x+sb,t+s)$. Hence $$ \frac{\partial}{\partial s}u(x+sb, t+s)) = \sum_{i=1}^n \frac{\partial}{\partial x_i}u(x+sb,t+s)\cdot b_i + \frac{\partial}{\partial t}u(x+sb,t+s)\cdot 1 $$ $$= \nabla u(x+sb,t+s)\cdot b + u_t(x+sb,t+s).$$