Partial derivatives and using the chain rule I have this simple question: 
In a course of differential equations I found this sentence: 
Let $u:\mathbb{R}^n\times \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $u(x,t)=g(x-t\overrightarrow{b})$ where $x,\overrightarrow{b} \in\mathbb{R}^n$, $t\in\mathbb{R}$ and $g$ is a given function.
My question is: 
Why will we have this (or how to use chaine rules to find this result) :
$$\frac{\partial u}{\partial t}=\sum_{i=1}^n -b_i\frac{\partial g}{\partial x_i}(x-t\overrightarrow{b}),$$
(all partial derivatives exist). 
Thank you very much!
 A: For a fixed $x \in \mathbb{R}^n$, $u(x,\cdot)$ is the composite function $\mathbb{R} \to \mathbb{R}^n \to \mathbb{R}$ given by $t \mapsto x - tb$ followed by $x \mapsto g(x)$. The derivative of the first map $\mathbb{R} \to \mathbb{R}^n$ is given by $$\begin{pmatrix} - b_1 \\ \vdots \\ - b_n\end{pmatrix}$$ and the derivative of the second map $g:\mathbb{R}^n \to \mathbb{R}$ is given by $$\begin{pmatrix}\frac{\partial g}{\partial x_1}, \ldots, \frac{\partial g}{\partial x_n}\end{pmatrix}.$$
Thus the derivative of the composite map at some $t$ is $$\begin{pmatrix}\frac{\partial g}{\partial x_1}(x-tb), \ldots, \frac{\partial g}{\partial x_n}(x-tb)\end{pmatrix}\begin{pmatrix} - b_1 \\ \vdots \\ - b_n\end{pmatrix} = \sum_{i=1}^n -b_i\frac{\partial g}{\partial x_i}(x-tb).$$
A: If $u(\vec x,t)=\left . g(\vec y)\right|_{\vec y=\vec x- \vec b t}$, then from the chain rule we can write
$$\begin{align}
\frac{\partial u(\vec x,t)}{\partial t}&=\frac{\partial g(\vec x - \vec b t)}{\partial t}\\\\
&=\sum_{k=1}^n \left. \left(\frac{\partial g(\vec y)}{\partial y_i}\frac{\partial y_i}{\partial t}\right)\right|_{y_i=x_i-b_it}\\\\
&=\sum_{k=1}^n \left. \left(-b_i\frac{\partial g(\vec y)}{\partial y_i}\right)\right|_{y_i=x_i-b_it}\\\\
&=\sum_{k=1}^n \left(-b_i\frac{\partial g(\vec x-\vec b t)}{\partial x_i}\right)
\end{align}$$
since $\frac{\partial  y_i}{\partial x_i}=1$.
