I want to apply the chain-rule to the function $z(s) := u(x + sb, t+s)$ where $x = (x_1, \cdots, x_n) \in \mathbb{R}^n$ and $b \in \mathbb{R}^n$, $s \in \mathbb{R}$ are constants. I am supposed to get this:
\begin{align} z(s) &= u(x+sb,t+s)\\[5pt] z'(s) &= u_{x_{1}}(x+sb,t+s)b_{1} + \dotsb + u_{x_{n}}(x+sb,t+s)b_{n} + u_{t}(x+sb,t+s)\cdot 1 \end{align}
but I do not know how this is done concretely.
Here is my attempt. I set $\xi(s) = x+sb$ and $\gamma(s) = t+s$. Then, $z(s) : = u(\xi(s), \gamma(s))$. The chain-rule then says: $$\frac{dz}{ds} = \frac{du}{d\xi}\frac{d\xi}{ds} + \frac{du}{d\gamma}\frac{d\gamma}{ds}.$$
What is $\frac{du}{d\xi}$ ?
Since $\xi$ is a vector then I think it is the sum $u_{\xi_1} + \cdots + u_{\xi_n}$. Then what is $u_{\xi_1}$?
This is my confusion that I hope someone can clear for me.
Based on the solution, it seems that $u(x + sb, t+s)$ is shorthand for $$u(x_{1}+sb_{1},x_{2} + sb_{2}, \dotsc, x_{n} + sb_{n},t+s).$$ So, we should really be defining $\xi_{i}(s) = x_{i} + sb_{i}$ and so $z(s)$ becomes $$z(s) = u(\xi_{1}(s), \xi_{2}(s),\dotsc,\xi_{n}(s),\gamma(s)).\tag{1}$$ Now, because $u\colon \mathbb{R}^{n+1}\to \mathbb{R}$ we think of $$u = u(x_{1},x_{2},\dotsc,x_{n},t).$$
That means that when we apply the chain rule to $(1)$ we get $$\frac{dz}{ds} = \frac{du}{dx_{1}}\frac{d\xi_{1}}{ds} + \frac{du}{dx_{2}}\frac{d\xi_{2}}{ds} + \dotsb + \frac{du}{dx_{n}}\frac{d\xi_{n}}{ds} + \frac{du}{dt}\frac{d\gamma}{ds}.$$
Then: \begin{align} \frac{du}{dx_{i}}\frac{d\xi_{i}}{ds} &= u_{x_{i}}(\xi_{1}(s), \xi_{2}(s),\dotsc,\xi_{n}(s),\gamma(s))\frac{d}{ds}(x_{i} + sb_{i})\\ &= u_{x_{i}}(\xi_{1}(s), \xi_{2}(s),\dotsc,\xi_{n}(s),\gamma(s))b_{i}\\ &=u_{x_{i}}(x_{1}+sb_{1},x_{2} + sb_{2}, \dotsc, x_{n} + sb_{n},t+s)b_{i}\\ &=u_{x_{i}}(x+sb,t+s)b_{i}. \end{align}