This is the book of Partial Differential Equation whose author is Lawrence C. Evans. I am reading the Linear Transport Partial Differential Equation of the First Order. In the formulation of the solution of it. The author is defining a function $z(s)$ where $s$ is a real number.I didn't understand how the derivative of $z$ is calculated with respect to the $s$.
To reduce confusion, note that: $$\frac{d}{{ds}}z\left( s \right) = \frac{d}{{ds}}u\left( {{\xi _1}\left( s \right), \ldots ,{\xi _n}\left( s \right),\tau \left( s \right)} \right)$$ where for $1 \leqslant i \leqslant n$: $${\xi _i}\left( s \right) = {x_i} + s{b_i}$$ and: $$\tau \left( s \right) = t + s$$ Using the chain rule, you could write: $$\frac{d}{{ds}}u\left( {{\xi _1}\left( s \right), \ldots ,{\xi _n}\left( s \right),\tau \left( s \right)} \right) = \frac{{\partial u}}{{\partial {\xi _1}}}\frac{{d{\xi _1}}}{{ds}} + \ldots + \frac{{\partial u}}{{\partial {\xi _1}}}\frac{{d{\xi _n}}}{{ds}} + \frac{{\partial u}}{{\partial \tau }}\frac{{d\tau }}{{ds}}$$ Now, using the definition of $\xi$ and $\tau$ and taking the derivative, one arrives at: $$\frac{d}{{ds}}z\left( s \right) = \frac{{\partial u}}{{\partial {\xi _1}}}{b_1} + \ldots + \frac{{\partial u}}{{\partial {\xi _1}}}{b_n} + \frac{{\partial u}}{{\partial \tau }}$$ (which by the definition of the $D$ operator is the same as the result mentioned in the textbook)