Galilean transformation and differentiation Given $x=x’-vt$ and $t=t’$, why is $\frac{\partial t}{\partial x’}=0$ instead of $1/v$? Maybe the answer has something to do with the fact that $dx’=dx$ in this Galilean transformation. Is $dx’=dx$ always the case for Galilean transformations?

Partial derivatives are only defined when you specify a convention regarding what's held constant, or that convention is obvious in context. For example, $\frac{\partial t}{\partial x^\prime}=0$ is derived from $t=t^\prime$ and assumes you're holding $t^\prime$ constant, and we can express this by writing $\left(\frac{\partial t}{\partial x^\prime}\right)_{t^\prime}=0$. By contrast, from $t=\frac{x^\prime-x}{v}$ we get $\left(\frac{\partial t}{\partial x^\prime}\right)_x=\frac{1}{v}$. (Of course, we can't define $\frac{\partial t}{\partial x^\prime}$ with a convention that holds either $t$ or $x^\prime$ constant.)