I am essentially (as in: essential singularity) confused by "change of variables" formula of two-variable function.
All functions below are $\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, meaning a functions of two variables. So let there be $f$, $g$, $u$ and $v$ of that form, and additionally impose $g$ by $g(u(x,y), v(x,y)) = f(x, y)$.
Often in physics one would then talk about the following objects, none of which I ever understood:
- $\frac{\partial g}{\partial x}$ - I would interpret this as $\partial_1 g$, but apparently one is allowed to write $$\frac{\partial g}{\partial x} = \frac{\partial g}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial g}{\partial v} \frac{\partial v}{\partial x} $$
- $\frac{\partial f}{\partial u}$ - $u$ is a function, not a label of an argument, and yet I am to believe that $$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u} $$ where again $x$ and $y$ are not functions, so $\partial_u x$, $\partial_v x$ have no meaning.
The matter is further complicated when one writes out the arguments: $$\frac{\partial g}{\partial u} (u(a, b), v(a,b)) \frac{\partial u}{\partial x} (a, b) + \frac{\partial g}{\partial v} (u(a, b), v(a,b)) \frac{\partial v}{\partial x} (a,b)$$ is supposed to be $\frac{\partial g}{\partial x}(a, b)$ or $\frac{\partial g}{\partial x}(u(a,b), v(a,b))$?
It always seemed to me that one tacitly promotes $\mathbb{R} \times \mathbb{R}$ to a vector space $(\mathbb{R}^2, +, \cdot)$ and all functions are then converted into functions of vector variables (e.g. $\tilde{f}(\vec{x}) = f(\vec{x} \cdot \vec{i}, \vec{x} \cdot \vec{j})$). But this scheme is doomed to fail in general if the domains of the three variables together do not form a vector space.
How should I interpret partial derivatives contructs such as those listed? A reference to a multivariable calculus book that is very strict about this would be helpful too.