From differential geometry I have learned that:
$$ \partial_{x + y} = \partial_x + \partial_y $$
Now trying to prove this property for partial derivatives as I know them from multivariable calculus I encounter conceptual problems.
I want to prove that $A_{x+y}(x,y) = A_x(x,y) + A_y(x,y)$
First naive approach was:
$A_{x+y} = A_x \frac{\partial x}{\partial (x+y)} + A_y \frac{\partial y}{\partial (x+y)} = \{t:=x+y\} = A_x \frac{\partial (t -y)}{\partial t} + A_y \frac{\partial (t - x)}{\partial t} = A_x + A_y$
This seems plausible at first, but I assumed here that $\frac{\partial y}{\partial t} = \frac{\partial x}{\partial t} = 0$, which must be wrong because: $1 = \frac{\partial t}{\partial t} = \frac{\partial (x+y)}{\partial t} = \frac{\partial x}{\partial t} + \frac{\partial y}{\partial t} = 0$
I tried keeping track of the complete set a variables that I define $A$ as of function of (e.g. $A(x,y) \rightarrow A(x(t),y)$, but this still gives me contradictions.
So what is really happening here? And how to rigorously prove this simple property using a similar approach to the one I made? (if it is even possible, but why souldn't it be?).
Any lengthy commentary is extremely welcome!
Partial derivatives are one of the most confusing mathematical subjects, mainly because our notation for them is so bad, in that it does not express one of their main aspects: that a variable is being considered as one of a set of variables, all others of which are being held constant while this one variable is varied.
Thus, there is no meaning to $\partial_{x+y}$ unless you specify what set of variables the variable $t=x+y$ is meant to be a part of. The natural choice for the second variable is $u=x-y$, and indeed for this choice you can prove an equation similar to yours, but with an additional factor of $2$:
\begin{eqnarray} \left.\frac{\partial}{\partial(x+y)}\right|_{x-y} &=&\left.\frac{\partial x}{\partial(x+y)}\right|_{x-y}\left.\frac{\partial}{\partial x}\right|_y+\left.\frac{\partial y}{\partial(x+y)}\right|_{x-y}\left.\frac{\partial}{\partial y}\right|_x\\ &=&\frac12\left.\frac{\partial((x+y)+(x-y))}{\partial(x+y)}\right|_{x-y}\left.\frac{\partial}{\partial x}\right|_y+\frac12\left.\frac{\partial ((x+y)-(x-y))}{\partial(x+y)}\right|_{x-y}\left.\frac{\partial}{\partial y}\right|_x\\ &=&\frac12\left.\frac{\partial}{\partial x}\right|_y+\frac12\left.\frac{\partial}{\partial y}\right|_x\;, \end{eqnarray}
where the subscript of the vertical bar indicates the variable being held fixed.
If you want to read more about confusions surrounding partial derivatives, and how to clear them up by using less ambiguous notation, you can check out some of my other answers to questions tagged partial-derivative.