partial derivative transformation rule

Question

If $\phi$ is a map between coordinates given by $$ \xi=\xi(x,y)\\ \eta=\eta(x,y)\\ $$ with inverse $\phi^{-1}$ $$ x=x(\xi,\eta)\\ y=y(\xi,\eta)\\ $$ the transformation rule for partial derivatives are $$ \partial_\xi=\partial_\xi x\partial_x+\partial_\xi y\partial_y\\ \partial_\eta=\partial_\eta x\partial_x+\partial_\eta y\partial_y $$ and $$ \partial_x=\partial_x \xi\partial_\xi+\partial_x \eta\partial_\eta\\ \partial_y=\partial_y \xi\partial_\xi+\partial_y \eta\partial_\eta.\\ $$ now my perplexity is the following: can I put one set of equation inside the other? It is possible to write $$ \partial_\xi=A\partial_\xi+B\partial_\eta\\ \partial_\eta=C\partial_\xi+D\partial_\eta\\ $$ and it seems that the combination of partial derivative must obey the following conditions $A=1$, $B=0$, $C=0$ and $D=1$. Is it true?

Answer 1

(Just for clarity, we'll use the $\cdot$-symbol to distinguish multiplication from a partial derivation operation.)

As peek-a-boo commented, we should anticipate that $\partial_\xi=1\cdot\partial_\xi+0\cdot\partial_\eta$ or else something is seriously awry.

So is this what happens?

$\begin{align}\partial_\xi &= \partial_\xi x\cdot\partial_x + \partial_\xi y\cdot\partial_y&&\text{chain rule}\\&=\partial_\xi x\cdot(\partial _x\xi\cdot\partial_\xi+\partial_x\eta\cdot\partial_\eta)+\partial_\xi y\cdot(\partial _y\xi\cdot\partial_\xi+\partial_y\eta\cdot\partial_\eta)&&\text{chain rule}\times 2\\&=(\partial_\xi x\cdot\partial_x\xi+\partial_\xi y\cdot\partial _y\xi)\cdot\partial_\xi+(\partial_\xi x\cdot\partial_x\eta+\partial_\xi y\cdot\partial _y\eta)\cdot\partial_\eta&&\text{multiplication distributes, commutes & associates}\\&=[\partial_\xi x\cdot\partial_x+\partial_\xi y\cdot\partial_y]\xi\cdot\partial_\xi +[\partial_\xi x\cdot\partial_x+\partial_\xi y\cdot\partial_y]\eta\cdot\partial_\eta&&\text{linear operation distributes}\\&=\partial_\xi\xi\cdot\partial_\xi+\partial_\xi\eta\cdot\partial_\eta&&\text{chain rule}\times 2\\&=1\cdot\partial_\xi+0\cdot\partial_\eta&&\text{partial differentiation, by definition}\\&=\partial_\xi&&\text{... as it should}\end{align}$

What a relief.