partial derivative with respect to itself

Question

Given some state function $$f(x,y,z)=0$$ I want to prove the reciprocity relationship $$\left(\dfrac{\partial x}{\partial y}\right)_z=\dfrac{1}{\left(\dfrac{\partial y}{\partial x}\right)_z}$$

I know how to do this the standard way (implicit function theorem, $f$ is exactly differentiable, etc.) but I've recently seen someone employ a somewhat trivial method of demonstrating this and I can't help but feel they've done something smelly but I lack the background to make a determination.

Beginning with the assumption that $z$ is held constant,

$$\dfrac{\partial x}{\partial y}=\dfrac{\partial x}{\partial y}\cdot\dfrac{\partial x}{\partial x}=\dfrac{\partial x}{\partial y}\cdot\dfrac{1}{\dfrac{\partial x}{\partial x}}=\dfrac{1}{\dfrac{\partial y}{\partial x}}$$

Is this valid? More importantly, how is $\frac{\partial x}{\partial x}$ to be interpreted? In physics we treat derivatives and partials like fractions and do algebraic operations on them all the time. What are the problems with this sort of treatment?

Answer 1

The short answer is no. I have read a lot of great posts extolling the virtues (or lack thereof) of the Leibniz notation (such as the top response of How is it that treating Leibniz notation as a fraction is fundamentally incorrect but at the same time useful?).

There is a way to make the abuse of differentials rigorous known as non-standard analysis, which you can read about here. Whether or not the manipulations you have written down constitute a formal proof in the context of non-standard analysis is beyond me, since I have never bothered learning it. Even if it was, to be truly rigorous you would need to build the framework of non-standard analysis from the ground up before you could quote such a proof.