If a function $f$ can be expressed in two different coordinate systems say $(x,y)$ and $(\bar x,\bar y)$, how would one take the partial differential of $f$ with respect to $\bar x$ with $\bar y$ held constant? I've seen it written:

$$\left( \dfrac{\partial f}{\partial \bar x}\right)_{\bar y} = \left(\dfrac{\partial f}{\partial x}\right)_y \left(\dfrac{\partial x}{\partial \bar x}\right)_{\bar y} + \left(\dfrac{\partial f}{\partial y}\right)_x \left(\dfrac{\partial y}{\partial \bar x}\right)_{\bar y}$$

But I do not understand where this comes from, specifically which variables should be held constant in each derivative. For example why couldn't the $\dfrac{\partial f}{\partial x}$ derivative have $\bar y$ held constant instead of $y$?


why couldn't the $\frac{\partial f}{\partial x}$ derivative have $\bar y$ held constant instead of $y$?

In principle, you could consider such a derivative, provided that it is possible to change $x$ while holding $\bar y$ constant. E.g., if $\bar y =2x$ you would not be able to do that, and then $\left(\frac{\partial f}{\partial x}\right)_{\bar y}$ would not make sense. In any event, the derivative would likely be different, and then the equality would not hold.

The chain rule that you stated is the usual multivariable chain rule, although it's normally not written with those subscripts. When subscripts are not given, readers are expected to understand that $(x,y)$ and $(\bar x,\bar y)$ are the coordinate pairs of interest here, and that when we take partial derivatives, the other member of the same pair is held constant. With this convention understood, one does not need subscripts in the formula you gave.

The proof of multivariable chain rule is not hard to find. I'll give a sketch in terms of differentials: $$dx = \left(\dfrac{\partial x}{\partial \bar x}\right)_{\bar y} d\bar x + \left(\dfrac{\partial x}{\partial \bar y}\right)_{\bar x} d\bar y$$ $$dy = \left(\dfrac{\partial y}{\partial \bar x}\right)_{\bar y} d\bar x + \left(\dfrac{\partial y}{\partial \bar y}\right)_{\bar x} d\bar y$$ Plug the above into $$df = \left(\dfrac{\partial f}{\partial x}\right)_{ y} d x + \left(\dfrac{\partial f}{\partial y}\right)_{x} d y$$ and the result follows.


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