How do we know that we only need three second derivatives to represent all the second derivatives of a thermodynamic system?

Question

In the topic of Maxwell relations of thermodynamics, my books says

Of all such derivatives, only three can be independent, and any given derivative can be expressed in terms of an arbitrarily chosen set of three basic derivatives. This set is conventionally chosen as ${c_p}$ , α, and ${κ_t}$. ... All first derivatives ( involving both extensive and intensive parameters) can be written in terms of second derivatives of the Gibbs potential, of which we have now seen that ${c_p}$ , α, and ${κ_t}$ constitute a complete independent set ( at constant mole numbers).

I was wondering if there is any methods (without calculation of derivative) ,finding that we only need three second derivatives to represent all the second derivatives of a thermodynamic system.

Answer 1

In the context that your thermodynamical system has 3 independent variables this is true. Because any other derivative along some direction in parameter space can be expressed as a linear combination of these. But in general, e.g. if you have like 200 particle species with all different chemical potential, I'm not sure if this statement is correct. (Or imagine other "thermodynamical forces")

EDIT: Adressing the comment: Suppose $f(x,y)$ depends on two independent variables $x$,$y$. Now you have a third one, $z(x,y)$ which can locally be inverted (This can be done if assumptions of the thm. of implicit inverse is satisfied). Then the derivative w.r.t. $z$ looks like:

$\frac{\partial f(z(x,y),x,y)}{\partial z}= \frac{\partial f(z(x,y),x,y)}{\partial x}\frac{\partial x(z,y)}{\partial z}+\frac{\partial f(z(x,y),x,y)}{\partial y}\frac{\partial y(z,x)}{\partial z}$

per Assumption, these things are all well defined.