<p>This is from Philip Morse's textbook called Thermal Physics. It's pretty old, the last edition was printed in 1968 or so. Pages 96-67 of the second edition.</p>
<p>As a note: (dX/dY)Z means the partial of X with respect to Y at a constant Z.</p>
<p>
[quote]
We have now displayed most of the techniques needed to work out relationships between one thermodynamic function and another, so that the thermal properties of the system can be explicitly given in terms of an experimentally determined heat capacity and various partials obtained from the equations of state. It may be useful to assemble these techniques in a sequence of recipes, which can be applied to any specific case. To keep the recipes simple we give them for a system in which (T,S) and one mechanical pair (Y,X) of variables are involved. The pair can be (-P,V) or (H,M), etc., with Y the intensive and X the extensive variable. Cases where more variables are simultaneously involved can be worked out from the equations already given in this chapter.</p>
<p>There's a paragraph with a cool description of how to make a diagram by which you can derive all of the Maxwell Relations and figure out which variables are the natural ones for a given potential, but it's a little hard to type up the box.</p>
<p>Using Euler's Eq. for this case, the four related potentials are</p>
<p>U = ST + YX + un
H = U - YX = TS + un
F = U - ST = YX + un
G = F - YX + un</p>
<p>Using the diagram we are now in a position to formulate a strategy for expressing any possible rate of change of a thermodynamic variable in terms of the immediately measurable quantities such as heat capacity and the partials (dX/dT)Y or (dX/dY)T, etc. coming from an equation of state relating X, Y, and T. Either Cx or Cy can be considered basic (Cp = Cy is the one usually measured)</p>
<p>Cx = (dU/dT)X = T(dS/dT)X (eq 8-18)
Cy = (dH/dT)Y = T(dS/dT)Y</p>
<p>The relation between them is given in terms fo partials from the equation of state [Eq 6-17]:</p>
<p>Cy = Cx - T(dY/dT)X (dX/dT)Y = Cx + T * [(dX/dT)Y]² / [(dX/dY)T]
Cy = Cx + T [(dY/dT)²x / (dY/dX)T]</p>
<p>The various tactics which can be used to express an unfamiliar partial in terms of an immediately measurable one are:</p>
<p>a. Replacing the partials of the potentials with respect to their adjoining variables in Figure 8-2 {the one I couldn't draw}, such as (dF/dT)X = -S or (dU/dS)X = T, etc.</p>
<p>b. Replacing a partial of a potential with respect to a nonadjacent variable, obtainable from its basic equation, such as dF = -S dT + Y dX, from which we get (dF/dX)S = -S(dT/dX)S + Y and (dF/dS)Y = -S(dT/dS)Y + Y(dX/dS)Y, etc.</p>
<p>c. Using one or more of the Maxwell relations, obtainable from Figure 8-2.</p>
<p>d. Using the basic properties fo partial derivatives, s displayed in Eqs. (3-9) and (3-10). {These are cyclic permutation and chain rule type things.}</p>
<p>In terms of these tactics, the appropriate strategies are:
1. If a potential is an independent variable in the given partial, make it the dependent variable by using (d) (this process is called bringing the potential into the numerator) and then using (a) or (b) to eliminate the potential.</p>
<ol>
<li><p>Next, if the entropy is an independent variable in the given partial or in the result of step 1, bring S into the numerator and eliminate it by using (c) or Eq. 8-18.</p></li>
<li><p>If measured equation-of-state partials have X in the numerator, bring X into the numerator of the result of steps 1 and 2, by using (d). If the equation fo state is in the form of Y = f(X,T), bring Y into the numerator.</p></li>
</ol>
<p>The result of applying these successive steps will be an expression for the partial of interest in terms of measured, or measurable, quantities.
[/quote]
</p>
<p>So, we can see that maxwell's relations are just like derivatives in that they're a useful mathematical tool to help us figure out ways to get real work done.</p>