Derivation of $(\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f=0$ in Chandler's Statistical mechanics book I am having trouble understanding something very introductory in Chandler's statistical mechanics book regarding the derivation of $(\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f=0$
The argument laid out in the book goes as follows.
For thermodynamic equilibrium states, entropy is an extensive function of state: S(E,X) where E is the internal energy and X is a mechanical extensive variable. Computing the differential we get:
$dS=(\partial S/\partial E)_{X}dE+(\partial S/\partial X)_{E}dX$
If we assume we have a reversible process:
$dE=(dQ)_{rev}+fdX$,
where f is an applied force. Combining the previous two equations we get for a reversible process:
$dS=(\partial S/\partial E)_{X}(dQ)_{rev}+((\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f)dX$
For processes which are both adiabatic and reversible, $dS$ and $(dQ)_{rev}$ are zero. Hence we have
$(\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f=0$
Till here I have no trouble. Then it is stated that since all quantities involved in this equation are functions of state, that implies the equality holds for adiabatic as well as non-adiabatic processes. I do not see why this should hold for non-adiabatic processes .
 A: From what I understand, he's just saying that since state functions are path-independent, that is, they describe equilibrium states irrespectively of how has the system has arrived to them (either by an adiabatic or non-adiabatic process, for instance), then an equality involving uniquely state functions must always hold. That wouldn't be the case for the preceding equality, which involves $dQ$, a path-dependent function.
A: Here is another way of proceeding.  Start with the equation
$$dS=(\partial S/\partial E)_{X}dE+(\partial S/\partial X)_{E}dX$$So, if we set dS = 0, it follows from partial differential mathematics that, in general, $$\left(\frac{\partial E}{\partial X}\right)_S=-\frac{(\partial S/\partial X)_{E}}{(\partial S/\partial E)_{X}}$$
But, we know that $$dE=TdS+fdX$$  So, $$\left(\frac{\partial E}{\partial X}\right)_S=f$$
So,$$f=-\frac{(\partial S/\partial X)_{E}}{(\partial S/\partial E)_{X}}$$
This gives:$$(\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f=0$$
I don't know where the extra minus sign came from, but I'm confident in what I did.
