Where does the relative sign come from in this chain rule application? I'm studying thermodynamics I came across this definition
\begin{equation}
p = \left( \frac{\partial u}{\partial x} \right)_{S}
\end{equation}
where $p$ is a generalised pressure, $u$ is the energy of the system, $x$ is an extensive parameter and $S$ is the entropy. I know this page is not about physics but neither is my doubt, which is:
The text then proceeds to write this pressure in terms of a derivative of entropy and apply to it the chain rule
\begin{equation}
p=\left( \frac{\partial u}{\partial x} \right)_{S}=-\left( \frac{\partial u}{\partial S} \right)_{x}\left( \frac{\partial S}{\partial x} \right)_{u}
\end{equation}
Where does the minus sign come from? What is the rule here? If I was asked to write this pressure I would apply the chain rule without the minus sign, so what am I missing here?
Thank you very much.
 A: Yes, thermodynamics people like to call this the cyclic rule for thermodynamics. If you have three variables, say, $x,u,S$, related by some functional equation $f(x,u,S)=0$, then the result is that
$$\left(\frac{\partial u}{\partial x}\right)_S \left(\frac{\partial S}{\partial u}\right)_x \left(\frac{\partial x}{\partial S}\right)_u = -1,$$
along with the rule that
$$\left(\frac{\partial u}{\partial x}\right)_S \left(\frac{\partial x}{\partial u}\right)_S = 1$$
(in the latter you're back to the usual single-variable calculus result that inverse functions have inverse derivatives, since $S$ is held constant in both).
To see where the surprising $-1$ comes from, say that $f(x,u,S)=0$ defines $x$ implicitly as a function of $u$ and $S$, and differentiate implicitly with respect to $u$ (fixing $S$, of course) to get that
$$\frac{\partial f}{\partial x}\left(\frac{\partial x}{\partial u} \right)_S+ \frac{\partial f}{\partial u} = 0, \quad\text{and so}\quad \left(\frac{\partial x}{\partial u}\right)_S = -\frac{\frac{\partial f}{\partial u}}{\frac{\partial f}{\partial x}}.$$
Doing this, analogously, for the other two quantities, will give you the product of three $-1$s and everything else cancels!
REMARK: If you have four variables $x,y,z,w$ related by some functional equation $f(x,y,z,w)=0$, then you should be able to see that the corresponding cyclic rule is
$$\left(\frac{\partial x}{\partial y}\right)_{z,w}\left(\frac{\partial y}{\partial z}\right)_{x,w}\left(\frac{\partial z}{\partial w}\right)_{x,y}\left(\frac{\partial w}{\partial x}\right)_{y,z}=+1.$$
