Suppose a function f(x, y, z) = 0, where x, y, and z are functions of each other. Write the total differentials of the variables $${\displaystyle dx=\left({\frac {\partial x}{\partial y}}\right)dy+\left({\frac {\partial x}{\partial z}}\right)dz}$$ $${\displaystyle dy=\left({\frac {\partial y}{\partial x}}\right)dx+\left({\frac {\partial y}{\partial z}}\right)dz}$$ Substitute dy into dx $${\displaystyle dx=\left({\frac {\partial x}{\partial y}}\right)\left[\left({\frac {\partial y}{\partial x}}\right)dx+\left({\frac {\partial y}{\partial z}}\right)dz\right]+\left({\frac {\partial x}{\partial z}}\right)dz}$$ By using the chain rule one can show the coefficient of dx on the right hand side is equal to one, thus the coefficient of dz must be zero $${\displaystyle \left({\frac {\partial x}{\partial y}}\right)\left({\frac {\partial y}{\partial z}}\right)+\left({\frac {\partial x}{\partial z}}\right)=0}$$ Subtracting the second term and multiplying by its inverse gives the triple product rule $${\displaystyle \left({\frac {\partial x}{\partial y}}\right)\left({\frac {\partial y}{\partial z}}\right)\left({\frac {\partial z}{\partial x}}\right)=-1.}$$

This section is based on chapter 5 of Pippard.^[2]

Suppose we are given four real variables ${\displaystyle (x,y,z,w)}$, restricted to move on a 2-dimensional ${\displaystyle C^{2}}$ surface in ${\displaystyle \mathbb {R} ^{4}}$. Then, if we know two of them, we can determine the other two uniquely (generically).

In particular, we may take any two variables as the independent variables, and let the other two be the dependent variables, then we can take all these partial derivatives.

Proposition: ${\displaystyle \left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}\left({\frac {\partial z}{\partial x}}\right)_{y}=-1}$

Proof. We can ignore ${\displaystyle w}$. Then locally the surface is just ${\displaystyle ax+by+cz+d=0}$. Then ${\displaystyle \left({\frac {\partial x}{\partial y}}\right)_{z}=-{\frac {b}{a}}}$, etc. Now multiply them.

The ideal gas law relates the state variables of pressure (P), volume (V), and temperature (T) via

${\displaystyle PV=nRT}$

which can be written as

${\displaystyle f(P,V,T)=PV-nRT=0}$

so each state variable can be written as an implicit function of the other state variables:

${\displaystyle {\begin{aligned}P&=P(V,T)={\frac {nRT}{V}}\\[1em]V&=V(P,T)={\frac {nRT}{P}}\\[1em]T&=T(P,V)={\frac {PV}{nR}}\end{aligned}}}$

From the above expressions, we have

${\displaystyle {\begin{aligned}-1&=\left({\frac {\partial P}{\partial V}}\right)\left({\frac {\partial V}{\partial T}}\right)\left({\frac {\partial T}{\partial P}}\right)\\[1em]&=\left(-{\frac {nRT}{V^{2}}}\right)\left({\frac {nR}{P}}\right)\left({\frac {V}{nR}}\right)\\[1em]&=\left(-{\frac {nRT}{PV}}\right)\\[1em]&=-{\frac {P}{P}}=-1\end{aligned}}}$

A geometric realization of the triple product rule can be found in its close ties to the velocity of a traveling wave

${\displaystyle \phi (x,t)=A\cos(kx-\omega t)}$

shown on the right at time t (solid blue line) and at a short time later t+Δt (dashed). The wave maintains its shape as it propagates, so that a point at position x at time t will correspond to a point at position x+Δx at time t+Δt,

${\displaystyle A\cos(kx-\omega t)=A\cos(k(x+\Delta x)-\omega (t+\Delta t)).}$

This equation can only be satisfied for all x and t if k Δx − ω Δt = 0, resulting in the formula for the phase velocity

${\displaystyle v={\frac {\Delta x}{\Delta t}}={\frac {\omega }{k}}.}$

To elucidate the connection with the triple product rule, consider the point p₁ at time t and its corresponding point (with the same height) p̄₁ at t+Δt. Define p₂ as the point at time t whose x-coordinate matches that of p̄₁, and define p̄₂ to be the corresponding point of p₂ as shown in the figure on the right. The distance Δx between p₁ and p̄₁ is the same as the distance between p₂ and p̄₂ (green lines), and dividing this distance by Δt yields the speed of the wave.

To compute Δx, consider the two partial derivatives computed at p₂,

${\displaystyle \left({\frac {\partial \phi }{\partial t}}\right)\Delta t={\text{rise from }}p_{2}{\text{ to }}{\bar {p}}_{1}{\text{ in time }}\Delta t{\text{ (gold line)}}}$

${\displaystyle \left({\frac {\partial \phi }{\partial x}}\right)={\text{slope of the wave (red line) at time }}t.}$

Dividing these two partial derivatives and using the definition of the slope (rise divided by run) gives us the desired formula for

${\displaystyle \Delta x=-{\frac {\left({\frac {\partial \phi }{\partial t}}\right)\Delta t}{\left({\frac {\partial \phi }{\partial x}}\right)}},}$

where the negative sign accounts for the fact that p₁ lies behind p₂ relative to the wave's motion. Thus, the wave's velocity is given by

${\displaystyle v={\frac {\Delta x}{\Delta t}}=-{\frac {\left({\frac {\partial \phi }{\partial t}}\right)}{\left({\frac {\partial \phi }{\partial x}}\right)}}.}$

For infinitesimal Δt, ${\displaystyle {\frac {\Delta x}{\Delta t}}=\left({\frac {\partial x}{\partial t}}\right)}$ and we recover the triple product rule

${\displaystyle v={\frac {\Delta x}{\Delta t}}=-{\frac {\left({\frac {\partial \phi }{\partial t}}\right)}{\left({\frac {\partial \phi }{\partial x}}\right)}}.}$