Editing Exact differential (section)

==Partial differential relations==

If a [[differentiable function]] <math>z(x,y)</math> is [[Injective function|one-to-one (injective)]] for each independent variable, e.g., <math>z(x,y)</math> is one-to-one for <math>x</math> at a fixed <math>y</math> while it is not necessarily one-to-one for <math>(x,y)</math>, then the following [[total differential]]s exist because each independent variable is a differentiable function for the other variables, e.g., <math>x(y,z)</math>.

:<math>d x = {\left ( \frac{\partial x}{\partial y} \right )}_z \, d y + {\left ( \frac{\partial x}{\partial z} \right )}_y \,dz</math>
:<math>d z = {\left ( \frac{\partial z}{\partial x} \right )}_y \, d x + {\left ( \frac{\partial z}{\partial y} \right )}_x \,dy.</math>

Substituting the first equation into the second and rearranging, we obtain

:<math>d z = {\left ( \frac{\partial z}{\partial x} \right )}_y \left [ {\left ( \frac{\partial x}{\partial y} \right )}_z d y + {\left ( \frac{\partial x}{\partial z} \right )}_y dz \right ] + {\left ( \frac{\partial z}{\partial y} \right )}_x dy,</math>
:<math>d z = \left [ {\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial y} \right )}_z + {\left ( \frac{\partial z}{\partial y} \right )}_x \right ] d y + {\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial z} \right )}_y dz,</math>
:<math>\left [ 1 - {\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial z} \right )}_y \right ] dz = \left [ {\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial y} \right )}_z + {\left ( \frac{\partial z}{\partial y} \right )}_x \right ] d y.</math>
Since <math>y</math> and <math>z</math> are independent variables, <math>d y</math> and <math>d z</math> may be chosen without restriction. For this last equation to generally hold, the bracketed terms must be equal to zero.<ref name="Cengel1998">{{cite book|last=Çengel|first=Yunus A.|title=Thermodynamics - An Engineering Approach|author2=Boles, Michael A.|last3=Kanoğlu|first3=Mehmet|publisher=McGraw-Hill Education|year=2019|isbn=978-1-259-82267-4|edition=9th|location=New York|pages=647–648|language=English|chapter=Thermodynamics Property Relations|orig-year=1989}}</ref> The left bracket equal to zero leads to the reciprocity relation while the right bracket equal to zero goes to the cyclic relation as shown below.

===Reciprocity relation===
Setting the first term in brackets equal to zero yields

:<math>{\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial z} \right )}_y = 1.</math>

A slight rearrangement gives a reciprocity relation,

:<math>{\left ( \frac{\partial z}{\partial x} \right )}_y = \frac{1}{{\left ( \frac{\partial x}{\partial z} \right )}_y}.</math>

There are two more [[permutations]] of the foregoing derivation that give a total of three reciprocity relations between <math>x</math>, <math>y</math> and <math>z</math>.

===Cyclic relation===
The cyclic relation is also known as the cyclic rule or the [[Triple product rule]]. Setting the second term in brackets equal to zero yields

:<math>{\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial y} \right )}_z = - {\left ( \frac{\partial z}{\partial y} \right )}_x.</math>

Using a reciprocity relation for <math>\tfrac{\partial z}{\partial y}</math> on this equation and reordering gives a cyclic relation (the [[triple product rule]]),

:<math>{\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.</math>

If, ''instead'', reciprocity relations for <math>\tfrac{\partial x}{\partial y}</math> and <math>\tfrac{\partial y}{\partial z}</math> are used with subsequent rearrangement, a [[Implicit function#Formula for two variables|standard form for implicit differentiation]] is obtained:

:<math>{\left ( \frac{\partial y}{\partial x} \right )}_z = - \frac { {\left ( \frac{\partial z}{\partial x} \right )}_y }{ {\left ( \frac{\partial z}{\partial y} \right )}_x }.</math>