# Exact differential

This article is about the concept from elementary differential calculus. For the generalized advanced mathematical concept from differential topology and differential geometry, see closed and exact differential forms.

In multivariate calculus, a differential is said to be exact (or perfect), as contrasted with an inexact differential, if it is of the form dQ, for some differentiable function Q.

## Overview

### Definition

We work in three dimensions, with similar definitions holding in any other number of dimensions. In three dimensions, a form of the type

$A(x,y,z) dx + B(x,y,z) dy + C(x,y,z) dz$

is called a differential form. This form is called exact on a domain $D \subset \mathbb{R}^3$ in space if there exists some scalar function $Q = Q(x,y,z)$ defined on $D$ such that

$\; dQ \; \equiv \; \left ( \frac{\partial Q}{\partial x} \right )_{\bold{y,z}} dx \quad + \quad \left ( \frac{\partial Q}{\partial y} \right )_{\bold{z,x}} dy \quad + \quad \left ( \frac{\partial Q}{\partial z} \right )_{\bold{x,y}} dz,$   $dQ = A dx + B dy + C dz$

throughout D. This is equivalent to saying that the vector field $(A, B, C)$ is a conservative vector field, with corresponding potential $Q$.

### One dimension

In one dimension, a differential form

$A(x) \, dx$

is exact as long as $A$ has an antiderivative; in this case let $Q$ be the antiderivative of $A$. Otherwise, if $A$ does not have an antiderivative, we cannot write $dQ = A(x) \, dx$ and so the differential form is inexact.

### Two and three dimensions

By symmetry of second derivatives, for any "nice" (non-pathological) function $Q$ we have

$\frac{\partial ^2 Q}{\partial x \partial y} = \frac{\partial ^2 Q}{\partial y \partial x}$

Hence, it follows that in a simply-connected region R of the xy-plane, a differential

$A(x, y)\,dx + B(x, y)\,dy$

is an exact differential if and only if the following holds:

$\left( \frac{\partial A}{\partial y} \right)_x = \left( \frac{\partial B}{\partial x} \right)_y$

For three dimensions, a differential

$dQ = A(x, y, z) \, dx + B(x, y, z) \, dy + C(x, y, z) \, dz$

is an exact differential in a simply-connected region R of the xyz-coordinate system if between the functions A, B and C there exist the relations:

$\left( \frac{\partial A}{\partial y} \right)_{x,z} \!\!\!= \left( \frac{\partial B}{\partial x} \right)_{y,z}$   ;   $\left( \frac{\partial A}{\partial z} \right)_{x,y} \!\!\!= \left( \frac{\partial C}{\partial x} \right)_{y,z}$   ;   $\left( \frac{\partial B}{\partial z} \right)_{x,y} \!\!\!= \left( \frac{\partial C}{\partial y} \right)_{x,z}$
Note: The subscripts outside the parenthesis indicate which variables are being held constant during differentiation. Due to the definition of the partial derivative, these subscripts are not required, but they are included as a reminder.

These conditions are equivalent to the following one: If G is the graph of this vector valued function then for all tangent vectors X,Y of the surface G then s(XY) = 0 with s the symplectic form.

These conditions, which are easy to generalize, arise from the independence of the order of differentiations in the calculation of the second derivatives. So, in order for a differential dQ, that is a function of four variables to be an exact differential, there are six conditions to satisfy.

In summary, when a differential dQ is exact:

• the function Q exists;
• $\int_i^f dQ=Q(f)-Q(i),$ independent of the path followed.

In thermodynamics, when dQ is exact, the function Q is a state function of the system. The thermodynamic functions U, S, H, A and G are state functions. Generally, neither work nor heat is a state function. An exact differential is sometimes also called a 'total differential', or a 'full differential', or, in the study of differential geometry, it is termed an exact form.

## Partial differential relations

If three variables, $x$, $y$ and $z$ are bound by the condition $F(x,y,z) = \text{constant}$ for some differentiable function $F(x,y,z)$, then the following total differentials exist[1]:667&669

$d x = {\left ( \frac{\partial x}{\partial y} \right )}_z \, d y + {\left ( \frac{\partial x}{\partial z} \right )}_y \,dz$
$d z = {\left ( \frac{\partial z}{\partial x} \right )}_y \, d x + {\left ( \frac{\partial z}{\partial y} \right )}_x \,dy.$

Substituting the first equation into the second and rearranging, we obtain[1]:669

$d z = {\left ( \frac{\partial z}{\partial x} \right )}_y \left [ {\left ( \frac{\partial x}{\partial y} \right )}_z d y + {\left ( \frac{\partial x}{\partial z} \right )}_y dz \right ] + {\left ( \frac{\partial z}{\partial y} \right )}_x dy,$
$d z = \left [ {\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial y} \right )}_z + {\left ( \frac{\partial z}{\partial y} \right )}_x \right ] d y + {\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial z} \right )}_y dz,$
$\left [ 1 - {\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial z} \right )}_y \right ] dz = \left [ {\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial y} \right )}_z + {\left ( \frac{\partial z}{\partial y} \right )}_x \right ] d y.$

Since $y$ and $z$ are independent variables, $d y$ and $d z$ may be chosen without restriction. For this last equation to hold in general, the bracketed terms must be equal to zero.[1]:669

### Reciprocity relation

Setting the first term in brackets equal to zero yields[1]:60฿฿฿70

${\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial z} \right )}_y = 1.$

A slight rearrangement gives a reciprocity relation,[1]:670

${\left ( \frac{\partial z}{\partial x} \right )}_y = \frac{1}{{\left ( \frac{\partial x}{\partial z} \right )}_y}.$

There are two more permutations of the foregoing derivation that give a total of three reciprocity relations between $x$, $y$ and $z$. Reciprocity relations show that the inverse of a partial derivative is equal to its reciprocal.

### 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[1]:670

${\left ( \frac{\partial z}{\partial x} \right )}_y {\left ( \frac{\partial x}{\partial y} \right )}_z = - {\left ( \frac{\partial z}{\partial y} \right )}_x.$

Using a reciprocity relation for $\tfrac{\partial z}{\partial y}$ on this equation and reordering gives a cyclic relation (the triple product rule),[1]:670

${\left ( \frac{\partial x}{\partial y} \right )}_z {\left ( \frac{\partial y}{\partial z} \right )}_x {\left ( \frac{\partial z}{\partial x} \right )}_y = -1.$

If, instead, a reciprocity relation for $\tfrac{\partial x}{\partial y}$ is used with subsequent rearrangement, a standard form for implicit differentiation is obtained:

${\left ( \frac{\partial y}{\partial x} \right )}_z = - \frac { {\left ( \frac{\partial z}{\partial x} \right )}_y }{ {\left ( \frac{\partial z}{\partial y} \right )}_x }.$

## Some useful equations derived from exact differentials in two dimensions

(See also Bridgman's thermodynamic equations for the use of exact differentials in the theory of thermodynamic equations)

Suppose we have five state functions $z,x,y,u$, and $v$. Suppose that the state space is two dimensional and any of the five quantities are exact differentials. Then by the chain rule

$(1)~~~~~ dz = \left(\frac{\partial z}{\partial x}\right)_y dx+ \left(\frac{\partial z}{\partial y}\right)_x dy = \left(\frac{\partial z}{\partial u}\right)_v du +\left(\frac{\partial z}{\partial v}\right)_u dv$

but also by the chain rule:

$(2)~~~~~ dx = \left(\frac{\partial x}{\partial u}\right)_v du +\left(\frac{\partial x}{\partial v}\right)_u dv$

and

$(3)~~~~~ dy= \left(\frac{\partial y}{\partial u}\right)_v du +\left(\frac{\partial y}{\partial v}\right)_u dv$

so that:

$(4)~~~~~ dz = \left[ \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial u}\right)_v + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial u}\right)_v \right]du$

$+ \left[ \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial v}\right)_u + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial v}\right)_u \right]dv$

which implies that:

$(5)~~~~~ \left(\frac{\partial z}{\partial u}\right)_v = \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial u}\right)_v + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial u}\right)_v$

Letting $v=y$ gives:

$(6)~~~~~ \left(\frac{\partial z}{\partial u}\right)_y = \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial u}\right)_y$

Letting $u=y$ gives:

$(7)~~~~~ \left(\frac{\partial z}{\partial y}\right)_v = \left(\frac{\partial z}{\partial y}\right)_x + \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial y}\right)_v$

Letting $u=y$, $v=z$ gives:

$(8)~~~~~ \left(\frac{\partial z}{\partial y}\right)_x = - \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial y}\right)_z$

using ($\partial a/\partial b)_c = 1/(\partial b/\partial a)_c$ gives the triple product rule:

$(9)~~~~~ \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x =-1$