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Exact differential
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<h2 id="overview">Overview</h2> <h3>Definition</h3> <p>Even if we work in three dimensions here, the definitions of exact differentials for other dimensions are structurally similar to the three dimensional definition. In three dimensions, a form of the type A ( x , y , z ) d x + B ( x , y , z ) d y + C ( x , y , z ) d z {\displaystyle A(x,y,z)\,dx+B(x,y,z)\,dy+C(x,y,z)\,dz} is called a <a href="/facts/Differential_form/T9gyHtMv">differential form</a>. This form is called <i>exact</i> on an open domain D ⊂ R 3 {\displaystyle D\subset \mathbb {R} ^{3}} in space if there exists some <a href="/facts/Differentiable_function/jQqTgLk1">differentiable</a> <a href="/facts/Scalar_function/8MtPTGob">scalar function</a> Q = Q ( x , y , z ) {\displaystyle Q=Q(x,y,z)} defined on D {\displaystyle D} such that d Q ≡ ( ∂ Q ∂ x ) y , z d x + ( ∂ Q ∂ y ) x , z d y + ( ∂ Q ∂ z ) x , y d z , d Q = A d x + B d y + C d z {\displaystyle dQ\equiv \left({\frac {\partial Q}{\partial x}}\right)_{y,z}\,dx+\left({\frac {\partial Q}{\partial y}}\right)_{x,z}\,dy+\left({\frac {\partial Q}{\partial z}}\right)_{x,y}\,dz,\quad dQ=A\,dx+B\,dy+C\,dz} throughout D {\displaystyle D} , where x , y , z {\displaystyle x,y,z} are <a href="/facts/Orthogonal_coordinates/LLoa2lAJ">orthogonal coordinates</a> (e.g., <a href="/facts/Cartesian_coordinate_system/lkTqvMmB">Cartesian</a>, <a href="/facts/Cylindrical/0wQPPr3k">cylindrical</a>, or <a href="/facts/Spherical_coordinate_system/5cVnNof3">spherical coordinates</a>). In other words, in some open domain of a space, a differential form is an <i>exact differential</i> if it is equal to the general differential of a differentiable function in an orthogonal coordinate system. </p><p>The subscripts outside the parenthesis in the above mathematical expression indicate which variables are being held constant during differentiation. Due to the definition of the <a href="/facts/Partial_derivative/JWHsBkAu">partial derivative</a>, these subscripts are not required, but they are explicitly shown here as reminders. </p> <h3>Integral path independence</h3> <p>The exact differential for a differentiable scalar function Q {\displaystyle Q} defined in an open domain D ⊂ R n {\displaystyle D\subset \mathbb {R} ^{n}} is equal to d Q = ∇ Q ⋅ d r {\displaystyle dQ=\nabla Q\cdot d\mathbf {r} } , where ∇ Q {\displaystyle \nabla Q} is the <a href="/facts/Gradient/TsR7uukj">gradient</a> of Q {\displaystyle Q} , ⋅ {\displaystyle \cdot } represents the <a href="/facts/Dot_product/tNz8MLjT">scalar product</a>, and d r {\displaystyle d\mathbf {r} } is the general differential displacement vector, if an orthogonal coordinate system is used. If Q {\displaystyle Q} is of differentiability class C 1 {\displaystyle C^{1}} (<a href="/facts/Smoothness/mffL4ch2">continuously differentiable</a>), then ∇ Q {\displaystyle \nabla Q} is a <a href="/facts/Conservative_vector_field/MoS3V9aV">conservative vector field</a> for the corresponding potential Q {\displaystyle Q} by the definition. For three dimensional spaces, expressions such as d r = ( d x , d y , d z ) {\displaystyle d\mathbf {r} =(dx,dy,dz)} and ∇ Q = ( ∂ Q ∂ x , ∂ Q ∂ y , ∂ Q ∂ z ) {\displaystyle \nabla Q=\left({\frac {\partial Q}{\partial x}},{\frac {\partial Q}{\partial y}},{\frac {\partial Q}{\partial z}}\right)} can be made. </p><p>The <a href="/facts/Gradient_theorem/EQehJw1t">gradient theorem</a> states </p> ∫ i f d Q = ∫ i f ∇ Q ( r ) ⋅ d r = Q ( f ) − Q ( i ) {\displaystyle \int _{i}^{f}dQ=\int _{i}^{f}\nabla Q(\mathbf {r} )\cdot d\mathbf {r} =Q\left(f\right)-Q\left(i\right)} <p>that does not depend on which integral path between the given path endpoints i {\displaystyle i} and f {\displaystyle f} is chosen. So it is concluded that <i>the integral of an exact differential is independent of the choice of an integral path between given path endpoints <a href="/facts/Conservative_vector_field/MoS3V9aV">(path independence)</a>.</i> </p><p>For three dimensional spaces, if ∇ Q {\displaystyle \nabla Q} defined on an open domain D ⊂ R 3 {\displaystyle D\subset \mathbb {R} ^{3}} is of <a href="/facts/Smoothness/mffL4ch2">differentiability class</a> C 1 {\displaystyle C^{1}} (equivalently Q {\displaystyle Q} is of C 2 {\displaystyle C^{2}} ), then this integral path independence can also be proved by using the <a href="/facts/Vector_calculus_identity/QWAaYA7t">vector calculus identity</a> ∇ × ( ∇ Q ) = 0 {\displaystyle \nabla \times (\nabla Q)=\mathbf {0} } and <a href="/facts/Stokes%2527_theorem/z0pAoj2x">Stokes' theorem</a>. </p> ∮ ∂ Σ ∇ Q ⋅ d r = ∬ Σ ( ∇ × ∇ Q ) ⋅ d a = 0 {\displaystyle \oint _{\partial \Sigma }\nabla Q\cdot d\mathbf {r} =\iint _{\Sigma }(\nabla \times \nabla Q)\cdot d\mathbf {a} =0} <p>for a simply closed loop ∂ Σ {\displaystyle \partial \Sigma } with the smooth oriented surface Σ {\displaystyle \Sigma } in it. If the open domain D {\displaystyle D} is <a href="/facts/Simply_connected_space/xFlta63J">simply connected open space</a> (roughly speaking, a single piece open space without a hole within it), then any irrotational vector field (defined as a C 1 {\displaystyle C^{1}} vector field v {\displaystyle \mathbf {v} } which curl is zero, i.e., ∇ × v = 0 {\displaystyle \nabla \times \mathbf {v} =\mathbf {0} } ) has the path independence by the Stokes' theorem, so the following statement is made; <i>In a simply connected open region, any</i> C 1 {\displaystyle C^{1}} <i>vector field that has the path-independence property (so it is a conservative vector field.) must also be irrotational and vice versa.</i> The equality of the path independence and conservative vector fields is shown <a href="/facts/Conservative_vector_field/MoS3V9aV">here</a>. </p> <h4>Thermodynamic state function</h4> <p>In <a href="/facts/Thermodynamics/q4hIx3yP">thermodynamics</a>, when d Q {\displaystyle dQ} is exact, the function Q {\displaystyle Q} is a <a href="/facts/State_function/JThJasbq">state function</a> of the system: a <a href="/facts/Function_(mathematics)/CdReEKCh">mathematical function</a> which depends solely on the current <a href="/facts/Thermodynamic_equilibrium/1D7nGm0d">equilibrium state</a>, not on the path taken to reach that state. <a href="/facts/Internal_energy/IYdsoxhR">Internal energy</a> U {\displaystyle U} , <a href="/facts/Entropy/s52IXeFz">Entropy</a> S {\displaystyle S} , <a href="/facts/Enthalpy/bBou8lCE">Enthalpy</a> H {\displaystyle H} , <a href="/facts/Helmholtz_free_energy/2AK0HpPe">Helmholtz free energy</a> A {\displaystyle A} , and <a href="/facts/Gibbs_free_energy/bvenDuzh">Gibbs free energy</a> G {\displaystyle G} are <a href="/facts/State_function/JThJasbq">state functions</a>. Generally, neither <a href="/facts/Work_(thermodynamics)/SjChajtw">work</a> W {\displaystyle W} nor <a href="/facts/Heat/FjDGssCs">heat</a> Q {\displaystyle Q} is a state function. (Note: Q {\displaystyle Q} is commonly used to represent heat in physics. It should not be confused with the use earlier in this article as the parameter of an exact differential.) </p> <h3>One dimension</h3> <p>In one dimension, a differential form </p> A ( x ) d x {\displaystyle A(x)\,dx} <p>is exact <a href="/facts/If_and_only_if/bYSxGJ66">if and only if</a> A {\displaystyle A} has an <a href="/facts/Antiderivative/5qDsudIv">antiderivative</a> (but not necessarily one in terms of elementary functions). If A {\displaystyle A} has an antiderivative and let Q {\displaystyle Q} be an antiderivative of A {\displaystyle A} so d Q d x = A {\displaystyle {\frac {dQ}{dx}}=A} , then A ( x ) d x {\displaystyle A(x)\,dx} obviously satisfies the condition for exactness. If A {\displaystyle A} does <i>not</i> have an antiderivative, then we cannot write d Q = d Q d x d x {\displaystyle dQ={\frac {dQ}{dx}}dx} with A = d Q d x {\displaystyle A={\frac {dQ}{dx}}} for a differentiable function Q {\displaystyle Q} so A ( x ) d x {\displaystyle A(x)\,dx} is inexact. </p> <h3>Two and three dimensions</h3> <p>By <a href="/facts/Symmetry_of_second_derivatives/cj7Nz7SE">symmetry of second derivatives</a>, for any "well-behaved" (non-<a href="/facts/Pathological_(mathematics)/KmKHjJ1H">pathological</a>) function Q {\displaystyle Q} , we have </p> ∂ 2 Q ∂ x ∂ y = ∂ 2 Q ∂ y ∂ x . {\displaystyle {\frac {\partial ^{2}Q}{\partial x\,\partial y}}={\frac {\partial ^{2}Q}{\partial y\,\partial x}}.} <p>Hence, in a <a href="/facts/Simply-connected/xFlta63J">simply-connected</a> region <i>R</i> of the <i>xy</i>-plane, where x , y {\displaystyle x,y} are independent,<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a> a differential form </p> A ( x , y ) d x + B ( x , y ) d y {\displaystyle A(x,y)\,dx+B(x,y)\,dy} <p>is an exact differential if and only if the equation </p> ( ∂ A ∂ y ) x = ( ∂ B ∂ x ) y {\displaystyle \left({\frac {\partial A}{\partial y}}\right)_{x}=\left({\frac {\partial B}{\partial x}}\right)_{y}} <p>holds. If it is an exact differential so A = ∂ Q ∂ x {\displaystyle A={\frac {\partial Q}{\partial x}}} and B = ∂ Q ∂ y {\displaystyle B={\frac {\partial Q}{\partial y}}} , then Q {\displaystyle Q} is a differentiable (smoothly continuous) function along x {\displaystyle x} and y {\displaystyle y} , so ( ∂ A ∂ y ) x = ∂ 2 Q ∂ y ∂ x = ∂ 2 Q ∂ x ∂ y = ( ∂ B ∂ x ) y {\displaystyle \left({\frac {\partial A}{\partial y}}\right)_{x}={\frac {\partial ^{2}Q}{\partial y\partial x}}={\frac {\partial ^{2}Q}{\partial x\partial y}}=\left({\frac {\partial B}{\partial x}}\right)_{y}} . If ( ∂ A ∂ y ) x = ( ∂ B ∂ x ) y {\displaystyle \left({\frac {\partial A}{\partial y}}\right)_{x}=\left({\frac {\partial B}{\partial x}}\right)_{y}} holds, then A {\displaystyle A} and B {\displaystyle B} are differentiable (again, smoothly continuous) functions along y {\displaystyle y} and x {\displaystyle x} respectively, and ( ∂ A ∂ y ) x = ∂ 2 Q ∂ y ∂ x = ∂ 2 Q ∂ x ∂ y = ( ∂ B ∂ x ) y {\displaystyle \left({\frac {\partial A}{\partial y}}\right)_{x}={\frac {\partial ^{2}Q}{\partial y\partial x}}={\frac {\partial ^{2}Q}{\partial x\partial y}}=\left({\frac {\partial B}{\partial x}}\right)_{y}} is only the case. </p><p>For three dimensions, in a simply-connected region <i>R</i> of the <i>xyz</i>-coordinate system, by a similar reason, a differential </p> d Q = A ( x , y , z ) d x + B ( x , y , z ) d y + C ( x , y , z ) d z {\displaystyle dQ=A(x,y,z)\,dx+B(x,y,z)\,dy+C(x,y,z)\,dz} <p>is an exact differential if and only if between the functions <i>A</i>, <i>B</i> and <i>C</i> there exist the relations </p> ( ∂ A ∂ y ) x , z = ( ∂ B ∂ x ) y , z {\displaystyle \left({\frac {\partial A}{\partial y}}\right)_{x,z}\!\!\!=\left({\frac {\partial B}{\partial x}}\right)_{y,z}} ; ( ∂ A ∂ z ) x , y = ( ∂ C ∂ x ) y , z {\displaystyle \left({\frac {\partial A}{\partial z}}\right)_{x,y}\!\!\!=\left({\frac {\partial C}{\partial x}}\right)_{y,z}} ; ( ∂ B ∂ z ) x , y = ( ∂ C ∂ y ) x , z . {\displaystyle \left({\frac {\partial B}{\partial z}}\right)_{x,y}\!\!\!=\left({\frac {\partial C}{\partial y}}\right)_{x,z}.} <p>These conditions are equivalent to the following sentence: If <i>G</i> is the graph of this vector valued function then for all tangent vectors <i>X</i>,<i>Y</i> of the <i>surface</i> <i>G</i> then <i>s</i>(<i>X</i>, <i>Y</i>) = 0 with <i>s</i> the <a href="/facts/Symplectic_form/o95tWKBc">symplectic form</a>. </p><p>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 <i>dQ</i>, that is a function of four variables, to be an exact differential, there are six conditions (the <a href="/facts/Combination/nmvrhMm1">combination</a> C ( 4 , 2 ) = 6 {\displaystyle C(4,2)=6} ) to satisfy. </p> <h2 id="partial-differential-relations">Partial differential relations</h2> <p>If a <a href="/facts/Differentiable_function/jQqTgLk1">differentiable function</a> z ( x , y ) {\displaystyle z(x,y)} is <a href="/facts/Injective_function/5ES6tqf5">one-to-one (injective)</a> for each independent variable, e.g., z ( x , y ) {\displaystyle z(x,y)} is one-to-one for x {\displaystyle x} at a fixed y {\displaystyle y} while it is not necessarily one-to-one for ( x , y ) {\displaystyle (x,y)} , then the following <a href="/facts/Total_differential/337XJVMC">total differentials</a> exist because each independent variable is a differentiable function for the other variables, e.g., x ( y , z ) {\displaystyle x(y,z)} . </p> d x = ( ∂ x ∂ y ) z d y + ( ∂ x ∂ z ) y d z {\displaystyle dx={\left({\frac {\partial x}{\partial y}}\right)}_{z}\,dy+{\left({\frac {\partial x}{\partial z}}\right)}_{y}\,dz} d z = ( ∂ z ∂ x ) y d x + ( ∂ z ∂ y ) x d y . {\displaystyle dz={\left({\frac {\partial z}{\partial x}}\right)}_{y}\,dx+{\left({\frac {\partial z}{\partial y}}\right)}_{x}\,dy.} <p>Substituting the first equation into the second and rearranging, we obtain </p> d z = ( ∂ z ∂ x ) y [ ( ∂ x ∂ y ) z d y + ( ∂ x ∂ z ) y d z ] + ( ∂ z ∂ y ) x d y , {\displaystyle dz={\left({\frac {\partial z}{\partial x}}\right)}_{y}\left[{\left({\frac {\partial x}{\partial y}}\right)}_{z}dy+{\left({\frac {\partial x}{\partial z}}\right)}_{y}dz\right]+{\left({\frac {\partial z}{\partial y}}\right)}_{x}dy,} d z = [ ( ∂ z ∂ x ) y ( ∂ x ∂ y ) z + ( ∂ z ∂ y ) x ] d y + ( ∂ z ∂ x ) y ( ∂ x ∂ z ) y d z , {\displaystyle 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]dy+{\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial z}}\right)}_{y}dz,} [ 1 − ( ∂ z ∂ x ) y ( ∂ x ∂ z ) y ] d z = [ ( ∂ z ∂ x ) y ( ∂ x ∂ y ) z + ( ∂ z ∂ y ) x ] d y . {\displaystyle \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]dy.} <p>Since y {\displaystyle y} and z {\displaystyle z} are independent variables, d y {\displaystyle dy} and d z {\displaystyle dz} may be chosen without restriction. For this last equation to generally hold, the bracketed terms must be equal to zero.<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> 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. </p> <h3>Reciprocity relation</h3> <p>Setting the first term in brackets equal to zero yields </p> ( ∂ z ∂ x ) y ( ∂ x ∂ z ) y = 1. {\displaystyle {\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial z}}\right)}_{y}=1.} <p>A slight rearrangement gives a reciprocity relation, </p> ( ∂ z ∂ x ) y = 1 ( ∂ x ∂ z ) y . {\displaystyle {\left({\frac {\partial z}{\partial x}}\right)}_{y}={\frac {1}{{\left({\frac {\partial x}{\partial z}}\right)}_{y}}}.} <p>There are two more <a href="/facts/Permutations/JSw9ukmv">permutations</a> of the foregoing derivation that give a total of three reciprocity relations between x {\displaystyle x} , y {\displaystyle y} and z {\displaystyle z} . </p> <h3>Cyclic relation</h3> <p>The cyclic relation is also known as the cyclic rule or the <a href="/facts/Triple_product_rule/yYkWby8v">Triple product rule</a>. Setting the second term in brackets equal to zero yields </p> ( ∂ z ∂ x ) y ( ∂ x ∂ y ) z = − ( ∂ z ∂ y ) x . {\displaystyle {\left({\frac {\partial z}{\partial x}}\right)}_{y}{\left({\frac {\partial x}{\partial y}}\right)}_{z}=-{\left({\frac {\partial z}{\partial y}}\right)}_{x}.} <p>Using a reciprocity relation for ∂ z ∂ y {\displaystyle {\tfrac {\partial z}{\partial y}}} on this equation and reordering gives a cyclic relation (the <a href="/facts/Triple_product_rule/yYkWby8v">triple product rule</a>), </p> ( ∂ x ∂ y ) z ( ∂ y ∂ z ) x ( ∂ z ∂ x ) y = − 1. {\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.} <p>If, <i>instead</i>, reciprocity relations for ∂ x ∂ y {\displaystyle {\tfrac {\partial x}{\partial y}}} and ∂ y ∂ z {\displaystyle {\tfrac {\partial y}{\partial z}}} are used with subsequent rearrangement, a <a href="/facts/Implicit_function/UJmQ2Cyk">standard form for implicit differentiation</a> is obtained: </p> ( ∂ y ∂ x ) z = − ( ∂ z ∂ x ) y ( ∂ z ∂ y ) x . {\displaystyle {\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}}}.} <h2 id="some-useful-equations-derived-from-exact-differentials-in-two-dimensions">Some useful equations derived from exact differentials in two dimensions</h2> <p>(See also <a href="/facts/Bridgman%2527s_thermodynamic_equations/1s3F66xx">Bridgman's thermodynamic equations</a> for the use of exact differentials in the theory of <a href="/facts/Thermodynamic_equations/B90xDQJs">thermodynamic equations</a>) </p><p>Suppose we have five state functions z , x , y , u {\displaystyle z,x,y,u} , and v {\displaystyle v} . Suppose that the state space is two-dimensional and any of the five quantities are differentiable. Then by the <a href="/facts/Chain_rule/aMLYxP0x">chain rule</a> </p> <table><tbody><tr><td> d z = ( ∂ z ∂ x ) y d x + ( ∂ z ∂ y ) x d y = ( ∂ z ∂ u ) v d u + ( ∂ z ∂ v ) u d v {\displaystyle 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} </td> <td></td> <td>1</td></tr></tbody></table> <p>but also by the chain rule: </p> <table><tbody><tr><td> d x = ( ∂ x ∂ u ) v d u + ( ∂ x ∂ v ) u d v {\displaystyle dx=\left({\frac {\partial x}{\partial u}}\right)_{v}du+\left({\frac {\partial x}{\partial v}}\right)_{u}dv} </td> <td></td> <td>2</td></tr></tbody></table> <p>and </p> <table><tbody><tr><td> d y = ( ∂ y ∂ u ) v d u + ( ∂ y ∂ v ) u d v {\displaystyle dy=\left({\frac {\partial y}{\partial u}}\right)_{v}du+\left({\frac {\partial y}{\partial v}}\right)_{u}dv} </td> <td></td> <td>3</td></tr></tbody></table> <p>so that (by substituting (2) and (3) into (1)): </p> <table><tbody><tr><td> d z = [ ( ∂ z ∂ x ) y ( ∂ x ∂ u ) v + ( ∂ z ∂ y ) x ( ∂ y ∂ u ) v ] d u + [ ( ∂ z ∂ x ) y ( ∂ x ∂ v ) u + ( ∂ z ∂ y ) x ( ∂ y ∂ v ) u ] d v {\displaystyle {\begin{aligned}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\end{aligned}}} </td> <td></td> <td>4</td></tr></tbody></table> <p>which implies that (by comparing (4) with (1)): </p> <table><tbody><tr><td> ( ∂ z ∂ u ) v = ( ∂ z ∂ x ) y ( ∂ x ∂ u ) v + ( ∂ z ∂ y ) x ( ∂ y ∂ u ) v {\displaystyle \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}} </td> <td></td> <td>5</td></tr></tbody></table> <p>Letting v = y {\displaystyle v=y} in (5) gives: </p> <table><tbody><tr><td> ( ∂ z ∂ u ) y = ( ∂ z ∂ x ) y ( ∂ x ∂ u ) y {\displaystyle \left({\frac {\partial z}{\partial u}}\right)_{y}=\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial u}}\right)_{y}} </td> <td></td> <td>6</td></tr></tbody></table> <p>Letting u = y {\displaystyle u=y} in (5) gives: </p> <table><tbody><tr><td> ( ∂ z ∂ y ) v = ( ∂ z ∂ x ) y ( ∂ x ∂ y ) v + ( ∂ z ∂ y ) x {\displaystyle \left({\frac {\partial z}{\partial y}}\right)_{v}=\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial y}}\right)_{v}+\left({\frac {\partial z}{\partial y}}\right)_{x}} </td> <td></td> <td>7</td></tr></tbody></table> <p>Letting u = y {\displaystyle u=y} and v = z {\displaystyle v=z} in (7) gives: </p> <table><tbody><tr><td> ( ∂ z ∂ y ) x = − ( ∂ z ∂ x ) y ( ∂ x ∂ y ) z {\displaystyle \left({\frac {\partial z}{\partial y}}\right)_{x}=-\left({\frac {\partial z}{\partial x}}\right)_{y}\left({\frac {\partial x}{\partial y}}\right)_{z}} </td> <td></td> <td>8</td></tr></tbody></table> <p>using ( ∂ a / ∂ b ) c = 1 / ( ∂ b / ∂ a ) c {\displaystyle \partial a/\partial b)_{c}=1/(\partial b/\partial a)_{c}} gives the <a href="/facts/Triple_product_rule/yYkWby8v">triple product rule</a>: </p> <table><tbody><tr><td> ( ∂ z ∂ x ) y ( ∂ x ∂ y ) z ( ∂ y ∂ z ) x = − 1 {\displaystyle \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} </td> <td></td> <td>9</td></tr></tbody></table> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Closed_and_exact_differential_forms/uvEUbzUK">Closed and exact differential forms</a> for a higher-level treatment</li> <li><a href="/facts/Differential_(mathematics)/YmSDWCol">Differential (mathematics)</a></li> <li><a href="/facts/Inexact_differential_equation/lf59GDf4">Inexact differential</a></li> <li><a href="/facts/Integrating_factor/5tKufbqg">Integrating factor</a> for solving non-exact differential equations by making them exact</li> <li><a href="/facts/Exact_differential_equation/DXI9tgkD">Exact differential equation</a></li></ul> <ul><li>Perrot, P. (1998). <i>A to Z of Thermodynamics.</i> New York: Oxford University Press.</li> <li>Dennis G. Zill (14 May 2008). <a href="https://books.google.com/books?id=BnArjLNjXuYC&q=%22exact+differential%22"><i>A First Course in Differential Equations</i></a>. Cengage Learning. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-495-10824-5.</li></ul> <h2 id="external-links">External links</h2> <ul><li><a href="http://mathworld.wolfram.com/InexactDifferential.html">Inexact Differential</a> – from Wolfram MathWorld</li> <li><a href="https://web.archive.org/web/20151016052646/http://www.chem.arizona.edu/~salzmanr/480a/480ants/e%26idiff/e%26idiff.html">Exact and Inexact Differentials</a> – University of Arizona</li> <li><a href="http://farside.ph.utexas.edu/teaching/sm1/lectures/node36.html">Exact and Inexact Differentials</a> – University of Texas</li> <li><a href="http://mathworld.wolfram.com/ExactDifferential.html">Exact Differential</a> – from Wolfram MathWorld</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>If the pair of independent variables ( x , y ) {\displaystyle (x,y)} is a (locally reversible) function of dependent variables ( u , v ) {\displaystyle (u,v)} , all that is needed for the following theorem to hold, is to replace the partial derivatives with respect to x {\displaystyle x} or to y {\displaystyle y} , by the partial derivatives with respect to u {\displaystyle u} and to v {\displaystyle v} involving their Jacobian components. That is: A ( u , v ) d u + B ( u , v ) d v , {\displaystyle A(u,v)du+B(u,v)dv,} is an exact differential, if and only if: ∂ A ∂ u ∂ u ∂ y + ∂ A ∂ v ∂ v ∂ y = ∂ B ∂ u ∂ u ∂ x + ∂ B ∂ v ∂ v ∂ x . {\displaystyle {\frac {\partial A}{\partial u}}{\frac {\partial u}{\partial y}}+{\frac {\partial A}{\partial v}}{\frac {\partial v}{\partial y}}={\frac {\partial B}{\partial u}}{\frac {\partial u}{\partial x}}+{\frac {\partial B}{\partial v}}{\frac {\partial v}{\partial x}}.} <a href="/wiki/Jacobian" target="_blank">/wiki/Jacobian</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Çengel, Yunus A.; Boles, Michael A.; Kanoğlu, Mehmet (2019) [1989]. "Thermodynamics Property Relations". Thermodynamics - An Engineering Approach (9th ed.). New York: McGraw-Hill Education. pp. 647–648. ISBN 978-1-259-82267-4. <a href="978-1-259-82267-4" target="_blank">978-1-259-82267-4</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> </ol>