Clairaut's Theorem

Core Definition

Let $f(x, y)$ be defined on an open disk $D$ containing the point $(a, b)$ . If the mixed second-order partial derivative functions $f_{xy}$ and $f_{yx}$ are both continuous throughout $D$ , then:

f_{xy}(a, b) = f_{yx}(a, b)

In Leibniz notation:

\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}

Furthermore, this symmetry extends to any order

n

for functions of

C^n

smoothness (having continuous

n^{\text{th}}

-order partial derivatives): the order in which partial derivatives are evaluated does not affect the final result (e.g.,

f_{xxy} = f_{xyx} = f_{yxx}

Concept Explanation

The Commutative Nature of Mixed Differentiation

In single-variable calculus, operations rarely commute. However, for sufficiently smooth surfaces $z = f(x, y)$ , measuring the rate of change of the $x$ -slope as you move in the $y$ -direction yields the exact same rate as measuring the change of the $y$ -slope as you move in the $x$ -direction. Geometrically, both expressions quantify the same local twisting (torsional) curvature of the surface patch over the $xy$ -plane.

                  f(x, y)
                 /       \
        d/dx    /         \    d/dy
               v           v
              f_x         f_y
               \         /
        d/dy    \       /      d/dx
                 v     v
              f_{xy} = f_{yx}  (Clairaut Symmetry)

The Continuity Requirement ( $C^2$ Smoothness)

Clairaut's Theorem is guaranteed for all elementary functions (polynomials, rational functions, exponentials, logarithms, and trigonometric functions) on their open domains of definition, because these functions have continuous partial derivatives of all orders ( $C^\infty$ ). Pathological counterexamples only occur when the mixed partials suffer jump or infinite discontinuities at isolated points (such as $(0, 0)$ in certain piecewise rational functions).

Physical Foundations & Conservation Laws

The mathematical symmetry of Clairaut's Theorem is the direct reason why physical nature exhibits reciprocity:

Thermodynamics (Maxwell Relations): The Helmholtz free energy $F(T, V)$ satisfies $dF = -S dT - P dV$ . Because $F$ is a state function with continuous mixed partials: $\frac{\partial^2 F}{\partial V \partial T} = \frac{\partial^2 F}{\partial T \partial V} \implies \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V$ Engineers use this to measure entropy changes $S$ purely through easily measurable pressure $P$ and temperature $T$ instruments.
Conservative Forces: A 2D force field $\vec{F} = \langle P, Q \rangle$ is conservative (meaning work is path-independent) if and only if it is the gradient of a potential function $f$ ( $\nabla f = \vec{F}$ ). By Clairaut's Theorem, $f_{xy} = f_{yx}$ , which forces the curl condition: $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$

Engineering Applications

Civil & Structural Engineering: In structural elasticity, Betti's Reciprocal Work Theorem states that the deflection at point A due to a load at point B equals the deflection at point B due to a load at point A. This symmetry is an immediate physical manifestation of Clairaut's equality of mixed partials of the strain energy potential (see Structural Column Strain Energy Stability (Civil & Mechanical Engineering)).
Thermal & Fluid Systems: Equations of state in power plant steam cycles use Maxwell relations to compute unmeasurable thermodynamic enthalpies from temperature and volume sensors.
Electrical Circuit Optimization: Hessian matrices in automated audio amplifier control loops are guaranteed to be symmetric by Clairaut's Theorem, cutting computation time in half during real-time matrix inversion (see Automated Audio Amplifier Loop Gain (Electrical & Computer Engineering)).

Example Problems

Example 1: Verifying Clairaut's Theorem for Transcendental Functions

Let $f(x, y) = e^{2x} \cos(3y) + x^4 y^3$ . Verify Clairaut's Theorem by computing both $f_{xy}$ and $f_{yx}$ .

Solution

Compute $f_x$ and $f_{xy}$ :
$f_x = \frac{\partial}{\partial x}[e^{2x} \cos(3y) + x^4 y^3] = 2e^{2x} \cos(3y) + 4x^3 y^3$ $f_{xy} = \frac{\partial}{\partial y}[2e^{2x} \cos(3y) + 4x^3 y^3] = 2e^{2x} (-3\sin(3y)) + 4x^3 (3y^2) = -6e^{2x}\sin(3y) + 12x^3 y^2$
Compute $f_y$ and $f_{yx}$ :
$f_y = \frac{\partial}{\partial y}[e^{2x} \cos(3y) + x^4 y^3] = -3e^{2x} \sin(3y) + 3x^4 y^2$ $f_{yx} = \frac{\partial}{\partial x}[-3e^{2x} \sin(3y) + 3x^4 y^2] = -3(2e^{2x}) \sin(3y) + 3(4x^3) y^2 = -6e^{2x}\sin(3y) + 12x^3 y^2$
Conclusion: Both expressions match identically:
$f_{xy} = f_{yx} = -6e^{2x}\sin(3y) + 12x^3 y^2$ Since $f(x, y)$ is a combination of smooth elementary functions, its second partials are continuous on $\mathbb{R}^2$ , confirming Clairaut's Theorem.

Example 2: Higher-Order Clairaut Symmetry ( $n = 3$ )

For $w = x^3 y^2 z^4$ , show that the mixed third-order partial derivatives $w_{xyz}$ , $w_{xzy}$ , and $w_{zyx}$ are all equal.