Questions: Mixed Partial Derivatives and Clairaut's Theorem

Question 1

A student claims: 'To compute ∂²f/∂x∂y, I must differentiate first with respect to x, then y — the notation shows the order.' For f(x,y) = x²y + e^(xy), what is actually correct?

• A. The student is right — ∂²f/∂x∂y means differentiate with respect to x first, then y
• B. The notation ∂²f/∂x∂y means differentiate with respect to y first (rightmost denominator variable), then x — but by Clairaut's theorem the result equals ∂²f/∂y∂x for smooth functions
• C. The order is arbitrary and neither convention is standard
• D. Clairaut's theorem only applies to polynomials, not to functions involving exponentials

Answer: (B)

In denominator notation, you differentiate right-to-left: ∂²f/∂x∂y means first ∂/∂y, then ∂/∂x. The student has the order backwards. The deeper point is Clairaut's theorem: since f(x,y) = x²y + e^(xy) is smooth (continuous mixed partials everywhere), the order doesn't matter — ∂²f/∂x∂y = ∂²f/∂y∂x. Differentiate in whichever order is computationally easier.

Question 2

Under what condition does Clairaut's theorem guarantee that ∂²f/∂x∂y = ∂²f/∂y∂x at a point?

• A. Whenever f is defined and differentiable at the point
• B. Whenever both mixed partial derivatives exist and are continuous at the point
• C. Only when f is a polynomial or trigonometric function
• D. Whenever f has no critical points in a neighborhood of the point

Answer: (B)

Clairaut's theorem (also called Schwarz's theorem) requires continuity of both mixed partials at the point, not merely their existence. Differentiability alone is insufficient — there exist differentiable functions whose mixed partials exist at a point but are discontinuous there, and for these the mixed partials can differ. The continuity condition is sufficient (not necessary), so any smooth function you encounter in practice will satisfy it.

Question 3

For any differentiable function f(x, y), the mixed partial derivatives ∂²f/∂x∂y and ∂²f/∂y∂x are typically equal.

• T. True
• F. False

Answer: False

Differentiability alone is not sufficient. Clairaut's theorem requires the mixed partials themselves to be continuous. A canonical counterexample: f(x,y) = xy(x²−y²)/(x²+y²) for (x,y) ≠ (0,0) and f(0,0) = 0, where ∂²f/∂x∂y at the origin equals 1 but ∂²f/∂y∂x at the origin equals −1. The function is differentiable but its mixed partials are discontinuous at the origin.

Question 4

Clairaut's theorem implies that the Hessian matrix of a smooth function f(x₁, ..., xₙ) is always a symmetric matrix.

• T. True
• F. False

Answer: True

The Hessian H(f) has entry (i,j) equal to ∂²f/∂xᵢ∂xⱼ. By Clairaut's theorem, ∂²f/∂xᵢ∂xⱼ = ∂²f/∂xⱼ∂xᵢ whenever the mixed partials are continuous, meaning H(i,j) = H(j,i) — the definition of a symmetric matrix. This symmetry gives the Hessian real eigenvalues and an orthogonal eigenbasis, which is what makes the second-partials test for classifying critical points work.

Question 5

Why does Clairaut's theorem matter practically? What would change about computing second derivatives if the theorem were false for smooth functions?

Model answer: Clairaut's theorem allows you to differentiate in whichever order is computationally convenient, knowing the result is the same. Without it, every mixed partial would need to be computed in both orders to confirm the result — doubling the work. More critically, the Hessian would not be symmetric, losing its real eigenvalues and making the second-partials test for critical points inapplicable.

The practical importance extends further: Lagrangian mechanics, optimization theory, and differential geometry all rely on commutativity of partial derivatives. Clairaut's theorem is one of those background guarantees so ubiquitous that its absence would make modern analysis deeply complicated.