A student computes ∂²f/∂x∂y for f(x,y) = x³y by differentiating with respect to x first (getting 3x²y, then 3x²), while another student differentiates with respect to y first (getting x³, then 3x²). Both get 3x². The first student concludes: 'The notation ∂²f/∂x∂y means differentiate x first, since x appears first in the denominator.' What is wrong with this reasoning?
AThe notation actually means differentiate with respect to y first, then x; they got the same answer only because Clairaut's theorem applies here
BThe student is correct — in Leibniz notation, you differentiate the leftmost variable first
CClairaut's theorem doesn't apply to polynomials, so both computations are coincidentally correct
DBoth computations are invalid — higher-order partials require the limit definition
In Leibniz notation, ∂²f/∂x∂y is read right to left: the variable closest to f (rightmost in the denominator, here y) is differentiated first. So the correct interpretation is: differentiate with respect to y, then differentiate the result with respect to x. The two students got the same answer only because f = x³y has continuous mixed partials, so Clairaut's theorem guarantees f_xy = f_yx. The notation convention, however, designates y-first — and for functions where mixed partials are not continuous, the order would matter.
Question 2 Multiple Choice
For a function f(x,y) with continuous second-order partial derivatives, how many *distinct* second-order partial derivatives does it have?
AFour — f_xx, f_xy, f_yx, and f_yy are all potentially different
BTwo — only the pure second derivatives f_xx and f_yy carry independent information
CThree — f_xx, f_yy, and f_xy, since Clairaut's theorem guarantees f_xy = f_yx
DOne — the Hessian determinant summarizes all second-order behavior in a single number
When the mixed partial derivatives are continuous, Clairaut's theorem guarantees f_xy = f_yx — so the four formally distinct second-order partial derivatives collapse to three independent ones: f_xx (pure second derivative in x), f_yy (pure second derivative in y), and f_xy = f_yx (the shared mixed partial). All three appear on the Hessian matrix, which is symmetric precisely because of this equality. Option A would be correct for a pathological function where mixed partials are not continuous.
Question 3 True / False
The Hessian matrix H of a function f(x,y) is generally symmetric.
TTrue
FFalse
Answer: False
The Hessian is symmetric only when Clairaut's theorem applies — that is, when the mixed partial derivatives f_xy and f_yx are continuous. Pathological functions exist where f_xy(0,0) ≠ f_yx(0,0), making the Hessian asymmetric at that point. In practice, virtually every function encountered in applied work has continuous mixed partials (and therefore a symmetric Hessian), but the theorem requires continuity as a hypothesis, not just existence of the mixed partials.
Question 4 True / False
In Leibniz notation, ∂²f/∂y∂x means: differentiate with respect to y first, then x.
TTrue
FFalse
Answer: False
The notation is read right to left: the variable closest to f in the denominator is differentiated first. In ∂²f/∂y∂x, x is closest to f (rightmost), so you differentiate with respect to x first, then y. This is the opposite of left-to-right reading. The easy mnemonic: peel off variables from right to left, just as you peel off operators in function composition. Note this is also opposite to subscript notation: f_yx means differentiate x first, then y — adding to the confusion between conventions.
Question 5 Short Answer
Clairaut's theorem states that mixed partial derivatives are equal when they are continuous. Why does the theorem require *continuity* of the mixed partials — isn't it enough that they simply exist?
Think about your answer, then reveal below.
Model answer: Existence alone is not sufficient. Pathological functions can be constructed where f_xy(0,0) and f_yx(0,0) both exist but are unequal. Continuity of the mixed partials at a point is what forces the two orders of differentiation to agree. The continuity condition ensures the limiting processes involved in computing f_xy and f_yx interact in a well-behaved way. Without it, the order of differentiation can affect the result.
A classic counterexample is f(x,y) = xy(x²−y²)/(x²+y²) for (x,y) ≠ (0,0), f(0,0) = 0. For this function, f_xy(0,0) = 1 and f_yx(0,0) = −1. Both mixed partials exist at the origin but are unequal — and they are discontinuous there. Clairaut's theorem is not violated because the hypothesis (continuity of the mixed partials) fails at that point. This shows that mere existence is genuinely weaker than continuity in this context.