Clairaut's theorem — mixed partials are equal (under conditions): Exercises

Quick reminder of the two symbols we lean on the whole way down:

The picture below is what all this means: a mixed partial is a twist of the surface, and Clairaut says the twist is the same whichever way you sneak up on it.

Figure — Clairaut's theorem — mixed partials are equal (under conditions)

Figure 1 — A saddle surface $z = x y$ over the $x y$ -plane. Walk East (blue) and the ground tilts; now drift North and watch that East-tilt rotate — that rotation is $f_{x y}$ . The green path (North first, then East) measures the very same twisting of the corner. Clairaut: the two twists agree when the surface has no crease.

Definition Two words we will keep using

$C^{2}$ (read "see-two") is shorthand for "all second partial derivatives exist and are continuous." A function that is $C^{2}$ on a region is smooth enough for Clairaut to apply there. Whenever this page says " $C^{2}$ ," mentally replace it with "the mixed partials are continuous here."
Open region = a patch of the plane with no boundary edge included (like the inside of a disk). Clairaut only needs an open region around the single point. A stronger idea, simply connected (an open region with no holes — you can shrink any loop to a point), is not needed for Clairaut itself; it only shows up later in L4 for exact equations. We flag the distinction now so it is never a surprise:
- Clairaut ( $f_{x y} = f_{y x}$ ): needs only continuity on an open region.
- Exactness has a potential: needs $M_{y} = N_{x}$ and a simply connected domain.

Level 1 — Recognition

L1.1 True or false, and say why: "For $f (x, y) = x^{4} y^{7}$ , we are guaranteed $f_{x y} = f_{y x}$ without computing either."

L1.2 In the symbol $f_{y x}$ , which variable is differentiated first? And in the Leibniz form $\frac{\partial ^{2} f}{\partial x \partial y}$ , which is first?

L1.3 State the one hypothesis that Clairaut's theorem requires, in your own words.

Recall Solution L1.1

True. $f (x, y) = x^{4} y^{7}$ is a polynomial. Every partial derivative of a polynomial is again a polynomial, and polynomials are continuous everywhere. So $f_{x y}$ and $f_{y x}$ are continuous on all of $R^{2}$ — i.e. $f$ is $C^{2}$ — Clairaut's hypothesis holds automatically — hence they must be equal. We didn't need the actual value to know they agree. (For the curious: $f_{x} = 4 x^{3} y^{7}$ , so $f_{x y} = 28 x^{3} y^{6}$ ; and $f_{y} = 7 x^{4} y^{6}$ , so $f_{y x} = 28 x^{3} y^{6}$ . ✅)

Recall Solution L1.2

$f_{y x}$ : subscripts read left to right, so $y$ is differentiated first, then $x$ .
$\frac{\partial ^{2} f}{\partial x \partial y}$ : in Leibniz notation the operator closest to $f$ acts first. Here that is $\partial y$ , so $y$ is first — the same order as $f_{y x}$ . The two notations agree; they just read in opposite directions.

Recall Solution L1.3

The mixed partials $f_{x y}$ and $f_{y x}$ must exist and be continuous on an open region containing the point (that is, $f$ is $C^{2}$ there). Continuity of the second mixed partials is the price of admission for swapping the order.

Level 2 — Application

L2.1 For $f (x, y) = x^{2} y^{3} + e^{x y}$ , compute $f_{x y}$ by differentiating $x$ first, then verify by computing $f_{y x}$ .

L2.2 For $f (x, y) = sin (x) cos (y) + x^{2} y$ , find both mixed partials and confirm they match.

L2.3 ("Pick the easy order.") For $f (x, y) = e^{x^{2}} ln (y) + y x^{5}$ , you want the value of the mixed partial. Compute it using whichever order is algebraically cleaner, and state which order you chose and why.

Recall Solution L2.1

$x$ first. Hold $y$ fixed: $f_{x} = \frac{\partial}{\partial x} (x^{2} y^{3} + e^{x y}) = 2 x y^{3} + y e^{x y} .$ Why the $y e^{x y}$ ? Chain rule: $\frac{d}{d x} e^{x y} = e^{x y} \cdot \frac{\partial}{\partial x} (x y) = e^{x y} \cdot y$ . Now differentiate in $y$ : $f_{x y} = \frac{\partial}{\partial y} (2 x y^{3} + y e^{x y}) = 6 x y^{2} + (e^{x y} + y \cdot x e^{x y}) = 6 x y^{2} + e^{x y} (1 + x y) .$ Why the second term? Product rule on $y e^{x y}$ : derivative of $y$ is $1$ (giving $e^{x y}$ ), plus $y$ times $\frac{\partial}{\partial y} e^{x y} = x e^{x y}$ .

Check with $y$ first. $f_{y} = 3 x^{2} y^{2} + x e^{x y}$ , then $f_{y x} = 6 x y^{2} + (e^{x y} + x \cdot y e^{x y}) = 6 x y^{2} + e^{x y} (1 + x y) .$ ✅ Identical.

Recall Solution L2.2

$f_{x} = cos (x) cos (y) + 2 x y$ , so $f_{x y} = - cos (x) sin (y) + 2 x .$ $f_{y} = - sin (x) sin (y) + x^{2}$ , so $f_{y x} = - cos (x) sin (y) + 2 x .$ ✅ Match. (The $2 x$ appears in both — the twist of the $x^{2} y$ term is $2 x$ either way.)

Recall Solution L2.3

The two terms behave differently, so handle the payoff order term-by-term. The clean order is $y$ first for the $e^{x^{2}} ln y$ piece (because $\partial_{y} ln y = 1/ y$ kills the messy $ln$ immediately), and either order is trivial for $y x^{5}$ .

$f_{y} = e^{x^{2}} \cdot \frac{1}{y} + x^{5}$ . Then $f_{y x} = \frac{\partial}{\partial x} (\frac{e ^{x^{2}}}{y} + x^{5}) = \frac{2 x e ^{x^{2}}}{y} + 5 x^{4} .$ Why this was easier: had we gone $x$ first, $f_{x} = 2 x e^{x^{2}} ln y + 5 y x^{4}$ carries the $ln y$ through an extra step. Clairaut guarantees the same final answer, so we chose the order that avoids dragging $ln y$ around. Final answer: $\frac{2 x e ^{x^{2}}}{y} + 5 x^{4} .$

Level 3 — Analysis

L3.1 For $f (x, y) = x^{2} y^{3}$ , compute $f_{xx y}$ , $f_{x y x}$ , and $f_{y xx}$ separately and confirm all three are equal. Explain why they must be, using Clairaut.

L3.2 A student claims $\frac{\partial ^{2} f}{\partial x \partial y}$ and $\frac{\partial ^{2} f}{\partial y \partial x}$ are "obviously the same by definition, no theorem needed." Explain precisely what Clairaut adds that a definition alone does not.

L3.3 Suppose you are told only that a function's Hessian matrix (the $2 \times 2$ table of second partials) is symmetric at every point. What does that tell you about $f_{x y}$ vs $f_{y x}$ , and what smoothness condition guarantees it?

Recall Solution L3.1

Recall from the top of the page: each subscript letter = one differentiation in reading order. We work each string left to right, justifying every single step. ( $f = x^{2} y^{3}$ .)

$f_{xx y}$ (" $x$ , then $x$ , then $y$ "):

$\partial_{x} (x^{2} y^{3}) = 2 x y^{3}$ . Why: $y^{3}$ is a constant factor here, and $\partial_{x} (x^{2}) = 2 x$ .
$\partial_{x} (2 x y^{3}) = 2 y^{3}$ . Why: again $y^{3}$ is constant; $\partial_{x} (2 x) = 2$ .
$\partial_{y} (2 y^{3}) = 6 y^{2}$ . Why: the $x$ 's are gone, $\partial_{y} (y^{3}) = 3 y^{2}$ , times the $2$ .

$f_{x y x}$ (" $x$ , then $y$ , then $x$ "):

$\partial_{x} (x^{2} y^{3}) = 2 x y^{3}$ . Why: same first step, $y^{3}$ constant.
$\partial_{y} (2 x y^{3}) = 6 x y^{2}$ . Why: now $2 x$ is the constant factor; $\partial_{y} (y^{3}) = 3 y^{2}$ , times $2 x$ .
$\partial_{x} (6 x y^{2}) = 6 y^{2}$ . Why: $y^{2}$ constant; $\partial_{x} (6 x) = 6$ .

$f_{y xx}$ (" $y$ , then $x$ , then $x$ "):

$\partial_{y} (x^{2} y^{3}) = 3 x^{2} y^{2}$ . Why: $x^{2}$ constant; $\partial_{y} (y^{3}) = 3 y^{2}$ , times $x^{2}$ .
$\partial_{x} (3 x^{2} y^{2}) = 6 x y^{2}$ . Why: $y^{2}$ constant; $\partial_{x} (3 x^{2}) = 6 x$ .
$\partial_{x} (6 x y^{2}) = 6 y^{2}$ . Why: $y^{2}$ constant; $\partial_{x} (6 x) = 6$ .

All three give $6 y^{2}$ . ✅

Why they must match. $f$ is a polynomial, so every partial is continuous everywhere ( $f$ is $C^{\infty}$ , hence certainly $C^{2}$ ). Clairaut lets you swap any two adjacent differentiations (that's the two-variable theorem applied to $f$ or to a first derivative of $f$ ). By repeatedly swapping adjacent letters — like sorting — you can rearrange $xx y$ , $x y x$ , $y xx$ into one another. Therefore only the count of each variable (here: two $x$ 's, one $y$ ) matters, not the order.

Recall Solution L3.2

By definition, $\frac{\partial ^{2} f}{\partial x \partial y} = f_{y x}$ (differentiate in $y$ , then $x$ ) and $\frac{\partial ^{2} f}{\partial y \partial x} = f_{x y}$ (in $x$ , then $y$ ). These are two genuinely different computations — different limit processes, evaluated in different orders. Nothing in the definition forces them to give the same number. The counterexample at the origin (L5 below) shows they really can differ.

What Clairaut adds: an extra hypothesis (continuity of the mixed partials, i.e. $C^{2}$ ) under which those two different computations are proven to land on the same value. So it is a theorem with content, not a restatement of a definition.

Recall Solution L3.3

The Hessian Matrix is $H = (f_{xx} f_{y x} f_{x y} f_{y y}) .$ "Symmetric" means the off-diagonal entries are equal: $f_{x y} = f_{y x}$ — which is exactly Clairaut's conclusion. So a symmetric Hessian at every point is the statement "mixed partials agree everywhere." The condition that guarantees it: the second partials are continuous on the region ( $f \in C^{2}$ ). Then Clairaut forces symmetry.

Level 4 — Synthesis

L4.1 (Link to exact ODEs.) A differential form $M (x, y) d x + N (x, y) d y$ is called exact when it equals $df$ for some potential $f$ , which forces $M = f_{x}$ and $N = f_{y}$ . Use Clairaut to derive the standard test $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ . Then apply it to $M = 2 x y + y^{2}$ , $N = x^{2} + 2 x y$ : is the form exact?

L4.2 (Reverse the proof.) Recall the symmetric second difference from the parent proof: $Δ (h, k) = f (a + h, b + k) - f (a + h, b) - f (a, b + k) + f (a, b) .$ Compute $Δ$ explicitly for $f (x, y) = x y$ at $(a, b) = (0, 0)$ , then divide by $hk$ . What do you get, and how does it confirm $f_{x y} = f_{y x} = 1$ for this $f$ ?

L4.3 Build a function whose mixed partial equals $12 x^{2} y$ . (Reverse-engineer: integrate.)

Recall Solution L4.1

Deriving the exactness test. If the form is exact, $M = f_{x}$ and $N = f_{y}$ for a potential $f$ . Then $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} f_{x} = f_{x y}, \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} f_{y} = f_{y x} .$ If $f$ is $C^{2}$ , Clairaut gives $f_{x y} = f_{y x}$ , hence $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} .$ This is the necessary condition used in Exact Differential Equations — it is literally Clairaut wearing different clothes.

Apply it. $M = 2 x y + y^{2} \Rightarrow M_{y} = 2 x + 2 y$ . $N = x^{2} + 2 x y \Rightarrow N_{x} = 2 x + 2 y$ . They are equal, so the form is exact — and since the domain here is all of $R^{2}$ (open and simply connected, no holes), a genuine potential really exists.

Recall Solution L4.2

For $f = x y$ : $f (a + h, b + k) = (a + h) (b + k)$ , etc. At $(0, 0)$ : $Δ (h, k) = (h) (k) - (h) (0) - (0) (k) + (0) (0) = hk .$ Divide by $hk$ : $\frac{Δ ( h , k )}{hk} = 1 for all h, k \neq = 0.$ The parent proof showed $Δ/ (hk)$ limits to $f_{x y}$ one way and $f_{y x}$ the other. Here it is a constant $1$ , so both limits are $1$ regardless of order: $f_{x y} (0, 0) = f_{y x} (0, 0) = 1.$ Direct check: $f_{x} = y \Rightarrow f_{x y} = 1$ ; $f_{y} = x \Rightarrow f_{y x} = 1$ . ✅ The symmetric difference literally is the mixed partial for this simple $f$ .

Recall Solution L4.3

We want $f_{x y} = 12 x^{2} y$ . Undo the " $x$ then $y$ " chain in reverse. First find an $f_{x}$ whose $y$ -derivative is $12 x^{2} y$ : integrate $12 x^{2} y$ with respect to $y$ , $f_{x} = \int 12 x^{2} y d y = 6 x^{2} y^{2} + C (x),$ where $C (x)$ is an arbitrary function of $x$ alone (it plays the role of the "constant of integration," but since we integrated in $y$ , anything with no $y$ survives as $\partial_{y} C (x) = 0$ ). Take the simplest choice $C (x) = 0$ , giving $f_{x} = 6 x^{2} y^{2}$ . Now integrate that in $x$ : $f = \int 6 x^{2} y^{2} d x = 2 x^{3} y^{2} + D (y),$ where $D (y)$ is again arbitrary in $y$ (it vanishes under $\partial_{x}$ ). So the family of solutions is $f = 2 x^{3} y^{2} + D (y) + (terms from C)$ ; the simplest is $f = 2 x^{3} y^{2}$ . Why the arbitrary pieces matter: any $D (y)$ leaves $f_{x}$ unchanged and hence $f_{x y}$ unchanged, so the potential is not unique — this is exactly the non-uniqueness you meet when reconstructing potentials in Exact Differential Equations. Check: $f = 2 x^{3} y^{2} \Rightarrow f_{x} = 6 x^{2} y^{2} \Rightarrow f_{x y} = 12 x^{2} y$ . ✅ (And $f_{y} = 4 x^{3} y \Rightarrow f_{y x} = 12 x^{2} y$ — Clairaut agrees, as it must for this polynomial.)

Level 5 — Mastery

L5.1 (The counterexample, hands-on.) Let $f (x, y) = ⎩ ⎨ ⎧ \frac{x y ( x ^{2} - y ^{2} )}{x ^{2} + y ^{2}}, 0, (x, y) \neq = (0, 0) (x, y) = (0, 0) .$ Show $f_{x y} (0, 0) = - 1$ while $f_{y x} (0, 0) = + 1$ , so Clairaut fails here — and identify exactly which hypothesis broke.

L5.2 Explain, using the figure below, why continuity of the mixed partial is the precise thing that fails in L5.1, connecting it back to the "unknown MVT points squeezing toward the origin" step of the parent proof.

Figure 2 — Along the $y$ -axis the first partial $f_{x} (0, y) = - y$ (blue) falls with slope $- 1$ ; along the $x$ -axis $f_{y} (x, 0) = + x$ (yellow) rises with slope $+ 1$ . Those two slopes ARE the mixed partials $f_{x y} (0, 0) = - 1$ and $f_{y x} (0, 0) = + 1$ . They disagree at the shared red point (the origin) — the visual signature of a mixed partial that is not continuous there.

Recall Solution L5.1

We need first partials away from the origin, then evaluate second partials at the origin using difference quotients.

Step 1 — $f_{x}$ for $(x, y) \neq = (0, 0)$ . With $f = \frac{x y ( x ^{2} - y ^{2} )}{x ^{2} + y ^{2}} = \frac{x ^{3} y - x y ^{3}}{x ^{2} + y ^{2}}$ , the quotient rule gives (after simplification) $f_{x} (x, y) = \frac{y ( x ^{4} + 4 x ^{2} y ^{2} - y ^{4} )}{( x ^{2} + y ^{2} ) ^{2}} .$ Its value on the $y$ -axis: put $x = 0$ : $f_{x} (0, y) = \frac{y ( - y ^{4} )}{y ^{4}} = - y$ for $y \neq = 0$ . Also $f_{x} (0, 0) = 0$ (from the definition: $f$ is $0$ along the $x$ -axis, so its $x$ -slope at the origin is $0$ ).

Step 2 — $f_{x y} (0, 0)$ , the change of $f_{x}$ in the $y$ -direction at the origin: $f_{x y} (0, 0) = lim_{k \to 0} \frac{f _{x} ( 0 , k ) - f _{x} ( 0 , 0 )}{k} = lim_{k \to 0} \frac{- k - 0}{k} = - 1.$

Step 3 — $f_{y}$ for $(x, y) \neq = (0, 0)$ , by symmetry ( $f$ is antisymmetric under $x \leftrightarrow y$ ): $f_{y} (x, y) = \frac{x ( x ^{4} - 4 x ^{2} y ^{2} - y ^{4} )}{( x ^{2} + y ^{2} ) ^{2}}, f_{y} (x, 0) = \frac{x \cdot x ^{4}}{x ^{4}} = x (x \neq = 0), f_{y} (0, 0) = 0.$

Step 4 — $f_{y x} (0, 0)$ , the change of $f_{y}$ in the $x$ -direction at the origin: $f_{y x} (0, 0) = lim_{h \to 0} \frac{f _{y} ( h , 0 ) - f _{y} ( 0 , 0 )}{h} = lim_{h \to 0} \frac{h - 0}{h} = + 1.$

Conclusion: $f_{x y} (0, 0) = - 1 \neq = + 1 = f_{y x} (0, 0)$ . The mixed partials exist but are not continuous at the origin — the exact hypothesis Clairaut requires (the $C^{2}$ condition fails). So the theorem's continuity condition is genuinely necessary. ✅

Recall Solution L5.2

Look at Figure 2. Along the axes, $f_{x} (0, y) = - y$ climbs one way and $f_{y} (x, 0) = + x$ climbs the other; approaching the origin from different directions makes the mixed partial's value head toward different numbers ( $- 1$ vs $+ 1$ ). That is precisely discontinuity of the mixed partial at $(0, 0)$ .

Recall the parent proof's finish: the MVT hands us unknown points $(c_{i}, d_{i})$ that only squeeze toward $(a, b)$ , and we needed " $f_{x y}$ at a nearby point $\to f_{x y} (a, b)$ " — that step is continuity. When continuity fails (as here), the squeezing points can approach along directions giving different limits, and the two orders no longer have to agree. The figure's clashing colored slopes on the two axes are the visual signature of that failure.

Recall One-line summary of the whole ladder

Recognise it (L1) → apply it both orders and pick the easy one (L2) → sort higher-derivative strings & read the Hessian (L3) → recognise it inside exactness tests and the proof's symmetric difference (L4) → and know the one function where it breaks, and why (L5): no continuity → no Clairaut.