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

The scenario matrix

Every problem about mixed partials falls into one of these cells. We work at least one example per cell.

#	Case class	What makes it tricky	Example that hits it
A	Polynomial (all signs, positive powers)	pure bookkeeping	Ex 1
B	Product with trig / chain rule	product + chain interact	Ex 2
C	"Which order is cheaper?" (the 80/20 payoff)	choose the easy path	Ex 3
D	Higher order ( $f_{xx y} = f_{y xx}$ )	count, don't order	Ex 4
E	Zero / degenerate input (a variable missing)	a partial collapses to $0$	Ex 5
F	Word problem (real-world twist, units)	translate physics ↔ math	Ex 6
G	Failure case — discontinuous mixed partial	Clairaut does not apply	Ex 7
H	Exam twist — "find a constant so equality holds"	reverse-engineer a function	Ex 8

Before we start, one reminder in plain words so no symbol is unearned:

Cell A — Pure polynomial (all signs)

Worked example Example 1 — signs and powers

Let $f (x, y) = - 2 x^{4} y^{3} + 5 x y^{2} - y^{5}$ . Verify $f_{x y} = f_{y x}$ .

Forecast: guess now — will the two orders match? (They must, since polynomials are smooth everywhere.)

$f_{x}$ (freeze $y$ ): $f_{x} = - 8 x^{3} y^{3} + 5 y^{2}$ . Why this step? Differentiate each term in $x$ ; the lone $- y^{5}$ has no $x$ , so it vanishes — our first taste of a term dropping out.
$f_{x y}$ (now differentiate that in $y$ ): $f_{x y} = - 24 x^{3} y^{2} + 10 y$ . Why this step? $\partial_{y} (- 8 x^{3} y^{3}) = - 24 x^{3} y^{2}$ and $\partial_{y} (5 y^{2}) = 10 y$ .
$f_{y}$ (freeze $x$ ): $f_{y} = - 6 x^{4} y^{2} + 10 x y - 5 y^{4}$ . Why this step? The other order — differentiate the original in $y$ first.
$f_{y x}$ : $f_{y x} = - 24 x^{3} y^{2} + 10 y$ . Why this step? $\partial_{x} (- 6 x^{4} y^{2}) = - 24 x^{3} y^{2}$ , $\partial_{x} (10 x y) = 10 y$ , and $- 5 y^{4}$ drops.

Verify: $f_{x y} = f_{y x} = - 24 x^{3} y^{2} + 10 y$ ✅. Sanity spot-check at $(1, 1)$ : both give $- 24 + 10 = - 14$ .

Cell B — Product with trig (chain rule inside)

Worked example Example 2 — the twist meets the chain rule

Let $f (x, y) = x^{2} cos (x y)$ . Show the two orders agree.

Forecast: with a product and a chain rule, it feels like the two orders should diverge — bet on your gut before reading.

$f_{x}$ : $f_{x} = 2 x cos (x y) + x^{2} \cdot (- sin (x y)) \cdot y = 2 x cos (x y) - x^{2} y sin (x y)$ . Why this step? Product rule on $x^{2}$ and $cos (x y)$ ; the chain rule sends $\partial_{x} cos (x y) = - y sin (x y)$ .
$f_{x y}$ : differentiate in $y$ .
- $\partial_{y} [2 x cos (x y)] = - 2 x^{2} sin (x y)$ .
- $\partial_{y} [- x^{2} y sin (x y)] = - x^{2} sin (x y) - x^{3} y cos (x y)$ (product $y \cdot sin$ , chain on $sin$ ).
- Sum: $f_{x y} = - 3 x^{2} sin (x y) - x^{3} y cos (x y)$ .
$f_{y}$ : $f_{y} = x^{2} \cdot (- sin (x y)) \cdot x = - x^{3} sin (x y)$ . Why this step? Freeze $x$ ; the $x^{2}$ is a constant multiplier now.
$f_{y x}$ : $\partial_{x} [- x^{3} sin (x y)] = - 3 x^{2} sin (x y) - x^{3} y cos (x y)$ . Why this step? Product rule on $x^{3}$ and $sin (x y)$ ; chain gives $\partial_{x} sin (x y) = y cos (x y)$ .

Verify: both equal $- 3 x^{2} sin (x y) - x^{3} y cos (x y)$ ✅. At $(1, 0)$ : $sin 0 = 0, cos 0 = 1$ , both give $0 - 1 \cdot 0 \cdot 1 = 0$ .

Cell C — Which order is cheaper? (80/20)

Worked example Example 3 — pick the lazy path on purpose

Find $f_{x y}$ for $f (x, y) = \frac{ln ( 1 + y ^{2} )}{x}$ , then confirm by the other order.

Forecast: which order avoids the messy $ln$ derivative longest — $x$ first or $y$ first?

Cheap path — $y$ first (because $x$ appears only as $1/ x$ , the tame part): $f_{y} = \frac{1}{x} \cdot \frac{2 y}{1 + y ^{2}} = \frac{2 y}{x ( 1 + y ^{2} )}$ . Why this step? $\partial_{y} ln (1 + y^{2}) = \frac{2 y}{1 + y ^{2}}$ by chain rule; $1/ x$ rides along.
Then $x$ : $f_{y x} = \partial_{x} [\frac{2 y}{1 + y ^{2}} \cdot x^{- 1}] = - \frac{2 y}{x ^{2} ( 1 + y ^{2} )}$ . Why this step? $\partial_{x} x^{- 1} = - x^{- 2}$ ; the $y$ -block is a constant here.
Expensive path — $x$ first (to show the two orders literally match): $f_{x} = \partial_{x} [ln (1 + y^{2}) \cdot x^{- 1}] = - \frac{ln ( 1 + y ^{2} )}{x ^{2}}$ . Why this step? Now the whole $ln (1 + y^{2})$ is a constant multiplier; only $x^{- 1}$ moves.
Then $y$ : $f_{x y} = \partial_{y} [- \frac{ln ( 1 + y ^{2} )}{x ^{2}}] = - \frac{1}{x ^{2}} \cdot \frac{2 y}{1 + y ^{2}} = - \frac{2 y}{x ^{2} ( 1 + y ^{2} )}$ . Why this step? Here we do pay the price — we differentiate the $ln$ on this path. It lands on the exact same expression as Step 2, confirming $f_{x y} = f_{y x}$ .

Verify: both paths give $- \frac{2 y}{x ^{2} ( 1 + y ^{2} )}$ ✅. At $(2, 1)$ : $- \frac{2 \cdot 1}{4 \cdot 2} = - \frac{1}{4}$ . Payoff: either order works, so in an exam pick the path (Step 1–2) that never touches the $ln$ derivative until forced.

Cell D — Higher order: count, don't order

Worked example Example 4 — three derivatives, three orders

For $f (x, y) = x^{3} y^{4}$ , show $f_{xx y} = f_{x y x} = f_{y xx}$ and give the value.

Forecast: all three should agree — the theorem says only how many $x$ 's and $y$ 's matters (here: two $x$ 's, one $y$ ).

$f_{xx y}$ (order $x, x, y$ ): first $f_{x} = \partial_{x} (x^{3} y^{4}) = 3 x^{2} y^{4}$ ; then $f_{xx} = \partial_{x} (3 x^{2} y^{4}) = 3 \cdot 2 x^{2 - 1} y^{4} = 6 x y^{4}$ ; then $f_{xx y} = \partial_{y} (6 x y^{4}) = 6 x \cdot 4 y^{3} = 24 x y^{3}$ . Why this step? Each $x$ -step lowers the $x$ -power by one and multiplies by the old power ( $x^{3} \to 3 x^{2} \to 6 x$ ), keeping $y^{4}$ frozen; only the final step spends the single $y$ -derivative ( $y^{4} \to 4 y^{3}$ ).
$f_{x y x}$ (order $x, y, x$ ): $f_{x} = 3 x^{2} y^{4}$ , then $f_{x y} = \partial_{y} (3 x^{2} y^{4}) = 12 x^{2} y^{3}$ , then $f_{x y x} = \partial_{x} (12 x^{2} y^{3}) = 24 x y^{3}$ . Why this step? Same first $x$ -step, but now we hit $y$ in the middle ( $y^{4} \to 4 y^{3}$ gives $12 x^{2} y^{3}$ ) and finish with $x$ ( $x^{2} \to 2 x$ gives $24 x y^{3}$ ).
$f_{y xx}$ (order $y, x, x$ ): $f_{y} = \partial_{y} (x^{3} y^{4}) = 4 x^{3} y^{3}$ , then $f_{y x} = \partial_{x} (4 x^{3} y^{3}) = 12 x^{2} y^{3}$ , then $f_{y xx} = \partial_{x} (12 x^{2} y^{3}) = 24 x y^{3}$ . Why this step? Hit $y$ first ( $y^{4} \to 4 y^{3}$ ), then differentiate in $x$ twice ( $x^{3} \to 3 x^{2}$ , then $x^{2} \to 2 x$ ).
The pattern: each route rearranges when we spend the single $y$ -derivative, but every route spends one $y$ and two $x$ 's — so by Clairaut they must land in the same place.

Verify: all three $= 24 x y^{3}$ ✅. At $(1, 1)$ : $24$ .

Cell E — Zero / degenerate input

Worked example Example 5 — when a variable is missing

Let $f (x, y) = 7 x^{2} + cos (3 x) - e^{2 y}$ . Compute $f_{x y}$ and $f_{y x}$ .

Forecast: notice something — no term contains both $x$ and $y$ . What should a mixed partial of a "separable" sum be?

$f_{x} = 14 x - 3 sin (3 x)$ — depends on $x$ only. Why this step? Freeze $y$ : the term $- e^{2 y}$ has no $x$ , so it drops; the $x$ -terms differentiate to $14 x$ and $- 3 sin (3 x)$ . Notice the result contains no $y$ — this is the seed of what happens next.
$f_{x y} = \partial_{y} (14 x - 3 sin (3 x)) = 0$ . Why this step? $f_{x}$ turned out to be a function of $x$ alone. An expression with no $y$ in it has zero rate of change as $y$ moves — so its $y$ -derivative is exactly $0$ .
$f_{y} = - 2 e^{2 y}$ — depends on $y$ only. Why this step? Freeze $x$ : every $x$ -only term ( $7 x^{2}$ , $cos 3 x$ ) is now a constant and drops; only $- e^{2 y}$ survives, giving $- 2 e^{2 y}$ . Again the result contains no $x$ .
$f_{y x} = \partial_{x} (- 2 e^{2 y}) = 0$ . Why this step? $f_{y}$ is a function of $y$ alone, so differentiating it in $x$ gives $0$ — mirror image of Step 2.

Verify: $f_{x y} = f_{y x} = 0$ ✅. General law you just discovered: if $f (x, y) = g (x) + h (y)$ (a separable sum), the first partial in one variable kills all dependence on the other, so every mixed partial is $0$ . That is the degenerate/zero cell of the matrix — Clairaut holds trivially ( $0 = 0$ ).

Cell F — Word problem with units

Worked example Example 6 — heat plate (real-world twist)

A metal plate's temperature is $T (x, y) = 40 + 3 x^{2} y - 2 x y^{2}$ (in $^{\circ} C$ , with $x, y$ in metres). The quantity $T_{x y}$ measures how the East–West temperature-slope changes as you walk North. Compute it two ways and interpret.

Forecast: the two orders describe physically different-sounding questions ("East-slope's Northward change" vs "North-slope's Eastward change"). Should they give the same number?

$T_{x} = 6 x y - 2 y^{2}$ — the East–West slope, in $^{\circ} C / m$ . Why this step? Freeze $y$ : the constant $40$ drops, $\partial_{x} (3 x^{2} y) = 6 x y$ , $\partial_{x} (- 2 x y^{2}) = - 2 y^{2}$ . This is the temperature gradient as you step East.
$T_{x y} = \partial_{y} (6 x y - 2 y^{2}) = 6 x - 4 y$ , in $^{\circ} C / m^{2}$ . Why this step? Now differentiate that East-slope in $y$ : $\partial_{y} (6 x y) = 6 x$ , $\partial_{y} (- 2 y^{2}) = - 4 y$ . It is the rate at which the East-slope changes per metre walked North.
Other order — $T_{y} = 3 x^{2} - 4 x y$ , in $^{\circ} C / m$ . Why this step? Freeze $x$ : $40$ drops, $\partial_{y} (3 x^{2} y) = 3 x^{2}$ , $\partial_{y} (- 2 x y^{2}) = - 4 x y$ . This is the North–South slope.
$T_{y x} = \partial_{x} (3 x^{2} - 4 x y) = 6 x - 4 y$ . Why this step? Differentiate the North-slope in $x$ : $\partial_{x} (3 x^{2}) = 6 x$ , $\partial_{x} (- 4 x y) = - 4 y$ . Because $T$ is a smooth polynomial, Clairaut forces this to equal Step 2 — and it does.

Verify: both $= 6 x - 4 y$ ✅. At the point $(2 m, 1 m)$ : $6 (2) - 4 (1) = 8^{\circ} C / m^{2}$ . The twist of the temperature field is $8$ regardless of which walk you take — the physical meaning of Clairaut.

Cell G — The FAILURE case (Clairaut does not apply)

Here we see the only place the two orders disagree — because the hypothesis (continuity of the mixed partial) breaks. The figure below shows the surface: read it before the algebra so you can see where the trouble lives.

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

Figure (Cell G): the surface $f (x, y) = x y (x^{2} - y^{2}) / (x^{2} + y^{2})$ . The teal curve is the slice along the $x$ -axis ( $y = 0$ ); the plum curve is the slice along the $y$ -axis ( $x = 0$ ). The orange dot is the origin $O$ , where the surface is pinched — the mixed partials fail to be continuous exactly there, and that pinch is what lets the two orders disagree.

Worked example Example 7 — mixed partials differ at the origin

$f (x, y) = ⎩ ⎨ ⎧ \frac{x y ( x ^{2} - y ^{2} )}{x ^{2} + y ^{2}}, 0, (x, y) \neq = (0, 0) (x, y) = (0, 0) .$ Show $f_{x y} (0, 0) = - 1$ but $f_{y x} (0, 0) = + 1$ .

Forecast: every earlier example matched. Predict whether this one can — and why the origin is special (look at the pinch in the figure).

Find $f_{x} (0, y)$ from the definition of a partial derivative. The definition is $f_{x} (0, y) = lim_{h \to 0} \frac{f ( 0 + h , y ) - f ( 0 , y )}{h} .$ Why this step? Near the pinch we cannot trust shortcut rules, so we return to the raw definition: "slope in $x$ = rise over run as the run $h$ shrinks to $0$ ."
Plug in the formula. For $y \neq = 0$ , $f (0, y) = 0$ (the numerator has a factor $x = 0$ ), and $f (h, y) = \frac{h y ( h ^{2} - y ^{2} )}{h ^{2} + y ^{2}} .$ So the difference quotient is $\frac{f ( h , y ) - 0}{h} = \frac{y ( h ^{2} - y ^{2} )}{h ^{2} + y ^{2}} .$ Why this step? Dividing by $h$ cancels the single power of $h$ in the numerator — that cancellation is why the limit will exist cleanly.
Take $h \to 0$ . As $h \to 0$ , $h^{2} \to 0$ , so $f_{x} (0, y) = \frac{y ( 0 - y ^{2} )}{0 + y ^{2}} = \frac{- y ^{3}}{y ^{2}} = - y .$ Why this step? Letting the run shrink to zero gives the exact slope; the messy quotient collapses to the tidy function $f_{x} (0, y) = - y$ .
Differentiate that in $y$ : $f_{x y} (0, 0) = \partial_{y} [f_{x} (0, y)]_{y = 0} = \partial_{y} (- y) = - 1$ . Why this step? $f_{x y}$ means "do $x$ first (Steps 1–3), then $y$ " — and $- y$ has $y$ -derivative $- 1$ .
Repeat with the roles of $x, y$ swapped — writing the limit out in full. By definition, $f_{y} (x, 0) = lim_{k \to 0} \frac{f ( x , 0 + k ) - f ( x , 0 )}{k} .$ For $x \neq = 0$ , $f (x, 0) = 0$ (numerator has a factor $y = 0$ ), and $f (x, k) = \frac{x k ( x ^{2} - k ^{2} )}{x ^{2} + k ^{2}}, so \frac{f ( x , k ) - 0}{k} = \frac{x ( x ^{2} - k ^{2} )}{x ^{2} + k ^{2}} .$ Letting $k \to 0$ (so $k^{2} \to 0$ ): $f_{y} (x, 0) = \frac{x ( x ^{2} - 0 )}{x ^{2} + 0} = \frac{x ^{3}}{x ^{2}} = + x .$ Then $f_{y x} (0, 0) = \partial_{x} [f_{y} (x, 0)]_{x = 0} = \partial_{x} (x) = + 1$ . Why this step? Swapping $x \leftrightarrow y$ turns the factor $x^{2} - y^{2}$ into $x^{2} - k^{2}$ with the opposite overall sign pattern — that built-in sign flip is exactly what makes $f_{y} (x, 0) = + x$ where $f_{x} (0, y) = - y$ , and hence the two orders come out with opposite signs.

Verify: $f_{x y} (0, 0) = - 1 \neq = f_{y x} (0, 0) = 1$ ✅. As the figure shows, the surface has a pinched twist at the origin — the mixed partial is not continuous there, so Clairaut's admission price is unpaid. See Continuity of Multivariable Functions for why that discontinuity is the real culprit.

Cell H — Exam twist: solve for a constant

Before the example, one piece of vocabulary so nothing is unearned:

Worked example Example 8 — reverse-engineer equality

For which constant $a$ is the pair $(P, Q) = (a x y^{2}, x^{2} y)$ an exact pair — i.e. the two slopes $P = f_{x}, Q = f_{y}$ of some single function $f (x, y)$ ?

Forecast: a parent $f$ exists only if the "twist" is consistent. Which theorem enforces that?

Apply the exactness test. If $P = f_{x}$ and $Q = f_{y}$ , Clairaut forces $P_{y} = f_{x y} = f_{y x} = Q_{x}$ . So we need $P_{y} = Q_{x}$ . Why this step? Clairaut is the gatekeeper: a valid parent $f$ can only exist when $\partial_{y} P = \partial_{x} Q$ .
Compute both sides: $P_{y} = \partial_{y} (a x y^{2}) = 2 a x y$ ; $Q_{x} = \partial_{x} (x^{2} y) = 2 x y$ . Why this step? Differentiate $P$ in $y$ ( $y^{2} \to 2 y$ ) and $Q$ in $x$ ( $x^{2} \to 2 x$ ) to get the two "twists" we must match.
Set equal and solve: $2 a x y = 2 x y$ for all $x, y \Rightarrow 2 a = 2 \Rightarrow a = 1$ . Why this step? The equation must hold for every $(x, y)$ , so match the coefficient of the identical monomial $x y$ : $2 a = 2$ .
Build the parent $f$ to confirm. With $a = 1$ we need $f_{x} = x y^{2}$ and $f_{y} = x^{2} y$ . The function $f = \frac{1}{2} x^{2} y^{2}$ works: $f_{x} = \partial_{x} (\frac{1}{2} x^{2} y^{2}) = x y^{2}$ ✅ and $f_{y} = \partial_{y} (\frac{1}{2} x^{2} y^{2}) = x^{2} y$ ✅. Why this step? Producing an actual parent $f$ proves the pair really is exact, not just that the test passed.

Verify: with $a = 1$ , $P_{y} = Q_{x} = 2 x y$ ✅ and the parent $f = \frac{1}{2} x^{2} y^{2}$ regenerates both slopes ✅. The answer is $a = 1$ . This is why Clairaut underpins both the Hessian Matrix being symmetric and the exactness test for Higher Order Partial Derivatives.

Recall One-line recap of every cell

Poly (A) — trig/chain (B) — cheaper order (C) — higher order counts (D) — separable sum gives $0$ (E) — physical twist with units (F) — discontinuous ⇒ $- 1 \neq = 1$ (G) — solve for $a$ via $P_{y} = Q_{x}$ (H).

Active recall

Ex 5 discovered: what are all mixed partials of a separable sum

g (x) + h (y)

All zero.

In the failure case, what are

f_{x y} (0, 0)

and

f_{y x} (0, 0)

- 1

and

+ 1

In Ex 8, which constant makes

(a x y^{2}, x^{2} y)

an exact pair?

a = 1

(from

P_{y} = Q_{x}

Why can Ex 7 break Clairaut even though both partials exist?

The mixed partials are not continuous at the origin.

What does it mean for

(P, Q)

to be an "exact" pair?

There is one function

f

with

P = f_{x}

and

Q = f_{y}

; the test is

P_{y} = Q_{x}