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

What is a mixed partial?

Definition Notation

For $f(x,y)$ :

$f_x = \dfrac{\partial f}{\partial x}$ : differentiate w.r.t. $x$ , hold $y$ fixed.
$f_{xy} = \dfrac{\partial}{\partial y}\!\left(\dfrac{\partial f}{\partial x}\right) = \dfrac{\partial^2 f}{\partial y\,\partial x}$ : first $x$ , then $y$ .
$f_{yx} = \dfrac{\partial}{\partial x}\!\left(\dfrac{\partial f}{\partial y}\right) = \dfrac{\partial^2 f}{\partial x\,\partial y}$ : first $y$ , then $x$ .

⚠️ Subscript order reads left to right ( $f_{xy}$ = $x$ first), but Leibniz order reads right to left ( $\partial^2 f/\partial y\,\partial x$ = $x$ first). They mean the same thing.

Deriving it from first principles

Step 1 — Build a symmetric second difference. Fix step sizes $h,k>0$ . Define $\Delta(h,k) = f(a+h,\,b+k) - f(a+h,\,b) - f(a,\,b+k) + f(a,\,b).$ Why this step? This combination is the discrete analogue of a mixed second derivative. Notice it is perfectly symmetric: swapping $h\leftrightarrow k$ and $a\leftrightarrow b$ (i.e. swapping the roles of $x$ and $y$ ) leaves $\Delta$ unchanged. That hidden symmetry is the whole trick.

Step 2 — Group it as an $x$ -difference of a $y$ -difference. Let $g(x) = f(x, b+k) - f(x, b)$ . Then $\Delta(h,k) = g(a+h) - g(a).$ Why this step? We've turned $\Delta$ into a difference of $g$ in the $x$ -direction, so we can apply the Mean Value Theorem in $x$ .

Step 3 — Apply MVT in $x$ . By the MVT there is $c_1$ between $a$ and $a+h$ with $g(a+h)-g(a) = h\,g'(c_1) = h\big[f_x(c_1,b+k) - f_x(c_1,b)\big].$ Why this step? $g'(x) = f_x(x,b+k)-f_x(x,b)$ , and MVT converts a difference into a derivative times the step $h$ .

Step 4 — Apply MVT again, now in $y$ . The bracket is a $y$ -difference of $f_x(c_1,\cdot)$ . By MVT there is $d_1$ between $b$ and $b+k$ with $f_x(c_1,b+k)-f_x(c_1,b) = k\,f_{xy}(c_1,d_1).$ So $\Delta(h,k) = h\,k\,f_{xy}(c_1,d_1).$ Why this step? Differentiating $f_x$ w.r.t. $y$ gives $f_{xy}$ — exactly the mixed partial we want.

Step 5 — Do Steps 2–4 in the other order. Now let $\phi(y) = f(a+h,y)-f(a,y)$ , so $\Delta(h,k)=\phi(b+k)-\phi(b)$ . The same $\Delta$ . Two MVTs (first in $y$ , then in $x$ ) give, for some $c_2,d_2$ , $\Delta(h,k) = h\,k\,f_{yx}(c_2,d_2).$ Why this step? The symmetry of $\Delta$ means we are free to peel it apart in either order.

Step 6 — Equate and take the limit. Both expressions equal the same $\Delta$ , so dividing by $hk$ : $f_{xy}(c_1,d_1) = f_{yx}(c_2,d_2).$ As $h,k\to 0$ , all four points $(c_1,d_1),(c_2,d_2)\to(a,b)$ . Because $f_{xy}$ and $f_{yx}$ are continuous, both sides converge to their values at $(a,b)$ : $\boxed{f_{xy}(a,b) = f_{yx}(a,b).}$ Why continuity is essential: the points $c_i,d_i$ are unknown; we only know they squeeze toward $(a,b)$ . Continuity is what lets "value at a nearby point" become "value at $(a,b)$ ." Without it, the limit can fail — see the famous counterexample below.

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

Worked examples

Worked example Example 1 — Direct verification

Let $f(x,y)=x^3 y^2 + \sin(xy)$ . Show $f_{xy}=f_{yx}$ .

$x$ first: $f_x = 3x^2y^2 + y\cos(xy)$ . Why? Treat $y$ as constant; $\frac{d}{dx}\sin(xy)=y\cos(xy)$ by chain rule. Then $f_{xy} = \partial_y(3x^2y^2 + y\cos(xy)) = 6x^2 y + \cos(xy) - xy\sin(xy)$ . Why? $\partial_y[y\cos(xy)] = \cos(xy) + y\cdot(-\sin(xy))\cdot x$ (product + chain rule).

$y$ first: $f_y = 2x^3 y + x\cos(xy)$ . Then $f_{yx} = \partial_x(2x^3y + x\cos(xy)) = 6x^2 y + \cos(xy) - xy\sin(xy)$ . ✅ Equal.

The counterexample (when it FAILS)

Common mistake "Mixed partials are ALWAYS equal." — Steel-manning the error

Why it feels right: every polynomial, exponential, trig function you ever differentiate satisfies it. The theorem is so reliable that students assume it's a law of algebra with no conditions. The fix — a function where $f_{xy}(0,0)\neq f_{yx}(0,0)$ : $f(x,y) = \begin{cases} \dfrac{xy(x^2-y^2)}{x^2+y^2}, & (x,y)\neq(0,0)\\[2mm] 0, & (x,y)=(0,0)\end{cases}$ Careful computation gives $f_{xy}(0,0) = -1$ but $f_{yx}(0,0)= +1$ . The mixed partials exist everywhere but are not continuous at the origin — exactly the hypothesis that fails. So Clairaut's continuity condition is genuinely necessary, not decorative.

Recall Feynman: explain to a 12-year-old

Imagine a stretchy trampoline. Standing on it, ask: "If I walk East, how fast does the ground tilt?" That's an $x$ -slope. Now ask "and as I shift North, how does that East-tilt change?" That's $f_{xy}$ — a twist. Clairaut says the twist you measure (East-then-North) is the same twist you'd measure North-then-East. It's like saying the corner of the trampoline is dented the same amount no matter which way you sneak up on it — as long as the trampoline is smooth with no sharp creases. If there's a pinch or crease (a point where the slopes jump), the two ways can disagree.

Active recall

What does Clairaut's theorem conclude?

f_{xy}=f_{yx}

— mixed second partials are equal.

What is the key HYPOTHESIS of Clairaut's theorem?

The mixed partials are continuous on an open region around the point.

f_{xy}

, which variable is differentiated first?

x

(subscripts read left to right).

In Leibniz form

\partial^2 f/\partial y\,\partial x

, which is first?

x

(operator nearest

f

acts first).

What single symmetric quantity is used to prove Clairaut's theorem?

The second difference

\Delta = f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)

Why does the proof need continuity of the mixed partials?

The MVT points are unknown; continuity lets their limit values become the value at

(a,b)

Give a function where mixed partials differ at the origin.

f=xy(x^2-y^2)/(x^2+y^2)

(and

0

at origin):

f_{xy}(0,0)=-1\neq f_{yx}(0,0)=1

For higher derivatives like

f_{xxy}

, what determines the value?

Only the count of each variable (here two

x

's, one

y

), not the order — under smoothness.

Compute

f_{xy}

for

f=x^2y^3

\partial_y(2xy^3)=6xy^2

What twice-applied tool drives the proof?

The Mean Value Theorem (once in

x

, once in

y

Connections

Partial Derivatives — the building blocks $f_x,f_y$ .
Mean Value Theorem — the engine of the proof.
Hessian Matrix — Clairaut makes the Hessian symmetric, key for the second-derivative test.
Exact Differential Equations — exactness condition $\partial_y M=\partial_x N$ relies on equal mixed partials.
Continuity of Multivariable Functions — the hypothesis that makes it all work.
Higher Order Partial Derivatives — where iterated Clairaut lets us reorder freely.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Clairaut's theorem ek bahut hi pyaari aur kaam ki cheez hai. Jab tumhare paas do variable wala function $f(x,y)$ hota hai, toh tum usko pehle $x$ se differentiate kar sakte ho, phir $y$ se — ya phir ulta, pehle $y$ phir $x$ . Theorem kehta hai: dono ka answer same aayega, yaani $f_{xy} = f_{yx}$ . Bas ek condition hai — second mixed partials continuous honi chahiye us point ke aas-paas. Continuity hi entry ticket hai.

Intuition kya hai? Socho ek trampoline (surface) hai. $f_x$ matlab "East walk karte hue zameen kitni tilt hoti hai." Ab $f_{xy}$ matlab "jab main North shift karta hoon toh wo East-tilt kitna badalta hai" — ye ek twist (marod) hai. Clairaut kehta hai ki agar surface smooth hai (koi sharp crease nahi), toh twist East-phir-North napo ya North-phir-East, same niklega. Practical fayda: exam mein jis order mein algebra aasan ho, usi order mein differentiate karo — answer wahi rahega (Example 2 dekho).

Proof ka asli trick ek symmetric quantity $\Delta$ banana hai — ek rectangle ke chaaron corner ka combination. Ye $\Delta$ $x$ aur $y$ mein bilkul symmetric hai. Phir Mean Value Theorem do baar lagao ek order mein toh $f_{xy}$ milta hai, dusre order mein toh $f_{yx}$ . Kyunki $\Delta$ ek hi hai, dono barabar. Limit lete waqt continuity use karke hum points ko exact $(a,b)$ pe le jaate hain.

Aur haan, ye theorem hamesha sach nahi hota! Ek famous counterexample hai $f = xy(x^2-y^2)/(x^2+y^2)$ jahan origin pe $f_{xy}=-1$ par $f_{yx}=+1$ — kyunki wahan mixed partials continuous nahi hain. Isliye yaad rakho: No continuity, no Clairaut. Yehi condition hi pura khel hai.

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

What is a mixed partial?

Deriving it from first principles

Worked examples

The counterexample (when it FAILS)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Multivariable Calculus

Connections