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

4.4.4 · D3 · Maths › Multivariable Calculus › Clairaut's theorem — mixed partials are equal (under conditi

Scenario matrix

Mixed partials ke baare mein har problem in cells mein se kisi ek mein aati hai. Hum har cell mein kam se kam ek example karte hain.

#	Case class	Tricky kya hai	Jo example hit karta hai
A	Polynomial (sab signs, positive powers)	sirf bookkeeping	Ex 1
B	Product with trig / chain rule	product + chain interact karte hain	Ex 2
C	"Kaun sa order sasta hai?" (80/20 payoff)	aasan raasta chuno	Ex 3
D	Higher order ( $f_{xx y} = f_{y xx}$ )	gino, order mat karo	Ex 4
E	Zero / degenerate input (ek variable missing)	ek partial $0$ ho jaata hai	Ex 5
F	Word problem (real-world twist, units)	physics ↔ math translate karo	Ex 6
G	Failure case — discontinuous mixed partial	Clairaut apply nahin hota	Ex 7
H	Exam twist — "ek constant dhundo taaki equality hold kare"	ek function ko reverse-engineer karo	Ex 8

Shuru karne se pehle, ek reminder seedhe shabdon mein taaki koi symbol unearned na rahe:

Cell A — Pure polynomial (sab signs)

Worked example Example 1 — signs aur powers

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

Forecast: abhi andaaza lagao — kya dono orders match karenge? (Karenge hi, kyunki polynomials har jagah smooth hote hain.)

$f_{x}$ ( $y$ freeze karo): $f_{x} = - 8 x^{3} y^{3} + 5 y^{2}$ . Yeh step kyun? Har term ko $x$ mein differentiate karo; akela $- y^{5}$ mein koi $x$ nahin hai, isliye woh vanish ho jaata hai — yeh pehli baar hai jab ek term drop out hoti hai.
$f_{x y}$ (ab us ko $y$ mein differentiate karo): $f_{x y} = - 24 x^{3} y^{2} + 10 y$ . Yeh step kyun? $\partial_{y} (- 8 x^{3} y^{3}) = - 24 x^{3} y^{2}$ aur $\partial_{y} (5 y^{2}) = 10 y$ .
$f_{y}$ ( $x$ freeze karo): $f_{y} = - 6 x^{4} y^{2} + 10 x y - 5 y^{4}$ . Yeh step kyun? Doosra order — pehle original ko $y$ mein differentiate karo.
$f_{y x}$ : $f_{y x} = - 24 x^{3} y^{2} + 10 y$ . Yeh step kyun? $\partial_{x} (- 6 x^{4} y^{2}) = - 24 x^{3} y^{2}$ , $\partial_{x} (10 x y) = 10 y$ , aur $- 5 y^{4}$ drop ho jaata hai.

Verify: $f_{x y} = f_{y x} = - 24 x^{3} y^{2} + 10 y$ ✅. $(1, 1)$ par sanity spot-check: dono $- 24 + 10 = - 14$ dete hain.

Cell B — Product with trig (chain rule andar)

Worked example Example 2 — twist meets chain rule

Maano $f (x, y) = x^{2} cos (x y)$ . Dikhao ki dono orders agree karte hain.

Forecast: ek product aur ek chain rule ke saath, lagta hai dono orders alag ho jaayenge — padhne se pehle apne gut pe bet lagao.

$f_{x}$ : $f_{x} = 2 x cos (x y) + x^{2} \cdot (- sin (x y)) \cdot y = 2 x cos (x y) - x^{2} y sin (x y)$ . Yeh step kyun? $x^{2}$ aur $cos (x y)$ par product rule; chain rule se $\partial_{x} cos (x y) = - y sin (x y)$ milta hai.
$f_{x y}$ : $y$ mein differentiate karo.
- $\partial_{y} [2 x cos (x y)] = - 2 x^{2} sin (x y)$ .
- $\partial_{y} [- x^{2} y sin (x y)] = - x^{2} sin (x y) - x^{3} y cos (x y)$ (product $y \cdot sin$ , chain on $sin$ ).
- Sum: $f_{x y} = - 3 x^{2} sin (x y) - x^{3} y cos (x y)$ .
$f_{y}$ : $f_{y} = x^{2} \cdot (- sin (x y)) \cdot x = - x^{3} sin (x y)$ . Yeh step kyun? $x$ freeze karo; $x^{2}$ ab ek constant multiplier hai.
$f_{y x}$ : $\partial_{x} [- x^{3} sin (x y)] = - 3 x^{2} sin (x y) - x^{3} y cos (x y)$ . Yeh step kyun? $x^{3}$ aur $sin (x y)$ par product rule; chain se $\partial_{x} sin (x y) = y cos (x y)$ milta hai.

Verify: dono $- 3 x^{2} sin (x y) - x^{3} y cos (x y)$ ke barabar hain ✅. $(1, 0)$ par: $sin 0 = 0, cos 0 = 1$ , dono $0 - 1 \cdot 0 \cdot 1 = 0$ dete hain.

Cell C — Kaun sa order sasta hai? (80/20)

Worked example Example 3 — jaanbujhkar aasaan raasta chuno

$f (x, y) = \frac{ln ( 1 + y ^{2} )}{x}$ ke liye $f_{x y}$ dhundo, phir doosre order se confirm karo.

Forecast: kaun sa order messy $ln$ derivative ko sabse zyada der tak avoid karta hai — pehle $x$ ya pehle $y$ ?

Sasta raasta — pehle $y$ (kyunki $x$ sirf $1/ x$ ke roop mein aata hai, jo tame part hai): $f_{y} = \frac{1}{x} \cdot \frac{2 y}{1 + y ^{2}} = \frac{2 y}{x ( 1 + y ^{2} )}$ . Yeh step kyun? $\partial_{y} ln (1 + y^{2}) = \frac{2 y}{1 + y ^{2}}$ chain rule se; $1/ x$ saath chalta hai.
Phir $x$ : $f_{y x} = \partial_{x} [\frac{2 y}{1 + y ^{2}} \cdot x^{- 1}] = - \frac{2 y}{x ^{2} ( 1 + y ^{2} )}$ . Yeh step kyun? $\partial_{x} x^{- 1} = - x^{- 2}$ ; $y$ -block yahan ek constant hai.
Mahanga raasta — pehle $x$ (yeh dikhane ke liye ki dono orders literally match karte hain): $f_{x} = \partial_{x} [ln (1 + y^{2}) \cdot x^{- 1}] = - \frac{ln ( 1 + y ^{2} )}{x ^{2}}$ . Yeh step kyun? Ab poora $ln (1 + y^{2})$ ek constant multiplier hai; sirf $x^{- 1}$ move karta hai.
Phir $y$ : $f_{x y} = \partial_{y} [- \frac{ln ( 1 + y ^{2} )}{x ^{2}}] = - \frac{1}{x ^{2}} \cdot \frac{2 y}{1 + y ^{2}} = - \frac{2 y}{x ^{2} ( 1 + y ^{2} )}$ . Yeh step kyun? Yahaan hum price pay karte hain — is raaste mein hum $ln$ differentiate karte hain. Yeh bilkul Step 2 jaise expression par land karta hai, confirm karta hai $f_{x y} = f_{y x}$ .

Verify: dono rastey $- \frac{2 y}{x ^{2} ( 1 + y ^{2} )}$ dete hain ✅. $(2, 1)$ par: $- \frac{2 \cdot 1}{4 \cdot 2} = - \frac{1}{4}$ . Payoff: koi bhi order kaam karta hai, isliye exam mein woh raasta chuno (Step 1–2) jo $ln$ derivative ko tab tak nahin chhuta jab tak mazburi na ho.

Cell D — Higher order: gino, order mat karo

Worked example Example 4 — teen derivatives, teen orders

$f (x, y) = x^{3} y^{4}$ ke liye dikhao ki $f_{xx y} = f_{x y x} = f_{y xx}$ aur value batao.

Forecast: teeno agree karenge — theorem kehta hai sirf kitne $x$ 's aur $y$ 's hain yeh matter karta hai (yahaan: do $x$ 's, ek $y$ ).

$f_{xx y}$ (order $x, x, y$ ): pehle $f_{x} = \partial_{x} (x^{3} y^{4}) = 3 x^{2} y^{4}$ ; phir $f_{xx} = \partial_{x} (3 x^{2} y^{4}) = 3 \cdot 2 x^{2 - 1} y^{4} = 6 x y^{4}$ ; phir $f_{xx y} = \partial_{y} (6 x y^{4}) = 6 x \cdot 4 y^{3} = 24 x y^{3}$ . Yeh step kyun? Har $x$ -step $x$ -power ko ek kam karta hai aur purani power se multiply karta hai ( $x^{3} \to 3 x^{2} \to 6 x$ ), $y^{4}$ ko frozen rakhte hue; sirf aakhri step single $y$ -derivative kharach karta hai ( $y^{4} \to 4 y^{3}$ ).
$f_{x y x}$ (order $x, y, x$ ): $f_{x} = 3 x^{2} y^{4}$ , phir $f_{x y} = \partial_{y} (3 x^{2} y^{4}) = 12 x^{2} y^{3}$ , phir $f_{x y x} = \partial_{x} (12 x^{2} y^{3}) = 24 x y^{3}$ . Yeh step kyun? Wahi pehla $x$ -step, lekin ab hum $y$ ko beech mein hit karte hain ( $y^{4} \to 4 y^{3}$ se $12 x^{2} y^{3}$ milta hai) aur $x$ se finish karte hain ( $x^{2} \to 2 x$ se $24 x y^{3}$ milta hai).
$f_{y xx}$ (order $y, x, x$ ): $f_{y} = \partial_{y} (x^{3} y^{4}) = 4 x^{3} y^{3}$ , phir $f_{y x} = \partial_{x} (4 x^{3} y^{3}) = 12 x^{2} y^{3}$ , phir $f_{y xx} = \partial_{x} (12 x^{2} y^{3}) = 24 x y^{3}$ . Yeh step kyun? $y$ ko pehle hit karo ( $y^{4} \to 4 y^{3}$ ), phir $x$ mein do baar differentiate karo ( $x^{3} \to 3 x^{2}$ , phir $x^{2} \to 2 x$ ).
Pattern: har raasta kab hum single $y$ -derivative kharach karte hain yeh rearrange karta hai, lekin har raasta ek $y$ aur do $x$ 's kharach karta hai — isliye Clairaut ke hisaab se sab ek hi jagah land karte hain.

Verify: teeno $= 24 x y^{3}$ ✅. $(1, 1)$ par: $24$ .

Cell E — Zero / degenerate input

Worked example Example 5 — jab ek variable missing ho

Maano $f (x, y) = 7 x^{2} + cos (3 x) - e^{2 y}$ . $f_{x y}$ aur $f_{y x}$ compute karo.

Forecast: kuch notice karo — koi bhi term dono $x$ aur $y$ contain nahin karta. Ek "separable" sum ka mixed partial kya hona chahiye?

$f_{x} = 14 x - 3 sin (3 x)$ — sirf $x$ par depend karta hai. Yeh step kyun? $y$ freeze karo: term $- e^{2 y}$ mein koi $x$ nahin hai, isliye drop ho jaata hai; $x$ -terms differentiate hokar $14 x$ aur $- 3 sin (3 x)$ bante hain. Dhyan do ki result mein koi $y$ nahin — yeh seed hai jo aage hoga.
$f_{x y} = \partial_{y} (14 x - 3 sin (3 x)) = 0$ . Yeh step kyun? $f_{x}$ sirf $x$ ka function nikla. Jis expression mein koi $y$ nahin hota, uski $y$ ke saath rate of change zero hoti hai — isliye uski $y$ -derivative exactly $0$ hai.
$f_{y} = - 2 e^{2 y}$ — sirf $y$ par depend karta hai. Yeh step kyun? $x$ freeze karo: har $x$ -only term ( $7 x^{2}$ , $cos 3 x$ ) ab constant ho jaati hai aur drop ho jaati hai; sirf $- e^{2 y}$ bachti hai, giving $- 2 e^{2 y}$ . Phir se result mein koi $x$ nahin.
$f_{y x} = \partial_{x} (- 2 e^{2 y}) = 0$ . Yeh step kyun? $f_{y}$ sirf $y$ ka function hai, isliye isse $x$ mein differentiate karne par $0$ milta hai — Step 2 ka mirror image.

Verify: $f_{x y} = f_{y x} = 0$ ✅. Tumne jo general law discover kiya: agar $f (x, y) = g (x) + h (y)$ (ek separable sum) hai, toh ek variable ka pehla partial doosre par saari dependence khatam kar deta hai, isliye har mixed partial $0$ hota hai. Yeh matrix ka degenerate/zero cell hai — Clairaut trivially hold karta hai ( $0 = 0$ ).

Cell F — Units ke saath word problem

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

Ek metal plate ka temperature $T (x, y) = 40 + 3 x^{2} y - 2 x y^{2}$ hai ( $^{\circ} C$ mein, $x, y$ metres mein). Quantity $T_{x y}$ measure karta hai East–West temperature-slope kitna change hota hai jab tum North walk karo. Isse do tarike se compute karo aur interpret karo.

Forecast: dono orders physically alag-alag lagte sawaal describe karte hain ("East-slope ka Northward change" vs "North-slope ka Eastward change"). Kya inhe same number dena chahiye?

$T_{x} = 6 x y - 2 y^{2}$ — East–West slope, $^{\circ} C / m$ mein. Yeh step kyun? $y$ freeze karo: constant $40$ drop ho jaata hai, $\partial_{x} (3 x^{2} y) = 6 x y$ , $\partial_{x} (- 2 x y^{2}) = - 2 y^{2}$ . Yeh temperature gradient hai jab tum East step karte ho.
$T_{x y} = \partial_{y} (6 x y - 2 y^{2}) = 6 x - 4 y$ , $^{\circ} C / m^{2}$ mein. Yeh step kyun? Ab us East-slope ko $y$ mein differentiate karo: $\partial_{y} (6 x y) = 6 x$ , $\partial_{y} (- 2 y^{2}) = - 4 y$ . Yeh woh rate hai jis par East-slope change hota hai per metre North chalne par.
Doosra order — $T_{y} = 3 x^{2} - 4 x y$ , $^{\circ} C / m$ mein. Yeh step kyun? $x$ freeze karo: $40$ drop ho jaata hai, $\partial_{y} (3 x^{2} y) = 3 x^{2}$ , $\partial_{y} (- 2 x y^{2}) = - 4 x y$ . Yeh North–South slope hai.
$T_{y x} = \partial_{x} (3 x^{2} - 4 x y) = 6 x - 4 y$ . Yeh step kyun? North-slope ko $x$ mein differentiate karo: $\partial_{x} (3 x^{2}) = 6 x$ , $\partial_{x} (- 4 x y) = - 4 y$ . Kyunki $T$ ek smooth polynomial hai, Clairaut force karta hai ki yeh Step 2 ke barabar ho — aur hota hai.

Verify: dono $= 6 x - 4 y$ ✅. Point $(2 m, 1 m)$ par: $6 (2) - 4 (1) = 8^{\circ} C / m^{2}$ . Temperature field ka twist $8$ hai chahe koi bhi walk karo — Clairaut ka physical meaning.

Cell G — FAILURE case (Clairaut apply nahin hota)

Yahaan hum woh ek jagah dekhte hain jahan dono orders disagree karte hain — kyunki hypothesis (mixed partial ki continuity) toot jaati hai. Neeche figure surface dikhata hai: isse algebra se pehle padho taaki tum dekh sako ki trouble kahaan hai.

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

Figure (Cell G): surface $f (x, y) = x y (x^{2} - y^{2}) / (x^{2} + y^{2})$ . Teal curve $x$ -axis ke saath slice hai ( $y = 0$ ); plum curve $y$ -axis ke saath slice hai ( $x = 0$ ). Orange dot origin $O$ hai, jahan surface pinched hai — mixed partials exactly wahan continuous hona fail karte hain, aur wahi pinch hai jo dono orders ko disagree karne deta hai.

Worked example Example 7 — origin par mixed partials differ karte hain

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

Forecast: har pehla example match hua. Andaaza lagao ki kya yeh hoga — aur origin kyun special hai (figure mein pinch dekho).

Definition se $f_{x} (0, y)$ dhundo. Definition hai: $f_{x} (0, y) = lim_{h \to 0} \frac{f ( 0 + h , y ) - f ( 0 , y )}{h} .$ Yeh step kyun? Pinch ke paas hum shortcut rules par trust nahin kar sakte, isliye hum raw definition par wapas aate hain: "slope in $x$ = rise over run jab run $h$ $0$ ki taraf shrink hota hai."
Formula plug in karo. $y \neq = 0$ ke liye, $f (0, y) = 0$ (numerator mein factor $x = 0$ hai), aur $f (h, y) = \frac{h y ( h ^{2} - y ^{2} )}{h ^{2} + y ^{2}} .$ Toh difference quotient hai: $\frac{f ( h , y ) - 0}{h} = \frac{y ( h ^{2} - y ^{2} )}{h ^{2} + y ^{2}} .$ Yeh step kyun? $h$ se divide karne par numerator mein $h$ ki single power cancel ho jaati hai — wahi cancellation hai jis wajah se limit cleanly exist karegi.
$h \to 0$ lo. Jab $h \to 0$ , $h^{2} \to 0$ , isliye $f_{x} (0, y) = \frac{y ( 0 - y ^{2} )}{0 + y ^{2}} = \frac{- y ^{3}}{y ^{2}} = - y .$ Yeh step kyun? Run ko zero tak shrink karne dene se exact slope milta hai; messy quotient tidy function $f_{x} (0, y) = - y$ mein collapse ho jaata hai.
Use ko $y$ mein differentiate karo: $f_{x y} (0, 0) = \partial_{y} [f_{x} (0, y)]_{y = 0} = \partial_{y} (- y) = - 1$ . Yeh step kyun? $f_{x y}$ ka matlab hai "pehle $x$ karo (Steps 1–3), phir $y$ " — aur $- y$ ki $y$ -derivative $- 1$ hai.
$x, y$ ki roles swap karke repeat karo — limit puri tarah likh ke. Definition se, $f_{y} (x, 0) = lim_{k \to 0} \frac{f ( x , 0 + k ) - f ( x , 0 )}{k} .$ $x \neq = 0$ ke liye, $f (x, 0) = 0$ (numerator mein factor $y = 0$ hai), aur $f (x, k) = \frac{x k ( x ^{2} - k ^{2} )}{x ^{2} + k ^{2}}, isliye \frac{f ( x , k ) - 0}{k} = \frac{x ( x ^{2} - k ^{2} )}{x ^{2} + k ^{2}} .$ $k \to 0$ lete hue (toh $k^{2} \to 0$ ): $f_{y} (x, 0) = \frac{x ( x ^{2} - 0 )}{x ^{2} + 0} = \frac{x ^{3}}{x ^{2}} = + x .$ Phir $f_{y x} (0, 0) = \partial_{x} [f_{y} (x, 0)]_{x = 0} = \partial_{x} (x) = + 1$ . Yeh step kyun? $x \leftrightarrow y$ swap karne par factor $x^{2} - y^{2}$ ka $x^{2} - k^{2}$ ban jaata hai opposite overall sign pattern ke saath — wahi built-in sign flip hai jo $f_{y} (x, 0) = + x$ banata hai jahan $f_{x} (0, y) = - y$ tha, aur isliye dono orders opposite signs ke saath aate hain.

Verify: $f_{x y} (0, 0) = - 1 \neq = f_{y x} (0, 0) = 1$ ✅. Jaise figure dikhata hai, surface mein origin par ek pinched twist hai — mixed partial wahan continuous nahin hai, isliye Clairaut ki admission price unpaid hai. Dekho Continuity of Multivariable Functions kyun woh discontinuity asli culprit hai.

Cell H — Exam twist: ek constant ke liye solve karo

Example se pehle, vocabulary ka ek piece taaki kuch unearned na rahe:

Worked example Example 8 — equality reverse-engineer karo

Kin constant $a$ ke liye pair $(P, Q) = (a x y^{2}, x^{2} y)$ ek exact pair hai — yaani kisi ek function $f (x, y)$ ke do slopes $P = f_{x}, Q = f_{y}$ ?

Forecast: parent $f$ tabhi exist karta hai jab "twist" consistent ho. Kaun sa theorem yeh enforce karta hai?

Exactness test apply karo. Agar $P = f_{x}$ aur $Q = f_{y}$ , toh Clairaut force karta hai $P_{y} = f_{x y} = f_{y x} = Q_{x}$ . Isliye hume $P_{y} = Q_{x}$ chahiye. Yeh step kyun? Clairaut ek gatekeeper hai: ek valid parent $f$ tabhi exist kar sakta hai jab $\partial_{y} P = \partial_{x} Q$ .
Dono sides compute karo: $P_{y} = \partial_{y} (a x y^{2}) = 2 a x y$ ; $Q_{x} = \partial_{x} (x^{2} y) = 2 x y$ . Yeh step kyun? $P$ ko $y$ mein differentiate karo ( $y^{2} \to 2 y$ ) aur $Q$ ko $x$ mein ( $x^{2} \to 2 x$ ) do "twists" pane ke liye jo match karne chahiye.
Equal set karo aur solve karo: $2 a x y = 2 x y$ sabhi $x, y$ ke liye $\Rightarrow 2 a = 2 \Rightarrow a = 1$ . Yeh step kyun? Equation har $(x, y)$ ke liye hold karni chahiye, isliye identical monomial $x y$ ka coefficient match karo: $2 a = 2$ .
Confirm karne ke liye parent $f$ banao. $a = 1$ ke saath hume $f_{x} = x y^{2}$ aur $f_{y} = x^{2} y$ chahiye. Function $f = \frac{1}{2} x^{2} y^{2}$ kaam karta hai: $f_{x} = \partial_{x} (\frac{1}{2} x^{2} y^{2}) = x y^{2}$ ✅ aur $f_{y} = \partial_{y} (\frac{1}{2} x^{2} y^{2}) = x^{2} y$ ✅. Yeh step kyun? Ek actual parent $f$ produce karna prove karta hai ki pair sach mein exact hai, na sirf ki test pass hua.

Verify: $a = 1$ ke saath, $P_{y} = Q_{x} = 2 x y$ ✅ aur parent $f = \frac{1}{2} x^{2} y^{2}$ dono slopes regenerate karta hai ✅. Answer hai $a = 1$ . Yahi wajah hai ki Clairaut Hessian Matrix ke symmetric hone aur Higher Order Partial Derivatives ke exactness test ko underpin karta hai.

Recall Har cell ka one-line recap

Poly (A) — trig/chain (B) — sasta order (C) — higher order counts (D) — separable sum $0$ deta hai (E) — units ke saath physical twist (F) — discontinuous ⇒ $- 1 \neq = 1$ (G) — $P_{y} = Q_{x}$ se $a$ ke liye solve karo (H).

Active recall

Ex 5 mein discover hua: separable sum

g (x) + h (y)

ke saare mixed partials kya hote hain?

Sab zero.

In the failure case,

f_{x y} (0, 0)

aur

f_{y x} (0, 0)

kya hain?

- 1

aur

+ 1

Ex 8 mein, kaun sa constant

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

ko exact pair banata hai?

a = 1

(

P_{y} = Q_{x}

se).

Ex 7 Clairaut kyun break kar sakta hai jabki dono partials exist karte hain?

Mixed partials origin par continuous nahin hain.

(P, Q)

ke "exact" pair hone ka kya matlab hai?

Ek function

f

exist karta hai jisme

P = f_{x}

aur

Q = f_{y}

; test hai

P_{y} = Q_{x}