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

4.4.4 · D4 · Maths › Multivariable Calculus › Clairaut's theorem — mixed partials are equal (under conditi

Poore page mein jo do symbols use honge, unka ek quick reminder:

Neeche di gayi picture ka matlab yahi hai: ek mixed partial surface ka ek twist hai, aur Clairaut kehta hai ki twist same hai chahe jis taraf se bhi aao.

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

Figure 1 — $x y$ -plane ke upar saddle surface $z = x y$ . East chalo (blue) toh zameen tilts; ab North drift karo aur dekho East-tilt rotate hoti hai — woh rotation hai $f_{x y}$ . Green path (pehle North, phir East) same corner ki twisting measure karta hai. Clairaut: jab surface mein koi crease nahi hoti tab dono twists agree karte hain.

Definition Do words jo baar baar aayenge

$C^{2}$ (padho "see-two") shorthand hai "saare second partial derivatives exist karte hain aur continuous hain." Jo function ek region pe $C^{2}$ hai woh Clairaut ke liye kaafi smooth hai. Jab bhi is page pe " $C^{2}$ " likha ho, mentally replace karo "mixed partials yahaan continuous hain."
Open region = plane ka ek patch jisme koi boundary edge include nahi hoti (jaise ek disk ka andar wala hissa). Clairaut ko sirf ek open region chahiye us single point ke aas paas. Ek aur zyada strong idea, simply connected (ek open region jisme koi holes nahi — tum kisi bhi loop ko ek point pe shrink kar sako), Clairaut ke liye zaruri nahi hai; woh sirf baad mein exact equations waale L4 mein aata hai. Hum yeh distinction abhi flag kar rahe hain taaki baad mein surprise na ho:
- Clairaut ( $f_{x y} = f_{y x}$ ): sirf continuity on an open region chahiye.
- Exactness ka potential hona: $M_{y} = N_{x}$ aur simply connected domain chahiye.

Level 1 — Recognition

L1.1 Sach ya jhooth, aur kyun: " $f (x, y) = x^{4} y^{7}$ ke liye, hum guarantee kar sakte hain ki $f_{x y} = f_{y x}$ bina kuch compute kiye."

L1.2 Symbol $f_{y x}$ mein, kaun sa variable pehle differentiate hota hai? Aur Leibniz form $\frac{\partial ^{2} f}{\partial x \partial y}$ mein, kaun sa pehle hai?

L1.3 Clairaut's theorem ki ek hypothesis apne words mein batao.

Recall Solution L1.1

Sach. $f (x, y) = x^{4} y^{7}$ ek polynomial hai. Polynomial ka har partial derivative phir se polynomial hota hai, aur polynomials har jagah continuous hote hain. Toh $f_{x y}$ aur $f_{y x}$ poore $R^{2}$ pe continuous hain — yaani $f$ $C^{2}$ hai — Clairaut ki hypothesis automatically hold hoti hai — isliye woh equal hone hi chahiye. Hume actual value jaane ki zarurat nahi thi ki woh agree karte hain. (Curious logon ke liye: $f_{x} = 4 x^{3} y^{7}$ , toh $f_{x y} = 28 x^{3} y^{6}$ ; aur $f_{y} = 7 x^{4} y^{6}$ , toh $f_{y x} = 28 x^{3} y^{6}$ . ✅)

Recall Solution L1.2

$f_{y x}$ : subscripts left to right padhte hain, toh pehle $y$ differentiate hota hai, phir $x$ .
$\frac{\partial ^{2} f}{\partial x \partial y}$ : Leibniz notation mein operator jo $f$ ke sabse kareeb hota hai woh pehle act karta hai. Yahaan woh $\partial y$ hai, toh $y$ pehle hai — same order jaise $f_{y x}$ . Dono notations agree karte hain; sirf opposite directions mein padhte hain.

Recall Solution L1.3

Mixed partials $f_{x y}$ aur $f_{y x}$ us point ke aas paas ek open region pe exist aur continuous hone chahiye (yaani $f$ wahaan $C^{2}$ ho). Second mixed partials ki continuity hi Clairaut ke liye entry price hai.

Level 2 — Application

L2.1 $f (x, y) = x^{2} y^{3} + e^{x y}$ ke liye, pehle $x$ se differentiate karke $f_{x y}$ compute karo, phir $f_{y x}$ compute karke verify karo.

L2.2 $f (x, y) = sin (x) cos (y) + x^{2} y$ ke liye, dono mixed partials nikalo aur confirm karo ki woh match karte hain.

L2.3 ("Easy order chuno.") $f (x, y) = e^{x^{2}} ln (y) + y x^{5}$ ke liye, tum mixed partial ki value chahte ho. Jo order algebraically clean ho use compute karo, aur batao tune kaun sa order choose kiya aur kyun.

Recall Solution L2.1

Pehle $x$ . $y$ ko fixed rakho: $f_{x} = \frac{\partial}{\partial x} (x^{2} y^{3} + e^{x y}) = 2 x y^{3} + y e^{x y} .$ $y e^{x y}$ kyun? Chain rule: $\frac{d}{d x} e^{x y} = e^{x y} \cdot \frac{\partial}{\partial x} (x y) = e^{x y} \cdot y$ . Ab $y$ mein differentiate karo: $f_{x y} = \frac{\partial}{\partial y} (2 x y^{3} + y e^{x y}) = 6 x y^{2} + (e^{x y} + y \cdot x e^{x y}) = 6 x y^{2} + e^{x y} (1 + x y) .$ Doosra term kyun? $y e^{x y}$ pe product rule: $y$ ka derivative $1$ hai (deta hai $e^{x y}$ ), plus $y$ times $\frac{\partial}{\partial y} e^{x y} = x e^{x y}$ .

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

Recall Solution L2.2

$f_{x} = cos (x) cos (y) + 2 x y$ , toh $f_{x y} = - cos (x) sin (y) + 2 x .$ $f_{y} = - sin (x) sin (y) + x^{2}$ , toh $f_{y x} = - cos (x) sin (y) + 2 x .$ ✅ Match. (Dono mein $2 x$ aata hai — $x^{2} y$ term ka twist dono taraf se $2 x$ hai.)

Recall Solution L2.3

Dono terms alag behave karte hain, toh payoff order term-by-term handle karo. Clean order hai pehle $y$ $e^{x^{2}} ln y$ piece ke liye (kyunki $\partial_{y} ln y = 1/ y$ immediately messy $ln$ ko khatam kar deta hai), aur $y x^{5}$ ke liye koi bhi order trivial hai.

$f_{y} = e^{x^{2}} \cdot \frac{1}{y} + x^{5}$ . Phir $f_{y x} = \frac{\partial}{\partial x} (\frac{e ^{x^{2}}}{y} + x^{5}) = \frac{2 x e ^{x^{2}}}{y} + 5 x^{4} .$ Yeh kyun easy tha: agar $x$ pehle karte, toh $f_{x} = 2 x e^{x^{2}} ln y + 5 y x^{4}$ mein $ln y$ ek extra step tak carry hota. Clairaut same final answer guarantee karta hai, toh humne woh order choose kiya jo $ln y$ ko idhar udhar ghisitane se bachata hai. Final answer: $\frac{2 x e ^{x^{2}}}{y} + 5 x^{4} .$

Level 3 — Analysis

L3.1 $f (x, y) = x^{2} y^{3}$ ke liye, $f_{xx y}$ , $f_{x y x}$ , aur $f_{y xx}$ alag alag compute karo aur confirm karo ki teeno equal hain. Explain karo kyun woh hone chahiye, Clairaut use karke.

L3.2 Ek student claim karta hai ki $\frac{\partial ^{2} f}{\partial x \partial y}$ aur $\frac{\partial ^{2} f}{\partial y \partial x}$ "obviously same hain by definition, koi theorem ki zarurat nahi." Precisely explain karo ki Clairaut kya add karta hai jo sirf definition nahi deti.

L3.3 Maan lo tumhe sirf yeh bataya gaya hai ki kisi function ki Hessian matrix (second partials ki $2 \times 2$ table) har point pe symmetric hai. Yeh $f_{x y}$ vs $f_{y x}$ ke baare mein kya batata hai, aur kaun si smoothness condition ise guarantee karti hai?

Recall Solution L3.1

Yaad karo page ke upar se: har subscript letter = reading order mein ek differentiation. Hum har string left to right work karte hain, har step justify karte hue. ( $f = x^{2} y^{3}$ .)

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

$\partial_{x} (x^{2} y^{3}) = 2 x y^{3}$ . Kyun: $y^{3}$ yahaan constant factor hai, aur $\partial_{x} (x^{2}) = 2 x$ .
$\partial_{x} (2 x y^{3}) = 2 y^{3}$ . Kyun: phir se $y^{3}$ constant hai; $\partial_{x} (2 x) = 2$ .
$\partial_{y} (2 y^{3}) = 6 y^{2}$ . Kyun: $x$ 's chale gaye, $\partial_{y} (y^{3}) = 3 y^{2}$ , times $2$ .

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

$\partial_{x} (x^{2} y^{3}) = 2 x y^{3}$ . Kyun: same pehla step, $y^{3}$ constant.
$\partial_{y} (2 x y^{3}) = 6 x y^{2}$ . Kyun: ab $2 x$ constant factor hai; $\partial_{y} (y^{3}) = 3 y^{2}$ , times $2 x$ .
$\partial_{x} (6 x y^{2}) = 6 y^{2}$ . Kyun: $y^{2}$ constant; $\partial_{x} (6 x) = 6$ .

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

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

Teeno $6 y^{2}$ dete hain. ✅

Kyun match karna zaroori hai. $f$ ek polynomial hai, toh har partial derivative har jagah continuous hai ( $f$ $C^{\infty}$ hai, toh certainly $C^{2}$ bhi). Clairaut tumhe koi bhi do adjacent differentiations swap karne deta hai (yeh do-variable theorem hai $f$ ya $f$ ki first derivative pe apply hota hai). Adjacent letters ko baar baar swap karke — jaise sorting — tum $xx y$ , $x y x$ , $y xx$ ko ek doosre mein badal sakte ho. Isliye sirf har variable ki count matter karti hai (yahaan: do $x$ 's, ek $y$ ), order nahi.

Recall Solution L3.2

Definition ke according, $\frac{\partial ^{2} f}{\partial x \partial y} = f_{y x}$ (pehle $y$ mein differentiate, phir $x$ ) aur $\frac{\partial ^{2} f}{\partial y \partial x} = f_{x y}$ ( $x$ mein pehle, phir $y$ ). Yeh do genuinely alag computations hain — alag limit processes, alag orders mein evaluate hote hain. Definition mein kuch bhi force nahi karta ki woh same number dein. Counterexample origin pe (L5 neeche) dikhata hai ki woh really differ kar sakte hain.

Clairaut kya add karta hai: ek extra hypothesis (mixed partials ki continuity, yaani $C^{2}$ ) jiske under woh do alag computations proven hain ki same value pe land karte hain. Toh yeh ek theorem hai jisme content hai, definition ka restatement nahi.

Recall Solution L3.3

Hessian Matrix hai $H = (f_{xx} f_{y x} f_{x y} f_{y y}) .$ "Symmetric" ka matlab hai off-diagonal entries equal hain: $f_{x y} = f_{y x}$ — jo bilkul wahi hai jo Clairaut conclude karta hai. Toh har point pe symmetric Hessian ka matlab hai "mixed partials har jagah agree karte hain." Jo condition ise guarantee karti hai: second partials region pe continuous hain ( $f \in C^{2}$ ). Tab Clairaut symmetry force karta hai.

Level 4 — Synthesis

L4.1 (Exact ODEs se link.) Ek differential form $M (x, y) d x + N (x, y) d y$ ko exact kehte hain jab yeh kisi potential $f$ ke liye $df$ ke equal ho, jo force karta hai $M = f_{x}$ aur $N = f_{y}$ . Clairaut use karke standard test $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ derive karo. Phir ise $M = 2 x y + y^{2}$ , $N = x^{2} + 2 x y$ pe apply karo: kya form exact hai?

L4.2 (Proof ko reverse karo.) Parent proof ki symmetric second difference yaad karo: $Δ (h, k) = f (a + h, b + k) - f (a + h, b) - f (a, b + k) + f (a, b) .$ $f (x, y) = x y$ ke liye $(a, b) = (0, 0)$ pe $Δ$ explicitly compute karo, phir $hk$ se divide karo. Kya milta hai, aur yeh kaise confirm karta hai ki $f_{x y} = f_{y x} = 1$ is $f$ ke liye?

L4.3 Ek aisa function banao jiska mixed partial $12 x^{2} y$ ho. (Reverse-engineer karo: integrate karo.)

Recall Solution L4.1

Exactness test derive karna. Agar form exact hai, toh $M = f_{x}$ aur $N = f_{y}$ kisi potential $f$ ke liye. Tab $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} f_{x} = f_{x y}, \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} f_{y} = f_{y x} .$ Agar $f$ $C^{2}$ hai, toh Clairaut deta hai $f_{x y} = f_{y x}$ , isliye $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} .$ Yahi woh necessary condition hai jo Exact Differential Equations mein use hoti hai — yeh literally Clairaut hi hai alag clothes mein.

Apply karo. $M = 2 x y + y^{2} \Rightarrow M_{y} = 2 x + 2 y$ . $N = x^{2} + 2 x y \Rightarrow N_{x} = 2 x + 2 y$ . Woh equal hain, toh form exact hai — aur kyunki domain yahaan poora $R^{2}$ hai (open aur simply connected, koi holes nahi), ek genuine potential really exist karta hai.

Recall Solution L4.2

$f = x y$ ke liye: $f (a + h, b + k) = (a + h) (b + k)$ , etc. $(0, 0)$ pe: $Δ (h, k) = (h) (k) - (h) (0) - (0) (k) + (0) (0) = hk .$ $hk$ se divide karo: $\frac{Δ ( h , k )}{hk} = 1 for all h, k \neq = 0.$ Parent proof ne dikhaya ki $Δ/ (hk)$ ek taraf $f_{x y}$ ko aur doosri taraf $f_{y x}$ ko limit karta hai. Yahaan yeh ek constant $1$ hai, toh dono limits $1$ hain order ke baad bhi: $f_{x y} (0, 0) = f_{y x} (0, 0) = 1.$ Direct check: $f_{x} = y \Rightarrow f_{x y} = 1$ ; $f_{y} = x \Rightarrow f_{y x} = 1$ . ✅ Symmetric difference is simple $f$ ke liye literally mixed partial hi hai.

Recall Solution L4.3

Hum chahte hain $f_{x y} = 12 x^{2} y$ . " $x$ phir $y$ " chain ko reverse order mein undo karo. Pehle ek $f_{x}$ dhundho jiska $y$ -derivative $12 x^{2} y$ ho: $12 x^{2} y$ ko $y$ ke respect se integrate karo, $f_{x} = \int 12 x^{2} y d y = 6 x^{2} y^{2} + C (x),$ jahaan $C (x)$ sirf $x$ ka ek arbitrary function hai ("constant of integration" ki role nibhata hai, lekin kyunki humne $y$ mein integrate kiya, $y$ se kuch bhi nahi rakha $\partial_{y} C (x) = 0$ ke roop mein survive karta hai). Sabse simple choice $C (x) = 0$ lo, deta hai $f_{x} = 6 x^{2} y^{2}$ . Ab use $x$ mein integrate karo: $f = \int 6 x^{2} y^{2} d x = 2 x^{3} y^{2} + D (y),$ jahaan $D (y)$ phir se $y$ mein arbitrary hai ( $\partial_{x}$ ke under yeh vanish ho jaata hai). Toh solutions ki family hai $f = 2 x^{3} y^{2} + D (y) + (terms from C)$ ; sabse simple hai $f = 2 x^{3} y^{2}$ . Arbitrary pieces kyun matter karte hain: koi bhi $D (y)$ $f_{x}$ ko unchanged chodta hai aur isliye $f_{x y}$ ko bhi, toh potential unique nahi hota — yahi woh non-uniqueness hai jo tumhe Exact Differential Equations mein potentials reconstruct karte waqt milti hai. Check: $f = 2 x^{3} y^{2} \Rightarrow f_{x} = 6 x^{2} y^{2} \Rightarrow f_{x y} = 12 x^{2} y$ . ✅ (Aur $f_{y} = 4 x^{3} y \Rightarrow f_{y x} = 12 x^{2} y$ — Clairaut agree karta hai, jaise karna chahiye is polynomial ke liye.)

Level 5 — Mastery

L5.1 (Counterexample, hands-on.) Maano $f (x, y) = ⎩ ⎨ ⎧ \frac{x y ( x ^{2} - y ^{2} )}{x ^{2} + y ^{2}}, 0, (x, y) \neq = (0, 0) (x, y) = (0, 0) .$ Dikhao ki $f_{x y} (0, 0) = - 1$ jabki $f_{y x} (0, 0) = + 1$ , toh Clairaut yahaan fail karta hai — aur exactly identify karo ki kaun si hypothesis tooti.

L5.2 Neeche di gayi figure use karke explain karo, kyun mixed partial ki continuity precisely woh cheez hai jo L5.1 mein fail hoti hai, ise parent proof ke "unknown MVT points origin ki taraf squeeze hote hain" waale step se connect karte hue.

Figure 2 — $y$ -axis ke along first partial $f_{x} (0, y) = - y$ (blue) slope $- 1$ ke saath girta hai; $x$ -axis ke along $f_{y} (x, 0) = + x$ (yellow) slope $+ 1$ ke saath badhta hai. Woh do slopes HI mixed partials $f_{x y} (0, 0) = - 1$ aur $f_{y x} (0, 0) = + 1$ hain. Woh shared red point (origin) pe disagree karte hain — yeh ek mixed partial ka visual signature hai jo wahaan continuous nahi hai.

Recall Solution L5.1

Hume pehle origin se door first partials chahiye, phir origin pe difference quotients se second partials evaluate karni hain.

Step 1 — $f_{x}$ for $(x, y) \neq = (0, 0)$ . $f = \frac{x y ( x ^{2} - y ^{2} )}{x ^{2} + y ^{2}} = \frac{x ^{3} y - x y ^{3}}{x ^{2} + y ^{2}}$ ke saath, quotient rule deta hai (simplification ke baad) $f_{x} (x, y) = \frac{y ( x ^{4} + 4 x ^{2} y ^{2} - y ^{4} )}{( x ^{2} + y ^{2} ) ^{2}} .$ Iska value $y$ -axis pe: $x = 0$ rakho: $f_{x} (0, y) = \frac{y ( - y ^{4} )}{y ^{4}} = - y$ for $y \neq = 0$ . Aur $f_{x} (0, 0) = 0$ bhi hai (definition se: $f$ $x$ -axis ke along $0$ hai, toh origin pe uska $x$ -slope $0$ hai).

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

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

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

Conclusion: $f_{x y} (0, 0) = - 1 \neq = + 1 = f_{y x} (0, 0)$ . Mixed partials exist karte hain lekin origin pe continuous nahi hain — exactly wahi hypothesis jo Clairaut require karta hai ( $C^{2}$ condition fail hoti hai). Toh theorem ki continuity condition genuinely necessary hai. ✅

Recall Solution L5.2

Figure 2 dekho. Axes ke along, $f_{x} (0, y) = - y$ ek taraf badhta hai aur $f_{y} (x, 0) = + x$ doosri taraf; origin ki taraf alag directions se aane pe mixed partial ki value alag numbers ki taraf jaati hai ( $- 1$ vs $+ 1$ ). Yahi precisely $(0, 0)$ pe mixed partial ki discontinuity hai.

Parent proof ka finish yaad karo: MVT humein unknown points $(c_{i}, d_{i})$ deta hai jo sirf $(a, b)$ ki taraf squeeze karte hain, aur hume chahiye tha "nearby point pe $f_{x y}$ $\to f_{x y} (a, b)$ " — woh step continuity hi hai. Jab continuity fail hoti hai (jaise yahaan), squeezing points alag directions se aate hue alag limits de sakte hain, aur dono orders ko agree karna zaroori nahi reh jaata. Figure ke dono axes pe clashing colored slopes usi failure ka visual signature hain.

Recall Poori ladder ka one-line summary

Pehchaano (L1) → dono orders se apply karo aur easy wala chuno (L2) → higher-derivative strings sort karo aur Hessian padho (L3) → ise exactness tests aur proof ke symmetric difference ke andar pehchaano (L4) → aur woh ek function jaano jahan yeh toot jaata hai, aur kyun (L5): no continuity → no Clairaut.