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

False. Ye tab agree karte hain jab mixed partials us point ke paas continuous hon; origin counterexample $f=xy(x^2-y^2)/(x^2+y^2)$ deta hai $f_{xy}(0,0)=-1\ne f_{yx}(0,0)=1$.

False. Sirf exist karna kaafi nahi — dono counterexample mein exist karte hain phir bhi differ karte hain. Jo equality force karta hai woh hai mixed partials ki continuity, unka existence nahi.

True. Continuity exactly Clairaut ki hypothesis hai, aur yeh disk ke har point par hold karti hai, toh conclusion poore disk mein pointwise hold karta hai.

False. Yeh ek analysis theorem hai jiska poora content continuity condition hai; iske bina equality genuinely fail ho sakti hai, toh hypothesis decorative nahi hai.

True. $f_x$ hai $x$-direction mein slope; isse $y$ ke saath differentiate karna measure karta hai ki woh slope kitna twist karta hai jab tum $y$ mein shift karo — yeh woh geometric "warping" hai (Picture 2) jise Clairaut dono orders mein compare karta hai.

False. Mixed partials ki equality ek point par full two-variable continuity of $f$ force nahi karta: partial derivatives (chahe equal bhi hon) sirf coordinate lines ke along probe karte hain, jabki $f$ diagonal direction se approach karne par phir bhi jump kar sakta hai.

True. Polynomials ke har order ke continuous partials hote hain, toh Clairaut har point par apply hota hai aur har mixed partial order-independent hai.

True (Schwarz's refined version). Kaafi hai ki $f_x,f_y$ us point ke paas exist karein aur ek mixed partial wahan continuous ho; doosra tab exist karta hai aur uske equal hota hai.

True. Hessian ke off-diagonal entries $f_{xy}$ aur $f_{yx}$ hain; Clairaut ki continuity condition ke under woh equal hain, Hessian Matrix ko symmetric banate hain — isliye iske eigenvalues real hote hain.

Error. Leibniz notation mein operator jo $f$ ke sabse paas hota hai pehle act karta hai, toh yeh $y$-then-$x$ hai. Subscript form $f_{yx}$ left-to-right $y$ pehle padhta hai; dono notations opposite directions mein jaate hain.

Error. Point $(c_1,d_1)$ unknown hai aur generally $(a,b)$ nahi hota; sirf $h,k\to 0$ ke baad hi woh $(a,b)$ ki taraf squeeze hota hai, aur sirf continuity hi iske function value ko $(a,b)$ par value ke paas laane deti hai.

Error. Origin par existence swapping guarantee nahi karta; wahan mixed partials discontinuous hain (Picture 3), toh hypothesis fail ho jaati hai aur $f_{xy}(0,0)\ne f_{yx}(0,0)$.

Error. Har order $\Delta$ ko MVT ke do applications se todta hai — ek $x$ mein, ek $y$ mein — ek taraf $hk{f}_{xy}(c_1,d_1)$ produce karta hai aur doosri taraf $hk{f}_{yx}(c_2,d_2)$; kul chaar interior points.

Error. Continuity ke under Clairaut poori string ko reorder karne deta hai; sirf har variable ki count matter karti hai, toh $f_{xxy}=f_{xyx}=f_{yxx}$.

Error. $f$ ki continuity iske second mixed partials ki continuity se bahut weak hai. Counterexample function khud har jagah continuous hai phir bhi origin par Clairaut fail karta hai.

Kyunki $\Delta$ visibly unchanged rehta hai jab $x$ aur $y$ ki roles swap hoti hain (Picture 1 mein symmetric corner-signs dekho), toh ek hi quantity ko ek taraf $f_{xy}$ tak aur doosri taraf $f_{yx}$ tak limit karte hue dikhaya ja sakta hai — shared symmetry dono limits ko agree karne par force karti hai.

Mean Value Theorem (MVT) hume unknown interior points $c_i,d_i$ par derivative values deta hai; continuity hi ek cheez hai jo "squeezing point par value" ko "$(a,b)$ par value" mein convert karti hai, toh isse hatao toh conclusion bhi chala jaata hai.

"Jab tum $y$ mein move karo toh $x$-slope kaise change hota hai" aur "jab tum $x$ mein move karo toh $y$-slope kaise change hota hai" dono surface ki same local warping quantify karte hain, toh ek smooth surface ko dono taraf same twist report karna chahiye.

Rational formula $xy(x^2-y^2)/(x^2+y^2)$ origin par $0/0$ hai, toh ek alag value wahan set ki jaati hai; yeh seam (Picture 3 mein pinch) exactly wahan hai jahan second mixed partials continuity kho dete hain aur dono orders alag ho jaate hain.

Kyunki koi bhi order same result deta hai, tum algebraically cleaner path choose kar sakte ho — jaise pehle simpler variable differentiate karo — aur bina hard order dobaara kiye answer par trust kar sakte ho.

Symmetry $f_{xy}=f_{yx}$ Clairaut ki continuity hypothesis se earned hoti hai; ek non-smooth function ke liye Hessian ke off-diagonal entries differ kar sakte hain, toh symmetry ek theorem hai, built-in property nahi.

Nahi. $f_{xy}$ ke liye $f_x$ ka ek neighbourhood mein exist karna zaroori hai isse $y$ mein differentiate karne se pehle; agar $f_x$ paas mein undefined hai, toh mixed partial defined nahi hai, toh equality question void hai.

Nahi, double limit ka order matter karta hai. Proof ko $\Delta$ chahiye $hk$ se divided, dono steps genuinely nonzero ke saath; $k=0$ set karna $\Delta$ ko $0$ mein collapse kar deta hai aur second difference ko koi bhi information extract hone se pehle destroy kar deta hai.

Crease par hypothesis fail hoti hai, toh equality wahan break ho sakti hai; crease se door, jahan partials continuous hain, Clairaut phir bhi locally $f_{xy}=f_{yx}$ guarantee karta hai.

Nahi. First-partial continuity kaafi weak hai; Clairaut us point par second mixed partials ki continuity demand karta hai, jo exactly missing hai.

Haan, same continuity condition ke under. Adjacent variables ki koi bhi pair swap ki ja sakti hai jab relevant mixed partials continuous hon, toh $x,y,z$ ka koi bhi permutation same third-order partial deta hai.

Trivially true. Saare partials $0$ hain aur continuous hain, toh $f_{xy}=f_{yx}=0$; theorem hold karta hai lekin koi information nahi carry karta — ek useful sanity check ki hypotheses satisfiable hain.