Differentiability in multiple variables: Question bank

4.4.6 · D5 · Maths › Multivariable Calculus › Differentiability in multiple variables

Shuru karne se pehle, ye pictures dimag mein rakho: ek partial derivative sirf due East ya due North chalti hai; differentiability demand karti hai ki ek single flat tilted board (tangent plane) har direction se ek saath fit ho. Neeche ke zyaadatar traps yehi distinction hai ek disguise mein.

True or false — justify

f

ke dono partial derivatives

a

par hain, isliye

f

a

par continuous hai.

False. Partials sirf do axis directions ko probe karti hain; function ek diagonal ke along jump kar sakta hai, jaise

x y / (x^{2} + y^{2})

f_{x} = f_{y} = 0

origin par hain phir bhi

y = x

ke along limit

\frac{1}{2}

hai.

Agar

f

a

par differentiable hai, toh

f

a

par continuous hai.

True.

f (a + h) - f (a) = \nabla f \cdot h + E

se, linear term aur

E = o (∥ h ∥)

dono

h \to 0

hone par vanish ho jaate hain, isliye

f (a + h) \to f (a)

Agar

f

a

par continuous hai, toh

f

a

par differentiable hai.

False. Arrow sirf ek direction mein chalta hai:

∣ x ∣ + ∣ y ∣

har jagah continuous hai lekin origin par koi tangent plane nahi hai (ek sharp corner hai).

Agar dono partials

a

ke paas continuous hain, toh

f

a

par differentiable hai.

True. Yahi exactly

C^{1}

sufficient condition hai — woh workhorse jo tumhe almost har function ke liye raw

o (∥ h ∥)

limit skip karne deta hai.

Agar

f

a

par differentiable hai, toh uski partials

a

par continuous hain.

False. Differentiable

C^{1}

se weaker hai. Ek function ek point par differentiable ho sakta hai jabki uski partials wahan discontinuously wiggle karti rahein (jaise pieced-together

x^{2} sin (1/ x)

-style constructions).

Agar

f

ke har directional derivative

a

par exist karte hain, toh

f

a

par differentiable hai.

False. "Har direction" bhi fail ho sakta hai: formula

D_{u} f = \nabla f \cdot u

break ho sakta hai, aur continuity phir bhi fail ho sakti hai, isliye har jagah directional derivatives hona differentiability se strictly weaker hai.

x

aur

y

mein ek polynomial har jagah differentiable hota hai.

True. Uski partials phir se polynomials hain, isliye har jagah continuous hain, toh yeh

C^{1}

hai aur isliye har point par differentiable hai.

Agar

f

a

par differentiable hai, toh ek hi possible linear approximation

f (a) + \nabla f (a) \cdot h

hai.

True. Har axis ke along approach karne se coefficients ko

f_{x}

aur

f_{y}

ke barabar hona force hota hai; koi aur linear map

o (∥ h ∥)

condition satisfy nahi kar sakta.

Agar

E (h) \to 0

jab

h \to 0

, toh

a + m \cdot h

tangent plane hai.

False.

E \to 0

sirf mere touching hai (error ki continuity). Tangent plane ke liye

E /∥ h ∥ \to 0

chahiye — error ko step size se faster marna hoga.

C^{1}

a

aur differentiable at

a

ek hi cheez hai.

False.

C^{1}

(continuous partials) differentiable imply karta hai lekin genuinely stronger hai; differentiable functions ke continuous partials hone zaroori nahi.

Spot the error

f_{x} (0, 0) = 0

aur

f_{y} (0, 0) = 0

, isliye

\nabla f (0, 0) = 0

aur tangent plane

z = 0

hai." (

f = x y / (x^{2} + y^{2})

ke liye)

Error yeh hai ki tangent plane exist karna assume kar liya. Kyunki

f

origin par discontinuous hai, yeh differentiable nahi hai, toh "

\nabla f

" sirf numbers ka ek pair hai, valid linear approximation nahi.

f

differentiable hai kyunki maine limit

x

-axis,

y

-axis, aur

y = x

ke along check ki, aur error ratio teeno par 0 par gaya."

Teen lines har direction nahi hoti.

o (∥ h ∥)

limit ko sab paths ke saath ek saath vanish karna hoga; ek chautha curved path abhi bhi betray kar sakta hai.

"Tangent plane dhundne ke liye mujhe raw

o (∥ h ∥)

limit chahiye, toh main ise

f = x^{2} y + sin y

ke liye compute karunga."

Yeh overkill aur error-prone hai: partials

2 x y

aur

x^{2} + cos y

continuous hain, isliye

f

C^{1}

hai aur plane

z = f (a) + \nabla f (a) \cdot h

automatically valid hai.

f (x, y) = x^{2} + y^{2}

nice functions ka sum aur root hai, isliye yeh har jagah differentiable hai."

Origin par galat:

f_{x} = x / x^{2} + y^{2}

wahan undefined hai (ek cone point), isliye

C^{1}

aur differentiability dono

(0, 0)

par fail hote hain chahe baaki har jagah hold karein.

"Error ratio

a

se har straight line ke along

0

par gaya, isliye

f

differentiable hai."

Straight lines bhi kaafi nahi hain. Kuch functions error ratio ko har line par toh 0 tak bhejte hain lekin parabolic paths jaise

y = x^{2}

ke along nahi, isliye yeh argument incomplete hai.

"Kyunki differentiable

\Rightarrow

continuous, aur mera

f

continuous hai,

f

differentiable hona chahiye."

Yeh implication ko ulta kar raha hai. Continuity differentiability ka consequence hai, cause nahi;

∣ x ∣ + ∣ y ∣

iska standing counterexample hai.

Why questions

Hum error

E (h)

∥ h ∥

se kyun divide karte hain sirf

E \to 0

require karne ki jagah?

Divide karna error ko higher-order force karta hai — step se bhi chhota — jo precisely ek genuine tangent plane ko kisi bhi random plane se alag karta hai jo sirf point ko touch kare.

Partial derivatives "diagonals ke andhe" kyun hain?

Har partial ek one-variable limit hai jo doosre variable ko frozen rakh ke li jaati hai, isliye yeh sirf axis ke parallel motion sample karti hai aur kabhi bhi

(h, h)

jaisa off-axis step nahi dekhti.

C^{1}

condition practice mein itni useful kyun hai jabki yeh sirf sufficient hai?

Almost har function jo tum miloge (polynomials,

sin

exp

, ratios apne zeros se door) ke continuous partials hote hain, isliye ek line ka "

C^{1}

hence differentiable" 99% time ek painful limit computation replace kar deta hai.

Linear map, agar exist karta hai, gradient

(f_{x}, f_{y})

ke equal kyun hona chahiye?

x

-axis ke along approach restrict karna definition ko

x

mein single-variable derivative par collapse kar deta hai, jo

m_{1} = f_{x}

pin karta hai;

y

-axis

m_{2} = f_{y}

pin karta hai.

Differentiability sab directional derivatives exist kyun karta hai aur

D_{u} f = \nabla f \cdot u

kyun obey karte hain?

Tangent plane ek single linear map hai; direction

u

ke along chalna bas us map ka slope us direction mein read karta hai, jo dot product

\nabla f \cdot u

hai.

Differentiability ko "2D requirement" aur partial ko "1D" kyun kehte hain?

Ek partial ek frozen slice check karta hai; differentiability demand karti hai ki ek single flat board surface ko surrounding disk ke har point se fit kare, jo ek saath sab directions couple karta hai.

Ek function ek point par differentiable kyun ho sakta hai phir bhi

C^{1}

nahi?

Differentiability sirf yeh maangti hai ki us point par ek acha linear fit exist kare; yeh kuch nahi kehta ki partials ek neighbourhood mein smoothly match up karein, jo

C^{1}

ki extra demand hai.

Edge cases

Kya ek constant function

f (x, y) = c

differentiable hai? Uska tangent plane kya hai?

Yes. Uski partials dono

0

hain (continuous), isliye yeh

C^{1}

hai; tangent plane har jagah flat plane

z = c

hai.

Ek isolated point par jahan

f

sirf defined hai (koi neighbourhood nahi), kya differentiability poochha bhi ja sakta hai?

No. Limit

h \to 0

ko surrounding disk mein har direction se approach karna require karta hai, isliye differentiability ko point ke around

f

ka open set par defined hona zaroori hai.

f = x^{2} + y^{2}

ko origin par lo: partials, continuity, differentiability?

Wahan continuous hai, lekin partials exist nahi karte (cone tip ka koi single slope nahi hai), isliye yeh origin par differentiability fail karta hai jabki baaki har jagah differentiable rehta hai.

Agar

\nabla f (a) = 0

lekin

f

differentiable hai, tangent plane kaisa dikhta hai?

Ek perfectly horizontal plane

z = f (a)

— ek critical point jahan surface first order tak flat hai (peak, valley, ya saddle).

Kya

a

par differentiability tumhe

f

ke baare mein

a

se door kuch batata hai?

No. Yeh purely local, point-by-point property hai;

f

a

par well-behaved ho sakta hai aur thodi door wildly discontinuous.

Kya ek function exactly ek point par differentiable ho sakta hai aur kahin nahi?

Yes.

f = (x^{2} + y^{2}) \cdot (kuch jagged)

jaisi constructions sirf origin par ek valid linear fit force kar sakti hain, jahan multiplying factor saari buri behaviour ko crush kar deta hai.

Recall Pure bank ka one-line summary

Yahan har trap ek hi trap hai: axis-information (partials) ya touching-information (continuity, $E \to 0$ ) ko all-directions, faster-than-linear fit samajh lena jo hai differentiability.

Question bank — Differentiability in multiple variables

True or false — justify

Spot the error

Why questions

Edge cases

Connections