Visual walkthrough — Differentiability in multiple variables

4.4.6 · D2 · Maths › Multivariable Calculus › Differentiability in multiple variables

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Step 1 — Do-variable function ka graph kya hota hai?

KYA. Ek function $f(x,y)$ do numbers lete aur ek number deta hai. Ek flat floor ke har point $(x,y)$ ko feed karo aur output $z=f(x,y)$ ko upar ki taraf height ke roop mein stack karo. Result ek surface hota hai — floor ke upar floating ek landscape.

KYUN. Isse pehle ki hum poochhein "kya is landscape ko ek flat tilted board touch kar raha hai?", humein agree karna hoga ki landscape hai kya. Neeche ki saari cheezein is surface par rehti hain.

PICTURE. Floor $xy$-plane hai. Point $(a,b)$ (red mein marked) ke upar surface ek single height $f(a,b)$ tak uthti hai. Woh red vertical stick woh value hai jise hum approximate karenge.

Dekho Partial Derivatives ki hum is surface ko ek axis ek time kaise probe karte hain.

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Step 2 — "Tangent plane" ka matlab kya hoga?

KYA. Ek plane ek perfectly flat, infinitely wide tilted board hota hai. Hum woh ek board chahte hain jo surface ko $\mathbf{a}$ par touch kare aur uske saath jitna possible ho utna flat lage — tangent line ka 2D cousin.

KYUN. Ek variable mein, "differentiable" ka matlab tha "tangent line hai — zoom in karo aur curve seedha dikhta hai." Do variables mein honest generalisation hai "zoom in karo aur surface flat dikhta hai." Flat cheez plane hoti hai. To differentiability = yeh plane exist karo aur woh sach mein surface se chipki raho.

PICTURE. Black surface curve karti hai. Red plane point $(a,b,f(a,b))$ se gujarti hai aur match karne ke liye tilt karti hai. $\mathbf{a}$ ke paas unka gap tiny hai; door jaane par badhta hai.

Plane ki equation ki poori treatment Tangent Planes and Linear Approximation mein hai.

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Step 3 — Candidate plane ko algebraically likho

KYA. Point $(\mathbf{a},f(\mathbf{a}))$ se gujarne wala koi bhi plane likha ja sakta hai $L(\mathbf{x})=\underset{\text{height at }\mathbf{a}}{\underbrace{f(\mathbf{a})}}+\underset{\text{tilt vector}}{\underbrace{\mathbf{m}}}\cdot \underset{\text{step across floor}}{\underbrace{(\mathbf{x}-\mathbf{a})}}.$

KYUN. Hum abhi correct tilt nahi jaante, isliye hum ise $\mathbf{m}=(m_1,m_2)$ kehte hain aur baad mein solve karte hain. Point $f(\mathbf{a})$ se start karna guarantee karta hai ki plane wahan surface ko touch karegi. Dot product $\mathbf{m}\cdot (\mathbf{x}-\mathbf{a})$ woh hai jis tarah ek plane floor-step ko rise mein convert karti hai.

PICTURE. Floor par ek step $\mathbf{h}=\mathbf{x}-\mathbf{a}$ lo (red arrow). Plane ki rise step ke East-part times $m_1$ plus North-part times $m_2$ hai. Steeper $\mathbf{m}$ ⇒ per step badi rise.

Yahan $\mathbf{h}=(h_1,h_2)$ step vector hai aur $\Vert \mathbf{h}\Vert =\sqrt{h_1^2+h_2^2}$ uski length hai (straight-line distance stepped). Dekho Gradient and Directional Derivatives.

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Step 4 — Error measure karo, phir demand karo ki woh sahi tarike se chhota ho

KYA. Surface kehti hai $\mathbf{a}+\mathbf{h}$ par true height $f(\mathbf{a}+\mathbf{h})$ hai. Plane predict karta hai $L(\mathbf{a}+\mathbf{h})=f(\mathbf{a})+\mathbf{m}\cdot \mathbf{h}$. Unka difference error hai: $E(\mathbf{h})=\underset{\text{true height}}{\underbrace{f(\mathbf{a}+\mathbf{h})}}-\underset{\text{plane’s guess}}{\underbrace{[f(\mathbf{a})+\mathbf{m}\cdot \mathbf{h}]}}.$

KYUN. Ek plane jo sirf touch kare ($E\to 0$ as $\mathbf{h}\to 0$) special nahi hai — infinitely many tilted boards ek point par surface ko touch karti hain. Tangent plane ko pin down karne ke liye humein chahiye ki error step se tezi se mit jaaye.

PICTURE. Floor-step $\mathbf{h}$ par, black surface aur red plane do heights hain. Unke beech ka vertical red segment $E(\mathbf{h})$ hai. Jaise $\mathbf{h}$ ko $0$ ki taraf slide karte ho, woh gap sirf close hi nahi hona chahiye — use horizontal distance se tezi se close hona chahiye.

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Step 5 — $\Vert \mathbf{h}\Vert$ se divide kyun karte hain? Ratio picture

KYA. Hum condition impose karte hain ${\lim }_{\mathbf{h}\to 0}\frac{E(\mathbf{h})}{\Vert \mathbf{h}\Vert }=0.$

KYUN. Vertical gap $E$ ko horizontal distance $\Vert \mathbf{h}\Vert$ se divide karna gap ko error ka slope bana deta hai. Agar yeh slope bhi zero ho jaaye, surface aur plane tangent ho jaate hain, sirf intersecting nahi. Concretely: $\mathbf{a}$ ke paas error $\Vert \mathbf{h}{\Vert }^2$ jaisa behave karta hai — ek bura "touching" plane error $\sim \Vert \mathbf{h}\Vert$ chodta hai, aur divide karna difference expose kar deta hai.

PICTURE. "Error vs step" ki do curves: ek touching plane ka error (upper) ek straight line ki tarah $0$ tak shrink hota hai — lekin $\Vert \mathbf{h}\Vert$ se divide karne par yeh nonzero number par flatten ho jaata hai. Tangent plane ka error (red, lower) $\Vert \mathbf{h}{\Vert }^2$ ki tarah curl down karta hai — $\Vert \mathbf{h}\Vert$ se divide karne par bhi yeh $0$ ki taraf dive karta hai.

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Step 6 — Tilt $\mathbf{m}$ solve karo axes par chalke

KYA. Condition har direction se approach karne par hold karni chahiye. Pehle easy wale choose karo. Purely East step karo, $\mathbf{h}=(h,0)$, so $\Vert \mathbf{h}\Vert =|h|$: $\frac{f(a+h,b)-f(a,b)-m_{1}h}{|h|}\underset{h\to 0}{\to }0⟹m_1={\lim }_{h\to 0}\frac{f(a+h,b)-f(a,b)}{h}=f_x(a,b).$ Purely North step karo, $\mathbf{h}=(0,h)$: milta hai $m_2=f_y(a,b)$.

KYUN. Humare paas ek unknown tilt $\mathbf{m}$ tha. Ek axis par approach karna doosre component ko kill kar deta hai, ek one-variable limit chhodta hai — jo exactly partial derivative ki definition hai. To agar tangent plane exist karta hai, uski tilt force ho jaati hai $(f_x,f_y)$ hone ke liye; koi freedom nahi.

PICTURE. Surface ke through do red slices: East–West slice (uska slope $\mathbf{a}$ par $f_x$ hai) aur North–South slice (slope $f_y$). Tangent plane mein dono slice-tangent-lines honi chahiye.

Yeh Gradient and Directional Derivatives ka gradient hai; matrix ke roop mein package hua yeh Total Derivative / Jacobian hai.

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Step 7 — Master formula assemble karo

KYA. Forced tilt wapas plug in karne par, differentiability kehti hai surface plane + tiny leftover mein split hoti hai: $\boxed{f(\mathbf{a}+\mathbf{h})=\underset{\text{base height}}{\underbrace{f(\mathbf{a})}}+\underset{\text{plane’s rise}}{\underbrace{\nabla f(\mathbf{a})\cdot \mathbf{h}}}+\underset{\text{negligible error}}{\underbrace{o(\Vert \mathbf{h}\Vert )}}}$

KYUN. Yeh poora point ek line mein hai: ek differentiable surface apna tangent plane hai plus ek error jo matter karne ke liye bahut chhota hai. Yeh instantly differentiable ⇒ continuous bhi prove karta hai — jaise $\mathbf{h}\to 0$, plane-rise $\to 0$ aur error $\to 0$, isliye $f(\mathbf{a}+\mathbf{h})\to f(\mathbf{a})$. Dekho Continuity in Multiple Variables.

PICTURE. Ek fixed small step par teen stacked bars: total height (black), plane dwara explain kiya hua hissa (red, almost sab kuch), aur leftover error ka chhota sliver (nothing tak shrink hota hua).

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Step 8 — Degenerate case: partials exist hain par koi plane hug nahi karta

KYA. Lo $f(x,y)=\{\begin{array}{ll}\frac{xy}{x^2+y^2}& (x,y)\ne (0,0)\\ 0& (0,0).\end{array}$ Har axis par $f\equiv 0$, isliye $f_x(0,0)=f_y(0,0)=0$ — dono partials exist karte hain. Phir bhi diagonal $y=x$ par, $f(x,x)=\frac{x\cdot x}{x^2+x^2}=\frac{1}{2}\ne 0.$

KYUN. Step 6 ka axis-walk sirf ek tilt propose karta tha; usne Step 5 ki diagonal directions check nahi ki. Yahan surface axes par flat cross hai lekin diagonal par ek raised ridge hai — koi single plane dono ko match nahi kar sakta. Kyunki $f$ $0$ par continuous bhi nahi hai (approach par $\frac{1}{2}$ hit karta hai), Step 7 ke hisaab se yeh differentiable nahi ho sakta.

PICTURE. Upar se neeche dekho: black axes par height $0$ hai; red diagonal par height $\frac{1}{2}$ tak jump karti hai. Origin se gujarne wala ek plane with zero East/North slope flat floor hoga — lekin red ridge usse poke karti hai. Koi hug possible nahi.

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Step 9 — Escape hatch: $C^1$ ⇒ differentiable

KYA. Step 5 limit ko har direction mein verify karna exhausting hai. Shortcut: agar $f_x$ aur $f_y$ $\mathbf{a}$ ke paas exist aur continuous hain, toh $f$ automatically wahan differentiable hai. Aisa $f$ $C^1$ kehlata hai.

KYUN. Partials ki continuity matlab hai ki jaisi tum move karte ho East aur North slopes jump nahi karti — isliye unhe ek plane mein glue karna peeche koi diagonal ridge nahi chhod sakta. Pathological Example (Step 8) exactly isliye fail karta hai kyunki uske partials origin par continuous nahi hain (woh oscillate karte hain). Uss possibility ko rule out karna trap ko rule out karta hai.

PICTURE. Hierarchy ka ek flowchart: $C^1$ top par baitha hai aur har arrow ek one-way implication hai, kabhi reversible nahi.

Dekho C1 and Smooth Functions aur Chain Rule (Multivariable) jahan $C^1$ kaam aata hai.

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Ek-picture summary

Sab ek saath: black surface, red tangent plane $\mathbf{a}$ par touch karta hua, floor par step $\mathbf{h}$, aur shrinking vertical error $E$ jo $\Vert \mathbf{h}\Vert$ se tezi se khatam ho jaani chahiye.

Recall Feynman retelling — plain words mein poora walk

Ek pahadi landscape imagine karo (Step 1). Ek jagah par khado aur ek perfectly flat tilted board ko slide karne ki koshish karo taaki woh exactly wahan zameen ko kiss kare (Step 2). Tum board ka rule likhte ho: apni height se shuru karo, phir har step ke liye kuch amount tilt karo (Step 3). Ab thoda chalo aur zameen ko apne board se compare karo — vertical gap error hai (Step 4). Sirf touch karna kaafi nahi: tum demand karte ho ki gap jitni door tum chale usse tezi se shrink ho, isliye gap ko distance se divide karo aur insist karo ki woh zero ho jaaye (Step 5). Sahi tilt dhundhne ke liye, pehle exactly East aur exactly North chalo — woh slopes partials hain, aur woh board ki tilt gradient hone ke liye force karte hain (Step 6). Sab ek saath rakho: zameen = board + ek sliver too small to matter, jo yeh bhi prove karta hai ki ek differentiable pahaadi mein cliffs nahi hain (Step 7). Lekin savdhaan raho — ek pahaad East aur North par perfectly flat ho sakta hai phir bhi diagonal par ridge chhupa sakta hai; tab koi board fit nahi hota, chahe partials exist karein (Step 8). Safe check: agar East/North slopes smoothly vary karein (pahaad $C^1$ hai), ek fitting board guaranteed hai — aur yeh almost har function ko cover karta hai jo tum kabhi miloge (Step 9).

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Recall

Error ko $\Vert \mathbf{h}\Vert$ se divide kyun karna chahiye? ::: Taaki error higher-order ho ($o(\Vert \mathbf{h}\Vert )$) — yeh unique tangent plane select karta hai, sirf koi bhi plane nahi jo touch kare. Agar $f$ differentiable hai, uski tilt $\mathbf{m}$ kya honi chahiye, aur kyun? ::: Gradient $(f_x,f_y)$, do axes par approach karne se forced (har ek partial derivative deta hai). Step 8 mein, $f_x=f_y=0$ hone ke bawajood $f$ differentiable kyun nahi hai? ::: Woh origin par continuous nahi hai ($y=x$ par $\to \frac{1}{2}$); differentiable ⇒ continuous fail hota hai. $C^1$ kya deta hai? ::: Continuous partials ⇒ automatically differentiable — koi limit check nahi chahiye.

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Connections

• 4.4.06 Differentiability in multiple variables (Hinglish)
• Partial Derivatives
• Gradient and Directional Derivatives
• Tangent Planes and Linear Approximation
• Continuity in Multiple Variables
• Total Derivative / Jacobian
• C1 and Smooth Functions
• Chain Rule (Multivariable)