Visual walkthrough — Chain rule for multivariable functions — all cases

4.4.7 · D2 · Maths › Multivariable Calculus › Chain rule for multivariable functions — all cases

Pehli line se pehle, teen seedhe saade words:

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Step 1 — Ek walker, ek clock

KYA. Landscape par khade rehne ki jagah, ek aisa walker imagine karo jiska position time $t$ ke tick karne ke saath move karta hai. Unka east coordinate ek function $x=x(t)$ hai aur north coordinate $y=y(t)$ hai. Jab $t$ change hota hai, point $(x,y)$ ground par ek path trace karta hai, aur unke paon ke neeche ki height $z$ upar-neeche hoti rehti hai.

KYU. Yahi Case 1 ka poora sawaal hai: clock tick karne ke saath height kitni tez change hoti hai? Hume ek number chahiye, $\frac{dz}{dt}$ — ek ordinary derivative ($d$, $\partial$ nahi), kyunki end mein $z$ sirf ek variable $t$ par depend karta hai.

PICTURE. Neeche wala curve walker ki trail hai map par. Height ground ka color hai. Baaki poora page is trail ke EK tiny green segment mein zoom karta hai.

• $t$ ::: clock; Case 1 mein ek maatra truly independent variable.
• $(x(t),y(t))$ ::: walker ki position — dono coordinates $t$ ke move karne ke saath move karte hain.

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Step 2 — Time mein ek tiny step lo

KYA. Clock ko thodi si amount se aage badhao, jise $\Delta t$ likha jaata hai (symbol $\Delta$, capital Greek "D", matlab "mein ek chhota sa change"). Walker move karta hai. Unka east coordinate $\Delta x$ se change hota hai, north coordinate $\Delta y$ se, aur height $\Delta z$ se change hoti hai.

KYU. Derivatives tiny changes se paida hote hain. Agar hum samjhein ki ek chhota $\Delta t$ $\Delta z$ kaise produce karta hai, toh hum exact rate paane ke liye divide aur shrink kar sakte hain. Yeh wahi idea hai jaise Single-variable Chain Rule, bas do coordinates ek saath move ho rahe hain.

PICTURE. Step 1 ke green box mein zoom karo. Time mein step, map par ek horizontal nudge $\Delta x$ aur ek vertical nudge $\Delta y$ mein split ho jaati hai.

• $\Delta t$ ::: clock ka tiny tick (cause).
• $\Delta x,\Delta y$ ::: east aur north nudges jo yeh produce karta hai (do "doors").
• $\Delta z$ ::: height mein resulting change (woh effect jo hume chahiye).

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Step 3 — Height ek nudge par kaise respond karti hai: do ramps

KYA. Ek moment ke liye time freeze karo aur poochho: nudges $\Delta x$ aur $\Delta y$ diye hue, height kitni change hogi? Kyunki $f$ differentiable hai, apne point ke paas curved ground flat lagti hai — ek tilted plane. Ek plane par height change bas yeh hai:

$\Delta z\approx \underset{\text{east steepness}}{\underbrace{\frac{\partial z}{\partial x}}}\underset{\text{east nudge}}{\underbrace{\Delta x}}+\underset{\text{north steepness}}{\underbrace{\frac{\partial z}{\partial y}}}\underset{\text{north nudge}}{\underbrace{\Delta y}}.$

YEH TOOL KYU. Ek plane kyun, aur do effects add kyun karte hain? Kyunki "differentiable" precisely yeh promise hai ki pass mein surface ko ek flat tilted sheet se well-approximate kiya ja sakta hai — yeh sheet Total Differential hai. Ek flat sheet par, east phir north move karne se height change hoti hai (east-steepness × east-distance) plus (north-steepness × north-distance). Independent directions independently contribute karte hain, isliye hum sum karte hain. Isliye chain rule add karta hai — addition local plane ki flatness se inherit hoti hai.

PICTURE. Point ke upar tilted plane. East chalna ek ramp climb karta hai; north chalna doosra; total rise dono ramp-rises ka sum hai.

• $\frac{\partial z}{\partial x}\Delta x$ ::: east door se rise = (east steepness) × (east nudge).
• $\frac{\partial z}{\partial y}\Delta y$ ::: north door se rise = (north steepness) × (north nudge).

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Step 4 — Honest version: leftover error ko naam do

KYA. $\approx$ ek chhota sa jhooth chupaata hai: ground exactly ek plane nahi hai. Differentiability hume truth exactly likhne deti hai leftover terms add karke:

$\Delta z=\frac{\partial z}{\partial x}\Delta x+\frac{\partial z}{\partial y}\Delta y+\underset{\text{leftover curvature}}{\underbrace{{\epsilon }_1\Delta x+{\epsilon }_2\Delta y}},$

jahan ${\epsilon }_1,{\epsilon }_2$ (Greek "epsilon", matlab "ek vanishingly small number") dono $0$ par shrink ho jaate hain jab nudges $0$ par shrink hote hain.

KYU. Hume cheat nahi karna chahiye. Error terms ko explicitly rakhne ka matlab hai ki hamara final rule exact hoga, approximate nahi. Poora trick yeh dikhana hoga ki ye leftovers limit mein mar jaate hain.

PICTURE. True curved surface aur flat plane ke beech ka gap leftover hai. Jab step shrink hota hai, gap step se tez shrink hota hai.

• ${\epsilon }_1,{\epsilon }_2$ ::: unit nudge per curvature error — dono $\to 0$ jab $(\Delta x,\Delta y)\to (0,0)$.

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Step 5 — Clock tick se divide karo

KYA. Hume rate per unit time chahiye, isliye exact equation ke har term ko $\Delta t$ se divide karo:

$\frac{\Delta z}{\Delta t}=\frac{\partial z}{\partial x}\frac{\Delta x}{\Delta t}+\frac{\partial z}{\partial y}\frac{\Delta y}{\Delta t}+{\epsilon }_1\frac{\Delta x}{\Delta t}+{\epsilon }_2\frac{\Delta y}{\Delta t}.$

KYU. $\frac{dz}{dt}$ literally matlab hai "$z$ mein change per change in $t$." Isliye hum sab kuch per tick $\Delta t$ measure karte hain. Notice karo $\frac{\Delta x}{\Delta t}$ = "walker ki east-speed" aur $\frac{\Delta y}{\Delta t}$ = "north-speed."

PICTURE. Har nudge ek speed ban jaata hai: map nudges ko usi clock tick se divide karo.

• $\frac{\Delta x}{\Delta t}\to \frac{dx}{dt}$ ::: walker kitni tez east jaata hai.
• $\frac{\Delta y}{\Delta t}\to \frac{dy}{dt}$ ::: walker kitni tez north jaata hai.

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Step 6 — Tick ko shrink karo: leftovers vanish hote hain, rule appear hota hai

KYA. Ab $\Delta t\to 0$ hone do. Jab tick shrink hota hai, nudges $\Delta x,\Delta y\to 0$ bhi (walker ka path continuous hai), isliye ${\epsilon }_1,{\epsilon }_2\to 0$. Meanwhile $\frac{\Delta x}{\Delta t}\to \frac{dx}{dt}$ finite rehta hai. Isliye do leftover terms hain (kuch $\to 0$) × (kuch finite) $\to 0$ aur mar jaate hain:

LIMIT KYU. Limit woh cheez hai jo approximate plane ko exact rate mein turn karti hai — curvature error sirf finite step size par matter karti thi, aur woh $\Delta t$ se tez evaporate ho jaati hai.

PICTURE. Do supply-roads output mein flow karti hain. Har road ek multiplication hai (steepness × speed); meeting point sum hai.

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Step 7 — Edge cases: pakde mat jaana

Har woh scenario jo reader encounter kar sakta hai:

(a) Ek frozen coordinate. Agar walker sirf east move karta hai, toh $\frac{dy}{dt}=0$ aur north path kuch contribute nahi karta — hum Single-variable Chain Rule $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}$ recover karte hain. Formula 1-D case ko contain karta hai; yeh kabhi contradict nahi karta.

(b) Zero steepness lekin tez chal rahe hain. Agar tum ek level contour ke along chalte ho (height us direction mein constant hai), toh matching steepness $0$ hai, isliye woh path $0$ add karta hai chahe tum kitni tez bhi chalo. Speed akele height nahi change karti — sirf speed times steepness karta hai.

(c) Do ultimate variables → $\partial$ use karo (Case 2). Agar $x=x(s,t)$ aur $y=y(s,t)$, toh $z$ end mein do variables $s,t$ par depend karta hai. $t$ freeze karo, $s$ wiggle karo: yeh exactly Step 6 hai ek slice par apply kiya gaya, isliye $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}.$ Ab hum $\partial$ likhte hain ($d$ nahi) kyunki ek companion variable $t$ survive karta hai.

(d) Sneaky direct path (implicit reappearance). Agar $w=f(x,y,t)$ aur $x,y$ $t$ par depend karte hain, toh $t$ ki taraf seedha ek teesra road hai: $\frac{dw}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial t}.$ Tree literally teen branches dikhata hai — direct waala miss karo aur tumhara answer galat hoga. Yeh Implicit Differentiation (multivariable) ke peeche ka engine hai.

PICTURE. Char mini-trees, ek per case, jo dikhate hain kaun se branches survive karte hain.

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Ek-picture summary

Ek single map par poori derivation: ek tiny wiggle $\Delta t$ do nudges mein split hoti hai, har nudge apna ramp climb karti hai, dono rises add hote hain, hum $\Delta t$ se divide karte hain, shrink karte hain, aur error wash out ho jaata hai — boxed rule reh jaata hai. Matrix language mein yeh Jacobian product $J_{z\circ x}=J_z{J}_x$ mein compress hota hai (dekho Gradient and Jacobian).

Recall Poore walkthrough ki Feynman retelling

Tum ek hilly landscape par hiking kar rahe ho, lekin tum directly control nahi karte ki tum kahan khade ho — ek clock karta hai. Har tick par, clock tumhe thoda east aur thoda north ek saath push karta hai. Yeh pata karne ke liye ki tumhari altitude per tick kitni tez change hoti hai: dekho hill east ki taraf kitni steep hai aur apni eastward speed se multiply karo — yahi east-move tumhe kitna lift karta hai. North ke liye bhi yahi karo. Hill perfectly flat nahi hai, isliye ek tiny curvature fudge hoti hai, lekin jab tick chhota hota hai toh fudge kuch nahi reh jaati. Dono lifts add karo aur tumhare paas altitude ki rate of change hai. Zyada doors (zyada intermediate variables) → zyada roads add karne ke liye; agar clock bhi landscape ko khud change karta hai, toh ek aur seedha road add karo. Yahi poora multivariable chain rule hai: paths ka sum karo, links ko multiply karo.

Recall Quick self-check

Error terms ${\epsilon }_1\Delta x/\Delta t$ vanish kyun hote hain? ::: Kyunki ${\epsilon }_1\to 0$ jab step shrink hota hai jabki $\Delta x/\Delta t\to dx/dt$ finite rehta hai; zero times finite zero hota hai. Case 2 mein $\partial z/\partial s$ kyun hai, $dz/ds$ nahi? ::: Kyunki ek doosra ultimate variable $t$ abhi bhi exist karta hai, isliye $z$ ek se zyada variable par depend karta hai. Ex-check: $z=x^2+y^2$, $x=r\cos \theta$, $y=r\sin \theta$ ke liye, $\partial z/\partial r$ kya hai? ::: $2r$ — chain rule direct answer $z=r^2$ se match karta hai.

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Connections

• Parent topic — cases ka full catalogue.
• Partial Derivatives — "steepness" ramps $\partial z/\partial x$.
• Total Differential — Step 3 ka tilted plane.
• Directional Derivative — Case 2c disguise mein ek directional derivative hai.
• Gradient and Jacobian — one-picture matrix compression.
• Implicit Differentiation (multivariable) — sneaky direct path (7d).
• Single-variable Chain Rule — one-road special case (7a).