Visual walkthrough — Chain rule — proof, composite function derivatives

4.1.16 · D2 · Maths › Calculus I — Limits & Derivatives › Chain rule — proof, composite function derivatives

Sab kuch ek pehle ke idea par tika hai: derivative as a limit of a difference quotient. Agar yeh phrase abhi bhi kuch nahi kehta, toh Step 1 isse zero se rebuild karta hai.

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Step 1 — "Rate" hota kya hai (ek tiny secant ki slope)

KYA HAI. Koi bhi function lo — ek machine jo ek number $x$ khaati hai aur ek number output karti hai. Input ko ek tiny amount se nudge karo jise hum $\Delta x$ ("delta-x", padho "$x$ mein ek chhoti si change") kehte hain. Output kuch amount se change hoti hai. Rate yeh hai:

$$\underset{\text{"rise over run"}}{\underbrace{\frac{\text{change in output}}{\text{change in input}}}}=\frac{f(x+\Delta x)-f(x)}{\Delta x}.$$

Term by term: $x+\Delta x$ nudged input hai; $f(x+\Delta x)-f(x)$ batata hai ki output kitna hila (rise); $\Delta x$ batata hai ki input kitna hila (run). Yeh fraction secant line ki slope hai jo do dots ko join karti hai.

KYUN. Hume ek honest number chahiye "output kitni tezi se input ke response mein badlta hai" ke liye. Slope = rise/run exactly wahi hai. Jaise-jaise $\Delta x$ $0$ ki taraf shrink hota hai, secant curve ke upar chipak jaata hai aur uski slope derivative $f^{\prime }(x)$ ban jaati hai — instantaneous rate. Woh shrinking limit ${\lim }_{\Delta x\to 0}$ likhi jaati hai.

PICTURE. Blue curve; do dots jo $\Delta x$ ki distance par hain; orange secant; run aur rise labelled. Dekho secant green tangent ki taraf tilt hoti hai jab dots paas aate hain.

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Step 2 — Ek composite matlab DO machines ek ke baad ek

KYA HAI. Ab do machines chain karo. Inner machine $g$, $x$ khaati hai aur ek middle value output karti hai jise hum $u=g(x)$ kehte hain ("$u$" sirf inner output ka ek nickname hai). Outer machine $f$ woh $u$ khaati hai aur final number $y=f(u)$ output karti hai. Poora pipeline hai

$$h(x)=f\left(\underset{=u}{\underbrace{g(x)}}\right).$$

Term by term: $g(x)$ pehle run karta hai aur $u$ par land karta hai; $f$ us $u$ par kaam karta hai; $h$ do-step journey $x\to u\to y$ ka naam hai.

KYUN. Zyaadatar real functions secretly aise hi stacked hote hain — $\sin (x^2)$, $e^{\sin x}$, $\sqrt{\cos 4x}$. Agar hum ek stack ko differentiate kar sakte hain, toh hum almost kuch bhi differentiate kar sakte hain. Hume pehle middle station $u$ ka naam dena hoga, kyunki poora proof is baare mein hai ki wiggle us se kaise travel karti hai.

PICTURE. Ek conveyor belt par do boxes: $x$, $g$-box mein enter hota hai, $u$ ke roop mein middle rail par exit hota hai, $f$-box mein enter hota hai, $y$ ke roop mein exit hota hai.

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Step 3 — Input ko wiggle karo; wiggle ko relay hote dekho

KYA HAI. Input ko nudge karo: $x\to x+\Delta x$. Isse middle station ek amount se move karne par majboor hota hai jise hum inner increment kehte hain

$$\Delta u=g(x+\Delta x)-g(x).$$

Term by term: $\Delta u$ batata hai ki middle rail kitna slide hua kyunki humne $x$ ko push kiya. Phir outer machine us $\Delta u$ ko final move $\Delta y=f(u+\Delta u)-f(u)$ mein badalta hai.

KYUN. Yahi "gear-ratio" ki kahani ka dil hai. Input par ek push ($\Delta x$) middle par ek push ($\Delta u$) ban jaati hai, jo output par ek push ($\Delta y$) ban jaati hai. Hum total ratio $\Delta y/\Delta x$ chahte hain, aur natural move hai ise middle se route karna.

PICTURE. Teen number-lines stack ki gayi hain: $x$-line ek chhote arrow $\Delta x$ ke saath; $u$-line ek bade arrow $\Delta u$ ke saath; $y$-line ek aur bade arrow $\Delta y$ ke saath. Har amplification ek machine hai.

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Step 4 — Naive split (middle se route karo)

KYA HAI. Woh rate likho jo hum chahte hain aur $\Delta u$ se multiply-and-divide karo:

$$\frac{\Delta y}{\Delta x}=\underset{\text{outer rate}\to f^{\prime }(u)}{\underbrace{\frac{f(u+\Delta u)-f(u)}{\Delta u}}}\cdot \underset{\text{inner rate}\to g^{\prime }(x)}{\underbrace{\frac{\Delta u}{\Delta x}}}.$$

Term by term: left fraction "final move per middle move" hai — jaise $\Delta u\to 0$ yeh $f^{\prime }(u)$ ban jaata hai. Right fraction "middle move per input move" hai — jaise $\Delta x\to 0$ yeh $g^{\prime }(x)$ ban jaata hai. Unka product total stretch hai.

KYUN. Yahi gear-ratio intuition algebraic roop mein hai: total speed-up $=$ (outer gear ratio) $\times$ (inner gear ratio). Yeh sahi answer $h^{\prime }(x)=f^{\prime }(u)g^{\prime }(x)$ deta hai.

PICTURE. Do meshed gears: chhota input gear (ratio $g^{\prime }(x)$) ek bade output gear (ratio $f^{\prime }(u)$) ko drive karta hai; ek caption likha hai "total $=$ product". $\Delta u$ ke denominator mein ek red warning tag latka hua hai.

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Step 5 — Hole: jab middle rail HILTI nahi

KYA HAI. Step 4 ka split $\Delta u$ se divide karta hai. Lekin $\Delta u=g(x+\Delta x)-g(x)$ exactly $0$ ho sakta hai jab bhi $\Delta x\ne 0$ ho. Do tarike se yeh hota hai:

• Flat patch: agar $g$ ek stretch par constant hai, toh har nudge $\Delta u=0$ deta hai.
• Wiggly zero-crosser: $g(x)=x^2\sin (1/x)$ $0$ ke paas $\Delta u=0$ infinitely baar hit karta hai.

$0$ se divide karna undefined hai, toh "one-line proof" abhi tak proof nahi hai.

KYUN. Ek theorem har case mein hold karna chahiye, sirf friendly wale cases mein nahi. Hum $\Delta u=0$ ko wave away nahi kar sakte; hume aisa machinery banana hoga jo isse survive kare. Step 6 ka poora reason yahi hai.

PICTURE. Left panel: $g$ ka ek flat plateau — do dots alag $x$ par lekin identical height par, toh $\Delta u=0$. Right panel: spiky $x^2\sin (1/x)$ origin ke paas baar baar zero cross karta hua, har crossing ek jagah hai jahan $\Delta u=0$.

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Step 6 — Repair: ek "error function" jo divide ki jagah multiply karti hai

KYA HAI. Kyunki $f$, $u$ par differentiable hai, ek bookkeeping function $E$ define karo jo maapti hai ki secant slope true tangent slope $f^{\prime }(u)$ se kitni door hai:

$$E(k)=\{\begin{array}{ll}\frac{f(u+k)-f(u)}{k}-f^{\prime }(u),& k\ne 0,\\ 0,& k=0.\end{array}$$

Term by term: $k$ "outer machine ka input nudge" ke liye ek stand-in hai (baad mein hum $k=\Delta u$ plug karenge). Pehla branch secant slope minus tangent slope hai — error. Doosra branch hole ko patch karta hai: hum simply define karte hain ki error $0$ hai jab $k=0$.

Kyunki $f^{\prime }(u)$ precisely us secant slope ki limit hai, ${\lim }_{k\to 0}E(k)=0=E(0)$, toh $E$ $0$ par continuous hai (koi jump nahi). Ab pehle branch ko ek aisi form mein rearrange karo jo sab $k$ ke liye valid ho (including $k=0$, jahan dono sides $0$ hain):

$$f(u+k)-f(u)=(f^{\prime }(u)+E(k))k.$$

KYUN. Dekho: $k$ ab upar hai, multiply hua — kabhi neeche nahi. Yahi magic hai. Identity tab bhi hold karti hai jab $k=0$ (yeh $0=0$ padhta hai). Toh hum safely troublesome $k=\Delta u$ substitute kar sakte hain.

PICTURE. Tangent line (green, slope $f^{\prime }(u)$) aur ek secant (orange). Unke beech ki vertical gap, $k$ ke per unit scale ki gayi, exactly $E(k)$ hai; jaise $k\to 0$ gap collapse hota hai — $E$, $0$ par continuous hai.

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Step 7 — Substitute karo, $\Delta x$ se divide karo (legal!), limit lo

KYA HAI. $k=\Delta u=g(x+\Delta x)-g(x)$ ko Step 6 ki multiply-form mein dalo:

$$f(g(x+\Delta x))-f(g(x))=(f^{\prime }(u)+E(\Delta u))\Delta u.$$

Ab $\Delta x$ se divide karo — allowed hai, kyunki limit mein $\Delta x\ne 0$:

$$\frac{h(x+\Delta x)-h(x)}{\Delta x}=(f^{\prime }(u)+E(\Delta u))\cdot \frac{\Delta u}{\Delta x}.$$

$\Delta x\to 0$ hone do aur har piece track karo:

• $g$ differentiable $\Rightarrow$ $g$ continuous $\Rightarrow$ $\Delta u\to 0$, toh $E(\Delta u)\to E(0)=0$. (Yeh differentiable ⇒ continuous fact hai.)
• $\frac{\Delta u}{\Delta x}\to g^{\prime }(x)$ — inner rate.

Toh poori cheez $(f^{\prime }(u)+0)g^{\prime }(x)$ ki taraf tend karti hai.

KYUN. Humne sirf $\Delta x$ se divide kiya (kabhi bhi $\Delta u$ se nahi), toh koi bhi step illegal nahi tha. Error term $E(\Delta u)$ politely vanish ho jaata hai kyunki $g$, $\Delta u$ ko $0$ ki taraf khenchta hai. Jo survive karta hai woh exactly do rates ka product hai.

PICTURE. Ek collapsing bar chart: $f^{\prime }(u)$, $E(\Delta u)$, $\Delta u/\Delta x$ ke liye teen bars; jaise $\Delta x\to 0$ middle bar $E$ kuch nahi reh jaata aur surviving product $f^{\prime }(u)g^{\prime }(x)$ par settle ho jaata hai.

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Step 8 — Ek real example par dekho: $h(x)=\sin (x^2)$

KYA HAI. Inner $g(x)=x^2$ toh $g^{\prime }(x)=2x$; outer $f(u)=\sin u$ toh $f^{\prime }(u)=\cos u$. Assemble karo:

$$h^{\prime }(x)=\underset{f^{\prime }(g(x))}{\underbrace{\cos (x^2)}}\cdot \underset{g^{\prime }(x)}{\underbrace{2x}}=2x\cos (x^2).$$

KYUN. Yeh abstract product ko concrete banata hai aur classic trap expose karta hai: yeh $\cos (x^2)$ hai (inside intact raha), nahi $\cos (2x)$. Inner rate $2x$ ek alag multiplied factor hai, cosine ke andar kabhi nahi ghusta.

PICTURE. Upar: $h(x)=\sin (x^2)$ jaise $x$ badhta hai apni oscillations ko speed up karta hai. Neeche: computed $h^{\prime }(x)=2x\cos (x^2)$ $h$ ki numerical slope par overlaid — woh ek doosre ke upar land karte hain, formula confirm karte hain.

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Ek-picture summary

Sab kuch ek canvas par: input nudge $\Delta x$ inner gear mein enter karta hai (ratio $g^{\prime }(x)$), middle nudge $\Delta u$ ban jaata hai, outer gear mein enter karta hai (ratio $f^{\prime }(u)$), output nudge $\Delta y$ ban jaata hai — aur total ratio product hai. Chhota error term $E$ jo proof ko honest banata tha, woh zero ki taraf fade hota hua drawn hai.

Recall Feynman retelling — poora walkthrough simple words mein

Ek composite do machines hain ek belt par. Input ko thoda push karo (Step 1 ka rate idea). Woh push ek middle rail $u=g(x)$ tak pahuncha aur use $\Delta u$ se slide kiya (Step 3). Middle push phir output tak pahuncha aur use $\Delta y$ se slide kiya. "Output-per-input" paane ke liye tum middle se route karte ho: (output-per-middle) × (middle-per-input) (Step 4) — gear ratios multiply karte hain. Lekin ek trap hai: middle rail ruk sakti hai ($\Delta u=0$) jab bhi tumne input push kiya ho, aur tum ek ruki hui rail se divide nahi kar sakte (Step 5). Fix: $\Delta u$ se divide karne ki jagah, outer machine ke response ko $(f^{\prime }(u)+E)\Delta u$ likho jahan $E$ ek chhoti si "error" hai jise hum define karte hain ki $0$ ho jab kuch nahi hilta (Step 6). Ab $\Delta u$ multiply hua hai, kabhi denominator mein nahi, toh safe hai. Sirf $\Delta x$ se divide karo, use shrink karo, aur $E$ fade ho jaata hai kyunki ek differentiable $g$ jump nahi kar sakta — toh $\Delta u$ bhi aaram se zero ho jaata hai (Step 7). Jo bachta hai woh exactly $f^{\prime }(g(x))g^{\prime }(x)$ hai. $\sin (x^2)$ par test karo: $\cos (x^2)\cdot 2x$, inside intact rakho (Step 8).

Connections

• Derivative — limit definition — Step 1 literally yahi hai; poora proof ek difference quotient ki limit hai.
• Continuity implies differentiability fails converse — Step 7 "differentiable $g$ ⇒ continuous $g$" use karta hai $\Delta u\to 0$ force karne ke liye.
• Product rule aur Quotient rule — sibling rules jo aksar chain rule ke saath stack hote hain.
• Implicit differentiation — chain rule $y(x)$ par apply hua.
• Related rates — wahi $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ multiplication ek time setting mein.
• Inverse function derivative — chain-rule $f(f^{-1}(x))=x$.