Product rule — proof — Hinglish notes

4.1.14 · Maths › Calculus I — Limits & Derivatives

WHAT — statement kya hai

WHY corner term vanish karta hai (Dual Coding)

Picture mein area $uv$ ek rectangle ke roop mein dikhaya gaya hai. $x \to x + h$ badhane par $u \to u + Δ u$ aur $v \to v + Δ v$ ho jaata hai. Naya area 4 pieces mein split hota hai:

Piece	Area
original	$uv$
right strip (wider)	$v Δ u$
top strip (taller)	$u Δ v$
corner (tiny)	$Δ u Δ v$

Area mein change hai $Δ (uv) = v Δ u + u Δ v + Δ u Δ v$ . Corner do chhoti cheezein ka product hai → second-order tiny → limit mein vanish ho jaata hai.

HOW — Scratch se Derivation (limit definition se)

Formula First principles se step-by-step

$f = uv$ ke derivative ki definition se start karo: $f^{'} (x) = lim_{h \to 0} \frac{u ( x + h ) v ( x + h ) - u ( x ) v ( x )}{h}$

Step 1 — Ek clever middle term add aur subtract karo. Numerator mein $- u (x + h) v (x) + u (x + h) v (x) = 0$ insert karo: $u (x + h) v (x + h) - u (x) v (x) = height changes u (x + h) [v (x + h) - v (x)] + width changes v (x) [u (x + h) - u (x)]$ Yeh step kyun? Yeh "strips split karo" ka algebra version hai. Hum force karte hain ki ek group mein $Δ v$ aur doosre mein $Δ u$ aaye taaki dono known derivatives ban jaayein.

Step 2 — $h$ se divide karo aur limit split karo. $f^{'} (x) = lim_{h \to 0} [u (x + h) \frac{v ( x + h ) - v ( x )}{h} + v (x) \frac{u ( x + h ) - u ( x )}{h}]$ Yeh step kyun? Har fraction exactly ek difference quotient hai — derivative ka seed.

Step 3 — Limits piece by piece lo.

$h \to 0 lim \frac{v ( x + h ) - v ( x )}{h} = v^{'} (x)$ (definition).
$h \to 0 lim \frac{u ( x + h ) - u ( x )}{h} = u^{'} (x)$ (definition).
$h \to 0 lim u (x + h) = u (x)$ — iske liye zaroori hai ki $u$ , $x$ par continuous ho, jo guaranteed hai kyunki $u$ differentiable hai (differentiable ⇒ continuous).
$v (x)$ , $h$ mein constant hai.

Sabko milaakar: $f^{'} (x) = u (x) v^{'} (x) + v (x) u^{'} (x) ■$

Worked examples

Flashcards

Product rule state karo.

(uv)^{'} = u^{'} v + u v^{'}

Product rule ke limit proof mein kaun sa trick use hota hai?

Numerator mein

u (x + h) v (x)

add aur subtract karo.

Proof mein

lim_{h \to 0} u (x + h) = u (x)

kyun valid hai?

Kyunki

u

differentiable hai, isliye continuous bhi hai.

(uv)^{'} = u^{'} v^{'}

kyun galat hai (counterexample do)?

u = v = x

ke saath:

(x^{2})^{'} = 2 x

par

u^{'} v^{'} = 1

Geometrically, product rule ka "corner term"

Δ u Δ v

kya represent karta hai?

Ek second-order tiny area jo limit mein vanish ho jaata hai.

x^{2} sin x

ko differentiate karo.

2 x sin x + x^{2} cos x

(uv w)^{'}

kya hai?

u^{'} v w + u v^{'} w + uv w^{'}

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho ek rectangle mein stickers hain — itne wide, itne tall. Ab dono width aur height badhne lagti hain. Thodi der baad tumne ek thin strip side mein aur ek thin strip upar add ki hai. Ek teeny corner square bhi hai jahan dono strips overlap hoti hain, par woh itna small hai ki hum ignore kar dete hain. Toh total stickers kitni tezi se badhte hain = (kitni tezi se wide hota hai)×height

width×(kitni tezi se tall hota hai). Yahi product rule hai!

Connections

Limit definition of the derivative — is proof ka engine.
Differentiability implies continuity — Step 3 mein hidden lemma.
Quotient rule — product rule + chain rule se derived.
Chain rule — compositions ke liye iske saath partner hai.
Power rule — product rule se bootstrap kiya gaya (Example 2).
Leibniz rule (nth derivative of a product) — grand generalisation.