Product rule — proof: Question bank — Hinglish deep dive

4.1.14 · D5 · Maths › Calculus I — Limits & Derivatives › Product rule — proof

Is page ke vocabulary (ek baar banao, har jagah use karo)

Kisi bhi trap se pehle, hum symbols fix kar lete hain taaki kuch undefined na lage.

True ya false — justify karo

Product rule hai $(uv)^{'} = u^{'} v + u v^{'}$ ; sirf "true/false" mat kaho — kyun bhi batao.

Har item mein tumse verdict defend karne ko kaha gaya hai, sirf state karne ko nahi.

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

sabhi differentiable

u, v

ke liye.

False — linearity sirf addition ko respect karti hai. Test karo

u = v = x

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

lekin

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

. Missing cross-terms

u^{'} v + u v^{'}

exactly wahi hain jo products ko sums se alag banate hain.

Agar

u

aur

v

dono differentiable hain, toh

uv

automatically differentiable hoga.

True — proof upar di gayi char pieces se

h \to 0 lim \frac{uv ( x + h ) - uv ( x )}{h}

construct karta hai aur dikhata hai ki woh exist karta hai,

u^{'} v + u v^{'}

ke equal; limit ka existence hi differentiability hai.

Product rule ko

u

aur

v

ka continuous hona chahiye, lekin differentiable nahi.

False — dono ko differentiable hona chahiye; continuity akele

u^{'}

aur

v^{'}

ko undefined chhod deti hai, toh

u^{'} v + u v^{'}

ka koi matlab nahi. (Differentiability continuity to deti hi hai free mein, dekho Differentiability implies continuity.)

Agar

u

constant hai, toh product rule collapse hokar

(c v)^{'} = c v^{'}

ban jaata hai.

True —

u = c

ke saath,

u^{'} = 0

, toh

u^{'} v + u v^{'} = 0 \cdot v + c v^{'} = c v^{'}

. Constant-multiple rule, product rule ka ek special case hai.

Corner term

Δ u Δ v

vanish hota hai kyunki

Δ u

aur

Δ v

literally zero hain.

False — woh chhote hain lekin nonzero; term isliye vanish hoti hai kyunki

\frac{Δ u Δ v}{h} = Δ u \cdot \frac{Δ v}{h} \to 0 \cdot v^{'} = 0

(spare

Δ u \to 0

use maar deta hai), ye ek second-order effect hai, exact zero nahi.

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

tab bhi hold karta hai agar interest ke point par

v (x) = 0

ho.

True — rule koi assumption nahi karta ki

u

v

nonzero ho; jis point par

v = 0

ho wahan ye simply padhta hai

(uv)^{'} = u^{'} \cdot 0 + u v^{'} = u v^{'}

u \cdot u = u^{2}

par product rule apply karne par

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

milta hai.

True —

v = u

set karo:

(uu)^{'} = u^{'} u + u u^{'} = 2 u u^{'}

. (Chain rule

\frac{d}{d x} u^{2} = 2 u \cdot u^{'}

aur

x^{2}

par Power rule same number dete hain — dena hi chahiye, kyunki teeno ek hi function describe karte hain — lekin yahan humne ise purely product rule se derive kiya.)

Teen factors ke liye,

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

False —

(uv) \cdot w

group karo aur do baar apply karo:

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

. Har factor exactly ek baar differentiate hone ki baari leta hai jabki baaki waise rehte hain.

Error dhundho

Har line mein ek flawed statement ya step hai; reveal us flaw ka naam batata hai.

"Proof mein hum

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

bina kisi justification ke likhte hain — ye obvious hai."

Flaw ye hai ki skip kar diya kyun: ye step use karta hai ki

u

x

par continuous hai, yaani

Δ u \to 0

, jo tabhi hota hai jab

u

wahan differentiable ho. Is dependence ko name karna hi proof ka pura subtle point hai.

"Product rule prove karne ke liye, numerator mein

u (x) v (x)

add aur subtract karo."

Galat middle term — tumhe

u (x + h) v (x)

add/subtract karna chahiye (ya

u (x) v (x + h)

u (x) v (x)

use karna sirf wahi term phir se daalna hai jo pehle se hai aur koi difference quotients

Δ u / h, Δ v / h

produce nahi karta.

"Kyunki

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

, usi logic se

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

Analogy galat hai: sum rule limits ke addition par distribute hone se aata hai; multiplication limit par us tarah distribute nahi hoti, aur counterexample

u = v = x

ise turant khatam kar deta hai.

Δ u Δ v

drop kiya jaata hai kyunki hume chhote numbers ignore karne ki permission hai."

Vague aur galat reasoning — hum ise "ignore" nahi karte, hum iska limit compute karte hain:

\frac{Δ u Δ v}{h} = Δ u \cdot \frac{Δ v}{h} \to 0 \cdot v^{'} = 0

. Rigor, hand-waving nahi, ise door bhejta hai.

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

kyunki tum product ko ek bade function ki tarah differentiate karte ho."

Sirf teen mein se ek term — ek product ek single black box nahi hai; teeno factors mein se har ek ko baari baari differentiate karna hoga, jo

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

deta hai.

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

Sirf doosra half sahi hai, aur woh bhi galat jagah hai. Correct:

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

. Writer ne dono factors ek saath differentiate kar diye, classic product-rule blunder.

"Proof har piece ka limit check karne se pehle limit ko do pieces mein split karta hai."

Limit ko sum/product mein split karna tab hi valid hai jab har piece ka limit exist kare (aur product

u (x + h) \cdot \frac{Δ v}{h}

split hota hai kyunki dono factors converge karte hain). Order matter karta hai: existence split ko justify karta hai.

Why questions

Hum same middle term add aur subtract kyun karte hain instead of koi naya quantity add karne ke?

Same value add aur subtract karne se kuch nahi badalta (net zero), toh expression equal rehta hai — lekin ise regroup karne par ek bracket mein

Δ v

aur doosre mein

Δ u

manufacture hota hai, dono difference quotients.

Middle term

u (x + h)

use kyun karta hai,

u (x)

nahi?

Taaki ek bracket

u (x + h) [v (x + h) - v (x)] = u (x + h) Δ v

ek clean

Δ v

ban sake; factor

u (x + h)

ko phir limit mein

u (x)

banne ke liye continuity chahiye — woh lemma jis par proof rely karta hai.

Product rule "rectangle picture" se itna exactly mirror kyun karta hai?

Do surviving strips

v Δ u

aur

u Δ v

(figure dekho) do difference-quotient groups ke geometric twins hain; vanishing corner dropped second-order term hai. Algebra aur area ek hi kahani sunate hain.

Differentiability, sirf continuity nahi, dono

u

aur

v

se kyun chahiye?

Final answer mein

u^{'}

aur

v^{'}

hain; agar koi bhi derivative exist karna band kar de, toh

(uv)^{'}

define karne wala limit exist karna zaroori nahi, toh formula ka koi matlab nahi hoga.

Hum Power rule ko product rule se bootstrap kyun kar sakte hain lekin doosra way definition ke roop mein nahi?

x^{n} = x \cdot x^{n - 1}

likhna aur induction hypothesis ke saath product rule apply karna

n x^{n - 1}

deta hai; lekin ye base case

(x)^{'} = 1

ko first principles se presuppose karta hai — product rule induction organize karta hai, seed replace nahi karta.

Quotient rule product rule se kyun "aata hai"?

u / v = u \cdot v^{- 1}

likho; product rule aur Chain rule (

v^{- 1}

differentiate karne ke liye) mil kar quotient rule banaate hain. Ye koi independent axiom nahi hai.

Corner term conceptually kyun matter karta hai jabki woh vanish ho jaata hai?

Ye honest reason hai ki products sums se alag kyun behave karte hain: do growing quantities ka cross-interaction. Uska first-order shadow

u^{'} v + u v^{'}

ke roop mein bachta hai; sirf uska second-order part marta hai.

Edge cases

Agar kisi point par

v^{'} (x) = 0

ho (v momentarily flat), toh rule kya deta hai?

(uv)^{'} = u^{'} v + u \cdot 0 = u^{'} v

— height is waqt stretch nahi kar rahi, toh saari area growth sirf widening strip se aa rahi hai.

Jis point par dono

u = 0

aur

v = 0

hoon, wahan

(uv)^{'}

kya hai?

(uv)^{'} = u^{'} \cdot 0 + 0 \cdot v^{'} = 0

— chahe

u^{'}, v^{'} \neq = 0

bhi hoon, product ki rate of change wahan zero hai kyunki rectangle ki dono "sides" ki length zero hai.

Kya product rule tab bhi hold karta hai agar

u

differentiable ho lekin

v

x

par sirf continuous ho (differentiable nahi)?

Nahi — formula

u^{'} v + u v^{'}

ke liye

v^{'} (x)

ka exist karna zaroori hai. Agar

v

wahan corner (kona) rakhti hai, toh

uv

differentiable ho bhi sakta hai ya nahi bhi, aur product rule apply nahi hota.

(uv)^{'}

ka kya hoga jab ek factor constant ki taraf jaaye, maano

u \to c

u^{'} \to 0

, toh

(uv)^{'} \to 0 \cdot v + c v^{'} = c v^{'}

— constant-multiple rule ko smoothly recover karta hai ek limiting/boundary case ke roop mein.

n

factors ke product mein kitne terms aate hain aur higher-order pieces kyun survive nahi karte?

Exactly

n

terms, har ek ek factor differentiate karta hai:

\sum_{k = 1}^{n} f_{1} \dots f_{k}^{'} \dots f_{n}

\prod (f_{k} + Δ f_{k}) - \prod f_{k}

expand karo, har term kuch

Δ f_{k}

's ka product hai;

h

se divide karne par, do ya zyada

Δ

's wali term (finite)

\times (Δ f_{j} \to 0)

padhti hai, toh woh mar jaati hai — sirf single

Δ

wali terms (ek derivative, baaki undisturbed) survive karti hain, bilkul do-factor case ke single corner ki tarah.

Kya

(uv)^{'}

us point par defined hai jahan

u

v

jump kare (discontinuity)?

Nahi — jump ka matlab hai woh factor continuous bhi nahi hai, isliye differentiable bhi nahi, toh na factor ka derivative exist karta hai na product ka.

Recall Har trap ki one-line summary

Product rule ke cross-terms ( $u^{'} v + u v^{'}$ , kabhi $u^{'} v^{'}$ nahi) rectangle ki do growing sides se aate hain; har strip ek difference quotient $Δ u / h$ ya $Δ v / h$ hai, jabki corner $Δ u Δ v$ marta hai kyunki ek spare $Δ \to 0$ . Proof differentiability ⇒ continuity par lean karta hai; har extra factor ek term add karta hai (ek factor differentiate karo ek baar, higher- $Δ$ terms vanish ho jaate hain), aur ye zeros, flats, aur constants par gracefully degenerate ho jaata hai.

Connections

Product rule — proof — woh parent jise yahan har trap refer karta hai.
Limit definition of the derivative — jahan difference quotients $Δ u / h$ rehte hain.
Differentiability implies continuity — woh hidden lemma jise kai traps target karte hain.
Quotient rule — ek "why" item: product rule + chain rule.
Chain rule — $v^{- 1}$ ko quotient rule mein turn karne ke liye chahiye.
Power rule — product rule se bootstrapped, replace nahi kiya gaya.
Leibniz rule (nth derivative of a product) — large- $n$ edge case generalised.