Quotient rule — proof: Question bank — Hinglish deep dive

4.1.15 · D5 · Maths › Calculus I — Limits & Derivatives › Quotient rule — proof

Throughout, hum $f = \frac{u}{v}$ (upar $u$ , neeche $v$ , jahan $v (x) \neq = 0$ ) ke liye rule likhte hain: $f^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}} .$ Yahan $u^{'}$ ka matlab hai "upar kitni tezi se badal raha hai" aur $v^{'}$ ka matlab hai "neeche kitni tezi se badal raha hai" — isse zyada exotic kuch nahi.

Jis proof ko yeh bank interrogate karta hai uske paanch moves hain; "Spot the error" items unhe naam se refer karte hain:

Recall Paanch proof steps (traps ke liye reference)

Step 1 — $f (x + h) - f (x) = \frac{u ( x + h )}{v ( x + h )} - \frac{u ( x )}{v ( x )}$ likho.
Step 2 — common denominator pe combine karo: top $= u (x + h) v (x) - u (x) v (x + h)$ , bottom $= v (x + h) v (x)$ .
Step 3 — top ke andar add-and-subtract zero ( $- u (x) v (x) + u (x) v (x)$ ) karo.
Step 4 — $h$ se factor aur divide karo, $u$ aur $v$ ke difference quotients expose karo.
Step 5 — $h \to 0$ jaane do ( $v$ ki continuity use karke) aur formula tak pahuncho.

True ya false — justify karo

Quotient rule bas product rule hai jo

u

aur

1/ v

pe apply hoti hai.

True —

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

likhke aur product rule plus

\frac{d}{d x} v^{- 1} = - v^{'} / v^{2}

use karke same formula milta hai. Product rule — proof aur Chain rule dekho.

Dono numerator terms ko swap karna,

\frac{u v ^{'} - u ^{'} v}{v ^{2}}

, same answer deta hai kyunki do cheezoon ka subtraction "symmetrically dikhta hai".

False — yeh poori derivative ka sign palat deta hai, increasing ko decreasing bana deta hai. Lo dHi pehle ka order zaroori hai.

Agar kisi point pe

v (x) = 0

ho, toh quotient rule wahan bhi apply hoti hai jab tak

u

aur

v

differentiable hain.

False — formula

v^{2}

se divide karta hai, isliye

v (x) \neq = 0

zaroori hai;

v

ke zero pe function

u / v

typically defined bhi nahi hota.

Quotient rule apply karne ke liye aapko alag se yeh assume karna padta hai ki

v

continuous hai, uske differentiable hone ke upar se.

False — aap continuity alag se kabhi assume nahi karte, kyunki differentiability pehle se hi continuity guarantee karta hai; proof sirf us free continuity ko

h \to 0

step mein use karta hai. Continuity dekho.

\frac{d}{d x} tan x = sec^{2} x

derive karna possible hai bina

cos^{2} x + sin^{2} x = 1

jaane.

False — Pythagorean identity hi numerator

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

1

mein collapse karta hai; uske bina aap sirf

\frac{1}{c o s ^{2} x}

tak pahunchte ho, jo waise bhi usi identity se aata hai.

Reciprocal rule

\frac{d}{d x} \frac{1}{v} = - \frac{v ^{'}}{v ^{2}}

ek alag theorem hai jo aapko apne aap memorise karna padega.

False — yeh quotient rule ka special case hai jahan

u = 1

u^{'} = 0

; rule mein yeh pehle se hi contained hai.

"Add and subtract zero" step numerator ki value change karta hai.

False — yeh

- u (x) v (x) + u (x) v (x) = 0

add karta hai, toh value unchanged rehti hai; sirf grouping badlti hai, jo do difference quotients reveal karta hai.

Agar dono

u

aur

v

constants hain, toh quotient rule correctly

f^{'} = 0

deta hai.

True —

u^{'} = v^{'} = 0

se numerator

0 \cdot v - u \cdot 0 = 0

ban jaata hai, jo match karta hai ki constant ratio ki slope zero hoti hai.

Error dhundo

(\frac{u}{v})^{'} = \frac{u ^{'}}{v ^{'}}

— top aur bottom differentiate karo."

Division term-wise nahi hoti. Test karo

f = \frac{x ^{2}}{x} = x

: sach yeh hai ki

1

hona chahiye, lekin

\frac{2 x}{1} = 2 x

aata hai. Yeh shortcut bilkul galat hai.

f^{'} = \frac{u ^{'} v - u v ^{'}}{v}

— neeche single

v

rehta hai jaise original fraction mein."

Galat — denominator

v^{2}

hona chahiye. Yeh Step 2 mein common-denominator product

v (x + h) v (x) \to v^{2}

se aata hai.

"Step 2 mein maine common denominator pe combine kiya aur

\frac{u ( x + h ) v ( x + h ) - u ( x ) v ( x )}{v ( x + h ) v ( x )}

mila."

Cross-multiplication galat hai: Step 2 mein top pe

u (x + h) v (x) - u (x) v (x + h)

hona chahiye. Har numerator term opposite denominator se pair karta hai (yaad rakho

h

chhota step hai:

x + h

x

"Kyunki

v

differentiable hai, limit mein

v (x + h) \to v (x + h)

, toh bottom

v (x + h)^{2}

ban jaata hai."

Jaise

h \to 0

v (x + h) \to v (x)

v (x + h)

nahi; bottom

v (x)^{2}

ki taraf tend karta hai.

h \to 0

apply karna har jagah karna hoga.

\frac{d}{d x} tan x = \frac{c o s x c o s x - s i n x s i n x}{c o s ^{2} x}

Sign error:

v^{'} = - sin x

, toh doosra term hai

- sin x \cdot (- sin x) = + sin^{2} x

. Sahi numerator hai

cos^{2} x + sin^{2} x = 1

, jo

sec^{2} x

deta hai.

\frac{1}{v}

ke liye maine quotient rule use ki aur

\frac{0 \cdot v - 1 \cdot v ^{'}}{v} = - \frac{v ^{'}}{v}

mila."

Denominator squared hona chahiye:

- \frac{v ^{'}}{v ^{2}}

. Square bhool jaana classic galti hai.

"Proof numerator ko

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

ke roop mein factor karta hai."

Factors swap hain: Step 4 deta hai

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

. Har change ko multiply karne wala constant-in-the-limit factor doosra function hota hai.

Why questions

Minus sign kyun hai, seedhe physical terms mein?

Kyunki fraction chhota hota hai jab uska neeche wala badh jaata hai; toh

v

ka badhna (

v^{'} > 0

)

f^{'}

ko neeche push karna chahiye, jo exactly subtracted

- u v^{'}

term hai.

Hum specifically

v^{2}

se kyun divide karte hain,

v

v^{3}

se kyun nahi?

1/ v

ko Chain rule se differentiate karne par

- v^{'} / v^{2}

milta hai;

v

ki ek power original fraction se rehti hai, ek aur

v^{- 1}

ki chain-rule derivative se aati hai.

Proof Limit definition of derivative se kyun shuru karta hai product rule quote karne ki jagah?

First principles se rule prove karne ke liye humein ek aisa argument chahiye jo doosre differentiation rules assume na kare; limit definition common engine hai.

Hum exactly

- u (x) v (x) + u (x) v (x)

hi kyun insert karte hain, koi aur zero kyun nahi?

Yeh particular pair hume allow karta hai ek bracket se

v (x)

aur doosre se

u (x)

factor out karne ke liye, wo difference quotients manufacture karta hai jo

u^{'}

aur

v^{'}

bante hain.

Last step mein

v

ki continuity (aur

u

ki nahi) kyun singled out hoti hai?

Sirf

v

denominator

v (x + h) v (x)

mein appear karta hai; humein

v (x + h) \to v (x)

chahiye taaki bottom nonzero

v (x)^{2}

ki taraf tend kare. Continuity dekho.

End mein

u^{'}

aur

v^{'}

ko alag limits ke roop mein kyun pull out kar sakte hain?

Har difference quotient converge karta hai (dono functions differentiable hain) aur surrounding factors finite numbers ki taraf converge karte hain, toh product/sum ka limit, product/sum of limits hota hai.

Mnemonic order "Lo dHi minus Hi dLo" kyun hai aur "Hi dLo minus Lo dHi" kyun nahi?

Kyunki sahi sign mein

u^{'} v

(Lo·dHi) positive hai; ise reverse karne se poori derivative negate ho jaati hai, increasing/decreasing ulta ho jaata hai.

Edge cases

Quotient rule mein kya hota hai jab

u = v

, yaani

f = \frac{v}{v} = 1

Numerator

v^{'} v - v v^{'} = 0

ban jaata hai, toh

f^{'} = 0

— jo sahi match karta hai ki constant

1

ki zero derivative hoti hai jahan bhi

v \neq = 0

ho.

Ek aisi jagah jahan

v^{'} (x) = 0

lekin

u^{'} (x) \neq = 0

, kya

f^{'}

zero hona zaroori hai?

Nahi — formula deta hai

f^{'} = \frac{u ^{'} v - 0}{v ^{2}} = \frac{u ^{'}}{v}

, generally nonzero; flat denominator fraction ko flat nahi karta.

f^{'}

kya hoga jahan numerator

u^{'} v - u v^{'}

zero ho lekin

v \neq = 0

Wahan

f^{'} = 0

: ratio ka ek critical point (horizontal tangent), jaise

\frac{x ^{2}}{x + 1}

mein

x = 0, - 2

Agar

v

ka odd order ka zero (sign change)

x = a

pe ho, toh wahan kya hoga?

f = u / v

vertical asymptote ke saath blow up karta hai aur fraction

a

ke across sign flip karta hai (dono sides pe opposite signs); rule $a$ pe apply nahi hoti kyunki

v (a) = 0

Agar

v

ka even order ka zero (no sign change)

x = a

pe ho, jaise

v = (x - a)^{2}

, toh yeh kaise different hai?

Phir bhi vertical asymptote hai, lekin

v > 0

dono sides pe, toh

f

dono taraf same sign ke saath blow up hota hai (dono

+ \infty

ya dono

- \infty

); yahan bhi rule $a$ pe fail hoti hai kyunki

v (a) = 0

Kya quotient rule galat answer de sakti hai agar

u / v

ko pehle simplify kiya ja sake (jaise

\frac{x ^{2}}{x} = x

Nahi — yeh removable point ke bahar same sahi result deti hai, lekin cancelled point (

x = 0

) pe original

u / v

undefined hai, toh agreement sirf wahan hota hai jahan ratio defined ho.

Rule kya deti hai jab

u^{'} = 0

ho (constant top)?

Yeh reduce hoti hai

- \frac{u v ^{'}}{v ^{2}}

mein, constant

u

se scale ki hui reciprocal rule — ek built-in sanity check.

Kya rule ki parwah hai ki

u

aur

v

trig, polynomial, ya exponential functions hain?

Nahi — yeh sirf use karta hai ki dono

x

pe differentiable hain aur

v (x) \neq = 0

;

u^{'}, v^{'}

ke specific formulas kahin aur se aate hain, jaise Derivatives of trig functions.

Recall Bank ke baad self-test

Bina dekhe: (1) $v^{2}$ kyun, $v$ kyun nahi, (2) minus kyun, (3) hum kaunsa zero add karte hain, (4) $v$ ki continuity kyun matter karti hai. Koi bhi miss ho → woh trap phir se dekho.

Connections

Quotient rule — proof — woh parent jise yeh bank interrogate karta hai.
Product rule — proof — twin "add-zero" trick.
Chain rule — $- v^{'} / v^{2}$ aur $v^{2}$ denominator ka source.
Limit definition of derivative — yahan har "why" ke peeche ka engine.
Continuity — last limit step ka quiet hero.
Derivatives of trig functions — jahan se $tan, sec, csc$ traps aate hain.