Quotient rule — proof: Worked examples — Hinglish deep dive

4.1.15 · D3 · Maths › Calculus I — Limits & Derivatives › Quotient rule — proof

Yeh page ek drill hai har tarah ke quotient ke liye jo tum encounter kar sakte ho. Hum sirf formula mein plug nahi karte — balki har case ko picture se build karte hain, answer forecast karte hain, step by step kaam karte hain, aur verify karte hain. Agar tumne abhi proof nahi dekha, toh pehle Quotient rule — proof padho; yahan hum result ka use karte hain

$(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}} .$

The scenario matrix

Har quotient problem in cells mein se kisi ek mein aata hai. Hamare examples (E1–E8) us cell ke saath label kiye hain jo woh cover karta hai, toh saath milke yeh har box ko touch karte hain.

Cell	Kya khaas baat hai	Covered by
A. Polynomial / polynomial	seedha algebra, slope ke real roots	E1
B. Trig quotient	$u^{'}, v^{'}$ trig hain; identities simplify karte hain	E2
C. Reciprocal ( $u = 1$ )	degenerate numerator, $u^{'} = 0$	E3
D. Constant on top ( $u = c$ )	$u^{'} = 0$ lekin $u \neq = 1$	E3 (note)
E. Sign of the slope at a point	behaviour ka quadrant: increasing vs decreasing	E4
F. Exponential over polynomial	naya $u^{'}$ type, limiting behaviour $x \to \infty$	E5
G. Point where the answer is zero	$f^{'}$ ka numerator vanish hota hai (flat tangent)	E1, E4
H. Degenerate: $v (x) = 0$	rule apply NAHI hota — flag karna zaroori	E6
I. Real-world word problem	concentration / rate of change	E7
J. Exam twist: nested / with chain rule	$v$ khud ek composite hai	E8

E1 — Cell A + G: polynomial over polynomial

Worked example

$f (x) = \frac{x ^{2}}{x + 1}$ differentiate karo aur batao slope kahan zero hai. Forecast: Guess karo — kya slope kabhi zero hoga? Upar parabola, neeche line... expect karo do flat points.

Step 1. Parts ko naam do: $u = x^{2}, v = x + 1$ , toh $u^{'} = 2 x, v^{'} = 1$ . Yeh step kyun? Rule ko char ingredients chahiye; unhe label karo taaki tum inhe mix na karo.

Step 2. "Lo·dHi − Hi·dLo over Lo²" mein plug karo: $f^{'} (x) = \frac{( 2 x ) ( x + 1 ) - ( x ^{2} ) ( 1 )}{( x + 1 ) ^{2}} .$ Yeh step kyun? Direct substitution — Lo ( $v$ ) pehle har product mein aata hai, sign ko protect karta hai.

Step 3. Numerator simplify karo: $= \frac{2 x ^{2} + 2 x - x ^{2}}{( x + 1 ) ^{2}} = \frac{x ^{2} + 2 x}{( x + 1 ) ^{2}} = \frac{x ( x + 2 )}{( x + 1 ) ^{2}} .$ Yeh step kyun? Top ko factor karne se slope ke zeros saamne aate hain: $x = 0$ aur $x = - 2$ (cell G).

Step 4. $x = 1$ par: $f^{'} (1) = \frac{1 \cdot 3}{2 ^{2}} = \frac{3}{4} > 0$ . Yeh step kyun? Sign check — positive matlab function wahan chadh raha hai.

Verify: $f = \frac{x ^{2}}{x + 1}$ . Rewrite karo $x^{2} = (x + 1) (x - 1) + 1$ , toh $f = x - 1 + \frac{1}{x + 1}$ , jisse $f^{'} = 1 - \frac{1}{( x + 1 ) ^{2}}$ milta hai. $x = 1$ par: $1 - \frac{1}{4} = \frac{3}{4}$ . ✓ Match karta hai.

E2 — Cell B: trig quotient ( $sec^{2} x$ aur $csc^{2} x$ derive karo)

Worked example Dikhao

$\frac{d}{d x} tan x = sec^{2} x$ aur $\frac{d}{d x} cot x = - csc^{2} x$ . Forecast: Dono "1 over a squared trig" hone chahiye, lekin signs alag honge — kyun? (Hint: kaun sa decrease karta hai?)

Step 1. $tan x = \frac{sin x}{cos x}$ ke liye: $u = sin x, v = cos x$ , toh $u^{'} = cos x, v^{'} = - sin x$ . Yeh step kyun? Hume trig pieces ke derivatives chahiye — dekho Derivatives of trig functions.

Step 2. Rule apply karo: $\frac{d}{d x} tan x = \frac{c o s x c o s x - s i n x ( - s i n x )}{c o s ^{2} x} = \frac{c o s ^{2} x + s i n ^{2} x}{c o s ^{2} x} .$ Yeh step kyun? Minus times minus se $- sin x$ flip hokar $+ sin^{2} x$ ban jaata hai — yahan identity kaam aayegi.

Step 3. $cos^{2} x + sin^{2} x = 1$ use karo: $= \frac{1}{cos ^{2} x} = sec^{2} x$ . Hamesha $\geq 1$ , hamesha positive — $tan$ kabhi decrease nahi karta.

Step 4. $cot x = \frac{cos x}{sin x}$ ke liye: $u = cos x, v = sin x$ , $u^{'} = - sin x, v^{'} = cos x$ : $\frac{- s i n x s i n x - c o s x c o s x}{s i n ^{2} x} = \frac{- ( s i n ^{2} x + c o s ^{2} x )}{s i n ^{2} x} = - \frac{1}{s i n ^{2} x} = - csc^{2} x .$ Minus kyun? Yahan numerator ke dono terms negative hain — $cot$ ek decreasing function hai jahan bhi defined hai.

Verify: $x = \frac{π}{4}$ par, $sec^{2} \frac{π}{4} = (2)^{2} = 2$ aur $- csc^{2} \frac{π}{4} = - 2$ . ✓

E3 — Cell C + D: reciprocal aur constant-on-top

E4 — Cell E + G: har region mein slope ka sign padhna

Worked example

$g (x) = \frac{x}{x ^{2} + 1}$ ke liye, $g^{'}$ nikalo, aur $x$ ke har region mein slope ka sign classify karo. Forecast: Yeh function pehle badhta hai phir girta hai (peak aata hai). Predict karo: chhote $∣ x ∣$ ke liye slope positive, bade $∣ x ∣$ ke liye negative, kahin zero.

Step 1. $u = x, v = x^{2} + 1$ , toh $u^{'} = 1, v^{'} = 2 x$ : $g^{'} (x) = \frac{1 \cdot ( x ^{2} + 1 ) - x \cdot 2 x}{( x ^{2} + 1 ) ^{2}} = \frac{x ^{2} + 1 - 2 x ^{2}}{( x ^{2} + 1 ) ^{2}} = \frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}} .$ Yeh step kyun? Simplify karne ke baad, denominator hamesha $> 0$ hai, toh $g^{'}$ ka sign poori tarah top $1 - x^{2}$ se decide hota hai.

Step 2. $1 - x^{2} = 0 \Rightarrow x = \pm 1$ solve karo (cell G, flat tangents). Yeh step kyun? Yeh number line ko teen regions mein baantte hain jo humne forecast kiya tha.

Step 3. Har region test karo (cell E — har sign):

Region	sample $x$	$1 - x^{2}$	slope
$x < - 1$	$- 2$	$- 3$	negative (falling)
$- 1 < x < 1$	$0$	$+ 1$	positive (rising)
$x > 1$	$2$	$- 3$	negative (falling)

Yeh step kyun? Yahi "quadrant coverage" idea hai: hum ek point par nahi rukते, har sign jo slope le sakta hai woh cover karte hain.

Verify: $g^{'} (0) = \frac{1}{1} = 1 > 0$ ✓; $g^{'} (2) = \frac{1 - 4}{5 ^{2}} = \frac{- 3}{25} < 0$ ✓; $g^{'} (1) = 0$ ✓ ( $x = 1$ par peak).

E5 — Cell F: exponential over polynomial + limiting behaviour

Worked example

$h (x) = \frac{e ^{x}}{x}$ differentiate karo ( $x \neq = 0$ ke liye) aur batao $h^{'}$ kaisi hoti hai jab $x \to + \infty$ . Forecast: $e^{x}$ bahut tezi se badhta hai, toh eventually top sab kuch jeet leta hai. Predict karo $h^{'}$ badi aur positive ho jaayegi.

Step 1. $u = e^{x}, v = x$ , toh $u^{'} = e^{x}, v^{'} = 1$ : $h^{'} (x) = \frac{e ^{x} \cdot x - e ^{x} \cdot 1}{x ^{2}} = \frac{e ^{x} ( x - 1 )}{x ^{2}} .$ Yeh step kyun? Numerator se $e^{x}$ factor out karne par sign-controlling factor $(x - 1)$ saamne aata hai.

Step 2. Sign map (ek aur all-signs sweep):

$0 < x < 1$ : $x - 1 < 0 \Rightarrow h^{'} < 0$ (falling),
$x > 1$ : $h^{'} > 0$ (rising, minimum at $x = 1$ ),
$x < 0$ : $x^{2} > 0$ , $e^{x} > 0$ , $x - 1 < 0 \Rightarrow h^{'} < 0$ (falling).

Yeh step kyun? Negative- $x$ branch bhi cover karta hai — missing point ke left waali region mat bhoolo.

Step 3. Limit $x \to + \infty$ : $h^{'} = e^{x} \cdot \frac{x - 1}{x ^{2}} \to + \infty$ kyunki $e^{x}$ kisi bhi power of $x$ ko dominate karta hai. Yeh step kyun? Yahi "limiting behaviour" cell hai — exponential polynomial ko peeche chhod deta hai.

Verify: $x = 1$ par, $h^{'} (1) = \frac{e ^{1} ( 0 )}{1} = 0$ ✓ (minimum). $x = 2$ par, $h^{'} (2) = \frac{e ^{2} ( 1 )}{4} = \frac{e ^{2}}{4} \approx 1.847 > 0$ ✓.

E6 — Cell H: degenerate case $v (x) = 0$

Worked example

$x = 2$ par $\frac{d}{d x} \frac{x - 3}{x - 2}$ kya hai? Forecast: Dhyan se — quotient rule ke liye zaroori hai $v (x) \neq = 0$ . $x = 2$ par bottom zero hai. Predict karo: derivative wahan exist nahi karta kyunki function khud wahan exist nahi karta.

Step 1. $u = x - 3, v = x - 2$ , toh $u^{'} = 1, v^{'} = 1$ . $x \neq = 2$ ke liye: $\frac{d}{d x} \frac{x - 3}{x - 2} = \frac{1 \cdot ( x - 2 ) - ( x - 3 ) \cdot 1}{( x - 2 ) ^{2}} = \frac{( x - 2 ) - ( x - 3 )}{( x - 2 ) ^{2}} = \frac{1}{( x - 2 ) ^{2}} .$ Yeh step kyun? $x = 2$ chhod kar har jagah valid hai.

Step 2. $x = 2$ par: $v (2) = 0$ , toh rule ki hypothesis fail hoti hai. Function ka wahan vertical asymptote hai; koi tangent line nahi, isliye koi derivative nahi. Yeh step kyun? Yahi honesty cell hai — ek rule jisme precondition hai. Kabhi blindly $x = 2$ ko $\frac{1}{( x - 2 ) ^{2}}$ mein plug karke "infinity" mat kaho; sahi statement hai undefined / does not exist.

Verify: $x \neq = 2$ ke liye, $\frac{1}{( x - 2 ) ^{2}} > 0$ hamesha — function dono branches par increasing hai. $x = 3$ par, $h^{'} (3) = \frac{1}{1} = 1 > 0$ ✓. $x = 2$ par: undefined (dono $f$ aur $f^{'}$ ). ✓

E7 — Cell I: real-world word problem (concentration)

Worked example Ek tank

$10$ L paani mein $2$ kg namak ke saath shuru hota hai. Fresh namak $t$ kg total aur paani $t$ L total add hota hai, toh concentration hai $C (t) = \frac{2 + t}{10 + t} (kg per L) .$ $t = 0$ par concentration kitni tezi se badal rahi hai? Forecast: Salt fraction $\frac{2}{10} = 0.2$ add ho rahi fraction ( $\frac{1}{1}$ ) se kam hai, toh concentration badhni chahiye — expect karo $C^{'} > 0$ .

Step 1. $u = 2 + t, v = 10 + t$ , toh $u^{'} = 1, v^{'} = 1$ : $C^{'} (t) = \frac{1 \cdot ( 10 + t ) - ( 2 + t ) \cdot 1}{( 10 + t ) ^{2}} = \frac{( 10 + t ) - ( 2 + t )}{( 10 + t ) ^{2}} = \frac{8}{( 10 + t ) ^{2}} .$ Yeh step kyun? Top mein $t$ cancel ho jaate hain, ek constant $8$ bachta hai — jo andar hai aur jo add ho raha hai unke beech ka "gap".

Step 2. $t = 0$ par: $C^{'} (0) = \frac{8}{1 0 ^{2}} = \frac{8}{100} = 0.08$ kg/L per unit time. Yeh step kyun? Positive, jaise forecast tha — concentration badhti hai kyunki hum current mix se zyada salty cheez add kar rahe hain.

Step 3. Jab $t \to \infty$ , $C^{'} \to 0$ aur $C \to 1$ : concentration incoming ratio ki taraf flatten ho jaati hai. Yeh step kyun? Limiting behaviour — tank apni starting condition "bhool" jaata hai.

Verify: $C (0) = 0.2$ , $C (1) = \frac{3}{11} \approx 0.2727$ ; average rate $\approx 0.0727$ , $t = 0$ par instantaneous $0.08$ ke kareeb (rate drop ho raha hai). Units: $\frac{kg/L}{time}$ ✓. $C^{'} (0) = 0.08$ ✓.

E8 — Cell J: exam twist (quotient + chain rule)

Worked example

$k (x) = \frac{x}{x ^{2} + 1}$ differentiate karo. Forecast: Bottom ek composite hai, toh $v$ differentiate karne ke liye hume Chain rule chahiye. Predict karo clean result aayega kyunki yeh ek jaana-maana "sigmoid-shaped" function hai.

Step 1. $u = x \Rightarrow u^{'} = 1$ . $v = (x^{2} + 1)^{1/2}$ ; chain rule se $v^{'} = \frac{1}{2} (x^{2} + 1)^{- 1/2} \cdot 2 x = \frac{x}{x ^{2} + 1}$ . Yeh step kyun? $v$ elementary nahi hai — bahar wali power differentiate karo, phir andar wale ke derivative $2 x$ se multiply karo.

Step 2. Quotient rule: $k^{'} (x) = \frac{1 \cdot x ^{2} + 1 - x \cdot \frac{x}{x ^{2} + 1}}{( x ^{2} + 1 ) ^{2}} = \frac{x ^{2} + 1 - \frac{x ^{2}}{x ^{2} + 1}}{x ^{2} + 1} .$ Yeh step kyun? Denominator squared root ko hata deta hai: $(x^{2} + 1)^{2} = x^{2} + 1$ .

Step 3. Top ko $x^{2} + 1$ ke upar combine karo: $numerator = \frac{( x ^{2} + 1 ) - x ^{2}}{x ^{2} + 1} = \frac{1}{x ^{2} + 1} .$ Toh $k^{'} (x) = \frac{1}{x ^{2} + 1} \cdot \frac{1}{x ^{2} + 1} = \frac{1}{( x ^{2} + 1 ) ^{3/2}} .$ Yeh step kyun? Hamesha positive — $k$ har jagah increasing hai, jo iske S-curve shape se match karta hai.

Verify: $x = 0$ par, $k^{'} (0) = \frac{1}{1 ^{3/2}} = 1$ ✓; $x = 1$ par, $k^{'} (1) = \frac{1}{2 ^{3/2}} = \frac{1}{2 2} \approx 0.3536$ ✓.

Recall Self-test: har problem kaun si cell hai?

Bina dekhe matrix se match karo. $\frac{e ^{x}}{x}$ ::: Cell F (exponential over polynomial, limiting behaviour). $\frac{1}{x ^{2} + 1}$ ::: Cell C (reciprocal, $u^{'} = 0$ ). $\frac{x - 3}{x - 2}$ at $x = 2$ ::: Cell H (degenerate, $v = 0$ , undefined). $\frac{x}{x ^{2} + 1}$ ::: Cell J (chain rule andar chahiye).

Connections

Quotient rule — proof — parent result jo hum poore mein apply karte hain.
Derivatives of trig functions — E2 mein $u^{'}, v^{'}$ supply karta hai.
Chain rule — E8 mein composite denominator ke liye zaroori hai.
Limit definition of derivative — jahan se rule khud aata hai.
Product rule — proof — twin rule; yaad karo $u / v = u \cdot v^{- 1}$ .
Continuity — kyun degenerate $v = 0$ case (E6) tangent ko tod deta hai.