Product rule — proof: Worked examples — Hinglish deep dive

4.1.14 · D3 · Maths › Calculus I — Limits & Derivatives › Product rule — proof

Pehle — rule ko dekho use karne se pehle

Is page ki har cheez ek picture par tikhi hai: ek rectangle ka area jiske dono sides grow kar rahe hain. Koi bhi algebra se pehle ise dekho.

Scenario matrix

Kuch bhi work karne se pehle, saare case classes lay out karte hain jo ek product aap pe throw kar sakta hai. Har row ko aise padho: "ek aisi situation jo differently behave karti hai aur jise kam se kam ek baar dikhana zaroori hai."

Cell	Case class	Kya cheez special banati hai	Covered by
A	Do plain "nice" factors	Baseline: bas $u^{'} v + u v^{'}$ mein slot karo	Ex 1
B	Ek factor with sign changes (trig)	Answer ka sign depend karta hai kahan evaluate karo	Ex 2
C	Ek factor us point pe zero equals karta hai	Terms drop out ho jaate hain — check karo ki kuch 0 se divide na ho	Ex 3
D	Degenerate: ek factor constant	Rule collapse hokar $(c v)^{'} = c v^{'}$ ban jaata hai	Ex 4
E	Dono factors $x$ ki same power	Product rule ko Power rule se agree karna chahiye	Ex 5
F	Teen factors	Rule do baar apply karo / all-but-one pattern	Ex 6
G	Real-world word problem (units)	Product ki rate = e.g. area growth rate	Ex 7
H	Exam twist: disguised / simplify karna padega	Pehle product spot karna padega	Ex 8

Ab hum har cell fill karenge.

Warm-up: ek recipe jo har jagah use hoti hai

Cell A — do plain factors

Worked example Example 1 (Cell A) —

$f (x) = (3 x + 1) (x^{2} - 4)$ Forecast: aage padhne se pehle guess karo: kya $f^{'}$ ek quadratic hai, aur uska leading coefficient kya hai? (Hint: $f$ expand hokar ek cubic banta hai, toh $f^{'}$ ek quadratic hogi.)

Step 1 — Dono factors ko naam do. Maan lo $u = 3 x + 1$ aur $v = x^{2} - 4$ . Yeh step kyun? Rule sirf $u, v, u^{'}, v^{'}$ ki bhaasha mein bolta hai; naam dene se pieces mix hone se bacha rahe hain.

Step 2 — Har factor ko akele differentiate karo. $u^{'} = 3$ , aur $v^{'} = 2 x$ . Yeh step kyun? $u^{'}$ aur $v^{'}$ ingredients hain; dono ek-ek line ki power-rule derivative hain.

Step 3 — $u^{'} v + u v^{'}$ assemble karo. $f^{'} = u^{'} 3 v (x^{2} - 4) + u (3 x + 1) v^{'} (2 x)$ Yeh step kyun? Yahi poora rule hai; pieces ko boxed rakho taaki koi ek drop na ho.

Step 4 — Tidy up karo. $3 x^{2} - 12 + 6 x^{2} + 2 x = 9 x^{2} + 2 x - 12$ .

Verify: Pehle expand karo: $f = 3 x^{3} - 12 x + x^{2} - 4$ , toh $f^{'} = 9 x^{2} + 2 x - 12$ . ✓ Same answer — yahi hona chahiye. $x = 1$ pe dono routes dete hain $9 + 2 - 12 = - 1$ .

Cell B — sign-changing factor (trig)

Trig factors classic "sign ka dhyan rakho" wala case hai: $cos$ aur $sin$ quadrants mein sign flip karte hain, isliye $f^{'}$ ki value ka character is baat par depend karta hai ki aap kahan khade ho.

Worked example Example 2 (Cell B) —

$f (x) = x cos x$ , ek poori sign table Forecast: kya $f^{'}$ kabhi sirf $cos x$ ke barabar hoti hai? Kis $x$ pe " $- x sin x$ " piece $f^{'}$ ko neeche push karta hai versus upar?

Step 1 — Factors ko naam do aur differentiate karo. $u = x \Rightarrow u^{'} = 1$ ; $v = cos x \Rightarrow v^{'} = - sin x$ . Yeh step kyun? $cos$ differentiate hokar $- sin$ banta hai — minus sign wahin rehta hai jahan sign-mistakes hoti hain.

Step 2 — Assemble karo. $f^{'} = (1) (cos x) + (x) (- sin x) = cos x - x sin x$ Yeh step kyun? Seedha application; note karo $-$ $v^{'}$ se aaya, rule se nahi.

Step 3 — Saare chaar quadrants mein ek complete sign table banao. Yahan "quadrant" matlab angle $x$ ka kaunsa quarter-turn hai: QI $= (0, \frac{π}{2})$ , QII $= (\frac{π}{2}, π)$ , QIII $= (π, \frac{3 π}{2})$ , QIV $= (\frac{3 π}{2}, 2 π)$ .

$x$	quadrant	$cos x$	$sin x$	$f^{'} = cos x - x sin x$	$f^{'}$ ka sign
$\frac{π}{4} \approx 0.785$	QI	$+$	$+$	$0.707 - 0.785 (0.707) \approx 0.152$	$+$
$\frac{3 π}{4} \approx 2.356$	QII	$-$	$+$	$- 0.707 - 2.356 (0.707) \approx - 2.373$	$-$
$\frac{5 π}{4} \approx 3.927$	QIII	$-$	$-$	$- 0.707 - 3.927 (- 0.707) \approx 2.069$	$+$
$\frac{7 π}{4} \approx 5.498$	QIV	$+$	$-$	$0.707 - 5.498 (- 0.707) \approx 4.594$	$+$

Yeh step kyun? Har quadrant mein ek point guarantee karta hai ki reader har sign combination of $cos$ aur $sin$ dekhe, taaki koi scenario skip na ho. Note karo QIII mein $f^{'}$ positive ho jaata hai kyunki $- x sin x$ mein do minus signs milkar positive dete hain.

Verify: $x = \frac{3 π}{4}$ pe, $f^{'} = cos \frac{3 π}{4} - \frac{3 π}{4} sin \frac{3 π}{4} = - \frac{2}{2} - \frac{3 π}{4} \cdot \frac{2}{2} \approx - 2.373$ (negative, jaise table mein hai). ✓

Cell C — ek factor jo us point pe zero hai

Worked example Example 3 (Cell C) —

$f (x) = (x^{2} - 9) e^{x}$ , $f^{'} (3)$ nikalo Forecast: kyunki $x^{2} - 9 = 0$ at $x = 3$ , guess karo product rule ke dono terms mein se kaunsa bachega.

Step 1 — Factors ko naam do. $u = x^{2} - 9, v = e^{x}$ . Yeh step kyun? $e^{x}$ apna khud ka derivative hai, toh $v^{'} = e^{x}$ — ek clean second factor.

Step 2 — Differentiate karo. $u^{'} = 2 x, v^{'} = e^{x}$ .

Step 3 — Assemble karo. $f^{'} = 2 x e^{x} + (x^{2} - 9) e^{x}$ Yeh step kyun? Abhi kuch special nahi — zero tab strike karta hai jab hum koi value plug in karte hain.

Step 4 — $x = 3$ pe evaluate karo. $f^{'} (3) = 2 (3) e^{3} + = 0 (9 - 9) e^{3} = 6 e^{3}$ Yeh step kyun? Doosra term mar jaata hai kyunki $u (3) = 0$ hai, sirf " $u^{'} v$ " piece bachti hai. Yeh ek genuine feature hai, shortcut nahi: $u$ ke root pe, sirf $u^{'} v$ contribute kar sakta hai.

Verify: $6 e^{3} \approx 6 (20.0855) \approx 120.51$ . $x = 3$ ke paas $f$ ka finite-difference slope match karta hai. ✓ (Note: yahan kuch bhi zero se divide nahi hua — rule safe hai chahe ek factor 0 ho.)

Cell D — degenerate case: ek factor constant

Worked example Example 4 (Cell D) —

$f (x) = 7 (x^{3} + 2 x)$ Forecast: aap pehle se constant-multiple rule se "jaante" ho answer. Kya product rule use reproduce karega, ya kuch extra junk add karega?

Step 1 — Factors ko naam do. $u = 7$ (ek constant), $v = x^{3} + 2 x$ . Yeh step kyun? Hum deliberately ek constant ko "factor" ki tarah treat kar rahe hain taaki degenerate cell test kar sakein.

Step 2 — Differentiate karo. $u^{'} = 0$ (constants change nahi hote), $v^{'} = 3 x^{2} + 2$ . Yeh step kyun? $u^{'} = 0$ hi exactly "extra" term ko kill karta hai.

Step 3 — Assemble karo. $f^{'} = = 0 0 \cdot (x^{3} + 2 x) + 7 (3 x^{2} + 2) = 21 x^{2} + 14$ Yeh step kyun? Product rule degenerate hokar constant-multiple rule $(c v)^{'} = c v^{'}$ ban jaata hai. Yeh ek sanity check hai ki general rule mein special rules contained hain.

Verify: Direct: $f = 7 x^{3} + 14 x$ , $f^{'} = 21 x^{2} + 14$ . ✓ Identical. $x = 1$ pe: $35$ .

Cell E — Power rule se agree karna zaroori

Cell F — teen factors (yahan se scratch se derive kiya)

Hum teen-factor rule borrow nahi karte — hum ise two-factor rule se build karte hain taaki yeh page apne aap mein complete rahe.

Cell G — real-world word problem (units!)

Worked example Example 7 (Cell G) — expanding photo frame

Ek rectangular photo ki width $w (t) = 4 + 0.5 t$ cm aur height $h (t) = 3 + 0.2 t$ cm hai, jahan $t$ seconds mein hai. $t = 10$ s pe area kitni tezi se badh raha hai?

Forecast: kya answer "sirf width ki growth × current height" se bada hoga ya chhota? (Height bhi grow kar rahi hai, toh ek doosra contribution bhi hai.)

Step 1 — Area ko product ki tarah likho. $A (t) = w (t) h (t)$ . Yeh step kyun? Area hai ek product, isliye uski rate of change ek textbook product-rule problem hai — yeh literally page ke upar se rectangle picture zinda ho gayi hai.

Step 2 — Har side ki rates. $w^{'} (t) = 0.5$ cm/s, $h^{'} (t) = 0.2$ cm/s. Yeh step kyun? Yeh har side ki "growth speeds" hain — $u^{'}$ aur $v^{'}$ ingredients.

Step 3 — Rule apply karo. $A^{'} (t) = w^{'} h + w h^{'} = 0.5 h (t) + 0.2 w (t)$ Yeh step kyun? Area ki rate = (widening speed × current height) + (current width × heightening speed).

Step 4 — $t = 10$ plug in karo. $w (10) = 4 + 5 = 9$ cm, $h (10) = 3 + 2 = 5$ cm. $A^{'} (10) = 0.5 (5) + 0.2 (9) = 2.5 + 1.8 = 4.3 cm^{2} / s$ Yeh step kyun? Dono strips contribute karti hain; doosri ko ignore karna undercount hoga.

Verify (units + expansion): Units: (cm/s)(cm) + (cm)(cm/s) $=$ cm²/s ✓ — ek area rate. Direct check: $A (t) = (4 + 0.5 t) (3 + 0.2 t) = 12 + 0.8 t + 1.5 t + 0.1 t^{2} = 12 + 2.3 t + 0.1 t^{2}$ , toh $A^{'} (t) = 2.3 + 0.2 t$ , deta hai $A^{'} (10) = 2.3 + 2 = 4.3$ . ✓

Cell H — exam twist (hidden product spot karo)

Worked example Example 8 (Cell H) —

$f (x) = x (x - 2)$ Twist: pehli nazar mein $x (x - 2)$ ek single messy expression lagta hai, lekin yeh secretly do factors ka product hai — aur square root ek disguised power hai. Dono spot karna exam skill hai jo test ho rahi hai.

Pehle domain. $x$ sirf $x \geq 0$ ke liye defined hai, aur uska derivative $\frac{1}{2} x^{- 1/2}$ ko $x > 0$ chahiye ( $x$ se division). Toh neeche sab kuch sirf $x > 0$ pe valid hai — hum edge $x \to 0^{+}$ ko limit ki tarah study karenge, $x = 0$ directly kabhi plug nahi karenge.

Forecast: $x$ $x$ ki kaunsi power hai, aur uska derivative kya hai? Guess karo kya $f^{'}$ blow up karta hai jaise $x \to 0^{+}$ .

Step 1 — Disguised factor ko power ki tarah rewrite karo. $x = x^{1/2}$ . Set karo $u = x^{1/2}, v = x - 2$ . Yeh step kyun? Exams products ko radicals ke peeche chhupate hain; $x^{1/2}$ mein convert karne se Power rule ise handle kar leta hai. $u^{'} = \frac{1}{2} x^{- 1/2}$ ( $x > 0$ ke liye valid).

Step 2 — Doosre factor ko differentiate karo. $v^{'} = 1$ .

Step 3 — Assemble karo. $f^{'} = \frac{1}{2} x^{- 1/2} (x - 2) + x^{1/2} (1) = \frac{x - 2}{2 x} + x$ Yeh step kyun? Direct product rule; negative exponent ko $1/ (2 x)$ ki tarah rakho clarity ke liye.

Step 4 — Ek denominator pe combine karo (exam-neat form). $f^{'} = \frac{x - 2}{2 x} + \frac{2 x}{2 x} = \frac{3 x - 2}{2 x} (x > 0)$ Yeh step kyun? Examiners ek single simplified fraction chahte hain; common denominator $2 x$ .

Step 5 — Limiting behaviour jaise $x \to 0^{+}$ . Domain ki left edge ke paas jaate hue: numerator $\to - 2$ , denominator $\to 0^{+}$ , toh $f^{'} \to - \infty$ : slope origin ke paas vertical aur steeply negative ho jaata hai. Yeh step kyun? Matrix limiting behaviour maangta hai — derivative unbounded ho sakta hai chahe $f$ khud finite ho ( $f (0) = 0$ ). Hum limit right se lete hain kyunki domain $x > 0$ hai.

Verify: $x = 1$ pe: $f^{'} = \frac{3 - 2}{2} = 0.5$ . Direct: $f = x^{3/2} - 2 x^{1/2}$ , toh $f^{'} = \frac{3}{2} x^{1/2} - x^{- 1/2}$ ; $x = 1$ pe: $1.5 - 1 = 0.5$ . ✓ $x = 4$ pe: $\frac{10}{4} = 2.5$ aur $\frac{3}{2} (2) - \frac{1}{2} = 3 - 0.5 = 2.5$ . ✓

Matrix completeness check

Recall Kya humne sach mein har cell cover ki? (Neeche har line "prompt ::: answer" hai — khud check karo)

A → Ex 1 (plain factors) ::: ✓ B → Ex 2 (trig, sign per quadrant, full table) ::: ✓ C → Ex 3 (ek factor zero ke barabar) ::: ✓ D → Ex 4 (constant factor, degenerate) ::: ✓ E → Ex 5 (power rule se agree karta hai) ::: ✓ F → Ex 6 (teen factors, yahan derive kiye) ::: ✓ G → Ex 7 (word problem with units) ::: ✓ H → Ex 8 (disguised product + limiting slope) ::: ✓

Rapid recall

Recall Yeh lines kaise padhen

Har line prompt ::: answer ki tarah likhi hai. ::: ke baad waala part cover karo, answer try karo, phir reveal karo. Yeh vault ka standard self-quiz format hai.

(3 x + 1) (x^{2} - 4)

differentiate karo.

9 x^{2} + 2 x - 12

x cos x

differentiate karo.

cos x - x sin x

f = (x^{2} - 9) e^{x}

ke liye,

x = 3

pe ek term kyun vanish hoti hai?

Kyunki

u (3) = x^{2} - 9 = 0

u v^{'}

term ko kill karta hai, sirf

u^{'} v = 6 e^{3}

bachta hai.

Jab ek factor constant

c

ho, product rule ban jaata hai…

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

(doosra term

0 \cdot v

hai).

Expanding rectangle area ke liye,

A^{'} =

w^{'} h + w h^{'}

(widening speed × height + width × heightening speed).

\frac{d}{d x} [x (x - 2)] =

\frac{3 x - 2}{2 x}

(

x > 0

ke liye valid)

Connections

Product rule — proof — parent jise yeh page drill karta hai.
Power rule — Examples 5 aur 8 ise lean karte hain.
Chain rule — zaroori hota hai jab ek factor khud ek composition ho.
Quotient rule — products ki jagah divisions ke liye sibling rule.
Differentiability implies continuity — original proof ke peeche chhupa hua lemma.
Leibniz rule (nth derivative of a product) — jahan teen-factor pattern generalise hota hai.
Limit definition of the derivative — har step ke neeche ka engine.