Product rule — proof: Exercises — Hinglish deep dive

4.1.14 · D4 · Maths › Calculus I — Limits & Derivatives › Product rule — proof

Yeh page Product rule — proof ke liye ek self-test ladder hai. Har rung thodi mushkil hoti jaati hai: pehle aap rule ko sirf pehchaante ho, phir apply karte ho, phir aur tricky structures ko analyse karte ho, phir doosre rules ke saath synthesise karte ho, aur finally mastery tak pahunchte ho jahan aap cheezein prove karte ho.

Yahan use kiye gaye har symbol ko parent note mein banaya gaya tha. Jo tool hum use karte hain uska quick reminder:

Level 1 — Recognition

Yahan tumhe sirf do factors $u, v$ dhundhne hain, $u^{'}, v^{'}$ nikalni hain, aur unhe slot karna hai. Koi simplifying tricks nahi.

L1.1 $f (x) = x^{3} cos x$ ko differentiate karo.

Recall Solution L1.1

KYA: factors ke naam lo. Maano $u = x^{3}$ aur $v = cos x$ . KYUN: yeh function do cheezein ka product hai jinhe hum pehle se differentiate karna jaante hain (Power rule for $x^{3}$ , standard fact for $cos x$ ). $u^{'} = 3 x^{2}, v^{'} = - sin x$ Ab $(uv)^{'} = u^{'} v + u v^{'}$ mein slot karo: $f^{'} (x) = 3 x^{2} cos x + x^{3} (- sin x) = 3 x^{2} cos x - x^{3} sin x$

L1.2 $g (x) = (2 x + 1) e^{x}$ ko differentiate karo.

Recall Solution L1.2

Maano $u = 2 x + 1$ , $v = e^{x}$ . Toh $u^{'} = 2$ , $v^{'} = e^{x}$ . $g^{'} (x) = 2 e^{x} + (2 x + 1) e^{x} = e^{x} (2 x + 3)$ Last step sirf common $e^{x}$ factor karta hai — optional hai par tidy lagta hai.

L1.3 $h (x) = x ln x$ ko $x > 0$ ke liye differentiate karo.

Recall Solution L1.3

Maano $u = x$ , $v = ln x$ . Toh $u^{'} = 1$ , $v^{'} = \frac{1}{x}$ . $h^{'} (x) = 1 \cdot ln x + x \cdot \frac{1}{x} = ln x + 1$

Level 2 — Application

Ab factors kam obvious hain, ya phir baad mein algebra clean up karni padegi.

L2.1 $f (x) = x^{2} e^{3 x}$ ko differentiate karo. (Tum $(e^{3 x})^{'} = 3 e^{3 x}$ Chain rule se quote kar sakte ho.)

Recall Solution L2.1

Maano $u = x^{2}$ , $v = e^{3 x}$ . Toh $u^{'} = 2 x$ , aur $v^{'} = 3 e^{3 x}$ (andar $3 x$ par chain rule). $f^{'} (x) = 2 x e^{3 x} + x^{2} \cdot 3 e^{3 x} = e^{3 x} (3 x^{2} + 2 x)$

L2.2 $g (x) = x sin x$ ko $x > 0$ ke liye differentiate karo. $x = x^{1/2}$ likho.

Recall Solution L2.2

Maano $u = x^{1/2}$ , $v = sin x$ . Power rule se $u^{'} = \frac{1}{2} x^{- 1/2}$ , aur $v^{'} = cos x$ . $g^{'} (x) = \frac{1}{2} x^{- 1/2} sin x + x^{1/2} cos x = \frac{s i n x}{2 x} + x cos x$

L2.3 Product rule use karke $y = x^{2} (x - 3)$ ka slope $x = 1$ par nikalo (pehle expand mat karo).

Recall Solution L2.3

Maano $u = x^{2}$ , $v = x - 3$ . Toh $u^{'} = 2 x$ , $v^{'} = 1$ . $y^{'} = 2 x (x - 3) + x^{2} \cdot 1 = 2 x^{2} - 6 x + x^{2} = 3 x^{2} - 6 x$ $x = 1$ par: $y^{'} (1) = 3 (1) - 6 (1) = - 3$ . Slope hai $- 3$ .

Level 3 — Analysis

Yahan tum structure decide karte ho: teen factors, ek quotient ke andar chhupa hua product, ya ek graph se rule padhna.

L3.1 $f (x) = x e^{x} sin x$ ko differentiate karo (teen factors).

Recall Solution L3.1

KYA: parent ke Example 3 ka three-factor pattern — ek factor at a time differentiate karo, baaki ko fixed rakho, teen results ko sum karo. $(uv w)^{'} = u^{'} v w + u v^{'} w + uv w^{'}$ $u = x, v = e^{x}, w = sin x$ ke saath ( $u^{'} = 1, v^{'} = e^{x}, w^{'} = cos x$ ): $f^{'} (x) = 1 \cdot e^{x} sin x + x e^{x} sin x + x e^{x} cos x$ $= e^{x} (sin x + x sin x + x cos x)$

L3.2 Ek function $F (x) = \frac{x ^{2}}{cos x}$ likha gaya hai. Ise product $x^{2} (cos x)^{- 1}$ ke roop mein likhkar product rule + chain rule use karke differentiate karo (yahi jagah hai jahan Quotient rule chhupe taur par aata hai).

Recall Solution L3.2

Maano $u = x^{2}$ , $v = (cos x)^{- 1}$ . Toh $u^{'} = 2 x$ . $v^{'}$ ke liye $(cos x)^{- 1}$ par chain rule use karo: outer power $- 1$ , inner $cos x$ jiska derivative $- sin x$ hai: $v^{'} = - 1 \cdot (cos x)^{- 2} \cdot (- sin x) = \frac{s i n x}{c o s ^{2} x}$ Product rule: $F^{'} (x) = 2 x (cos x)^{- 1} + x^{2} \cdot \frac{s i n x}{c o s ^{2} x} = \frac{2 x}{c o s x} + \frac{x ^{2} s i n x}{c o s ^{2} x}$ Common denominator par: $F^{'} (x) = \frac{2 x cos x + x ^{2} sin x}{cos ^{2} x}$ — bilkul wahi jo quotient rule deta hai.

L3.3 (geometric) Picture mein growing-rectangle model dikhaya gaya hai. Width $u (x)$ aur height $v (x)$ dono badhte hain. Dikhaye gaye instant par, $u = 4, v = 3, u^{'} = 2, v^{'} = 5$ (rates per unit $x$ ). Area $A = uv$ kitni tezi se badh raha hai?

Recall Solution L3.3

YEH KAISA DIKHTA HAI: figure mein area ek right strip se rate $v u^{'}$ par aur ek top strip se rate $u v^{'}$ par badhta hai; corner $Δ u Δ v$ second-order tiny hai aur instantaneous rate mein contribute nahi karta. $A^{'} = u^{'} v + u v^{'} = (2) (3) + (4) (5) = 6 + 20 = 26$ Area $26$ square units per unit $x$ ki rate se badh raha hai.

Level 4 — Synthesis

Product rule ko doosre rules aur problem-solving (tangent lines, roots) ke saath combine karo.

L4.1 $y = x^{2} e^{x}$ par $x = 0$ par tangent line ki equation nikalo.

Recall Solution L4.1

Point: $y (0) = 0^{2} \cdot e^{0} = 0$ , toh point hai $(0, 0)$ . Slope: $u = x^{2}, v = e^{x}$ ke saath: $y^{'} = 2 x e^{x} + x^{2} e^{x} = e^{x} (2 x + x^{2})$ . $x = 0$ par: $y^{'} (0) = e^{0} (0 + 0) = 0$ . Tangent line: $(0, 0)$ se guzarti hui slope $0$ wali line hai $y = 0$ ( $x$ -axis). Curve wahan flat touch karta hai.

L4.2 $f (x) = (x^{2} - 1) e^{x}$ ke liye, saare $x$ nikalo jahan $f^{'} (x) = 0$ .

Recall Solution L4.2

Maano $u = x^{2} - 1, v = e^{x}$ . Toh $u^{'} = 2 x$ , $v^{'} = e^{x}$ . $f^{'} (x) = 2 x e^{x} + (x^{2} - 1) e^{x} = e^{x} (x^{2} + 2 x - 1)$ Kyunki $e^{x} > 0$ hamesha, humein $x^{2} + 2 x - 1 = 0$ chahiye. Quadratic formula se: $x = \frac{- 2 \pm 4 + 4}{2} = - 1 \pm 2$ Toh $x = - 1 + 2 \approx 0.4142$ aur $x = - 1 - 2 \approx - 2.4142$ .

L4.3 Maano $f (x) = u (x) v (x)$ jahan $x = 2$ par hum jaante hain $u (2) = 3, u^{'} (2) = - 1, v (2) = 5, v^{'} (2) = 4$ . $f^{'} (2)$ nikalo — tumhe $u, v$ ke formulas nahi diye gaye, sirf unki values.

Recall Solution L4.3

Product rule ko sirf chaar numbers chahiye, formulas nahi: $f^{'} (2) = u^{'} (2) v (2) + u (2) v^{'} (2) = (- 1) (5) + (3) (4) = - 5 + 12 = 7$

Level 5 — Mastery

Prove karo aur generalise karo. Ab tum rule wield kar rahe ho, woh rule nahi jo tumhe test kar raha ho.

L5.1 Product rule use karke power rule $(x^{4})^{'} = 4 x^{3}$ prove karo, $x^{4} = x \cdot x^{3}$ likh kar aur $(x^{3})^{'} = 3 x^{2}$ assume karke.

Recall Solution L5.1

Maano $u = x, v = x^{3}$ , toh $u^{'} = 1, v^{'} = 3 x^{2}$ (assumed fact). $(x^{4})^{'} = (1) (x^{3}) + (x) (3 x^{2}) = x^{3} + 3 x^{3} = 4 x^{3}$ Yeh Example 2 ka induction step hai: product rule Power rule ko ek degree at a time bootstrap karta hai.

L5.2 General three-factor rule $(uv w)^{'} = u^{'} v w + u v^{'} w + uv w^{'}$ prove karo, sirf two-factor product rule se shuru karke.

Recall Solution L5.2

KYA: $uv w$ ko do blocks $(uv)$ aur $w$ ke product ki tarah treat karo. KYUN: do-factor rule woh sab kuch hai jo hum use kar sakte hain; group karo taaki woh ek baar apply ho, phir inner product par dobara. $(uv w)^{'} = ((uv) w)^{'} = (uv)^{'} w + (uv) w^{'} (two-factor rule)$ Ab inner $(uv)^{'} = u^{'} v + u v^{'}$ expand karo: $= (u^{'} v + u v^{'}) w + uv w^{'} = u^{'} v w + u v^{'} w + uv w^{'} ■$

L5.3 Maano $f (x) = [g (x)]^{2}$ jahan $g$ differentiable hai. Sirf product rule use karke ( $[g]^{2} = g \cdot g$ likho), dikhao ki $f^{'} (x) = 2 g (x) g^{'} (x)$ . Yeh $[g]^{2}$ ke liye Chain rule hai, bina use invoke kiye derive kiya gaya.

Recall Solution L5.3

$f = u \cdot v$ likho jahan $u = g$ aur $v = g$ , toh $u^{'} = g^{'}$ aur $v^{'} = g^{'}$ . $f^{'} (x) = u^{'} v + u v^{'} = g^{'} \cdot g + g \cdot g^{'} = 2 g (x) g^{'} (x) ■$ $g (x) = x^{3}$ ke saath quick sanity check: $f = x^{6}$ , aur $2 g g^{'} = 2 x^{3} \cdot 3 x^{2} = 6 x^{5} = (x^{6})^{'}$ . ✓

L5.4 (Leibniz preview) $f = uv$ ke liye, pehla derivative hai $u^{'} v + u v^{'}$ . Dobara differentiate karo aur $f^{''} = (uv)^{''}$ $u, v$ aur unke derivatives ke terms mein nikalo. Woh binomial-coefficient pattern notice karo jo Leibniz rule (nth derivative of a product) ki taraf le jaata hai.

Recall Solution L5.4

$f^{'} = u^{'} v + u v^{'}$ ko term by term differentiate karo, har term par product rule apply karke: $(u^{'} v)^{'} = u^{''} v + u^{'} v^{'}, (u v^{'})^{'} = u^{'} v^{'} + u v^{''}$ Unhe joddo: $f^{''} = u^{''} v + u^{'} v^{'} + u^{'} v^{'} + u v^{''} = u^{''} v + 2 u^{'} v^{'} + u v^{''}$ Coefficients $1, 2, 1$ binomial coefficients $(0 2), (1 2), (2 2)$ hain — bilkul Leibniz rule pattern $f^{(n)} = k = 0 \sum n (k n) u^{(n - k)} v^{(k)}$ .

Recall Self-check summary (khatam karne ke baad click karo)

L1.1 ::: $3 x^{2} cos x - x^{3} sin x$ L1.2 ::: $e^{x} (2 x + 3)$ L1.3 ::: $ln x + 1$ L2.1 ::: $e^{3 x} (3 x^{2} + 2 x)$ L2.2 ::: $\frac{sin x}{2 x} + x cos x$ L2.3 slope at $x = 1$ ::: $- 3$ L3.1 ::: $e^{x} (sin x + x sin x + x cos x)$ L3.3 area rate ::: $26$ L4.1 tangent ::: $y = 0$ L4.2 roots ::: $x = - 1 \pm 2$ L4.3 ::: $7$ L5.4 ::: $u^{''} v + 2 u^{'} v^{'} + u v^{''}$

Connections

Product rule — proof — woh parent jise yeh page drill karta hai.
Power rule — L5.1 mein bootstrap kiya gaya.
Chain rule — factors ke andar zaroori hai (L2.1, L3.2) aur L5.3 mein rederive kiya gaya.
Quotient rule — L3.2 mein product + chain se reconstruct kiya gaya.
Limit definition of the derivative — yahan har rule ke peeche ka engine.
Differentiability implies continuity — parent ke proof mein chhupa hua lemma.
Leibniz rule (nth derivative of a product) — L5.4 mein preview kiya gaya.