Sum, difference, constant multiple rules: Exercises

4.1.13 · D4 · Maths › Calculus I — Limits & Derivatives › Sum, difference, constant multiple rules

Intuition Is page ko kaise use karein

Neeche har problem ek idea test karta hai Sum, difference, constant multiple rules se: differentiation ek linear operation hai — slopes add hote hain, constants saath chalte hain, aur subtraction bas negative add karna hai. Har problem ko iske solution ko kholne se pehle solve karo. Difficulty paanch levels mein badhti hai; har level ko ek trap-buster [!mistake] se band kiya gaya hai taaki tum seekh sako ki cliff edge kahan hai.

Yahan sirf yahi tools allowed hain, jo parent note mein bane hain:

Derivative ki definition — dekho Definition of the derivative (limit of difference quotient).
Teen linearity rules: $\frac{d}{d x} [f + g] = f^{'} + g^{'}$ , $\frac{d}{d x} [f - g] = f^{'} - g^{'}$ , $\frac{d}{d x} [c f] = c f^{'}$ .
Power rule: $\frac{d}{d x} x^{n} = n x^{n - 1}$ (kisi bhi real $n$ ke liye).
Kuch known single-function derivatives: $sin^{'} = cos$ , $cos^{'} = - sin$ , $(e^{x})^{'} = e^{x}$ , $ln^{'} = 1/ x$ .

Shuru karne se pehle, ek figure jo " $^{'}$ " ka matlab fix karta hai: yeh ek slope hai, curve ki steepness ek point par. Is page par har cheez slopes ki bookkeeping hai.

Figure — Sum, difference, constant multiple rules

Red tangent line dekho: iski steepness hi number $f^{'} (x)$ hai. Jab hum "slopes add karte hain," hum literally do aise red lines ki steepness add kar rahe hote hain.

Level 1 — Recognition

Yahan tumhe bas yeh spot karna hai ki kaunsa rule apply hoga aur answer read off karna hai. Koi algebra tricks nahi.

Recall Solution 1.1

KYA: Do powers ka sum. WOH rule KYUN: sum rule kehta hai sum ko split karo, phir power rule se har piece ko. $y^{'} = \frac{d}{d x} x^{7} + \frac{d}{d x} x^{3} = 7 x^{6} + 3 x^{2} .$

Recall Solution 1.2

KYA: Ek constant $5$ times ek power. KYUN: constant multiple rule $5$ ko bahar kheench leta hai; power rule $x^{2}$ handle karta hai. $y^{'} = 5 \cdot \frac{d}{d x} x^{2} = 5 \cdot 2 x = 10 x .$

Recall Solution 1.3

KYA: Ek scaled power minus ek constant. $y^{'} = 4 \cdot 3 x^{2} - \frac{d}{d x} (9) = 12 x^{2} - 0 = 12 x^{2} .$ Constant $9$ ek flat line hai — zero slope — isliye yeh $0$ contribute karta hai.

Level 2 — Application

Ab tum ek problem mein kai rules apply karte ho, jisme power dekhne se pehle ek rewrite bhi karna padta hai.

Recall Solution 2.1

Har term ko split karo (sum/difference), constants bahar nikalo, har ek par power-rule lagao: $y^{'} = 3 \cdot 4 x^{3} - 5 \cdot 2 x + 6 \cdot 1 - 0 = 12 x^{3} - 10 x + 6.$

Recall Solution 2.2

Step 1 — powers mein rewrite karo. KYUN: power rule ko $x^{n}$ form chahiye; fraction aur root chhupe hue powers hain. $\frac{2}{x} = 2 x^{- 1}, x = x^{1/2} .$ Step 2 — differentiate karo. $y^{'} = 2 \cdot (- 1) x^{- 2} + \frac{1}{2} x^{- 1/2} = - \frac{2}{x ^{2}} + \frac{1}{2 x} .$

Recall Solution 2.3

Linearity trig terms ko bina interact kiye saath rehne deta hai. Yaad karo $sin^{'} = cos$ , $cos^{'} = - sin$ : $f^{'} (x) = 2 cos x - 3 (- sin x) + 0 = 2 cos x + 3 sin x .$ Cosine term par double sign flip note karo: $3 cos x$ subtract karna aur $cos^{'} = - sin$ milke $+ 3 sin x$ ban jaata hai.

Level 3 — Analysis

Yahan signs aur grouping pareshaan karte hain. Tumhe reason karna hoga ki minus kaise distribute hota hai.

Recall Solution 3.1

KYA: Ek poore multi-term function $g (x) = 2 x^{2} - 5 x + 1$ ka subtraction. KYUN dhyan dein: minus actually $(- 1) \cdot g$ hai, isliye yeh $g$ ke har term ko multiply karta hai. Pehle conceptually distribute karo: $y = x^{3} - 2 x^{2} + 5 x - 1$ . $y^{'} = 3 x^{2} - 4 x + 5 - 0 = 3 x^{2} - 4 x + 5.$

Recall Solution 3.2

Step 1 — scaled powers mein rewrite karo. $2$ se divide karna $\frac{1}{2}$ se multiply karna hai, aur $x = x^{1/2}$ : $g (x) = \frac{1}{2} x^{2} + 3 x^{1/2} .$ Step 2 — term by term differentiate karo. $g^{'} (x) = \frac{1}{2} \cdot 2 x + 3 \cdot \frac{1}{2} x^{- 1/2} = x + \frac{3}{2 x} .$

Recall Solution 3.3

Rewrite karo: $\frac{5}{x ^{3}} = 5 x^{- 3}$ aur $\frac{x}{7} = \frac{1}{7} x$ . $y^{'} = 5 (- 3) x^{- 4} - 4 \cdot \frac{1}{2} x^{- 1/2} + \frac{1}{7} = - \frac{15}{x ^{4}} - \frac{2}{x} + \frac{1}{7} .$

Level 4 — Synthesis

Ab linearity ko ek extra condition ke saath combine karo — derivative evaluate karo, slope nikalo, ya koi point dhundho.

Recall Solution 4.1

Step 1 — differentiate karo (linearity + power rule): $y^{'} = 6 x^{2} - x .$ Step 2 — evaluate karo $x = 2$ par: $y^{'} (2) = 6 (4) - 2 = 24 - 2 = 22.$ Slope hai $22$ .

Recall Solution 4.2

KYUN derivative: "horizontal tangent" ka matlab slope $= 0$ hai, aur slope $f^{'} (x)$ hai. $f^{'} (x) = 3 x^{2} - 12 = 0 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2.$ Dono cases matter karte hain: $x = 2$ aur $x = - 2$ . Negative root kabhi mat chhodna.

Recall Solution 4.3

Differentiate karo: $f^{'} (x) = 2 a x + b$ .

$f^{'} (0) = b = 3$ .
$f^{'} (1) = 2 a + b = 7 \Rightarrow 2 a = 4 \Rightarrow a = 2$ . Toh $a = 2$ , $b = 3$ .

Level 5 — Mastery

Prove karo aur generalize karo. Yahan tum linearity ko uske deeper home se connect karte ho: Linear operators aur Definition of the derivative (limit of difference quotient).

Recall Solution 5.1

Maano $h (x) = 3 f (x) - 2 g (x)$ . Definition se: $h^{'} (x) = lim_{Δ x \to 0} \frac{[ 3 f ( x + Δ x ) - 2 g ( x + Δ x ) ] - [ 3 f ( x ) - 2 g ( x ) ]}{Δ x} .$ Regroup karo $f$ -terms ko $f$ -terms ke saath, $g$ ko $g$ ke saath (addition commutative hai): $= lim_{Δ x \to 0} [3 \cdot \frac{f ( x + Δ x ) - f ( x )}{Δ x} - 2 \cdot \frac{g ( x + Δ x ) - g ( x )}{Δ x}] .$ Limit laws apply karo (sum split hota hai; constants bahar aa jaate hain — legal kyunki dono limits exist karti hain): $= 3 lim_{Δ x \to 0} \frac{f ( x + Δ x ) - f ( x )}{Δ x} - 2 lim_{Δ x \to 0} \frac{g ( x + Δ x ) - g ( x )}{Δ x} = 3 f^{'} (x) - 2 g^{'} (x) . ■$ Yahi bilkul ek-line ki linearity statement hai $\frac{d}{d x} [a f + b g] = a f^{'} + b g^{'}$ jisme $a = 3, b = - 2$ .

Recall Solution 5.2

Lo $f (x) = g (x) = x$ . Tab $f (x) g (x) = x^{2}$ , isliye $\frac{d}{d x} [f g] = 2 x$ . Lekin $f^{'} (x) g^{'} (x) = 1 \cdot 1 = 1$ . Kyunki $2 x \neq = 1$ saare $x$ ke liye (yeh sirf $x = \frac{1}{2}$ par agree karte hain), naive product-split galat hai. Products ko Product rule chahiye: $(f g)^{'} = f^{'} g + f g^{'} = 1 \cdot x + x \cdot 1 = 2 x$ . ✅

Recall Solution 5.3

General form. Linearity sum ko split karta hai aur har coefficient bahar nikaalta hai; power rule har $x^{k}$ handle karta hai: $P^{'} (x) = \sum_{k = 1}^{n} k a_{k} x^{k - 1} .$ $k = 0$ term ( $a_{0}$ , constant) vanish ho jaata hai. Apply karo $4 x^{5} - 3 x^{2} + 9$ par: $P^{'} (x) = 20 x^{4} - 6 x .$ Evaluate karo $x = 1$ par: $P^{'} (1) = 20 - 6 = 14.$

Connections

Sum, difference, constant multiple rules — parent jisko yeh drill train karta hai.
Definition of the derivative (limit of difference quotient) — Exercise 5.1 mein use hua.
Limit laws (sum, scalar, product of limits) — us proof ke peeche ka engine.
Power rule — yahan lagbhag har problem mein linearity ke saath pair hua.
Product rule — non-linear sibling, Exercise 5.2 mein isolate hua.
Linear operators — Exercise 5.1 ki structure ka abstract ghar.
Linearity of integration — usi idea ka integral twin.