Power rule — proof for integer, rational exponents: Exercises

4.1.12 · D4 · Maths › Calculus I — Limits & Derivatives › Power rule — proof for integer, rational exponents

Neeche har jagah ek hi rule test ho raha hai: $\frac{d}{d x} x^{n} = n x^{n - 1} .$

Iske peeche ke do engines (parent note se): Limit definition of the derivative aur, whole-number powers ke liye, Binomial theorem.

Level 1 — Recognition

Bas "drop and drop" apply karo: $n$ ko front pe le aao, phir exponent se ek ghataao.

Q1. $f (x) = x^{7}$ ko differentiate karo.

Q2. $g (x) = x^{- 4}$ ko differentiate karo.

Q3. $h (x) = 3 x$ (yaani $x^{1/3}$ ) ko differentiate karo.

Recall Solution — Q1, Q2, Q3

Q1. Yahan $n = 7$ hai. KIYA KYA: $n x^{n - 1}$ apply karo. KYUN: $x^{7}$ ek variable base hai jiska exponent constant hai — exactly wahi shape jisko power rule handle karta hai. $f^{'} (x) = 7 x^{7 - 1} = 7 x^{6} .$

Q2. Yahan $n = - 4$ hai (ek negative integer — proof ka Stage 3 ise cover karta hai). $g^{'} (x) = - 4 x^{- 4 - 1} = - 4 x^{- 5} = - \frac{4}{x ^{5}} .$ KYUN beech mein minus aaya: negative exponent se ek ghataane par woh aur zyada negative ho jaata hai.

Q3. Pehle root ko power ke roop mein likhte hain: $3 x = x^{1/3}$ , toh $n = \frac{1}{3}$ . $h^{'} (x) = \frac{1}{3} x^{1/3 - 1} = \frac{1}{3} x^{- 2/3} = \frac{1}{3 x ^{2/3}} = \frac{1}{3 3 x ^{2}} .$ KYUN rewrite karein: power rule ko ek exponent chahiye, root symbol nahi — same number hai, bas cleaner form hai.

Level 2 — Application

Ab exponents thode zyada messy hain, aur tum power rule ko is fact ke saath combine karte ho ki constants multiply ho ke saath aate hain aur sums term-by-term differentiate hote hain.

Q4. $f (x) = 5 x^{3} - \frac{2}{x} + 6 x$ ko differentiate karo.

Q5. $g (x) = \frac{x ^{2} + 3 x}{x}$ ko differentiate karo. (Hint: pehle fraction ko split karo.)

Q6. $y = x^{5/2}$ ka slope $x = 4$ par nikalo.

Recall Solution — Q4, Q5, Q6

Q4. KIYA KYA: har piece ko ek clean power ke roop mein likho. $\frac{2}{x} = 2 x^{- 1}$ aur $6 x = 6 x^{1/2}$ . Ab term by term differentiate karo (constant multiple saath chalti rehti hai): $f^{'} (x) = 5 \cdot 3 x^{2} - 2 \cdot (- 1) x^{- 2} + 6 \cdot \frac{1}{2} x^{- 1/2} .$ $f^{'} (x) = 15 x^{2} + \frac{2}{x ^{2}} + \frac{3}{x} .$ KYUN beech wala term positive ho gaya: $- 2 \cdot (- 1) = + 2$ — do minus signs.

Q5. KIYA KYA: differentiate karne se pehle split karo (yahan quotient ko directly power-rule nahi kar sakte). $g (x) = \frac{x ^{2}}{x ^{1/2}} + \frac{3 x}{x ^{1/2}} = x^{3/2} + 3 x^{1/2} .$ KYUN split karein: same base ki powers ko divide karne par exponents subtract hote hain — ek messy fraction do simple powers mein badal jaata hai. Ab differentiate karo: $g^{'} (x) = \frac{3}{2} x^{1/2} + 3 \cdot \frac{1}{2} x^{- 1/2} = \frac{3}{2} x + \frac{3}{2 x} .$

Q6. KIYA KYA: differentiate karo, phir plug in karo. $n = \frac{5}{2}$ : $y^{'} = \frac{5}{2} x^{3/2} .$ $x = 4$ par: $4^{3/2} = (4)^{3} = 2^{3} = 8$ , toh $y^{'} (4) = \frac{5}{2} \cdot 8 = 20.$ KYUN $4^{3/2} = 8$ : exponent $3/2$ ka matlab hai "square root lo phir cube karo" — $4 = 2$ lo, phir cube karo.

Level 3 — Analysis

Yahan tum rule ke baare mein reason karte ho, sirf use nahi karte — ise ulta karte ho, geometrically use karte ho, aur dekhte ho kahan derivative strange behave karta hai.

Q7. Ek curve ka slope har jagah $y^{'} = 12 x^{3}$ hai aur woh origin ki height rule $y = x^{4}$ -type family se guzarti hai. Kaun sa power function $y = x^{n}$ hai jiska derivative exactly $12 x^{3}$ hai? ( $n$ aur leading coefficient dhoondho agar $y = c x^{n}$ ho.)

Q8. $y = x$ ki tangent line ki equation us point par nikalo jahan $x = 9$ ho.

Q9. $f (x) = x^{1/3}$ ke liye — derivative ka use karke — explain karo ki graph ka $x = 0$ par vertical tangent kyun hai. (Step figure dekho.)

Recall Solution — Q7, Q8, Q9

Q7. KIYA KYA: power rule ko ulta chalaao. Agar $y = c x^{n}$ toh $y^{'} = c n x^{n - 1}$ . Ise $12 x^{3}$ ke barabar karna hai. KYUN pieces match karein: do power functions tab hi equal hote hain jab unke exponents aur coefficients dono match karein.

Exponent: $n - 1 = 3 \Rightarrow n = 4$ .
Coefficient: $c n = 12 \Rightarrow c \cdot 4 = 12 \Rightarrow c = 3$ .

Toh $y = 3 x^{4}$ (check: $y^{'} = 3 \cdot 4 x^{3} = 12 x^{3}$ ✓).

Q8. KIYA KYA: slope nikalo, point nikalo, line assemble karo $y - y_{0} = m (x - x_{0})$ .

Point: $y (9) = 9 = 3$ . Toh $(x_{0}, y_{0}) = (9, 3)$ .
Slope: $y^{'} = \frac{1}{2 x}$ , toh $m = \frac{1}{2 9} = \frac{1}{6}$ . $y - 3 = \frac{1}{6} (x - 9) ⟹ y = \frac{1}{6} x + \frac{3}{2} .$ KYUN point-slope: ek point aur ek slope se line banane ka sabse fast tarika yahi hai.

Q9. KIYA KYA: $f^{'} (x)$ compute karo aur dekhte raho jab $x \to 0$ . $f^{'} (x) = \frac{1}{3} x^{- 2/3} = \frac{1}{3 x ^{2/3}} .$ KYUN blow up karta hai: $x^{2/3}$ ek positive number hai jo $x \to 0$ hone par $0$ ki taraf shrink karta hai, toh $1/ (3 x^{2/3})$ bina bound ke badhta hai: $f^{'} (x) \to + \infty$ . Infinite slope ka matlab hai tangent line bilkul seedhi khadi hai — vertical. Dono taraf ( $x \to 0^{+}$ aur $x \to 0^{-}$ ) $x^{2/3} > 0$ dete hain, toh slope dono taraf se $+ \infty$ ki taraf rocket karta hai. Neeche figure mein near-vertical tangent dekho.

Figure — Power rule — proof for integer, rational exponents

Level 4 — Synthesis

Ab tum rule ko khud se rebuild karte ho — Limit definition of the derivative se special cases prove karte ho aur Chain rule / Implicit differentiation ko jodke laate ho.

Q10. Limit definition se prove karo (Stage-1 style) ki $\frac{d}{d x} x^{3} = 3 x^{2}$ , har cancellation dikhate hue.

Q11. First principles se prove karo ki $\frac{d}{d x} x^{- 1} = - x^{- 2}$ , common-denominator trick use karke (Stage 3).

Q12. Implicit differentiation (Stage 4) use karke prove karo ki $\frac{d}{d x} x^{2/3} = \frac{2}{3} x^{- 1/3}$ .

Recall Solution — Q10, Q11, Q12

Q10. KIYA KYA: limit-definition quotient likho aur $(x + h)^{3}$ ko Binomial theorem se expand karo. $f^{'} (x) = lim_{h \to 0} \frac{( x + h ) ^{3} - x ^{3}}{h} .$ Expand karo: $(x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ . $x^{3}$ subtract karo (leading term cancel ho jaata hai): $(x + h)^{3} - x^{3} = 3 x^{2} h + 3 x h^{2} + h^{3} .$ KYUN $h$ se divide karein: har surviving term mein kam se kam ek $h$ hai, toh divide karna legal hai aur pata chal jaata hai kaunse term mein $h$ bacha nahi: $\frac{( x + h ) ^{3} - x ^{3}}{h} = 3 x^{2} + 3 x h + h^{2} .$ $h \to 0$ bhejo: aakhri do terms mein $h$ hai aur woh vanish ho jaate hain. $f^{'} (x) = 3 x^{2} . ✓$

Q11. KIYA KYA: $f (x) = x^{- 1} = \frac{1}{x}$ . Difference quotient banao aur common denominator par combine karo.

= \lim_{h\to0}\frac{1}{h}\cdot\frac{x - (x+h)}{(x+h)x}.$$ KYUN common denominator: do fractions ko ek mein collapse kar deta hai jiska top simplify ho jaata hai. Top hai $x-(x+h) = -h$: $$= \lim_{h\to0}\frac{-h}{h\,(x+h)x} = \lim_{h\to0}\frac{-1}{(x+h)x}.$$ KYUN $h$ cancel hota hai: wahi ek cheez thi jo hume $h=0$ set karne se rok rahi thi. Ab $h\to0$ jaane do: $$f'(x) = \frac{-1}{x\cdot x} = -\frac{1}{x^2} = -x^{-2}. \checkmark$$ **Q12.** KIYA KYA: $y = x^{2/3}$ set karo aur fraction clear karne ke liye dono taraf cube karo (yahan $p=2,\,q=3$): $$y^3 = x^2.$$ KYUN cube karein: fractional exponent do *integer* powers mein badal jaata hai jinhein hum pehle prove kar chuke hain. Dono sides ko $x$ ke respect mein differentiate karo; left side ko [[Chain rule]] chahiye kyunki $y$ depends karta hai $x$ par: $$3y^2\,\frac{dy}{dx} = 2x.$$ $\frac{dy}{dx}$ ke liye solve karo: $$\frac{dy}{dx} = \frac{2x}{3y^2}.$$ $y = x^{2/3}$ substitute karo, toh $y^2 = x^{4/3}$: $$\frac{dy}{dx} = \frac{2x}{3\,x^{4/3}} = \tfrac23\,x^{1 - 4/3} = \tfrac23\,x^{-1/3}. \checkmark$$

Level 5 — Mastery

Boundary cases, wrong-rule detection, aur rule ka limit-of-the-rule sanity check — jahan deep understanding mechanical use se alag hoti hai.

Q13. Do look-alikes: differentiate karo (a) $y = x^{e}$ aur (b) $y = e^{x}$ . Kaun sa power rule use karta hai, kaun sa nahi, aur kyun? Dono derivatives do.

Q14. Zero/degenerate case. $f (x) = x^{0}$ consider karo ( $x \neq = 0$ ke liye defined). $f^{'} (x)$ do tareekon se compute karo: (i) directly, aur (ii) power-rule formula $n x^{n - 1}$ se jahan $n = 0$ ho. Confirm karo ki dono agree karte hain, aur explain karo ki formula ka $n$ ka factor hume kya "protect" kar raha hai.

Recall Solution — Q13, Q14

Q13. Deciding sawaal: kya exponent constant hai aur base variable, ya base constant hai aur exponent variable?

(a) $y = x^{e}$ . Base $x$ vary karta hai; exponent $e \approx 2.718$ ek constant number hai. Yeh power rule hai — upar baitha number $e$ koi special nahi hai: $y^{'} = e x^{e - 1} .$
(b) $y = e^{x}$ . Base $e$ constant hai; exponent $x$ vary karta hai. Yeh ek exponential hai, power nahi. Iska rule (dekho Derivative of exponential functions) hai $\frac{d}{d x} a^{x} = a^{x} ln a$ , aur $ln e = 1$ hone se: $y^{'} = e^{x} .$ KYUN dono alag hain: power rule $(x + h)^{n}$ ko $x$ mein expand karke prove kiya gaya tha; woh machinery variable exponent ko touch nahi kar sakti. Alag structure ⇒ alag tool.

Q14. (i) Directly. $x \neq = 0$ ke liye, $x^{0} = 1$ , ek constant hai. Kisi bhi constant ka derivative $0$ hota hai (limit definition se, $\frac{1 - 1}{h} = 0$ sabhi $h$ ke liye): $f^{'} (x) = 0.$ (ii) Formula se. $n = 0$ deta hai: $f^{'} (x) = n x^{n - 1} = 0 \cdot x^{- 1} = 0.$ Dono agree karte hain. KIYA KYA $n$ ka factor protect karta hai: bhagle bhi $x^{- 1}$ $x = 0$ par undefined ho, leading factor $n = 0$ isse $0$ se pehle multiply kar deta hai, toh woh kabhi matter hi nahi karta. Formula chupke se encode karta hai "ek constant ka slope zero hota hai" — $n = 0$ wala front factor exactly wahi hai jo otherwise awkward $x^{- 1}$ ko khatam karta hai. KYUN yeh promised consistency hai: proof ka Stage 2 aur general formula same answer dete hain, toh charon stages ek saath seamlessly jud jaate hain.

Recall Self-test scoreboard

L1–L2 bina help ke kiya ::: Tum rule ko reliably apply kar sakte ho. L3 bina help ke kiya ::: Tum slopes aur tangents ke baare mein reason kar sakte ho, vertical wale bhi. L4 bina help ke kiya ::: Tum rule ko limit definition aur implicitly se rebuild kar sakte ho. L5 bina help ke kiya ::: Tum power-vs-exponential trap pakad lete ho aur degenerate cases handle karte ho — mastery.

Connections

Power rule — proof for integer, rational exponents — woh parent theorem jinka yeh drill hai.
Limit definition of the derivative — Q10, Q11, Q14(i) mein use hua.
Binomial theorem — Q10 mein $(x + h)^{3}$ expand karta hai.
Chain rule — Q12 mein $\frac{d y}{d x}$ jodta hai.
Implicit differentiation — Q12 ka method.
Quotient rule — Q5 mein warn kiya gaya tempting-but-wrong path.
Derivative of exponential functions — Q13(b) ke liye sahi rule.