Power rule — proof for integer, rational exponents

4.1.12 · Maths › Calculus I — Limits & Derivatives

WHAT we are proving

Stage 1 — Positive integer $n$ (binomial proof)

HOW — scratch se derive karo: $f^{'} (x) = lim_{h \to 0} \frac{( x + h ) ^{n} - x ^{n}}{h} .$ Numerator expand karo: $(x + h)^{n} - x^{n} = x^{n} + n x^{n - 1} h + (2 n) x^{n - 2} h^{2} + \dots + h^{n} - x^{n} .$ $h$ se divide karo: $\frac{( x + h ) ^{n} - x ^{n}}{h} = n x^{n - 1} + har term mein h hai (2 n) x^{n - 2} h + \dots + h^{n - 1} .$ $h \to 0$ lo: brace mein har term mein $h$ hai, toh woh mar jaata hai. $f^{'} (x) = n x^{n - 1} .$

Stage 2 — Exponent $n = 0$

Stage 3 — Negative integer $n = - m$ , $m > 0$ (quotient/limit proof)

HOW — first principles se: $f (x) = x^{- m} = \frac{1}{x ^{m}}, f^{'} (x) = lim_{h \to 0} \frac{1}{h} (\frac{1}{( x + h ) ^{m}} - \frac{1}{x ^{m}}) .$ Common denominator: $= lim_{h \to 0} \frac{x ^{m} - ( x + h ) ^{m}}{h ( x + h ) ^{m} x ^{m}} .$ Ye step kyun? Fractions combine karne se $x^{m} - (x + h)^{m}$ expose hota hai, jise hum pehle se handle karna jaante hain. Stage 1 use karo: $\frac{( x + h ) ^{m} - x ^{m}}{h} \to m x^{m - 1}$ , toh $\frac{x ^{m} - ( x + h ) ^{m}}{h} \to - m x^{m - 1}$ . Denominator wala factor $(x + h)^{m} x^{m} \to x^{2 m}$ . Isliye $f^{'} (x) = \frac{- m x ^{m - 1}}{x ^{2 m}} = - m x^{m - 1 - 2 m} = - m x^{- m - 1} .$ $n = - m$ ke saath: $f^{'} (x) = n x^{n - 1} .$ ✓

Stage 4 — Rational exponent $n = p / q$ (implicit differentiation)

HOW — derive karo: Maano $y = x^{p / q} \Rightarrow y^{q} = x^{p}$ (dono integer exponents). Dono sides ko $x$ ke w.r.t. differentiate karo (left side pe chain rule use karo — jo khud Stage 1 ka consequence hai): $q y^{q - 1} \frac{d y}{d x} = p x^{p - 1} .$ Ye step kyun? Left side: $\frac{d}{d x} y^{q} = q y^{q - 1} \frac{d y}{d x}$ . Solve karo: $\frac{d y}{d x} = \frac{p}{q} \cdot \frac{x ^{p - 1}}{y ^{q - 1}} .$ $y = x^{p / q}$ substitute karo, toh $y^{q - 1} = x^{p (q - 1) / q} = x^{p - p / q}$ : $\frac{d y}{d x} = \frac{p}{q} x^{p - 1 - (p - p / q)} = \frac{p}{q} x^{p / q - 1} .$ $n = p / q$ ke saath: $\frac{d y}{d x} = n x^{n - 1} .$ ✓

Figure — Power rule — proof for integer, rational exponents

Common mistake Classic traps ko steel-man karo

Trap 1: "Binomial proof sab $n$ ke liye kaam karta hai." Kyun sahi lagta hai: algebra "general lagta hai." Flaw: $(x + h)^{n}$ ka finite expansion tabhi hota hai jab $n$ ek non-negative integer ho. $n = 1/2$ ya $n = - 3$ ke liye expansion ek infinite series hai — aap cleanly divide-and-cancel nahi kar sakte. Fix: fraction trick use karo (negatives ke liye) ya implicit differentiation (rationals ke liye).

Trap 2: $x^{n}$ ko $n^{x}$ ki tarah treat karna. $\frac{d}{d x} 2^{x} \neq = x 2^{x - 1}$ ! Power rule ko variable base, constant exponent chahiye. Fix: exponentials mein $\frac{d}{d x} a^{x} = a^{x} ln a$ use hota hai.

Trap 3: Stage 4 mein chain rule bhool jaana. $\frac{d}{d x} y^{q} = q y^{q - 1}$ likhna (bina $\frac{d y}{d x}$ ke). Kyun sahi lagta hai: ye "power rule jaisa lagta hai." Fix: $y$ , $x$ ka function hai, isliye chain rule $\frac{d y}{d x}$ attach karta hai.

Active recall

Power rule ka statement kya hai?

\frac{d}{d x} x^{n} = n x^{n - 1}

Positive-integer proof mein kaun sa theorem kaam aata hai?

Binomial theorem jo

(x + h)^{n}

expand karta hai.

(x + h)^{n} - x^{n}

expand karke

h

se divide karne ke baad,

h \to 0

pe kaun sa term bachta hai?

n x^{n - 1}

wala term (baaki sab mein

h

ka factor hota hai).

Binomial proof

n = 1/2

n = - 3

ke liye kyun kaam nahi karta?

Un exponents se ek infinite series milti hai, na ki finite cancellable expansion.

Negative integers ke liye rule prove karne ki trick?

x^{- m} = 1/ x^{m}

likho, common denominator lo, positive-integer result reuse karo.

Rational

p / q

ke liye rule prove karne ki trick?

y = x^{p / q}

set karo, power

q

se raise karo taaki

y^{q} = x^{p}

mile, phir implicitly differentiate karo.

y^{q}

x

ke w.r.t. differentiate karne se kya milta hai?

q y^{q - 1} \frac{d y}{d x}

(chain rule).

\frac{d}{d x} x

\frac{1}{2 x}

Kya

\frac{d}{d x} 2^{x} = x 2^{x - 1}

hai?

Nahi — ye galat rule hai;

\frac{d}{d x} 2^{x} = 2^{x} ln 2

\frac{d}{d x} x^{- 2}

- 2 x^{- 3}

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek tower hai blocks ka jahan har level $x$ times wide hai, $n$ levels oonchi — woh hai $x^{n}$ . Agar tum $x$ ko thoda sa bada karo, tower kitni tezi se badhega? Answer hai: tumhe $n$ copies milenge ek-level-chote towers ( $x^{n - 1}$ ) ke, kyunki $n$ levels mein se har ek "feel" karta hai us stretch ko. Isliye tum $n$ ko aage slide karte ho aur height ek se ghata dete ho. Roots ke liye (jaise $x$ ) hum ek flip-game khelate hain: isko square karo taaki root khatam ho, easy rule use karo, phir flip back karo.

Connections

Limit definition of the derivative — Stages 1 & 3 ka engine.
Binomial theorem — finite expansion deta hai.
Chain rule — rational-exponent proof mein chahiye.
Implicit differentiation — Stage 4 ke peeche ka method.
Derivative of exponential functions — contrast: $a^{x}$ mein power rule NAHI lagta.
Quotient rule — negative exponents ke liye alternate path.