Power rule — proof for integer, rational exponents: Question bank

4.1.12 · D5 · Maths › Calculus I — Limits & Derivatives › Power rule — proof for integer, rational exponents

Definition Is PAGE pe use hone wale symbols aur stages (inhe pehle build karo)

Exponent $n$ — woh chota number jo upar-right mein baithta hai $x^{n}$ mein: yeh count hai ki $x$ ki kitni copies multiply hoti hain ( $x^{3} = x \cdot x \cdot x$ , toh $n = 3$ ). Yeh power ke top-right mein rehta hai, jab $x$ vary karta hai tab yeh constant hota hai, aur poora page ek hi formula $\frac{d}{d x} x^{n} = n x^{n - 1}$ ke baare mein hai. Jaise $n$ number line pe range karta hai — positive integer, $0$ , negative integer, fraction, irrational — hume har range ke liye ek alag tool chahiye, yahi wajah hai ki $n$ ko neeche stages mein sort kiya gaya hai.
Increment $h$ — $x$ mein add ki gayi ek choti si nudge. Har derivative yahan limit definition $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ se aati hai: hum $x$ ko $x + h$ tak push karte hain, change measure karte hain, nudge se divide karte hain, phir $h$ ko $0$ tak shrink karte hain. Figure 1 mein do coral dots ke beech ka horizontal gap hi $h$ hai, butter arrow height ka change hai, aur coral dashed line unka ratio hai (secant slope) jo true slope ban jaata hai jab $h$ shrink hota hai.
Integer $m$ — negative-exponent case mein hum $n = - m$ likhte hain jahan $m > 0$ ek positive integer hai, toh $x^{n} = x^{- m} = 1/ x^{m}$ . $m$ ko naam dene se hum $x^{m}$ pe positive-integer result reuse kar sakte hain.
Chaar stages Figure 2 mein labelled boxes hain: Stage 1 positive integer $n$ (binomial expansion), Stage 2 $n = 0$ (constant), Stage 3 negative integer $n = - m$ (common denominator), Stage 4 rational $n = p / q$ (implicit differentiation). Ek paanchwa idea — irrational $n$ — continuity ke zariye reach hota hai.

Figure — Power rule — proof for integer, rational exponents

Intuition Figure 2 padho — proof stages ke beech kaise

flow karta hai Figure 2 ek road map hai. Stage 1 (lavender, top-left) foundation hai: positive integer $n$ ke liye $(x + h)^{n}$ ki binomial expansion. Iske bahar arrows follow karo: ek arrow Stage 3 (butter) ko feed karta hai — kyunki common denominator lene ke baad, Stage 3 ka numerator $x^{m} - (x + h)^{m}$ literally positive integer $m$ mein ek Stage-1 difference hai. Doosra arrow Stage 4 (coral) ko feed karta hai — kyunki chain-rule step $\frac{d}{d x} y^{q} = q y^{q - 1} \frac{d y}{d x}$ hi Stage 1 hai inner function $y$ pe apply kiya hua. Stage 2 (mint) trivial constant case hai. Stage 4 se ek "rationals dense" arrow irrational- $n$ box tak pahunchta hai: irrational exponent ke paas crowd karte rationals lo, har ek pe Stage 4 apply karo, phir limit le lo. Saare arrows finally ek boxed result $\frac{d}{d x} x^{n} = n x^{n - 1}$ pe converge karte hain. Yeh picture hi proof ka dependency graph hai.

True or false — justify karo

Power rule

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

har real

n

ke liye hold karta hai.

Ek result ke taur pe True hai, lekin koi ek proof saare

n

ko cover nahi karta — chaar stages rational

n

handle karte hain, aur irrational

n

baad mein ek continuity/limiting argument se reach hota hai (edge cases dekho), yahi wajah hai ki parent ise pieces mein tod ta hai.

Binomial theorem akela saare exponents ke liye power rule prove karta hai.

False.

(x + h)^{n}

ka ek finite, cleanly-cancelling expansion tab hi hota hai jab

n

ek non-negative integer ho;

n = \frac{1}{2}

n = - 3

ke liye yeh infinite series ban jaata hai jisme koi last term nahi hota jo divide ho sake.

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

False. Yeh power rule (variable base, constant exponent) ko exponential (constant base, variable exponent) se mix up karta hai; sahi answer

2^{x} ln 2

hai, dekho Derivative of exponential functions.

n = 0

ke liye, formula

n x^{n - 1}

deta hai

0 \cdot x^{- 1}

, jo

x = 0

pe undefined hai, toh rule wahan fail karta hai.

False. Derivative definition se lo:

\frac{( x + h ) ^{0} - x ^{0}}{h} = \frac{1 - 1}{h} = 0

har

h \neq = 0

aur har

x

ke liye (including

x = 0

), toh limit har jagah

0

hai — "

0 \cdot x^{- 1}

" actually kabhi evaluate hi nahi hota; leading factor

0

poore product ko annihilate karta hai

x \neq = 0

ke liye, aur definition khud

x = 0

0

supply karta hai.

Stage 4 (rational exponent) mein tum

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

likh sakte ho kyunki yeh "power rule jaisa lagta hai."

False —

y

$x$ ka ek function hai, toh chain rule

\frac{d y}{d x}

attach karta hai: correct left side hai

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

Negative-integer proof (Stage 3,

n = - m

ke saath) secretly positive-integer proof pe rely karta hai.

True. Common denominator lene ke baad yeh

\frac{( x + h ) ^{m} - x ^{m}}{h} \to m x^{m - 1}

use karta hai, jo Stage 1 ka result hai positive integer

m

pe apply kiya hua — stages deliberately stacked hain.

y = x

ke dono sides ko square karke

y^{2} = x

get karna function ko change karta hai, toh jo derivative hum compute karte hain woh

x

ki derivative nahi hai.

False. Squaring domain

x \geq 0

pe ek valid, reversible bridge hai; yeh hume integer powers use karne deta hai, aur

\frac{d y}{d x}

solve karne ke baad hum substitute back karte hain, toh result genuinely

\frac{d}{d x} x

hai.

Irrational exponent jaise

n = π

ke liye power rule ko rational stages se bilkul alag ek brand-new proof chahiye.

False. Hum

π

ko rationals ki ek sequence se approximate karte hain (har ek Stage 4 se handle hota hai) aur limit pe jaate hain; kyunki

x^{n}

aur

n x^{n - 1}

$n$ mein continuous hain (fixed

x > 0

ke liye,

x^{n} = e^{n l n x}

n

mein continuous hai), formula limit ke baad bhi survive karta hai — koi alag machinery nahi chahiye.

Error dhundo

(x + h)^{1/2} = x^{1/2} + \frac{1}{2} x^{- 1/2} h

, toh

h

se divide karne par

\frac{1}{2} x^{- 1/2}

milta hai." Flaw kahan hai?

Expansion sirf ek infinite series ke pehle do terms hain, exact equality nahi — tum legitimately ek non-integer exponent ke liye finite expansion cancel nahi kar sakte, yahi wajah hai ki hum implicit differentiation use karte hain.

x^{- 2}

ke liye,

(x + h)^{- 2}

ko binomial theorem se expand karo aur Stage 1 ki tarah cancel karo." Kya toot ta hai?

Binomial theorem ke finite form ko non-negative integer exponent chahiye;

- 2

ek infinite series produce karta hai, toh kuch bhi cleanly cancel nahi hota — Stage 3 ka common-denominator trick use karo (

- 2 = - m

likho jahan

m = 2

y^{3} = x^{2}

differentiate karo: left side

3 y^{2}

hai, right side

2 x

hai, toh

\frac{d y}{d x} = \frac{2 x}{3 y ^{2}}

." Mistake dhundo.

Left side mein chain-rule factor missing hai:

\frac{d}{d x} y^{3} = 3 y^{2} \frac{d y}{d x}

, jisse

3 y^{2} \frac{d y}{d x} = 2 x

aur

\frac{d y}{d x} = \frac{2 x}{3 y ^{2}}

milta hai — final answer ittefaq se match karta hai kyunki writer ne factor omit kiya phir "recover" kiya; reasoning galat hai.

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

, toh

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

." Kya galat hai?

x^{x}

mein variable base AUR variable exponent dono hain — na power rule akela apply hota hai, na exponential rule; tumhe logarithmic/implicit differentiation chahiye. (Sahi answer

x^{x} (ln x + 1)

hai.)

"Stage 1 mein, nudge

h

se divide karne ke baad hume milta hai

n x^{n - 1} + (2 n) x^{n - 2} h + \dots

; finish karne ke liye

h = 0

set karo." Kya "

h = 0

set karo" sahi phrase hai?

Strictly hum limit

h \to 0

lete hain,

h = 0

plug nahi karte — original

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

mein

h = 0

plug karna

\frac{0}{0}

deta hai; poora point yahi hai ki cancellation ke baad limit exist karta hai aur

n x^{n - 1}

ke barabar hota hai.

Why questions

Positive integers ke liye specifically binomial theorem kyun invoke kiya jaata hai, koi aur expansion kyun nahi?

Kyunki yeh first-order-in- $h$ term

n x^{n - 1} h

ko saare higher-

h

terms se alag expose karta hai; nudge

h

se divide karke aur limit lene ke baad, sirf woh first-order coefficient survive karta hai — binomial theorem woh tool hai jo ise isolate karta hai.

h \to 0

ke saath

n x^{n - 1}

ke alawa saare terms kyun vanish ho jaate hain?

Divided expansion ke har doosre term mein ab bhi kam se kam nudge

h

ka ek factor hota hai (

(2 n) x^{n - 2} h

wala, etc.), aur koi bhi quantity times shrinking

h

0

ki taraf tend karta hai.

Stage 3 directly expand karne ki bajay common denominator kyun leta hai?

\frac{1}{( x + h ) ^{m}} - \frac{1}{x ^{m}}

combine karne se numerator

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

banta hai, jo exactly ek Stage-1-shaped difference hai (positive integer

m

mein) jisko limit karna hum pehle se jaante hain — yeh ek naye problem ko ek solved problem mein convert karta hai.

Rational exponent

n = p / q

mein

q \neq = 0

kyun hona chahiye?

0

ka denominator

p / q

ko undefined banata hai, aur trick

y^{q} = x^{p}

collapse kar jaati hai (

y^{0} = 1

koi information nahi rakhta), toh differentiate karne ke liye koi equation hi nahi bachti.

Step by step dikhao ki

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

kyun hai — aur woh step "just Stage 1" kyun hai?

Outer power ko

g (u) = u^{q}

likho jahan

u = y (x)

. Stage 1 (integer power rule) deta hai

g^{'} (u) = q u^{q - 1}

. Chain rule kehta hai

g (y (x))

ka rate of change hai kitna fast $g$ respond karta hai $u$ se times kitna fast $u = y$ respond karta hai $x$ se:

\frac{d}{d x} g (y) = g^{'} (y) \cdot \frac{d y}{d x} = q y^{q - 1} \frac{d y}{d x}

. Toh Stage 1

q y^{q - 1}

supply karta hai aur chain rule inner rate

\frac{d y}{d x}

se multiply karta hai; agar

y

sirf

x

hota toh

\frac{d y}{d x} = 1

aur hume plain Stage 1 mil jaata.

Exponent

n

mein thoda sa change,

x^{n}

mein sirf thoda sa change kyun produce karta hai (woh fact jis pe continuity lean karti hai)?

Fixed base

x > 0

ke liye,

x^{n} = e^{n l n x}

, aur

e^{\cdot}

continuous hai; toh agar

n_{k} \to n

toh

x^{n_{k}} \to x^{n}

aur similarly

n_{k} x^{n_{k} - 1} \to n x^{n - 1}

—

n

mein small wobble, dono sides mein small wobble, aur yahi cheez formula ko irrational limit tak pass karne deti hai.

x

ki derivative

x = 0

ke paas kyun blow up karti hai?

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

, aur jaise

x \to 0^{+}

denominator

0

tak shrink karta hai, toh slope

+ \infty

ki taraf tend karta hai — jo graph ki origin pe near-vertical steepness se match karta hai.

Edge cases

n = 1

ke saath

x = 0

\frac{d}{d x} x^{n}

kya hai?

f (x) = x

f^{'} (x) = 1 \cdot x^{0} = 1

har jagah hai, including

x = 0

; yahan

x^{0} = 1

by convention hai, toh "

x^{n - 1}

" factor harmless hai.

Kya power rule

x^{1/2}

ke liye

x = 0

pe derivative deta hai?

Nahi —

\frac{1}{2 x}

x = 0

pe undefined hai; function

x

wahan vertical tangent hai, toh derivative genuinely exist nahi karta bhale hi

0

theek hai.

x = 0

x^{- 1}

ke baare mein kya?

x^{- 1} = 1/ x

x = 0

pe define hi nahi hai (wahan vertical asymptote hai), toh na function exist karta hai na iska derivative

- x^{- 2}

— rule sirf domain ke points ke baare mein baat karta hai.

Kya

x^{2/3}

x = 0

pe differentiable hai?

Nahi —

\frac{2}{3} x^{- 1/3} \to \pm \infty

jaise

x \to 0

, ek sharp cusp deta hai; function

0

pe continuous hai lekin koi finite slope nahi hai.

Negative-exponent formula

- m x^{- m - 1}

ek limit argument ke taur pe kahan break down karta hai?

x = 0

pe, kyunki working

(x + h)^{m} x^{m} \to x^{2 m}

se divide karta hai, jo

x = 0

0

hota hai — derivation throughout assume karta hai

x \neq = 0

, jo

x^{- m}

ke wahan undefined hone ke consistent hai.

Kya rational-exponent proof

x < 0

ke liye valid hai jab

q

even ho (e.g.

x^{1/2}

Nahi —

x^{1/2}

x < 0

ke liye real nahi hai, toh equation

y^{q} = x^{p}

ka koi real

y

nahi hai; proof sirf wahan hold karta hai jahan

x^{p / q}

actually defined hai (yahan

x \geq 0

Kya rational-exponent proof

x < 0

ke liye valid hai jab

q

odd ho (e.g.

x^{1/3}

Haan

x < 0

ke liye, kyunki

x^{1/3}

real hai (e.g.

(- 8)^{1/3} = - 2

); lekin

x = 0

pe dhyan do, jahan

\frac{1}{3} x^{- 2/3} \to + \infty

vertical tangent deta hai — toh formula

0

ke dono taraf hold karta hai lekin

0

pe nahi.

x^{n}

irrational

n

ke liye defined kaise hai, aur kin bases pe?

Sirf

x > 0

ke liye: hum

x^{n} = e^{n l n x}

define karte hain, aur

ln x

x > 0

chahiye.

x < 0

ke liye irrational power jaise

(- 2)^{π}

ki koi real value nahi hai (koi consistent odd/even root trick irrational exponent ke saath survive nahi karti), toh irrational

n

ke liye continuity argument strictly positive-base domain

x > 0

pe rehta hai;

x = 0

exclude hai kyunki

ln 0

undefined hai.

Irrational exponent jaise

x^{π}

tak power rule kaise extend hota hai, aur rational approximations ko converge kya karta hai?

Rationals

r_{k} \to π

chuno (e.g.

π = 3.14159 \dots

3, \frac{31}{10}, \frac{314}{100}, \dots

tak truncate karke); Stage 4 deta hai

\frac{d}{d x} x^{r_{k}} = r_{k} x^{r_{k} - 1}

har ek ke liye. Kyunki (fixed

x > 0

ke liye) dono

x^{n} = e^{n l n x}

aur

n x^{n - 1}

n

mein continuous hain,

r_{k} \to π

lene se

r_{k} x^{r_{k} - 1} \to π x^{π - 1}

force hota hai — inputs

r_{k} \to π

ka convergence outputs pe transfer hota hai, deta hai

\frac{d}{d x} x^{π} = π x^{π - 1}

Kya hum

x^{0}

pe power rule use karke yeh conclude kar sakte hain ki kisi bhi constant ki derivative

0

hai?

Directly nahi —

x^{0} = 1

ek specific constant hai; ek general constant

c

x

ki power nahi hai, lekin iska derivative

0

hai usi underlying wajah se: limit definition mein

\frac{c - c}{h} = 0

har

h \neq = 0

ke liye.

Recall Traps ki ek-line summary

Exponent class ke liye wrong tool, chain-rule factor drop karna, $x^{n}$ ko $n^{x}$ se confuse karna, yeh bhool jaana ki "rule applies" ka matlab hai point domain mein hona chahiye, aur yeh sochna ki irrational exponents ko new machinery chahiye jab continuity kaam tamam kar deti hai. ::: Ye sab woh jagah hain jahan staged proof structure tumhe bachata hai — agar tum naam le sako ki kaunsa stage (ya continuity step) kisi exponent ko handle karta hai, tum shayad hi kisi trap mein giro.

Connections

Power rule — proof for integer, rational exponents — woh parent jise yeh bank drill karta hai.
Limit definition of the derivative — kyun nudge $h \to 0$ ek limit hai, plug-in nahi.
Binomial theorem — Stage 1 ke peeche finite-vs-infinite distinction.
Chain rule — woh factor jo har koi Stage 4 mein drop karta hai.
Implicit differentiation — rational-exponent method.
Derivative of exponential functions — $n^{x}$ contrast trap.
Quotient rule — negative exponents ke liye ek alternative.