Power rule — proof for integer, rational exponents: Worked examples

4.1.12 · D3 · Maths › Calculus I — Limits & Derivatives › Power rule — proof for integer, rational exponents

Symbols se pehle, ek reminder ki har symbol ka kya matlab hai — sab ko achhi tarah samjho:

Definition Woh symbols jo hum har jagah use karte hain

$x$ — input, ek aisi number jise hum thoda badal sakte hain (graph par horizontal position).
$x^{n}$ — " $x$ apne aap se multiply hota hai, $n$ ke pattern ke hisaab se". Agar $n = 3$ hai to yeh $x \cdot x \cdot x$ hai; agar $n = - 2$ hai to yeh $\frac{1}{x \cdot x}$ hai; agar $n = \frac{1}{2}$ hai to yeh $x$ hai.
$h$ — ek tiny nudge jo hum $x$ mein add karte hain. Slope ko measure kiya jaata hai $x + h$ par height aur $x$ par height ke beech compare karke, phir $h$ ko $0$ ki taraf shrink karke (yeh hai Limit definition of the derivative). Jab bhi " $h \to 0$ " dikhe, socho nudge gayab ho raha hai.
$\frac{d}{d x} ()$ — "jo kuch bhi andar hai uske graph ka slope, position $x$ par". Slope hota hai rise over run: curve kitne units upar chadh jaati hai ek unit seedhe jaane par. Bada slope = steep hill; slope $0$ = flat zameen; negative slope = neeche utarna.

Power rule us slope ka shortcut hai: exponent ko front par drop karo, phir exponent ko ek se kam karo.

Scenario matrix

Is poore topic mein har problem neeche ke cells mein se ek hoti hai. Jo examples aate hain woh us cell ke saath tagged hain jo woh cover karte hain.

Cell	Case class	Kya tricky hai isme	Example
A	Positive integer exponent	kuch nahi — "easy" base case	Ex 1
B	Exponent $= 0$ / plain constant	flat line ka slope $0$ hota hai	Ex 2
C	Negative integer exponent	answer ka sign flip hota hai; $x = 0$ par undefined	Ex 3
D	Rational exponent / root	fraction arithmetic; $x = 0$ par cusp	Ex 4
E	"Hidden" power — pehle algebra chahiye	$3 x$ , $\frac{1}{x ^{5}}$ , $\frac{x ^{2}}{x}$ ko $x^{n}$ ke roop mein rewrite karna hoga	Ex 5
F	Ek point par slope evaluate karo (sign of slope)	kya curve wahan upar ja raha hai ya neeche?	Ex 6
G	Degenerate / undefined input	slope blow up hota hai jab $x \to 0$	Ex 7
H	Real-world rate (with units)	word problem; units sahi rakhni hain	Ex 8
I	Irrational exponent	kya $x^{π}$ ke liye shortcut kaam karta hai?	Ex 9
J	Exam twist — power rule jaisa lagta hai lekin hai nahi	$2^{x}$ exponential hai, power nahi	Ex 10

Example 1 — Cell A: positive integer

Worked example

$f (x) = x^{7}$ ko differentiate karo. Forecast: answer cover karo aur guess karo. $7$ drop karo, exponent kam karo... kya milta hai?

Step 1 — $n$ identify karo. Yahan $n = 7$ hai, ek positive whole number. Yeh step kyun? Rule $\frac{d}{d x} x^{n} = n x^{n - 1}$ ke liye pehle $n$ ko name karna zaroori hai, taaki pata chale kya drop karna hai aur kya subtract karna hai.

Step 2 — exponent ko front par drop karo. Isse $7$ ka factor milta hai. Yeh step kyun? Yeh mnemonic ka pehla "Drop" hai.

Step 3 — exponent ko ek se kam karo. $7 - 1 = 6$ . $f^{'} (x) = 7 x^{6} .$ Yeh step kyun? Nudge $h$ yaad karo: jab $(x + h)^{7} - x^{7}$ expand karte ho aur $h$ se divide karte ho, to $7 x^{6}$ ke siwa har term mein $h$ ka factor rehta hai, isliye woh terms $0$ ho jaati hain jab nudge gayab hota hai. Sirf $7 x^{6}$ bachta hai.

Valid kahan? Har jagah — polynomial sabhi $x$ ke liye smooth hoti hai.

Verify: $x = 1$ par, $f^{'} (1) = 7$ . First principles se slope check karo tiny $h = 0.001$ ke saath: $\frac{1.00 1 ^{7} - 1}{0.001} \approx 7.02$ — $7$ ke kaafi paas. ✓

Example 2 — Cell B: exponent zero / constant

Worked example

$g (x) = x^{0}$ differentiate karo, aur alag se $c (x) = 5$ . Forecast: $x^{0}$ equals $1$ hota hai (koi bhi nonzero number ko zero power $1$ hota hai). $y = 1$ ka graph ek flat horizontal line hai. Flat zameen ka slope kya hota hai?

Step 1 — dono ko constants recognize karo. $x^{0} = 1$ aur $5$ dono sirf numbers hain; unke graphs horizontal lines hain. Yeh step kyun? Horizontal line kabhi rise nahi karti, isliye uska rise-over-run har jagah $0$ hai.

Step 2 — $x^{0}$ par rule apply karo. $n = 0$ ke saath: $0 \cdot x^{- 1} = 0.$ $g^{'} (x) = 0, c^{'} (x) = 0.$ Yeh step kyun? Jo factor hum "front par drop" karte hain woh $0$ hai, jo poori cheez zero kar deta hai — flat-line intuition se match karta hai.

Valid kahan? $c (x) = 5$ ke liye, har jagah. $g (x) = x^{0}$ ke liye, $x = 0$ ko chhod kar har jagah ( $0^{0}$ defined nahi hai) — lekin slope $0$ hai har us point par jahan yeh defined hai.

Verify: kisi bhi horizontal line ka slope $0$ hota hai. ✓ (Koi units nahi, koi motion nahi — kuch nahi badlta.)

Example 3 — Cell C: negative integer

Worked example

$f (x) = x^{- 5}$ differentiate karo. Forecast: answer ke aage negative number hoga. Guess karo: $- 5$ drop karo, phir $- 5 - 1 = ?$

Step 1 — $n$ identify karo. $n = - 5$ . Yeh step kyun? Rule kisi bhi real $n$ ke liye kaam karta hai; parent note ne negative-integer case ko $x^{- 5} = 1/ x^{5}$ likh kar aur common denominator lekar prove kiya tha, isliye hum shortcut seedha apply kar sakte hain.

Step 2 — $n$ ko front par drop karo. $- 5$ ka factor.

Step 3 — exponent ko ek se kam karo. $- 5 - 1 = - 6$ . Trap yeh hai ki log add kar dete hain: yeh subtract hai, isliye yeh aur zyada negative hota jaata hai. $f^{'} (x) = - 5 x^{- 6} = \frac{- 5}{x ^{6}} .$ Yeh step kyun? Negative-integer exponent ka general pattern hai $\frac{d}{d x} x^{- k} = - k x^{- k - 1}$ (yahan whole number $k = 5$ hai).

Valid kahan? Sabhi $x \neq = 0$ ke liye. Original function $x^{- 5} = 1/ x^{5}$ khud $x = 0$ par undefined hai (zero se divide nahi kar sakte), isliye na function exist karta hai na slope — formula $- 5/ x^{6}$ sahi tarah se blow up karta hai jab $x \to 0$ .

Verify: $f = 1/ x^{5}$ positive hai aur $x > 0$ ke liye decreasing hai (bada $x$ ⇒ chhoti value), isliye slope negative hona chahiye — aur $- 5/ x^{6}$ wahan negative hi hai. ✓

Example 4 — Cell D: rational exponent / root

Worked example

$f (x) = x^{2/3}$ differentiate karo. Forecast: $n = \frac{2}{3}$ . Front factor $\frac{2}{3}$ hai; naya exponent $\frac{2}{3} - 1$ hai. $\frac{2}{3} - 1$ kya hota hai?

Step 1 — $n = \frac{2}{3}$ identify karo (yeh $3 x^{2}$ hai, $x^{2}$ ka cube root). Yeh step kyun? Hum $x^{2/3}$ ko nudge $h$ ke saath whole powers ki tarah expand nahi kar sakte, isliye parent note ne alag raasta liya: $y = x^{2/3}$ rakho, dono sides cube karo to $y^{3} = x^{2}$ milta hai — ab dono exponents whole numbers hain, jinhe hum differentiate karna jaante hain. Us relation ko differentiate karne par (Implicit differentiation, kyunki $y$ secretly $x$ par depend karta hai) exactly wahi shortcut $n x^{n - 1}$ milta hai. Toh "drop and lower" fractions ke liye bhi kaam karta hai.

Step 2 — $\frac{2}{3}$ ko front par drop karo.

Step 3 — exponent se $1$ subtract karo. $1 = \frac{3}{3}$ likho, to $\frac{2}{3} - \frac{3}{3} = - \frac{1}{3}$ . $f^{'} (x) = \frac{2}{3} x^{- 1/3} = \frac{2}{3 3 x} .$ Yeh step kyun? Common mistake $\frac{2}{3} - 1$ mein fumble karna hai; pehle $1$ ko $\frac{3}{3}$ mein convert karo.

Valid kahan? Kyunki cube root negative numbers ke liye bhi defined hai, $f (x) = x^{2/3}$ sabhi $x$ ke liye exist karta hai — lekin slope $\frac{2}{3 3 x}$ $x = 0$ par undefined hai. Wahan curve mein ek cusp hai (upar define kiya gaya — sharp point jahan tangent flip hota hai): left se aate waqt tangent steeply up-left point karta hai, right se steeply up-right, aur dono agree nahi karte, isliye $x = 0$ par koi single slope exist nahi karta (figure mein sharp tip dekho). $x < 0$ ke liye formula abhi bhi apply hota hai (jaise $x = - 8$ par, $3 - 8 = - 2$ , to $f^{'} (- 8) = \frac{2}{3 \cdot ( - 2 )} = - \frac{1}{3}$ : curve left branch par neeche jaati hai).

Verify: $x = 8$ par: $f^{'} (8) = \frac{2}{3} \cdot 8^{- 1/3} = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$ . Numerically $\frac{8.00 1 ^{2/3} - 8 ^{2/3}}{0.001} \approx 0.333$ . Aur $x = - 8$ par: $f^{'} (- 8) = - \frac{1}{3}$ . ✓

Example 5 — Cell E: hidden powers (pehle rewrite karo!)

Worked example

$f (x) = \frac{x ^{2}}{x}$ differentiate karo. Forecast: power rule ko seedha fraction par apply nahi kar sakte. Pehle guess karo: kya isko single $x^{something}$ ke roop mein rewrite kar sakte hain?

Step 1 — ek power ke roop mein rewrite karo. $x = x^{1/2}$ , aur divide karne se exponents subtract hote hain: $\frac{x ^{2}}{x ^{1/2}} = x^{2 - 1/2} = x^{3/2} .$ Yeh step kyun? Power rule sirf $x^{n}$ dekhta hai. Expression ko ek clean $x^{3/2}$ mein badalna hi poori ladaai hai; phir differentiation trivial ho jaata hai.

Step 2 — $n = \frac{3}{2}$ ke saath rule apply karo. $\frac{3}{2}$ drop karo; naya exponent $\frac{3}{2} - 1 = \frac{1}{2}$ . $f^{'} (x) = \frac{3}{2} x^{1/2} = \frac{3}{2} x .$ Yeh step kyun? Yahan "differentiate karne se pehle rewrite karo" tumhe Quotient rule se bachata hai, jo yahan slower aur error-prone hoti.

Valid kahan? $x > 0$ ke liye (original mein $x$ ko non-negative input chahiye, aur division ke liye $x \neq = 0$ ).

Verify: $x = 4$ par: $f^{'} (4) = \frac{3}{2} 4 = \frac{3}{2} \cdot 2 = 3$ . Numeric slope $\frac{f ( 4.001 ) - f ( 4 )}{0.001} \approx 3.0$ . ✓

Example 6 — Cell F: ek point par slope ka sign (geometry)

Worked example

$f (x) = x^{3}$ ke liye, $x = - 2$ aur $x = 2$ par slope nikalo. Kya curve wahan upar ja rahi hai ya neeche? Forecast: $x^{3}$ ek connected curve hai. Guess karo: kya yeh kabhi downhill jaati hai?

Step 1 — differentiate karo. $f^{'} (x) = 3 x^{2}$ (Cell A: $n = 3$ ).

Step 2 — dono points mein plug in karo. $f^{'} (- 2) = 3 (- 2)^{2} = 3 \cdot 4 = 12, f^{'} (2) = 3 (2)^{2} = 12.$ Yeh step kyun? $(- 2)^{2} = + 4$ — squaring sign khatam kar deta hai. Toh slope dono points par $+ 12$ hai. Tangent lines equally steep hain aur dono uphill point karti hain.

Step 3 — geometry padho (neeche figure dekho: red tangent lines dono upar chadh rahi hain). $Dono x = \pm 2 par: slope = 12 > 0 \Rightarrow curve rise kar rahi hai.$ Yeh step kyun? Kyunki $3 x^{2} \geq 0$ sabhi $x$ ke liye, cubic kabhi downhill nahi jaati — yeh fact sign check kiye bina miss ho jaata.

Valid kahan? Har jagah — smooth polynomial.

Verify: $f^{'} (x) = 3 x^{2}$ sirf $x = 0$ par zero hai aur baki jagah positive; $\pm 2$ par slopes dono $12$ hain. ✓

Example 7 — Cell G: degenerate input (slope blow up karta hai)

Worked example

$f (x) = x = x^{1/2}$ ke liye, jab $x \to 0^{+}$ to slope ka kya hota hai? Forecast: $x = 0$ ke paas $x$ curve almost vertical lagti hai. Guess karo: kya slope finite rehti hai ya infinity tak jaati hai?

Step 1 — differentiate karo. $n = \frac{1}{2}$ , to $f^{'} (x) = \frac{1}{2} x^{- 1/2} = \frac{1}{2 x}$ (Cell D).

Step 2 — limit probe karo. Jab $x \to 0^{+}$ , $x \to 0$ , to $\frac{1}{2 x} \to + \infty$ . $lim_{x \to 0^{+}} f^{'} (x) = + \infty.$ Yeh step kyun? Formula valid rehta hai, lekin $x = 0$ par khud slope undefined hai — tangent vertical hai.

Step 3 — doosra end. Jab $x \to \infty$ , $\frac{1}{2 x} \to 0$ : curve flat ho jaati hai. Yeh step kyun? Dono limiting directions cover karna hi is cell ka point hai — $0$ ke paas steep, door flat (figure dekho).

Valid kahan? $x > 0$ ke liye. $x = 0$ par tangent vertical hai (koi finite slope nahi); $x$ $x < 0$ ke liye defined nahi hai.

Verify: $f^{'} (0.01) = \frac{1}{2 0.01} = \frac{1}{0.2} = 5$ (bada); $f^{'} (100) = \frac{1}{2 \cdot 10} = 0.05$ (chhota). ✓

Example 8 — Cell H: units ke saath real-world rate

Worked example Ek balloon ka volume

$V (r) = \frac{4}{3} π r^{3}$ (cm $^{3}$ , radius $r$ cm mein) hai. Jab $r = 5$ cm ho to volume har cm radius ke saath kitni tezi se badhta hai? Forecast: "volume kitni tezi se badhta hai per cm radius" ek slope hai: $\frac{d V}{d r}$ . Guess karo $r$ ki kaunsi power bachti hai.

Step 1 — differentiate karo, $\frac{4}{3} π$ ko constant multiplier maanke. $n = 3$ ke saath: $\frac{d}{d r} r^{3} = 3 r^{2}$ . $\frac{d V}{d r} = \frac{4}{3} π \cdot 3 r^{2} = 4 π r^{2} .$ Yeh step kyun? Front par constant bas saath aa jaata hai — power rule sirf $r^{3}$ par act karta hai. (Dhyan do $4 π r^{2}$ surface area hai — ek pyaara sanity signpost.)

Step 2 — $r = 5$ par evaluate karo. $\frac{d V}{d r}_{r = 5} = 4 π (25) = 100 π \approx 314.16 cm^{3} per cm .$ Yeh step kyun? Question ne specific radius poochhi; algebra ke baad ant mein plug in karo.

Valid kahan? $r > 0$ ke liye (physical radius positive hoti hai).

Verify: units hain $\frac{cm ^{3}}{cm} = cm^{2}$ (ek area rate — surface area se match karta hai). Numerically $\frac{V ( 5.001 ) - V ( 5 )}{0.001} \approx 314.2$ . ✓

Example 9 — Cell I: irrational exponent

Worked example

$f (x) = x^{π}$ differentiate karo (haan, exponent woh number $π \approx 3.14159$ hai). Forecast: kya "drop and lower" tab bhi kaam karta hai jab exponent fraction bhi nahi hai?

Step 1 — check karo ki rule abhi bhi apply hoti hai. $π$ ek real number hai, aur power rule har real exponent ke liye prove kiya gaya tha — $\frac{22}{7}$ jaise fraction isko maante hain, aur $π$ aise fractions ka limit hai, isliye shortcut bina kisi badlaav ke carry over karta hai. Yeh step kyun? Hum $(x + h)^{π}$ ko nudge se expand nahi kar sakte (non-whole power ke liye koi finite expansion exist nahi karta), isliye hum is baat par rely karte hain ki rule sabhi real $n$ tak extend hota hai — bas $π$ ko "just a number" ki tarah treat karo.

Step 2 — $π$ ko front par drop karo, exponent ek se kam karo. $f^{'} (x) = π x^{π - 1} .$ Yeh step kyun? Hamesha ki tarah same mechanical move; bas oddity yeh hai ki naya exponent $π - 1 \approx 2.14159$ khud irrational hai — koi baat nahi.

Valid kahan? $x > 0$ ke liye ( $x^{π}$ jaise general real power sirf positive bases ke liye defined hoti hai).

Verify: $x = 1$ par exact answer $f^{'} (1) = π \cdot 1^{π - 1} = π \approx 3.1416$ hai. Ise Limit definition of the derivative se confirm karne ke liye, tiny nudge use karo taaki secant slope (do nearby points ke beech rise over run) tangent slope ke paas aaye: $\frac{f ( 1 + h ) - f ( 1 )}{h}$ jab $h = 0.001$ to $\frac{1.00 1 ^{π} - 1 ^{π}}{0.001} = \frac{1.003146 - 1}{0.001} \approx 3.146$ milta hai. Yeh step kyun? Jab $h \to 0$ to yeh secant slope true derivative $π \approx 3.1416$ ki taraf converge karta hai, aur $3.146$ already kaafi close hai (bada $h$ overshoot karta kyunki curve bend karti hai). ✓

Example 10 — Cell J: exam twist (power rule NAHI hai)

Worked example

$f (x) = 2^{x}$ differentiate karo. (Dhyan se — kya yeh power rule problem hai?) Forecast: $x \cdot 2^{x - 1}$ likhne ka mann karta hai. Ruko: variable kaun hai — base ya exponent?

Step 1 — structure diagnose karo. $x^{n}$ mein base variable hai, exponent fixed hai. $2^{x}$ mein base fixed hai ( $2$ ) aur exponent variable hai. Yeh dono ulti cheezein hain. Yeh step kyun? Power rule variable base, constant exponent ke liye prove kiya gaya tha. Yeh yahan bilkul apply nahi hoti — ise use karna ek category error hai.

Step 2 — iske bajaay exponential rule use karo (dekho Derivative of exponential functions): $\frac{d}{d x} a^{x} = a^{x} ln a \Rightarrow \frac{d}{d x} 2^{x} = 2^{x} ln 2.$ Yeh step kyun? Yeh sahi tool hai; $ln 2 \approx 0.693$ base $2$ ka "growth constant" hai.

Valid kahan? Har jagah — $2^{x}$ sabhi real $x$ ke liye defined aur smooth hai.

Verify: $x = 3$ par: sahi answer $2^{3} ln 2 = 8 \cdot 0.693 \approx 5.545$ ; galat power-rule guess $x 2^{x - 1} = 3 \cdot 4 = 12$ — alag hai, confirm karta hai ki dono same rule nahi hain. ✓

Recall Self-test (pehle predict karo, phir reveal karo)

$\frac{1}{3 x}$ ko scratch se differentiate karo, aur batao yeh kahan valid hai. ::: $x^{- 1/3}$ ke roop mein rewrite karo (Cell E). Phir $n = - \frac{1}{3}$ : $f^{'} = - \frac{1}{3} x^{- 1/3 - 1} = - \frac{1}{3} x^{- 4/3} = \frac{- 1}{3 3 x ^{4}}$ . $x \neq = 0$ ke liye valid hai ( $x = 0$ par undefined). ✓

Active recall

\frac{d}{d x} x^{7}

kya hai?

7 x^{6}

\frac{d}{d x} x^{0}

aur

\frac{d}{d x} 5

kya hain?

Dono

0

hain (flat lines ka slope zero hota hai).

\frac{d}{d x} x^{- 5}

kya hai, aur yeh kahan valid hai?

- 5 x^{- 6} = - 5/ x^{6}

;

x \neq = 0

ke liye valid.

\frac{d}{d x} x^{2/3}

kya hai, aur

x = 0

par kya hota hai?

\frac{2}{3} x^{- 1/3}

;

x = 0

par cusp hai, isliye wahan koi slope exist nahi karta.

Cusp kya hota hai?

Curve par ek sharp point jahan tangent direction flip karta hai, isliye wahan koi single slope exist nahi karta.

\frac{x ^{2}}{x}

differentiate karne se pehle kya karna chahiye?

Ek power ke roop mein rewrite karo:

x^{3/2}

x^{3}

ka slope kabhi negative kyun nahi hota?

f^{'} = 3 x^{2} \geq 0

sabhi

x

ke liye (squares non-negative hote hain).

\frac{d}{d x} x

ka kya hota hai jab

x \to 0^{+}

Yeh

+ \infty

tak jaata hai (vertical tangent;

x = 0

par undefined).

\frac{d}{d x} x^{π}

kya hai?

π x^{π - 1}

(rule har real exponent ke liye kaam karta hai;

x > 0

ke liye valid).

V = \frac{4}{3} π r^{3}

ke liye

\frac{d V}{d r}

kya hai?

4 π r^{2}

(surface area).

Kya

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

sahi hai?

Nahi —

2^{x}

exponential hai; answer hai

2^{x} ln 2

Connections

Power rule — proof for integer, rational exponents — woh parent note jise yeh page drill karata hai.
Limit definition of the derivative — nudge $h$ ka matlab; har numeric slope-check isko echo karta hai.
Binomial theorem — integer cases ke peeche (Ex 1, 6).
Implicit differentiation — isliye shortcut rational exponents tak extend hota hai (Ex 4).
Chain rule — tab chahiye jab functions power ke andar nest ho jaayein.
Quotient rule — woh slower alternative jise hum Ex 5 mein avoid kiya.
Derivative of exponential functions — Ex 10 ke liye sahi tool.