Power rule — proof for integer, rational exponents

WHAT we are proving

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

HOW — derive from scratch: $f'(x)=\lim_{h\to0}\frac{(x+h)^n - x^n}{h}.$ Expand the numerator: $(x+h)^n - x^n = \cancel{x^n} + nx^{n-1}h + \binom{n}{2}x^{n-2}h^2+\cdots+h^n - \cancel{x^n}.$ Divide by $h$ : $\frac{(x+h)^n-x^n}{h}= nx^{n-1} + \underbrace{\binom{n}{2}x^{n-2}h+\cdots+h^{n-1}}_{\text{every term has } h}.$ Take $h\to0$ : each term in the brace contains $h$ , so it dies. $\boxed{f'(x)=nx^{n-1}.}$

Stage 2 — Exponent $n = 0$

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

HOW — from first principles: $f(x)=x^{-m}=\frac1{x^m},\qquad f'(x)=\lim_{h\to0}\frac1h\left(\frac{1}{(x+h)^m}-\frac{1}{x^m}\right).$ Common denominator: $=\lim_{h\to0}\frac{x^m-(x+h)^m}{h\,(x+h)^m x^m}.$ Why this step? Combining fractions exposes $x^m-(x+h)^m$ , which we already know how to handle. Use Stage 1: $\dfrac{(x+h)^m-x^m}{h}\to mx^{m-1}$ , so $\dfrac{x^m-(x+h)^m}{h}\to -mx^{m-1}$ . The denominator factor $(x+h)^m x^m \to x^{2m}$ . Hence $f'(x)=\frac{-mx^{m-1}}{x^{2m}}=-m\,x^{m-1-2m}=-m\,x^{-m-1}.$ With $n=-m$ : $\;f'(x)=n\,x^{n-1}.$ ✓

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

HOW — derive: Let $y = x^{p/q}\;\Rightarrow\; y^q = x^p$ (both integer exponents). Differentiate both sides w.r.t. $x$ (use chain rule on the left — itself a consequence of Stage 1): $q\,y^{q-1}\frac{dy}{dx} = p\,x^{p-1}.$ Why this step? Left side: $\frac{d}{dx}y^q = qy^{q-1}\frac{dy}{dx}$ . Solve: $\frac{dy}{dx}=\frac{p}{q}\cdot\frac{x^{p-1}}{y^{q-1}}.$ Substitute $y=x^{p/q}$ , so $y^{q-1}=x^{p(q-1)/q}=x^{p - p/q}$ : $\frac{dy}{dx}=\frac{p}{q}\,x^{p-1-(p-p/q)}=\frac{p}{q}\,x^{p/q-1}.$ With $n=p/q$ : $\;\boxed{\dfrac{dy}{dx}=n\,x^{n-1}.}$ ✓

Figure — Power rule — proof for integer, rational exponents

Common mistake Steel-manning the classic traps

Trap 1: "The binomial proof works for all $n$ ." Why it feels right: the algebra "looks general." The flaw: $(x+h)^n$ only has a finite expansion when $n$ is a non-negative integer. For $n=1/2$ or $n=-3$ the expansion is an infinite series — you can't divide-and-cancel cleanly. Fix: use the fraction trick (negatives) or implicit differentiation (rationals).

Trap 2: Treating $x^n$ like $n^x$ . $\frac{d}{dx}2^x \ne x\,2^{x-1}$ ! The power rule needs a variable base, constant exponent. Fix: exponentials use $\frac{d}{dx}a^x=a^x\ln a$ .

Trap 3: Forgetting the chain rule in Stage 4. Writing $\frac{d}{dx}y^q=qy^{q-1}$ (no $\frac{dy}{dx}$ ). Why it feels right: it "looks like" the power rule. Fix: $y$ is a function of $x$ , so the chain rule attaches $\frac{dy}{dx}$ .

Active recall

Statement of the power rule?

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

Which theorem powers the positive-integer proof?

The binomial theorem expanding

(x+h)^n

After expanding

(x+h)^n - x^n

and dividing by

h

, which single term survives as

h\to0

The

nx^{n-1}

term (all others still contain a factor of

h

Why can't the binomial proof handle

n=1/2

n=-3

Those exponents give an infinite series, not a finite cancellable expansion.

Trick to prove the rule for negative integers?

Write

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

, take a common denominator, reuse the positive-integer result.

Trick to prove the rule for rational

p/q

Set

y=x^{p/q}

, raise to power

q

to get

y^q=x^p

, then implicitly differentiate.

Differentiating

y^q

w.r.t.

x

gives?

q\,y^{q-1}\,\frac{dy}{dx}

(chain rule).

\frac{d}{dx}\sqrt{x}

\frac{1}{2\sqrt{x}}

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

No — that's the wrong rule;

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

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

-2x^{-3}

Recall Feynman: explain to a 12-year-old

Imagine a tower made of blocks where each level is $x$ times wider, $n$ levels high — that's $x^n$ . If you make $x$ a tiny bit bigger, how fast does the tower grow? The answer is: you get $n$ copies of one-level-shorter towers ( $x^{n-1}$ ) added on, because each of the $n$ levels "feels" the stretch. That's why you slide the $n$ to the front and drop the height by one. For roots (like $\sqrt x$ ) we play a flip-game: square it to undo the root, use the easy rule, then flip back.

Connections

Limit definition of the derivative — the engine for Stages 1 & 3.
Binomial theorem — supplies the finite expansion.
Chain rule — needed in the rational-exponent proof.
Implicit differentiation — the method behind Stage 4.
Derivative of exponential functions — contrast: $a^x$ is NOT power rule.
Quotient rule — alternative path for negative exponents.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Power rule ka matlab simple hai: $\frac{d}{dx}x^n = n x^{n-1}$ — exponent ko neeche le aao (multiply karo) aur exponent mein se 1 ghata do. Lekin ratta maarne ki zaroorat nahi, kyunki ise hum derivative ki definition se derive kar sakte hain. Positive integer ke liye hum $(x+h)^n$ ko binomial theorem se kholte hain. Jab $x^n$ cancel ho jaata hai aur $h$ se divide karte hain, sirf ek hi term bina $h$ ke bachta hai — wahi $nx^{n-1}$ hai. Baaki sab terms mein $h$ hota hai, jo $h\to0$ pe gayab ho jaata hai.

Negative integer (jaise $x^{-2}$ ) ke liye binomial seedha kaam nahi karta, isliye usse $1/x^m$ likh kar common denominator lete hain, aur positive integer wala result reuse karte hain. Rational exponent (jaise $\sqrt{x}=x^{1/2}$ ) ke liye ek mast trick hai: $y=x^{p/q}$ likho, dono taraf $q$ power lagao to fraction khatam — $y^q=x^p$ . Ab dono integer powers hain, to implicit differentiation (chain rule ke saath) lagao aur answer nikal aata hai $\frac{p}{q}x^{p/q-1}$ .

Sabse common galti: socho ki binomial proof har $n$ ke liye chalta hai — nahi! Sirf non-negative integer ke liye finite expansion milta hai. Doosri galti: $2^x$ ko power rule se differentiate karna — galat, kyunki yahan base constant aur exponent variable hai (uska rule $2^x\ln 2$ hai). Yeh rule itna important hai kyunki har dusra derivative isi pe khada hai — ek baar samajh lo, zindagi bhar kaam aayega.

Power rule — proof for integer, rational exponents

WHAT we are proving

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

Stage 2 — Exponent $n = 0$

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

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

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus I — Limits & Derivatives

Connections