Visual walkthrough — Power rule — proof for integer, rational exponents

4.1.12 · D2 · Maths › Calculus I — Limits & Derivatives › Power rule — proof for integer, rational exponents

Step 0 — Derivative ka matlab ek picture mein kya hai?

PICTURE: Curve par do points baithe hain. Unke beech ka horizontal gap $h$ hai. Vertical gap $f(x+h)-f(x)$ hai. Unhe jodhne wali seedhi line (the secant) ka slope hai $\underset{\text{rise}÷\text{run}}{\underbrace{\frac{f(x+h)-f(x)}{h}}}.$ Jaise hum doosre point ko pehle ki taraf slide karte hain ($h\to 0$), secant tip hokar tangent ban jaati hai — woh line jo curve ko bas thoda sa chhoo kar nikalti hai.

Har step neeche is ek machine mein ek specific $f$ daalta hai. Dekho Limit definition of the derivative.

Step 1 — $x^2$ ke liye area picture (kyun "2 neeche aata hai")

Square kyun? Kyunki "$x$ squared" ek square hai, aur square ko grow karna kuch aisa hai jo hum literally draw kar sakte hain. Jab side $x$ se $x+h$ tak badhti hai, nayi area $(x+h{)}^2$ hoti hai. Extra area ek patla L-shaped border hai.

PICTURE: Border teen pieces mein banta hai:

• ek strip upar ki taraf: width $x$, height $h$ → area $xh$,
• ek strip side ki taraf: width $h$, height $x$ → area $xh$,
• ek chhota sa corner square: $h\times h$ → area $h^2$.

$\underset{\text{extra area}}{\underbrace{(x+h{)}^2-x^2}}=\underset{\text{do lambi strips}}{\underbrace{xh+xh}}+\underset{\text{chhota corner}}{\underbrace{h^2}}=2xh+h^2.$

Run $h$ se divide karo: $\frac{2xh+h^2}{h}=2x+h.$ Ab $h\to 0$ shrink karo: tiny corner term $h$ gayab ho jaata hai, aur bacha $f^{\prime }(x)=2x$. "$2$" do lambi strips se aaya — do isliye hain kyunki square ki do dimensions hoti hain. Yeh power rule $n{x}^{n-1}$ hai $n=2$ ke saath: dekha, memorize nahi kiya.

Step 2 — General shape $(x+h{)}^n$ (kyun binomial theorem chahiye)

Binomial theorem kyun? Yeh exactly woh tool hai jo $(x+h{)}^n$ ko terms ki ek sorted list mein expand karta hai, kitne factors of $h$ hain uske hisaab se organize kiya hua. Yahi sorting poora point hai: hum woh pieces chahte hain jisme exactly ek $h$ ho, kyunki yahi pieces $h$ se divide karne ke baad aur phir $h\to 0$ par survive karti hain.

PICTURE: $n=3$ ke liye (side $x$ ka ek cube), jab side $h$ se badhti hai toh extra volume hai 3 flat faces (har ek $x^2\cdot h$ → total $3x^{2}h$), plus 3 patli edges ($x{h}^2$), plus 1 corner ($h^3$). Teen faces accent hain — yeh $n{x}^{n-1}h$ term hai $n=3$ ke saath.

Step 3 — Machine mein daalo (positive integer $n$)

Kyun collapse hota hai: front ka $x^n$, $-f(x)$ ke $-x^n$ ko cancel kar deta hai; $h$ se divide karne ke baad sirf ek term hai jisme koi leftover $h$ nahi hai.

$\frac{(x+h{)}^n-x^n}{h}=\frac{\cancel{x^n}+n{x}^{n-1}h+\left(\genfrac{}{}{0ex}{}{n}{2}\right)x^{n-2}{h}^2+\cdots -\cancel{x^n}}{h}=n{x}^{n-1}+\underset{\text{har term mein abhi bhi ek }h\text{ hai}}{\underbrace{\left(\genfrac{}{}{0ex}{}{n}{2}\right)x^{n-2}h+\cdots }}.$

PICTURE: $h$ se divide karne ke baad terms ke liye ek number line. Akela term $n{x}^{n-1}$ bina kisi $h$ ke akela baitha hai (accent). Baaki sab terms mein kam se kam ek $h$ hai aur jaise arrow $h\to 0$ karta hai yeh $0$ ki taraf slide ho jaate hain.

$\boxed{f^{\prime }(x)=n{x}^{n-1}}(n=1,2,3,\dots )$

Step 4 — Flat edge case $n=0$

PICTURE: $y=1$ ka graph: ek level line. $x$ ko $h$ se nudge karo; height nahi badlti, isliye rise $=0$, isliye slope $=0$. Formula bhi agree karta hai: $0\cdot x^{-1}=0$. ✓

Yeh koi special exception nahi hai — yahi rule hai jo keh raha hai "grow karne ke liye $0$ faces hain."

Step 5 — Negative exponent $n=-m$ (fraction picture)

Ek nayi move kyun? Binomial theorem sirf positive whole powers expand karta hai. Isliye hum $x^{-m}$ expand nahi kar sakte. Iske bajaye hum ise fraction ke roop mein likhte hain aur common denominator par combine karte hain, jo positive-power difference $x^m-(x+h{)}^m$ ko phir se expose karta hai jise hum already handle karna jaante hain.

$\frac{1}{h}\left(\frac{1}{(x+h{)}^m}-\frac{1}{x^m}\right)=\frac{\stackrel{\text{Step 3 yeh handle karta hai}}{\overbrace{x^m-(x+h{)}^m}}}{h\underset{\to x^{2m}}{\underbrace{(x+h{)}^m{x}^m}}}.$

• top: Step 3 jaisi hi difference, lekin sign flip hai, isliye yeh $-m{x}^{m-1}$ ki taraf jaata hai.
• bottom: $h\to 0$ par, $(x+h{)}^m{x}^m\to x^{2m}$.

$f^{\prime }(x)=\frac{-m{x}^{m-1}}{x^{2m}}=-m{x}^{-m-1}=n{x}^{n-1}(n=-m).$

PICTURE: $y=1/x$ ka curve ($m=1$) jisme ek point par tangent line drawn hai — uska slope ek downward (negative) ramp hai, jo $-1\cdot x^{-2}<0$ se match karta hai.

Step 6 — Rational exponent $n=p/q$ (flip-game)

Implicit differentiation kyun? Agar hum height ko $y=x^{p/q}$ naam dein aur dono sides ko power $q$ tak raise karein, fraction khatam ho jaata hai: $y^q=x^p$. Ab dono sides integer powers hain — woh territory jo hum Steps 3–5 mein already conquer kar chuke hain. Hum dono sides differentiate karte hain, left side par Chain rule use karte hue kyunki $y$ khud $x$ ka function hai. Dekho Implicit differentiation.

$\frac{d}{dx}{y}^q=\frac{d}{dx}{x}^p⟹\underset{\text{chain rule: }y\text{ depends on }x}{\underbrace{q{y}^{q-1}\frac{dy}{dx}}}=\underset{\text{plain power rule}}{\underbrace{p{x}^{p-1}}}.$

Slope $\frac{dy}{dx}$ ke liye solve karo, phir $y=x^{p/q}$ substitute karo (toh $y^{q-1}=x^{p-p/q}$):

$\frac{dy}{dx}=\frac{p}{q}\cdot \frac{x^{p-1}}{x^{p-p/q}}=\frac{p}{q}{x}^{p/q-1}=n{x}^{n-1}.$

PICTURE: $y=\sqrt{x}$ ka graph. $0$ ke paas steeply rise karta hai aur $x$ badhne par flatten ho jaata hai — uski tangent slopes $\frac{1}{2\sqrt{x}}$ se match karti hain, jo $0$ ke paas huge hai aur door chhoti.

Ek picture mein summary

Har stage usi limit machine mein jaata hai; sirf ratio simplify karne ki algebra badlti hai, aur har baad wala stage pichle wale ko reuse karta hai.

Active recall