Derivatives of sin x, cos x — proofs from first principles: Visual walkthrough

4.1.17 · D2 · Maths › Calculus I — Limits & Derivatives › Derivatives of sin x, cos x — proofs from first principles

Step 1 — "Derivative" kya pooch raha hai (ek slope jise hum shrink karte hain)

KYA. Sine curve par ek point chuno. Thoda sa daayein, horizontal gap $h$ ke faasle par, ek doosra point banao. Unhe ek seedhi line se jodo. Us line ki ek steepness hoti hai — kitna climb kiya ( $Δ y$ ) divided by kitna run kiya ( $h$ ).

KYUN. Slope hi ek honest tarika hai yeh kehne ka ki "yahan yeh curve kitni tezi se badh rahi hai?" Ek akele point ka khud ka koi slope nahi hota, isliye hum $h$ distance par ek paas ke dost ki madad lete hain, unke beech ka average climb measure karte hain, phir dost ko andar slide karte hain ( $h \to 0$ ) jab tak dono points lagbhag chhoo na len. Slope jis value par settle karta hai wahi derivative hai.

PICTURE. Neecha amber secant line kam aur kam tilt hoti hai jaise daaya point baayein khisalta hai; woh cyan tangent line ki taraf ja rahi hai.

Yahan $f (x) = sin x$ hai, isliye rise hai $sin (x + h) - sin x$ . Neecche har step us rise ko tame karne ke baare mein hai.

Step 2 — Pehle ek small-angle fact kyun chahiye

KYA. $sin (x + h)$ ko chhune se pehle, hum poore proof ke engine ko isolate karte hain: jab $h$ bahut chhota ho toh $\frac{sin h}{h}$ ka kya hota hai?

KYUN. Aage dekhein: expand karne ke baad answer mein $\frac{sin h}{h}$ aayega. Agar woh number pata nahi, toh proof ruk jaata hai. Toh hum ise abhi geometrically settle karte hain, ek unit circle (radius $1$ ka circle) use karke.

PICTURE. Unit circle par, ek chhota angle $h$ sweep karo (radians mein measure kiya — arc length khud). Teen regions ek doosre ke andar nest karti hain: ek patla triangle, pie-slice (sector), aur ek bada triangle.

Inner region sector ke andar hai jo outer triangle ke andar hai — picture hi inequality hai.

Step 3 — Ratio ko 1 par squeeze karo

KYA. Poori inequality ko positive quantity $\frac{1}{2} sin h$ se divide karo, phir flip karo, aur $h \to 0$ jaane do.

KYUN. Hum chahte hain ki $\frac{sin h}{h}$ do aise cheezon ke beech trap ho jo dono ek hi limit ki taraf ja rahe hon. Exactly yahi Squeeze Theorem ko chahiye: agar koi quantity do aise cheezon ke beech pin ho jo dono $1$ ki taraf ja rahe hain, toh woh bhi $1$ ki taraf jaayega.

PICTURE. Jaise $h$ shrink hota hai, dono seedhe sides aur curved arc alag nahi kiye ja sakte — ratio $\frac{sin h}{h}$ ko $1$ par squeeze kiya ja raha hai.

Trapped ratio ko flip karne se show ka star milta hai: $h \to 0 lim \frac{sin h}{h} = 1$ Yeh $h < 0$ ke liye bhi hold karta hai, kyunki $\frac{sin h}{h}$ unchanged rehta hai jab $h \to - h$ (ek even ratio).

Step 4 — Cosine companion limit

KYA. Doosra number establish karo jis proof ko chahiye: $lim_{h \to 0} \frac{cos h - 1}{h}$ .

KYUN. $sin (x + h)$ expand karne ke baad ek $cos h - 1$ piece aayega. Hum uski limit guess nahi kar sakte — yeh $\frac{0}{0}$ shape hai. Hum ise known $\frac{s i n h}{h}$ mein convert karte hain conjugate trick use karke (upar aur neecha $cos h + 1$ se multiply karo).

PICTURE. Geometrically $1 - cos h$ circle point ka right edge se chhota horizontal sag hai — yeh $h$ se tezi se shrink hota hai, isliye ratio zero ho jaata hai.

$h \to 0 lim \frac{cos h - 1}{h} = 0$

Step 5 — Addition formula se rise expand karo

KYA. Ab actual derivative par attack karo. $sin (x + h)$ ko angle-addition formula use karke replace karo.

KYUN. Rise $sin (x + h) - sin x$ mein $x$ (fixed) aur $h$ (vanishing) ek hi sine ke andar mix hain. Hum unhe alag karna chahte hain taaki $h$ -parts Steps 3–4 ki limits se mil sakein. Addition formula woh tool hai jo angle-sum ko aise pieces mein split karta hai jahaan $x$ aur $h$ alag khade hon.

PICTURE. Socho ki ek unit vector ko angle $x$ se $h$ extra rotate karo; iski nayi height $sin (x + h)$ purani height aur width se $cos h, sin h$ se scale karke banti hai.

Step 6 — Dono known limits mein group karo

KYA. $h$ se divide karo, phir $sin x$ terms ko saath mein gather karo.

KYUN. Hum chahte hain ki expression $sin x \cdot (Step 4 limit) + cos x \cdot (Step 3 limit)$ jaisi dikhe. Grouping pure bookkeeping hai taaki pehchaane jaane waale pieces surface par aayen.

PICTURE. Do labelled "slots" — ek $0$ value ka wait kar raha hai, ek $1$ value ka.

Step 7 — Limit lo aur answer padho

KYA. $h \to 0$ jaane do. Constants $sin x, cos x$ limit ke bahar rehte hain; dono slots $0$ aur $1$ se bhar jaate hain.

KYUN. Sum ki limit, limits ke sum ke barabar hoti hai (Limits — definition and properties se), aur constant factors seedhe pass through ho jaate hain. Toh har slot bas apni known value ban jaata hai.

PICTURE. Dono slots $0$ aur $1$ par snap karte hain; sirf $cos x$ term bachta hai.

Cosine ka mirror proof bilkul waisa hi hai $cos (x + h) = cos x cos h - sin x sin h$ ke saath, jisse milta hai $\frac{d}{d x} cos x = cos x \cdot 0 - sin x \cdot 1 = - sin x$ .

Step 8 — Degenerate & edge cases (inhe kabhi skip mat karo)

KYA. Woh corners check karo jo pictures ne quietly assume kar liye the.

KYUN. Proof tab tak complete nahi hota jab tak har input cover na ho.

PICTURE. Jahaan sine zero cross karta hai upar jaate hue, uska slope maximum hota hai ( $cos = + 1$ ); peak par, slope zero hota hai.

Charon quadrants mein har $x$ ke liye ek slope milta hai jo cosine wave se match karta hai, aur $h$ ke dono signs agree karte hain — result watertight hai.

Ek-picture summary

Neecche: sine curve (cyan) apne slope-arrows ke saath, aur seedha neecche cosine curve (amber) jo un slopes ko trace karti hai. Proof yeh statement hai ki yeh dono frames ek doosre se locked hain.

Recall Feynman retelling — ek dost ko batao

Hum sine wave ka slope jaanna chahte the. Slope ka matlab hai "rise over run", toh maine ek doosra dot thoda gap $h$ daayein rakha aur climb measure ki, phir use andar slide kiya. Climb nikalne ke liye mujhe pehle ek magic fact chahiye tha: chhote angle ke liye, angle ka sine basically angle hi hota hai — maine ise ek circle par ek pie-slice ko do triangles ke beech squeeze karke prove kiya, jisne $\frac{s i n h}{h} \to 1$ force kiya. Ek cousin trick (conjugate se multiply karo) ne dikhaya ki bacha hua cosine piece $\frac{c o s h - 1}{h}$ bas $0$ ho jaata hai. Phir maine $sin (x + h)$ ko addition formula se uske $x$ -part aur $h$ -part mein tod diya, terms ko do slots mein sort kiya, $0$ aur $1$ daala — aur $cos x$ nikal aaya. Pictures se sanity check: jahaan sine zero cross karta hai woh sabse tezi se climb karta hai (cosine $= 1$ ), peak par flat hai (cosine $= 0$ ). Slope-wave hi cosine wave hai, ek quarter turn shift ki hui. Cosine bhi usi tarah kaam karta hai lekin minus sign le leta hai, kyunki $x = 0$ par cosine apne peak par hai aur girne wala hai.

Connections

Derivatives of sin x, cos x — Proofs from First Principles — parent proof jise yeh page illustrate karta hai
Squeeze Theorem — Step 3 ka engine
Trigonometric addition formulas — Step 5 ka splitter
Limits — definition and properties — Step 7 mein sum/constant rules
Radian measure — kyun sector area $\frac{1}{2} h$ hai
Chain Rule — ise $sin (g (x))$ tak extend karta hai
Derivatives of tan, sec, csc, cot — in dono se bane hain
Taylor series of sin and cos — wahi fact ek alag angle se