Derivatives of all six trig functions: Visual walkthrough

4.1.18 · D2 · Maths › Calculus I — Limits & Derivatives › Derivatives of all six trig functions

Hum ek central result bilkul zero se hasil karenge: $\frac{d}{d x} sin x = cos x .$ Neeche har symbol ko use karne se pehle ek saral-shabdon mein meaning aur ek picture di jaayegi.

Step 1 — " $sin x$ " asal mein kya hai: ek ghoomte wheel par ek height

KYA. Radius $1$ ka ek circle banao (ek unit circle — "unit" ka matlab sirf itna hai ki radius ek ke barabar hai). Uske edge par ek point $P$ rakho. $x$ woh angle hai jo sabse daaye wale point se counter-clockwise maapa jaata hai. Do words:

$cos x$ = centre se $P$ ki horizontal doori.
$sin x$ = centre ke upar $P$ ki vertical height.

KYUN. Hum kisi cheez ko differentiate nahi kar sakte jo hum picture nahi kar sakte. $sin x$ ko "ek ghoomte point ki height" ke roop mein fix karna ek abstract symbol ko ek concrete arrow mein badal deta hai jise hum move hote dekh sakte hain.

KYUN radians, degrees nahi. Hum angle $x$ ko arc length se maapte hain — yani point rim par kitna actually chala hai. Chunki radius $1$ hai, ek quarter turn circumference ka quarter hai $= \frac{π}{2}$ . Angle ko ek length ke roop mein maapna exactly wohi hai jo neeche ki pictures ko line up karta hai; degrees secretly sab kuch rescale kar dete (parent note ki "silent killer" mistake dekho).

PICTURE.

Figure — Derivatives of all six trig functions

Step 2 — " $sin$ ka derivative" poochh kya raha hai

KYA. Derivative ek rate hai: agar angle $x$ thodi si matra se badhe, toh height $sin x$ kitni tezi se badlegi? Hum yeh sawaal kuch yun likhte hain: $\frac{d}{d x} sin x = lim_{h \to 0} \frac{s i n ( x + h ) - s i n x}{h} .$

KYUN yahi exact expression. Yeh $sin$ par lagaya gaya Limit definition of the derivative hai. Har symbol ko wahan padhte hain jahan woh baitha hai:

$h$ — angle mein ek tiny nudge jo hum add karte hain. Socho yeh $0$ ki taraf shrink ho raha hai.
$sin (x + h)$ — nudge karne ke baad nayi height.
$sin (x + h) - sin x$ — height mein badlaav (rise).
$h$ se divide karna — angle-nudge ki per unit height ka badlaav (rise over run, ek slope).
$lim_{h \to 0}$ — yeh sawaal ki "woh slope kis par settle hota hai jab nudge gayab ho jaata hai?"

PICTURE. Sine wave par, yeh $x$ aur $x + h$ ke beech tiny chord ki slope hai.

Recall Sirf ek fraction kyun nahi, limit kyun?

Kisi chote $h$ ke liye rise-over-run kyun calculate nahi karte? ::: Ek fixed $h$ ek interval par ek average slope deta hai; instantaneous slope sirf tab milti hai jab $h \to 0$ ki limit lete hain. Wohi limit derivative ka poora essence hai.

Step 3 — Wheel mein zoom karo: tiny nudge ek tiny arc hai

KYA. Circle par wapas jaao. Angle ko $x$ se $x + h$ tak nudge karna point $P$ ko rim par thoda sa ek naye point $Q$ tak move karta hai. Kyunki radius $1$ hai, $P$ se $Q$ tak woh arc ki length exactly $h$ hai.

KYUN. Hum $sin (x + h) - sin x$ samajhna chahte hain — height mein badlaav. Woh badlaav $P$ se $Q$ tak chote step ka vertical part hai. Toh us step ko bahut gehraai se dekho.

PICTURE. Zoom karne par, arc $P Q$ lagbhag length $h$ ka ek straight segment hai, aur yeh radius $O P$ ke tangent (perpendicular) ko point karta hai.

Step 4 — Chota right triangle: step ka rise

KYA. $P$ se $Q$ tak tiny step (length $h$ ) ek chote right triangle ka hypotenuse hai. Uska vertical leg height mein badlaav hai, $Δ y = sin (x + h) - sin x$ . Uska horizontal leg width mein badlaav hai, $Δ x = cos (x + h) - cos x$ (jo negative hota hai jab hum upar-baayen move karte hain).

KYUN. Humein $\frac{Δ y}{h}$ chahiye. Agar hum triangle se vertical leg padh sakein, toh kaam khatam. Ab punch line yeh hai: kyunki step radius $O P$ ke perpendicular hai, chota triangle radius aur axes se bane bade triangle ke similar (same shape, same angles) hai.

Bada radius horizontal ke saath angle $x$ banata hai. Step, radius se $9 0^{\circ}$ ghuma hua hai, isliye step ke chote triangle ka angle $x$ , vertical ke against maapa jaata hai. Result: $\frac{vertical leg}{hypotenuse} = cos x, isliye Δ y \approx h cos x .$

Term by term padhte hain:

$Δ y$ — height kitni badhi.
$h$ — poore tiny step ki length.
$cos x$ — us step ka woh fraction jo upar gaya, kyunki step vertical se angle $x$ par tilted hai.

PICTURE. Chota triangle, jisme uska vertical leg $h cos x$ label kiya gaya hai.

Step 5 — Wohi picture, algebra se (yeh exact kyun hai)

KYA. " $\approx$ " wala word tumhe nervous karna chahiye. Aao exact Angle addition formulas se confirm karte hain: $sin (x + h) = sin x cos h + cos x sin h .$

KYUN. Algebra "approximately" ko ek exact limit mein badalta hai, aur dikhata hai triangle ke kaunse pieces $cos x$ ban jaate hain. Substitute karo aur simplify karo:

= \sin x\cdot\underbrace{\frac{\cos h-1}{h}}_{\to\, 0} \;+\; \cos x\cdot\underbrace{\frac{\sin h}{h}}_{\to\, 1}.$$ Term by term padhte hain: - $\dfrac{\cos h - 1}{h}$ — step ka **horizontal** part kaise scale hota hai; yeh vanish ho jaata hai kyunki sideways motion second-order tiny hoti hai. - $\dfrac{\sin h}{h}$ — **vertical** part kaise scale hota hai; yeh $1$ ki taraf jaata hai, isliye $\cos x$ untouched bach jaata hai. **PICTURE.** Yeh exactly woh do foundation limits hain — geometric squeeze jo $\frac{\sin h}{h}\to 1$ deta hai, aur conjugate trick jo $\frac{\cos h-1}{h}\to 0$ deta hai. ![[deepdives/dd-maths-4.1.18-d2-s05.png]] Limit lete hain: $$\frac{d}{dx}\sin x = \sin x\cdot 0 + \cos x\cdot 1 = \boxed{\cos x}.$$ > [!intuition] Ek limit $0$ kyun hai aur doosri $1$ kyun > Ek shrinking arc ke bottom ke paas, tum almost **seedha upar** move karte ho — poora vertical progress ($\frac{\sin h}{h}\to 1$) aur essentially **zero sideways** progress ($\frac{\cos h-1}{h}\to 0$). Triangle picture aur algebra ek hi kahani ko *do baar* sunate hain. --- ## Step 6 — Har quadrant: kya $\cos x$ sach mein slope predict karta hai? **KYA.** Humne claim kiya tha ki sine wave ki slope *$\cos$ ki value hai*. Isko sabhi chaar quadrants mein check karte hain — koi scenario nahi chhoota. **KYUN.** Ek formula jo sirf ek point par test kiya gaya woh guess hai. $\cos x$ ka sign yeh match karna chahiye ki wave har jagah rise ho rahi hai ya fall. | angle range | $\cos x$ ka sign | sine wave hai... | match karta hai? | |---|---|---|---| | $0$ se $\tfrac{\pi}{2}$ | $+$ | rising | ✅ | | $\tfrac{\pi}{2}$ se $\pi$ | $-$ | falling | ✅ | | $\pi$ se $\tfrac{3\pi}{2}$ | $-$ | falling | ✅ | | $\tfrac{3\pi}{2}$ se $2\pi$ | $+$ | rising | ✅ | **Degenerate points.** Wave ke bilkul top par, $x=\tfrac{\pi}{2}$: point $P$ purely **sideways** move kar raha hai, isliye height change nahi ho rahi — slope $=0$, aur sach mein $\cos\tfrac{\pi}{2}=0$. $x=0$ par point purely **upar** move karta hai — steepest rise, slope $=1=\cos 0$. Dono extreme cases fit hote hain. **PICTURE.** ![[deepdives/dd-maths-4.1.18-d2-s06.png]] > [!example] $x=\tfrac{\pi}{3}$ par Numeric spot-check > $x=\tfrac{\pi}{3}$ par $\sin$ ki slope $\cos\tfrac{\pi}{3}=\tfrac12$ honi chahiye. $h=0.001$ se nudge karo: > $$\frac{\sin(\tfrac{\pi}{3}+0.001)-\sin\tfrac{\pi}{3}}{0.001}\approx 0.49957\ldots \approx \tfrac12.\ \checkmark$$ --- ## Step 7 — Twin: $\cos' = -\sin$, usi wheel se **KYA.** Bilkul same triangle argument se, $P$ ki **horizontal** motion $-\,h\sin x$ hai (minus isliye kyunki upar jana matlab *baayen* jaana hai). Isliye $$\frac{d}{dx}\cos x = -\sin x.$$ **KYUN.** Width $\cos x$ *ghatti* hai jab hum pehle quadrant mein chadhte hain, isliye uski rate negative hai. Yahi hai poore "co's are negative" rule ka origin jis par parent note lean karta hai — yeh literally step ka leftward tilt hai. **PICTURE.** Wohi chota triangle, ab horizontal leg padh rahe hain. ![[deepdives/dd-maths-4.1.18-d2-s07.png]] Jab tumhare paas $\sin'=\cos$ aur $\cos'=-\sin$ aa jaate hain, toh $\tan=\sin/\cos$, $\sec=1/\cos$, etc. par [[Quotient Rule]] tumhe baaki chaar de deta hai (parent note dekho). Aur yeh do facts [[Simple Harmonic Motion]] aur [[Derivatives of inverse trig functions]] ke peeche ka engine hain. --- ## Ek-picture summary ![[deepdives/dd-maths-4.1.18-d2-s08.png]] Yeh single figure sab kuch stack karta hai: ghoomta point $\sin x$ (blue height wave) trace karta hai, aur **us wave ki slope har moment par** neeche pink $\cos x$ wave hai. Jahan blue sabse tezi se upar hai, wahan pink apne peak par hai; jahan blue flat crest karta hai, wahan pink zero cross karta hai. > [!recall]- Poore walkthrough ki Feynman retelling > Derivation ko bina symbols ke explain karo. ::: Ek dot ko radius one ke circle par ghoomao. Middle ke upar uski height wohi hai jo hum $\sin$ kehte hain. Dot ko thoda aur aage push karo. Woh thodi si doori chalega, aur woh choti doori tilted hai — thodi upar, thodi sideways. Us push ka *upar wala* fraction, kisi bhi moment par, exactly utna hi hai jitna dot abhi middle se wide hai, jo $\cos$ hai. Isliye "height kitni tezi se badhti hai" equals "dot kitna wide hai" — yahi reason hai ki $\sin$ ki slope $\cos$ hai. Sideways fraction width ko shrink karta hai, aur woh fraction height hi hai — isliye $\cos$ ki slope *minus* $\sin$ hai. Do facts, ek tilted chote step ki picture. > [!mnemonic] Chaar shabdon mein picture > **Up-fraction is cosine.** Tiny circular step ka rise $h\cos x$ hai; $h$ se divide karo aur derivative $\cos x$ nikal aata hai. --- ## Connections - [[Limit definition of the derivative]] — Step 2 ka rise/run limit - [[The Squeeze Theorem]] — Step 5 mein $\frac{\sin h}{h}\to 1$ prove karta hai - [[Angle addition formulas]] — Step 5 mein exact expansion - [[Pythagorean identity sin^2 + cos^2 = 1]] — quotient-rule sequels ko power deta hai - [[Quotient Rule]] · [[Product Rule]] · [[Chain Rule]] — baaki chaar functions banate hain - [[Simple Harmonic Motion]] — physics jahan yeh slope ek velocity hai - [[Derivatives of inverse trig functions]] — reverse direction - [[Derivatives of all six trig functions|Parent topic par wapas jaao]]