Visual walkthrough — Derivatives of inverse trig functions — all six

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

Shuru karne se pehle, ek wada: hum koi bhi symbol tab tak use nahi karenge jab tak uska ek plain-word meaning aur drawing mein ek jagah na ho. Chalo cast se milte hain.

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Step 1 — "arcsin" asal mein poochta kya hai

KYA HAI. Ek machine jiska naam $\sin$ hai, ek angle khaati hai aur ek ratio — ek plain number — ugalti hai. Iska inverse, jo $\arcsin$ likha jaata hai, machine ko ulta chalaata hai: tum ratio dete ho, woh tumhe batata hai kaun sa angle tha jisne woh ratio produce kiya.

KYUN. Hum $\arcsin$ ko seedha differentiate nahi kar sakte — hamare paas "$x$ jiska sine hai woh angle" ke liye koi rule nahi hai. Lekin forward machine $\sin$ ko hum andar-bahar jaante hain. Toh poori strategy yeh hai: backward question ko ek aisa forward question banao jo hum pehle se handle kar sakein.

PICTURE. Left side pe, angle in → ratio out. Right side pe, arrow ulta: ratio in → angle out.

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Step 2 — Backward question ko forward mein flip karo

KYA HAI. $y=\arcsin x$ ke dono sides pe forward machine $\sin$ lagao. Kyunki $\sin$ aur $\arcsin$ ek doosre ko undo karte hain, right side collapse ho jaati hai: $\sin y=x.$

KYUN. Dekho abhi kya hua. Equation $\sin y=x$ mein sirf forward function hai. Yeh $y$ aur $x$ ko aapas mein jodne wala ek honest relationship hai — koi bhi inverse nahi dikhta. Hum us cheez se nikal aaye jo hum differentiate nahi kar sakte the.

PICTURE. Dono machines ek key aur uski lock ki tarah cancel ho jaate hain; jo bachta hai woh clean relation $\sin y=x$ hai.

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Step 3 — Relation ko ek right triangle ki tarah draw karo

KYA HAI. Statement $\sin y=x$ asal mein $\sin y=\frac{x}{1}$ hai. Yaad karo ki ek right triangle mein kisi angle ke liye, $\sin y=\frac{\text{opposite side}}{\text{hypotenuse}}.$ Toh hum ek triangle bana sakte hain jo hamare equation ko follow kare: opposite side $=x$ aur hypotenuse $=1$ rakho.

KYUN. Triangle ek memory device hai. Jab hum do sides rakh dete hain, toh Pythagoras teesri side free mein de deta hai — aur woh teesri side exactly wahi quantity hai jo humein thodi der baad chahiye hogi.

PICTURE. Kone pe angle $y$, vertical (opposite) side labelled $x$, tilti hui hypotenuse labelled $1$, aur horizontal (adjacent) side abhi unknown — ise $a$ kehte hain.

Ab $y$ ka cosine seedha triangle se padha ja sakta hai: $\cos y=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}.$

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Step 4 — Honest relation ko differentiate karo

KYA HAI. $\sin y=x$ ke dono sides ka $\frac{d}{dx}$ lo. Right side easy hai: $x$ ka $x$ ke respect mein derivative $1$ hai. Left side ko Chain Rule chahiye kyunki $y$ khud $x$ ka function hai (jab hum $x$ slide karte hain toh yeh badalta hai): $\underset{\text{deriv of }\sin }{\underbrace{\cos y}}\cdot \underset{\text{deriv of the inside}}{\underbrace{\frac{dy}{dx}}}=1.$

KYUN. Yeh Implicit Differentiation hai: humne $y$ ko explicitly kabhi solve nahi kiya, phir bhi poore relation ko differentiate karke hum baad mein slope $\frac{dy}{dx}$ solve kar sakte hain. Yeh tool exactly isliye exist karta hai kyunki hum $\arcsin$ ko cleanly isolate nahi kar sakte the.

PICTURE. Ratio mein ek tiny nudge $dx$ se angle mein ek tiny nudge $dy$ hoti hai; factor $\cos y$ unke beech ki "gearing" hai jo chain rule expose karta hai.

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Step 5 — Slope solve karo, phir $x$ mein wापस translate karo

KYA HAI. Dono sides ko $\cos y$ se divide karo: $\frac{dy}{dx}=\frac{1}{\cos y}.$ Lekin ek derivative $x$ ka function hona chahiye, hidden angle $y$ ka nahi. Toh hum $\cos y$ ko Step 3 ki triangle value se swap karte hain, $\cos y=\sqrt{1-x^2}$: $\boxed{\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}}$

KYUN. Triangle ek bridge hai. Yeh angle-language answer $\frac{1}{\cos y}$ ko ratio-language answer $\frac{1}{\sqrt{1-x^2}}$ mein convert karta hai jo sirf $x$ mention karta hai.

PICTURE. $\arcsin$ curve ka slope ratio $x$ pe triangle ki adjacent side ke reciprocal ke barabar hai — answer seedha drawing se padho.

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Step 6 — Positive root kyun, aur domain $-1<x<1$ kyun

KYA HAI. Step 3 mein humne $\cos y=+\sqrt{1-x^2}$ liya, negative option phek diya. Hume ise justify karna hoga, aur yeh bhi batana hoga ki formula kahaan tak allowed hai.

PLUS SIGN KYUN. Angle $y=\arcsin x$ ko $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ mein rehna padta hai (Step 1 ki restriction). Us poore band mein, cosine kabhi negative nahi hoti — yeh $0$ se upar $1$ tak jaati hai aur wapas $0$ pe aati hai. Toh $\cos y\ge 0$, aur positive root hi sahi choice hai.

OPEN INTERVAL KYUN. Dekho ends pe kya hota hai:

• $x=\pm 1$ par: $\sqrt{1-x^2}=\sqrt{0}=0$, toh $\frac{1}{\sqrt{1-x^2}}$ blow up karta hai — $\arcsin$ curve ki tangent vertical ho jaati hai. Derivative wahan simply exist nahi karta.
• $|x|>1$ se aage: $1-x^2<0$, toh square root real number nahi hai, aur na hi $\arcsin x$ khud. Koi angle nahi hai jiska sine $1$ se zyada ho.

PICTURE. Arcsin band ke upar cosine graph axis par ya uske upar baitha rehta hai (kabhi negative nahi jaata), jabki derivative ki spikes exactly $x=\pm 1$ par infinity ki taraf shoot karti hain.

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Step 7 — Edges ke paas shape ka sanity check karo

KYA HAI. Formula ka behaviour test karo, sirf uski algebra nahi. Jab $x\to 0$, slope $\frac{1}{\sqrt{1-0}}=1$ hai. Jab $x\to \pm 1$, slope $\to +\infty$.

KYUN. $0$ ke ratio ke paas angle ratio ke saath same pace se badalta hai (slope $1$). $\pm 1$ ke ratio ke paas sine curve almost flat hoti hai, toh ratio mein tiny change ke liye angle mein bada swing chahiye — isliye slope explode karta hai. Formula exactly yahi predict karta hai, jo iska sahi hone ka strong evidence hai.

PICTURE. $\arcsin$ curve beech mein gentle hai (slope $\approx 1$) aur $x=\pm 1$ ke paas jaate jaate almost vertically upar uthti hai.

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Ek-picture summary

Upar sab kuch, compressed: backward question $\sin y=x$ ban jaata hai, phir triangle ban jaata hai, phir adjacent side se padha gaya slope ban jaata hai.

Recall Feynman retelling — poori movie plain words mein

Hum jaanna chahte the ki angle ka answer kitni tezi se badalta hai jab woh ratio jo hum feed karte hain thoda sa wiggle karta hai. Hum usse directly attack nahi kar sakte the, toh humne flip kiya: hamare angle ka sine bas ratio ke barabar hai, $\sin y=x$. Woh sentence ek triangle hai — opposite side $x$, hypotenuse $1$ — aur Pythagoras aakhri side $\sqrt{1-x^2}$ se fill kar deta hai. $\sin y=x$ ke dono sides ko nudge karne se milta hai $\cos y\cdot (\text{angle wiggle})=(\text{ratio wiggle})$, toh angle wiggle ek divided by $\cos y$ hai — aur $\cos y$ bas adjacent side $\sqrt{1-x^2}$ hai. Answer: $\frac{1}{\sqrt{1-x^2}}$. Yeh positive hai kyunki haara angle wahan rehta hai jahan cosine kabhi negative nahi jaati; yeh $x=\pm 1$ par explode karta hai kyunki wahan sine curve flat ho jaati hai aur angle ko race karna padta hai; aur yeh $x=\pm 1$ se aage meaningless hai kyunki kisi angle ka sine ek se bada nahi hota.

Recall

$\arcsin x$ differentiate karne ke liye pehla move ::: $\sin y=x$ likho aur implicitly differentiate karo $\sqrt{1-x^2}$ kahan se aata hai ::: us triangle ki adjacent side se jisme opposite $x$ hai, hypotenuse $1$ hai $\cos y$ ke liye positive root kyun ::: arcsin ka angle $[-\pi /2,\pi /2]$ mein rehta hai jahan $\cos y\ge 0$ $x=\pm 1$ par slope ka kya hota hai ::: yeh $+\infty$ tak blow up karta hai; tangent vertical ho jaati hai $x=\frac{1}{2}$ par slope ki value ::: $\frac{2}{\sqrt{3}}$

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Connections

• Inverse Functions and their Domains — range $[-\frac{\pi }{2},\frac{\pi }{2}]$ jo sign fix karti hai
• Implicit Differentiation — woh engine jo $\sin y=x$ ko slope mein badalta hai
• Chain Rule — isliye left side pe trailing $\frac{dy}{dx}$ aata hai
• Pythagorean Identities — adjacent side $\sqrt{1-x^2}$ supply karta hai
• Derivatives of Trig Functions — forward $\frac{d}{dx}\sin y=\cos y$ jo hum invert karte hain
• Integration by Inverse Trig — yeh derivative ulta padha gaya