Derivatives of inverse trig functions — all six

4.1.21 · Maths › Calculus I — Limits & Derivatives

Master method (sirf EK baar seekho, chhe baar reuse karo)

YEH KYU KAAM KARTA HAI: inverse equation $\sin y=x$ ek sachcha relation hai $x$ aur $y$ ke beech. Abhi hume $\frac{dy}{dx}$ nahi pata, lekin hum relation ko differentiate karke iske liye solve kar sakte hain. Yahi trick "main $\arcsin$ ko directly differentiate nahi kar sakta" ko "main $\sin$ ko differentiate kar sakta hoon, koi problem nahi" mein badal deti hai.

$\frac{d}{dx}\arcsin x$ ko scratch se derive karna

$\sin y=x$

Dono sides differentiate karo ($\frac{d}{dx}$), left side pe chain rule use karte hue: $\cos y\cdot \frac{dy}{dx}=1$

Yeh step kyun? Left side $\sin (\text{kuch})$ hai, toh uski derivative $\cos (\text{kuch})\times (\text{kuch ki derivative})$ hogi, aur woh kuch $y$ hai.

Solve karo: $\frac{dy}{dx}=\frac{1}{\cos y}$

Ab hume $\cos y$ ko $x$ ke terms mein likhna hoga, $y$ mein nahi (derivative $x$ ka function honi chahiye). ${\sin }^{2}y+{\cos }^{2}y=1$ use karo: $\cos y=\sqrt{1-{\sin }^{2}y}=\sqrt{1-x^2}$

+ root kyun? Kyunki $\arcsin$ ka range $[-\frac{\pi }{2},\frac{\pi }{2}]$ hai, jahan $\cos y\ge 0$ hota hai. Toh hum positive square root lete hain.

$\boxed{\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}},-1<x<1}$

Baaki paanch — same recipe, alag trig identity

$\arccos$: $\cos y=x\Rightarrow -\sin y\frac{dy}{dx}=1\Rightarrow \frac{dy}{dx}=\frac{-1}{\sin y}=\frac{-1}{\sqrt{1-x^2}}$ (range $[0,\pi ]$ hai, toh $\sin y\ge 0$).

$\arctan$: $\tan y=x\Rightarrow {\sec }^{2}y\frac{dy}{dx}=1\Rightarrow \frac{dy}{dx}=\frac{1}{{\sec }^{2}y}=\frac{1}{1+{\tan }^{2}y}=\frac{1}{1+x^2}.$ ${\sec }^{2}y=1+{\tan }^{2}y$ kyun? Yeh Pythagorean identity ko ${\cos }^2$ se divide karne pe milta hai. Koi square root nahi chahiye — isliye $\arctan$ ki derivative itni clean hai.

$\mathrm{arcsec}$: $\sec y=x\Rightarrow \sec y\tan y\frac{dy}{dx}=1\Rightarrow \frac{dy}{dx}=\frac{1}{\sec y\tan y}$. Ab $\tan y=\pm \sqrt{{\sec }^{2}y-1}=\pm \sqrt{x^2-1}$ hai, aur range ko carefully handle karne ke baad $\frac{1}{|x|\sqrt{x^2-1}}$ milta hai. $|x|$ wahi hai jo derivative ko positive rakhta hai jahan bhi arcsec increasing hai.

Worked examples

Flashcards

Connections

• Implicit Differentiation — har derivation ke peeche yahi engine hai
• Chain Rule — composite inverse-trig functions ke liye zaroori hai
• Pythagorean Identities — $\cos y,\sec y$ ko wapas $x$ mein convert karte hain
• Inverse Functions and their Domains — range restrictions signs fix karti hain
• Integration by Inverse Trig — yeh formulas ulte padhe hue
• Derivatives of Trig Functions — woh forward versions jo hum invert karte hain

Concept Map