Exercises — Derivatives of inverse trig functions — all six

$\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}}.$ Minus kyun: jaise-jaise ratio $x$ badhta hai, angle $\arccos x$ ghatta hai (arccos $\pi$ se $0$ tak jaata hai). Jo function decrease karta hai uski slope negative hoti hai. $\sqrt{1-x^2}$ ki shape arcsin jaisi hai, sign ulta hai.

$\frac{d}{dx}\arctan x=\frac{1}{1+x^2},\frac{d}{dx}\mathrm{arccot}x=-\frac{1}{1+x^2}.$ Bilkul bhi square root nahi — yeh saaf-suthra result identity ${\sec }^{2}y=1+{\tan }^{2}y$ (dekho Pythagorean Identities) se aata hai, jisme solve karte waqt koi square root nahi chahiye. Co-partner bas sign flip kar deta hai.

Sirf $\mathrm{arcsec}$ aur $\mathrm{arccsc}$ mein $|x|$ hai: $\frac{d}{dx}\mathrm{arcsec}x=\frac{1}{|x|\sqrt{x^2-1}}.$ $x=2$ par: $|2|=2$ aur $\sqrt{4-1}=\sqrt{3}$, toh value hai $\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\approx \mathrm{0.2887.}$

Andar $u=5x$ hai, toh $u^{\prime }=5$. $f^{\prime }(x)=\frac{1}{1+(5x{)}^2}\cdot 5=\frac{5}{1+25x^2}.$ Ek extra factor andar ka derivative hai — isse miss karo toh $5$ ka factor choot jaata hai.

Andar $u=\sqrt{x}=x^{1/2}$ hai, toh $u^{\prime }=\frac{1}{2}{x}^{-1/2}=\frac{1}{2\sqrt{x}}$. Aur $u^2=x$. $g^{\prime }(x)=\frac{1}{\sqrt{1-(\sqrt{x}{)}^2}}\cdot \frac{1}{2\sqrt{x}}=\frac{1}{\sqrt{1-x}}\cdot \frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x(1-x)}}.$ Dhyan do kaise $(\sqrt{x}{)}^2$ collapse hokar $x$ ban jaata hai — answer likhne se pehle hamesha inside-squared ko simplify karo.

Andar $u=e^x$ hai, toh $u^{\prime }=e^x$. Arccos ek minus sign laata hai. $h^{\prime }(x)=-\frac{1}{\sqrt{1-(e^x{)}^2}}\cdot e^x=\frac{-e^x}{\sqrt{1-e^{2x}}}.$ Valid sirf wahan hai jahan $e^x<1$ ho, yani $x<0$ — usse bahar ratio $1$ se zyada ho jaata hai aur $\arccos$ undefined ho jaata hai.

Andar $u=x^3$, $u^{\prime }=3x^2$, aur $u^2=x^6$. $p^{\prime }(x)=\frac{3x^2}{1+x^6},p^{\prime }(1)=\frac{3}{1+1}=\frac{3}{2}.$

Product rule: $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$ jahan $u=x,v=\arctan x$. $f^{\prime }(x)=1\cdot \arctan x+x\cdot \frac{1}{1+x^2}=\arctan x+\frac{x}{1+x^2}.$ Aage koi collapse nahi — dono pieces genuinely alag tarah ke terms hain (ek angle aur ek rational function).

Andar $u=1/x=x^{-1}$ hai, toh $u^{\prime }=-x^{-2}=-\frac{1}{x^2}$, aur $u^2=\frac{1}{x^2}$. $g^{\prime }(x)=\frac{1}{1+\frac{1}{x^2}}\cdot \left(-\frac{1}{x^2}\right)=\frac{x^2}{x^2+1}\cdot \left(-\frac{1}{x^2}\right)=-\frac{1}{1+x^2}.$ Yeh exactly $\frac{d}{dx}\mathrm{arccot}x$ ke barabar hai — yeh reassuring hai, kyunki $\arctan (1/x)=\mathrm{arccot}x$ for $x>0$.

$\frac{1}{\sqrt{1-x^2}}-\left(-\frac{1}{\sqrt{1-x^2}}\right)=\frac{2}{\sqrt{1-x^2}}.$ Dono derivatives add ho jaate hain (kyunki negative ko subtract karne se flip ho jaata hai), arcsin derivative ka double milta hai.

$\frac{2x}{1-x^2}$ ko tangent ka double-angle formula samjho: agar $x=\tan \theta$ toh $\frac{2x}{1-x^2}=\tan (2\theta )$, isliye $|x|<1$ par $y=2\arctan x$ hai. Phir $\frac{dy}{dx}=2\cdot \frac{1}{1+x^2}=\frac{2}{1+x^2}.$ (Brute-force quotient-rule differentiation bhi same $\frac{2}{1+x^2}$ deta hai — identity bas algebra bachaa leti hai.)

$\frac{d}{dx}(\arctan x+\mathrm{arccot}x)=\frac{1}{1+x^2}+\left(-\frac{1}{1+x^2}\right)=0.$ Ek interval par zero derivative ka matlab hai function constant hai (ek flat graph). $x=1$ par evaluate karo: $\arctan 1+\mathrm{arccot}1=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}.$ Toh $\arctan x+\mathrm{arccot}x=\frac{\pi }{2}$ sabhi $x$ ke liye.

Kyunki $\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$ hai, antiderivative $\arctan x$ hai. ${\int }_0^1\frac{dx}{1+x^2}=[\arctan x{]}_0^1=\arctan 1-\arctan 0=\frac{\pi }{4}-0=\frac{\pi }{4}.$

Integrand $\frac{d}{dx}\arcsin x$ hai, isliye ${\int }_0^{1/2}\frac{dx}{\sqrt{1-x^2}}=[\arcsin x{]}_0^{1/2}=\arcsin \frac{1}{2}-\arcsin 0=\frac{\pi }{6}-0=\frac{\pi }{6}.$

Andar $u=x/a$, $u^{\prime }=1/a$, $u^2=x^2/a^2$. $\frac{d}{dx}\arctan \left(\frac{x}{a}\right)=\frac{1}{1+\frac{x^2}{a^2}}\cdot \frac{1}{a}=\frac{a^2}{a^2+x^2}\cdot \frac{1}{a}=\frac{a}{a^2+x^2}.$ Toh $\frac{1}{a^2+x^2}$ ka antiderivative $\frac{1}{a}\arctan \left(\frac{x}{a}\right)$ hai. Phir ${\int }_0^a\frac{dx}{a^2+x^2}=\frac{1}{a}[\arctan \frac{x}{a}{]}_0^a=\frac{1}{a}(\arctan 1-\arctan 0)=\frac{1}{a}\cdot \frac{\pi }{4}=\frac{\pi }{4a}.$

Dono laal curve pieces dekho. $x>1$ ke liye arcsec curve $0$ se $\frac{\pi }{2}$ ki taraf upar jaata hai; $x<-1$ ke liye yeh $\frac{\pi }{2}$ se $\pi$ ki taraf upar jaata hai. Dono pieces upar jaate hain, isliye dono par slope positive hai. Raw algebra $\frac{1}{x\sqrt{x^2-1}}$ deta hai, jo $x<-1$ ke liye negative hoga (kyunki wahan $x<0$ hai). $x$ ko $|x|$ mein wrap karne se denominator dono taraf positive rehta hai, isliye derivative wahan positive rehti hai jahan arcsec badh raha hai: $\frac{d}{dx}\mathrm{arcsec}x=\frac{1}{|x|\sqrt{x^2-1}}.$ $x=-2$ par check karo: $\frac{1}{|-2|\sqrt{3}}=\frac{1}{2\sqrt{3}}>0$ ✓ (bare-$x$ version galti se $-\frac{1}{2\sqrt{3}}$ deta).

$\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}.$ Jaise $x\to 1^{-}$ hota hai, quantity $1-x^2\to 0^{+}$ hoti hai, toh $\sqrt{1-x^2}\to 0^{+}$ aur poora fraction $\to +\infty$ ho jaata hai. Figure mein arcsin curve $x=1$ par vertically upar uthta hai: tangent line ki slope blow up ho jaati hai. Geometrically, ratio ke $1$ ke paas hone par angle, ratio mein thodi si change ke liye bahut zyada change hota hai — yahi steepness is exploding derivative hai. Yahi cheez (mirror image mein) $x\to -1^{+}$ par bhi hoti hai.

Andar $u=x^2$, $u^{\prime }=2x$, aur $u^2=x^4$. Kyunki $u=x^2\ge 0$, hame $|u|=x^2$ milta hai. $y^{\prime }=\frac{1}{|x^2|\sqrt{(x^2{)}^2-1}}\cdot 2x=\frac{2x}{x^2\sqrt{x^4-1}}=\frac{2}{x\sqrt{x^4-1}}.$ Domain: arcsec ko $|u|\ge 1$ chahiye, yani $x^2\ge 1$, toh $|x|\ge 1$. Aur derivative ko $x^4>1$ chahiye, yani strictly $|x|>1$ (equality par vertical tangent hoti hai). Dhyan do $\frac{2}{x\sqrt{x^4-1}}$ genuinely $x$ ka sign rakhta hai: $x>1$ ke liye positive aur $x<-1$ ke liye negative — yeh sahi hai kyunki $\mathrm{arcsec}(x^2)$ dono branches par pehle badhta phir ghatta hai.

Maan lo $u=\frac{x-1}{x+1}$. Quotient rule: $u^{\prime }=\frac{(1)(x+1)-(x-1)(1)}{(x+1{)}^2}=\frac{(x+1)-(x-1)}{(x+1{)}^2}=\frac{2}{(x+1{)}^2}.$ Ab $1-u^2$ ki tricky simplification: $1-u^2=1-\frac{(x-1{)}^2}{(x+1{)}^2}=\frac{(x+1{)}^2-(x-1{)}^2}{(x+1{)}^2}=\frac{4x}{(x+1{)}^2}.$ (Difference of squares: $(x+1{)}^2-(x-1{)}^2=4x$.) Toh $\sqrt{1-u^2}=\frac{2\sqrt{x}}{|x+1|}$ (valid jahan $x>0$ ho, giving $x+1>0$ toh $|x+1|=x+1$). Arccos ke minus sign ke saath: $y^{\prime }=-\frac{1}{\sqrt{1-u^2}}\cdot u^{\prime }=-\frac{x+1}{2\sqrt{x}}\cdot \frac{2}{(x+1{)}^2}=-\frac{1}{\sqrt{x}(x+1)}.$ Clean result: $y^{\prime }=-\frac{1}{\sqrt{x}(x+1)}$, valid for $x>0$.

Recognition sign rule ::: sin/tan/sec positive, unke co-partners negative $\arctan (5x)$ ke liye chain-rule extra factor ::: $u^{\prime }=5$ se multiply karo giving $\frac{5}{1+25x^2}$ ${\int }_0^1\frac{dx}{1+x^2}$ ::: $\arctan 1-\arctan 0=\frac{\pi }{4}$ Arcsec derivative mein $|x|$ kyun ::: dono branches par slope positive rakhta hai $\frac{d}{dx}\arcsin x$ ka limit jaise $x\to 1^{-}$ ::: $+\infty$ (vertical tangent)