Derivatives of inverse trig functions — all six: Question bank

4.1.21 · D5 · Maths › Calculus I — Limits & Derivatives › Derivatives of inverse trig functions — all six

True ya false — justify karo

Yahan har symbol ka matlab hai: $y = arcsin x$ etc., toh $y$ ek angle hai aur $x$ ek ratio hai jo us angle ka trig function produce karta hai.

\frac{d}{d x} arccos x = \frac{1}{1 - x ^{2}}

False.

arccos

decrease karta hai — ek bada ratio

x

ek chhote angle ko map karta hai — toh uski slope negative honi chahiye:

- \frac{1}{1 - x ^{2}}

arcsin x

aur

arccos x

ka derivative sign ke alaawa bilkul same hai

True. Dono

\frac{1}{1 - x ^{2}}

share karte hain;

arccos

sirf minus carry karta hai. Isi wajah se

arcsin x + arccos x

ka derivative zero hai aur yeh constant

(\frac{π}{2})

hai.

arctan x

ka derivative har real

x

ke liye defined hai

True. Iska denominator

1 + x^{2}

kabhi zero nahi hota, toh

\frac{1}{1 + x ^{2}}

saare

x

ke liye exist karta hai — unlike

arcsin

ka, jo

x = \pm 1

par khatam ho jaata hai.

\frac{d}{d x} arcsec x = \frac{1}{x x ^{2} - 1}

False.

∣ x ∣

zaroori hai.

arcsec

dono branches (

x < - 1

aur

x > 1

) par increase karta hai, toh iska derivative har jagah positive hona chahiye; sirf

∣ x ∣

(jo hamesha

\geq 0

hai) yeh enforce karta hai.

\frac{d}{d x} arcsin x

x = 0

ke paas sabse zyada (steepest) hai

False.

x = 0

par denominator

1 - x^{2} = 1

hai, toh slope

1

milti hai. Jaise

x \to \pm 1

, denominator

\to 0

aur slope

+ \infty

tak blow up ho jaati hai — graph edges par sabse steep hai, beech mein sabse flat.

arctan x

aur

arccot x

ke graphs reflections hain, toh unke derivatives sign mein opposite hain

True.

arccot

decrease karta hai jabki

arctan

increase karta hai; unke derivatives

\pm \frac{1}{1 + x ^{2}}

hain, ek doosre ke exact negatives.

Kyunki

sin y = x

hai, hum hamesha

cos y = 1 - x^{2}

positive root ke saath likh sakte hain

True — lekin sirf $arcsin$ ke liye. Iska range

[- \frac{π}{2}, \frac{π}{2}]

cos y \geq 0

rakhta hai. Kisi doosre inverse ke liye jo alag range rakhe, sign dobara check karna padega; positive root earned hai, automatic nahi.

Error dhundo

Har line mein kisi ka "solution" likha hai. Galti dhundho aur theek karo.

\frac{d}{d x} arctan (3 x) = \frac{1}{1 + 9 x ^{2}}

Chain-rule factor missing. Formula

\frac{1}{1 + u ^{2}}

arctan (u)

par apply hota hai; yahan

u = 3 x

hai toh

u^{'} = 3

ka hisaab bhi dena hoga. Sahi answer:

\frac{3}{1 + 9 x ^{2}}

\frac{d}{d x} arcsin (x^{2}) = \frac{1}{1 - x ^{2}}

Do galtiyan hain. Andar

u = x^{2}

hai, toh root ke neeche

u^{2} = x^{4}

hoga (

x^{2}

nahi), aur

u^{'} = 2 x

se multiply bhi karna hoga. Sahi answer:

\frac{2 x}{1 - x ^{4}}

sec^{2} y = 1 + tan^{2} y

hone ki wajah se,

tan y = x

differentiate karne par end mein ek square root chahiye."

Koi root nahi chahiye.

\frac{d y}{d x} = \frac{1}{sec ^{2} y} = \frac{1}{1 + tan ^{2} y} = \frac{1}{1 + x ^{2}}

sec^{2} y

already

1 + x^{2}

hai — yahi clean cancellation hai jis wajah se

arctan

ke derivative mein square root nahi hai.

\frac{d}{d x} arcsin x

x = 2

equals

\frac{1}{1 - 4} = \frac{1}{- 3}

Domain violation.

arcsin

sirf

∣ x ∣ \leq 1

ke liye exist karta hai; koi angle nahi hai jiska sine

2

ho. Yeh sawaal hi meaningless hai, sirf arithmetic nahi.

\frac{d}{d x} arccos x = - \frac{1}{1 - x ^{2}}

, aur maine

cos y = x

differentiate kiya toh

- sin y \frac{d y}{d x} = 1

mila, isliye

sin y = - 1 - x^{2}

Root ka sign galat hai.

arccos

ka range

[0, π]

hai, jahan

sin y \geq 0

hai, toh

sin y = + 1 - x^{2}

. Phir

\frac{d y}{d x} = \frac{- 1}{sin y} = \frac{- 1}{1 - x ^{2}}

— minus algebra se aata hai, root se nahi.

\int \frac{d x}{1 + x ^{2}} = \frac{1}{2 x} + C

by the power rule on

(1 + x^{2})^{- 1}

Power rule apply nahi hota

1/ (1 + x^{2})

par kyunki andar

x^{2}

hai. Isko ulta padho: yeh exactly

\frac{d}{d x} arctan x

hai, toh integral

arctan x + C

hai. Dekho Integration by Inverse Trig.

\frac{d}{d x} arcsec x

x = \frac{1}{2}

\frac{1}{\frac{1}{2} \frac{1}{4} - 1}

Domain violation.

arcsec

∣ x ∣ \geq 1

chahiye; koi angle nahi hai jiska secant

\frac{1}{2}

ho (secant kabhi strictly

- 1

aur

1

ke beech nahi aata). Root ke neeche ka expression ek wajah se negative hai.

Why questions

Derivations

arcsin x

ko directly differentiate karne ki jagah

sin y = x

kyun likhte hain?

Kyunki hum

arcsin

ko kisi known rule se differentiate nahi kar sakte, lekin

sin

ko kar sakte hain.

sin y = x

likhna ek unknown ko ek known relation mein badal deta hai jise hum implicitly differentiate karke solve kar sakte hain. Dekho Implicit Differentiation.

arctan

ko tidy

\frac{1}{1 + x ^{2}}

kyun milta hai jabki

arcsin

ko ugly square root?

Kyunki

x

mein wapas convert karne wali identity alag hai.

arcsin

use karta hai

cos y = 1 - sin^{2} y

(ek root), jabki

arctan

use karta hai

sec^{2} y = 1 + tan^{2} y

— ek polynomial, koi root nahi — isliye iska derivative rational rahta hai.

Har "co-" function ka derivative negative kyun hota hai?

Har co-function (

arccos

arccot

arccsc

) apne range par decrease karta hai: jaise ratio badhta hai, angle chhota hota hai. Ek decreasing function ki slope har jagah negative hoti hai jahan wo differentiable hai.

∣ x ∣

sirf

arcsec

aur

arccsc

mein kyun aata hai, baaki chaar mein nahi?

Kyunki

tan y = \pm sec^{2} y - 1

ek sign carry karta hai jo secant ke dono branches ke beech flip hota hai, phir bhi

arcsec

dono par increase karta hai.

∣ x ∣

us sign flip ko absorb karta hai taaki derivative dono branches par positive rahe. Baaki chaar ek hi monotone piece par rehte hain, toh absolute value ki zaroorat nahi.

arcsin x

ki slope

x \to \pm 1

par infinity tak kyun explode hoti hai?

Geometrically,

\pm 1

ke ratio ke paas ratio mein thodi si change angle mein bahut badi change correspond karti hai (triangle lagbhag degenerate ho jaata hai). Algebraically, adjacent side

1 - x^{2} \to 0

, toh

\frac{1}{1 - x ^{2}} \to \infty

Inverse function ki range restriction ki zaroorat kyun hai?

Kyunki

sin

(aur har trig function) repeat karta hai, toh "

sin y = x

" ke infinitely many solutions

y

hain. Range restrict karna ek branch pick karta hai, jo phir

cos y

sin y

, ya

tan y

ka sign pin karta hai — aur isliye derivative ka sign bhi. Dekho Inverse Functions and their Domains.

Yeh chhe derivatives jaante hi integrals ki ek poori family kyun mil jaati hai?

Ek derivative rule ulta padha jaaye toh antiderivative ban jaata hai. Agar

\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}}

hai, toh definition se

\int \frac{d x}{1 + x ^{2}} = arctan x + C

hai. Har formula apna ek integral donate karta hai.

Edge cases

\frac{d}{d x} arcsin x

exactly

x = \pm 1

par kya hai?

Undefined (vertical tangent). Wahan denominator

1 - x^{2} = 0

hai, toh slope infinite hai.

arcsin

\pm 1

par continuous hai lekin differentiable nahi — iska graph vertical tangent rakhta hai.

Kya

\frac{d}{d x} arctan x

kabhi zero ya negative hota hai?

Kabhi nahi.

\frac{1}{1 + x ^{2}}

saare real

x

ke liye strictly positive hai; yeh

x \to \pm \infty

par

0

ke paas aata hai lekin kabhi pahunchta nahi. Toh

arctan

strictly increasing hai lekin tails par flat hota jaata hai.

x = 0

par, kya

arcsin

aur

arctan

ki slope same hai?

Haan, dono $1$ ke barabar hain.

\frac{1}{1 - 0} = 1

aur

\frac{1}{1 + 0} = 1

. Origin ke paas dono curves line

y = x

ke saath chipke rehte hain; yeh sirf

∣ x ∣

badhne par alag hote hain.

\frac{d}{d x} arcsec x

ka kya hoga jab

∣ x ∣ \to \infty

Yeh

\to 0^{+}

ho jaata hai.

∣ x ∣

aur

x^{2} - 1

dono without bound badhte hain, toh

\frac{1}{∣ x ∣ x ^{2} - 1} \to 0

—

arcsec

apne horizontal asymptotes ki taraf level off ho jaata hai.

Kya

\frac{d}{d x} arcsec x

x = 1

par evaluate kar sakte hain?

Nahi. Wahan

x^{2} - 1 = 0

hai, toh derivative infinite hai (vertical tangent).

x = \pm 1

boundary points hain jahan dono branches shuru hoti hain.

Kya

arcsin x + arccos x = \frac{π}{2}

saare valid

x

ke liye sach hai, aur calculus yeh kaise prove karta hai?

Haan, saare $x \in [- 1, 1]$ ke liye. Unke derivatives exact negatives hain, toh sum ka derivative

0

hai aur yeh constant hai.

x = 0

par evaluate karne se

0 + \frac{π}{2} = \frac{π}{2}

milta hai, jo constant fix karta hai.

arccot

ke derivative ka sign kya hai, aur kya yeh apne domain mein kabhi change hota hai?

Hamesha negative, kabhi nahi badalta.

- \frac{1}{1 + x ^{2}} < 0

har real

x

ke liye hai, toh

arccot

apne poore domain par strictly decreasing hai, koi sign flip nahi.

Recall Har trap ki ek-line summary

Sign traps aate hain monotonicity se (co-functions decrease karte hain), root-sign traps aate hain range restrictions se, $∣ x ∣$ traps aate hain two-branch functions se, chain-rule traps aate hain bhool gaye inner derivatives se, aur domain traps aate hain aise ratios plug in karne se jo function produce nahi kar sakta. Category ka naam lo aur fix khud samajh aayega.

Connections

Derivatives of inverse trig functions — all six — woh parent jise yeh bank drill karta hai
Implicit Differentiation — kyun har derivation $sin y = x$ se shuru hoti hai
Pythagorean Identities — har sign decision ka source
Inverse Functions and their Domains — range restrictions jo signs pin karti hain
Chain Rule — bhool jaane wale inner-derivative traps
Integration by Inverse Trig — rules ko ulta padhna
Derivatives of Trig Functions — woh forward maps jinhe hum invert karte hain