4.1.21 · D3Derivatives of inverse trig functions — all six
Intuition Yeh page kis liye haiParent note ne tumhe chhe formulas diye aur kaise bante hain, yeh bataya. Yeh page hai drill floor : hum is topic ke har tarah ke problem ko dhundhte hain aur ek-ek solve karte hain — signs, domains, chain-rule traps, degenerate inputs, blow-up points, ek word problem, aur ek exam twist. Iske baad tumhe koi aisa case nahi milna chahiye jo tumne pehle defeat hote na dekha ho.
Yeh do "engine" formulas apne paas rakhlo; neeche ka almost sab kuch inhi mein se ek hai plus Chain Rule :
d x d arctan x = 1 + x 2 1 , d x d arcsin x = 1 − x 2 1 .
Kuch bhi solve karne se pehle, is topic ke har alag tarah ke situation ki list banate hain. Har row ek "cell" hai. Har cell ko neeche kam se kam ek worked example cover karta hai.
Cell
Ise alag kya banata hai
Example jo ise hit karta hai
C1 Plain formula, positive inputsirf derivative quote karo, koi chain nahi
Ex 1
C2 Chain rule, linear insideextra constant factor aata hai
Ex 2
C3 Chain rule, nonlinear insideandar squared/shifted, domain dekho
Ex 3
C4 Ek "co-" function (sign trap)minus sign carry karna zaroori
Ex 4
C5 arcsec/arccsc with $x
$
C6 Negative input / slope ka signcheck karo slope positive/negative theory ke mutabik rehti hai
Ex 6
C7 Limiting / blow-up value (x → ± 1 derivative → ∞
Ex 7
C8 Degenerate / out-of-domain input"koi jawab exist nahi karta" pehchano
Ex 8
C9 Word problem (rate of change)inverse trig ek real angle model karta hai
Ex 9
C10 Exam twist (identity + product)do rules combine karo, cleverly simplify karo
Ex 10
C11 arccot (chhetha function)chhe mein se aakhri, uska apna minus sign
Ex 11
Pehle ek picture lagata hoon taaki neeche ka har "slope" claim kuch point kar sake.
Red curve hai arcsin x sirf x = − 1 x = 1 (dashed walls), yeh hamesha rise karti hai (slope hamesha positive), aur dono edges par vertical ho jaati hai — yahi hai 1 − x 2 → 0 C6, C7, C8 sab bas yahi hain ki "main is red curve par kahan khada hoon?"
f ( x ) = arctan x x = 1 Forecast: padhne se pehle guess karo — x = 1 1
Step 1. Engine formula quote karo. Yeh step kyun? Andar x Chain Rule correction nahi deni.
f ′ ( x ) = 1 + x 2 1 .
Step 2. x = 1 Yeh step kyun? "Ek point par slope" matlab derivative wahan evaluate karo.
f ′ ( 1 ) = 1 + 1 2 1 = 2 1 .
Verify: arctan x π /2 1 kam hona chahiye. Humne 2 1 < 1
g ( x ) = arcsin ( 5 x ) Forecast: kya answer 1 − 25 x 2 1
Step 1. Inside u = 5 x Yeh step kyun? Formula d x d arcsin u = 1 − u 2 u ′ u u ′ = 5
Step 2. Formula apply karo, andar x u = 5 x Yeh step kyun? 1 − □ 2 pura andar wala squared hota hai.
g ′ ( x ) = 1 − ( 5 x ) 2 1 ⋅ 5 = 1 − 25 x 2 5 .
Verify: domain ab ∣5 x ∣ < 1 ∣ x ∣ < 5 1 arcsin narrow window, jo samajh aata hai kyunki 5 x ± 1 5 Trap A ). ✓
h ( x ) = arctan ( x 2 + 1 ) Forecast: andar kabhi 0 1 + ( inside ) 2
Step 1. Inside u = x 2 + 1 u ′ = 2 x Yeh step kyun? Inside ko alag se differentiate karna padega — yahi poora Chain Rule hai.
Step 2. d x d arctan u = 1 + u 2 u ′ Yeh step kyun? arctan h ′ ( x ) = 1 + ( x 2 + 1 ) 2 2 x .
Step 3. Optionally ( x 2 + 1 ) 2 = x 4 + 2 x 2 + 1 1 + x 4 + 2 x 2 + 1 = x 4 + 2 x 2 + 2 Yeh step kyun? Baad ke integration checks ke liye cleaner.
h ′ ( x ) = x 4 + 2 x 2 + 2 2 x .
Verify: denominator x 4 + 2 x 2 + 2 ≥ 2 > 0 sab real x defined hai — yeh match karta hai ki arctan x = 0 h ′ = 0 arctan ( x 2 + 1 ) x = 0
p ( x ) = arccos ( 2 x ) x = 0 Forecast: positive slope ya negative? Yaad karo arccos downhill jaata hai.
Step 1. Yaad karo d x d arccos u = − 1 − u 2 u ′ minus note karo. Yeh step kyun? arccos ek "co-" function hai; yeh decrease karta hai (bada ratio → Trap B ).
Step 2. Inside u = 2 x u ′ = 2 Yeh step kyun? Chain rule upar factor u ′ = 2 p ′ ( x ) = − 1 − 4 x 2 2 .
Step 3. x = 0 Yeh step kyun? "x = 0 p ′ x = 0 p ′ ( 0 ) = − 1 2 = − 2.
Verify: slope negative hai, exactly jaisa arccos ki downhill shape demand karti hai. Agar tumhe + 2
q ( x ) = arcsec ( x ) x = 2 x = − 2 arccsc ( x ) Forecast: arcsec dono branches par increase karta hai — toh kya x = 2 x = − 2 same sign hona chahiye? Aur kya tumhe lagta hai arccsc ka slope opposite sign hoga?
Step 1. d x d arcsec x = ∣ x ∣ x 2 − 1 1 ∣ x ∣ x x 2 − 1 1 negative hoga — lekin arcsec wahan bhi rise karta hai, isliye derivative positive rehni chahiye. ∣ x ∣ Trap C ).
Step 2. x = 2 Yeh step kyun? Point par derivative ka direct evaluation.
q ′ ( 2 ) = ∣2∣ 4 − 1 1 = 2 3 1 = 6 3 .
Step 3. x = − 2 Yeh step kyun? "Dono branches par same sign" ka claim test karne ke liye, left branch actual mein compute karna padega.
q ′ ( − 2 ) = ∣ − 2∣ ( − 2 ) 2 − 1 1 = 2 3 1 = 6 3 .
Step 4. Companion. Yeh step kyun? Cell hai "arcsec /arccsc ", isliye csc case bhi dena hoga. arccsc, arcsec ka "co-" partner hai, isliye parent ke pattern se uska derivative arcsec ka negative hai:
d x d arccsc x = − ∣ x ∣ x 2 − 1 1 . x = 2 − 6 3 x = − 2 − 6 3 negative slope, kyunki arccsc har jagah decrease karta hai.
Verify: arcsec dono branches par equal aur positive nikla (+ 6 3 ≈ 0.2887 ∣ x ∣ x equal aur negative nikla (− 6 3
arcsin x x = − 2 1 Forecast: figure mein red curve har jagah rise karti hai — toh negative x positive honi chahiye, hai na?
Step 1. Derivative likho. Yeh step kyun? "arcsin x f ′ ( x ) = 1 − x 2 1 .
Step 2. x = − 2 1 Yeh step kyun? "x = − 2 1 par slope" matlab wahan derivative evaluate karo; pehle 1 − x 2 1 − x 2 = 1 − 4 1 = 4 3 ⟹ f ′ ( − 2 1 ) = 3/4 1 = 3 2 = 3 2 3 .
Verify: answer positive hai chahe x < 0 x squared appear karta hai, x 3 2 3 ≈ 1.1547 > 1 4 5 ∘
x → 1 − d x d arcsin x Forecast: finite number, ya infinite?
Step 1. x → 1 − 1 − x 2 Yeh step kyun? Slope ka poora behavior us square root ke shrink hone se control hota hai.
1 − x 2 → 1 − 1 = 0 + ⟹ 1 − x 2 → 0 + .
Step 2. Fixed numerator ko us vanishing denominator se divide karo. Yeh step kyun? Constant 1 0 + lim x → 1 − 1 − x 2 1 = + ∞.
Verify (geometry): neeche picture mein red curve x = ± 1 vertical ho jaati hai. Vertical tangent = infinite slope. Parent mein [!recall] ne kaha tha: "± 1
Intuition Yeh figure kyun matter karta haiYeh doosri picture Ex 7 ka eyeball se proof hai. Black curve wohi arcsin x x = 1 vertical ki taraf tilt ho rahi hai. Jab graph vertical hota hai, "rise over run" ka run zero ki taraf jaata hai — yahi 1 − x 2 → 0 → + ∞
x = 2 d x d arcsin x Forecast: koi answer guess karne ki koshish karo — phir check karo ki point exist bhi karta hai ya nahi.
Step 1. Pehle domain test karo. Yeh step kyun? arcsin ∣ x ∣ ≤ 1 Inverse Functions and their Domains ).
Step 2. x = 2 1 − x 2 Yeh step kyun? Formula ke square root ko 1 − x 2 ≥ 0 1 − x 2 = 1 − 4 = − 3 < 0 ⟹ 1 − x 2 = − 3 is not real .
Conclusion: x = 2 exist nahi karta — wahan red curve par koi point hi nahi hai (yeh dashed walls ke bahar hai). Yahi sahi answer hai , koi number nahi (Trap D ).
Verify: figures ki red curve x = 1 x = 2
Worked example Ek insaan ek wall se10 m top edge eye level se x θ = arctan ( 10 x ) x = 10 m θ Forecast: x = 10 4 5 ∘
Step 1. θ x u = 10 x u ′ = 10 1 Yeh step kyun? "Height ke har metre par θ d x d θ d x d θ = 1 + ( 10 x ) 2 1 ⋅ 10 1 = 10 1 ⋅ 1 + 100 x 2 1 = 100 + x 2 10 . Algebra kyun? Inner fraction clear karne ke liye upar aur neeche 100
Step 2. x = 10 Yeh step kyun? Question us particular height par rate poochh raha hai, isliye x = 10 d x d θ x = 10 = 100 + 100 10 = 200 10 = 20 1 = 0.05 rad/m .
Verify (units + sense): units hain radians per metre ✓. Value 0.05 ≈ 2. 9 ∘ 4 5 ∘
r ( x ) = x arctan x − 2 1 ln ( 1 + x 2 ) Forecast: aise exam questions usually kisi choti cheez par collapse hote hain. Guess karo r ′ ( x )
Step 1. r product rule se differentiate karo. Yeh step kyun? x arctan x x x arctan x d x d ( uv ) = u ′ v + u v ′ u = x , v = arctan x d x d ( x arctan x ) = u ′ 1 ⋅ arctan x + x ⋅ v ′ 1 + x 2 1 = arctan x + 1 + x 2 x . v ′ = 1 + x 2 1 arctan
Step 2. Doosra piece − 2 1 ln ( 1 + x 2 ) logarithm–chain rule se differentiate karo. Yeh step kyun? Kisi cheez ka ln d x d ln ( w ) = w w ′ w = 1 + x 2 w ′ = 2 x d x d ( − 2 1 ln ( 1 + x 2 ) ) = − 2 1 ⋅ 1 + x 2 2 x = − 1 + x 2 x . − 2 1
Step 3. Dono derivatives add karo aur cancellation dekho. Yeh step kyun? r ′ 1 + x 2 x r ′ ( x ) = ( arctan x + 1 + x 2 x ) + ( − 1 + x 2 x ) = arctan x + 1 + x 2 x − 1 + x 2 x = arctan x .
Verify: extra fractions annihilate ho jaate hain, clean r ′ ( x ) = arctan x r ( x ) arctan x antiderivative hai — ek favourite Integration by Inverse Trig result ulta padha. x = 0 r ′ ( 0 ) = arctan 0 = 0 0 + 0 − 0 = 0
s ( x ) = arccot ( 4 x ) x = 0 Forecast: arccot, arctan ka "co-" partner hai, isliye uska base derivative minus carry karta hai. x = 0
Step 1. Chhetha engine formula yaad karo d x d arccot x = − 1 + x 2 1 Minus kyun? arccot har jagah decrease karta hai (bada ratio → Trap B disguise mein).
Step 2. Inside u = 4 x u ′ = 4 Yeh step kyun? Chain Rule upar factor u ′ = 4 1 + □ 2 s ′ ( x ) = − 1 + ( 4 x ) 2 1 ⋅ 4 = − 1 + 16 x 2 4 .
Step 3. x = 0 Yeh step kyun? "x = 0 s ′ x = 0 s ′ ( 0 ) = − 1 + 0 4 = − 4.
Verify: slope negative hai (− 4 + 1 + x 2 1 arccot x
Recall Kaun sa cell kaun sa tha?Plain formula ::: Ex 1 (C1)
Chain rule with a linear inside gives an extra constant factor ::: Ex 2 (C2)
Nonlinear inside, domain stays all reals for arctan ::: Ex 3 (C3)
"co-" function needs the minus sign ::: Ex 4 (C4)
arcsec has ∣ x ∣ x → + ∞ x → 1 − d x d θ = 100 + x 2 10 0.05 x = 10 d x d ( x arctan x − 2 1 ln ( 1 + x 2 ) ) = arctan x d x d arccot ( 4 x ) = − 1 + 16 x 2 4 − 4 x = 0
Mnemonic Differentiate karne se pehle teen sawaal"Domain? Inside? Sign?" — (1) Kya input allowed hai (Trap D , C8)? (2) Kya andar koi function hai jise Chain Rule chahiye (Trap A , C2, C3, C11)? (3) Kya yeh ek "co-" function hai jise minus dena hai (Trap B , C4, C11)?
Multiply by inside derivative
