Chapter 1: Further Differentiation — Exercise 2.3
Q1. Tangent line ka equation nikalo (y = sin⁻¹(x/2) + cos x, at x = π/2)
Concept: Tangent line nikalne ke liye hume 3 cheezein chahiye: slope (derivative se), point ka x aur y coordinate.
Step 1: Pehle dy/dx nikalo. sin⁻¹(x/2) ki derivative formula se (1/√(1-u²))·u’ aati hai, aur cos x ki derivative -sin x hoti hai.
Step 2: x = π/2 par slope nikalo — is point par sirf -sin(π/2) wala part reh jata hai kyunke sin⁻¹ wala part bhi hai lekin final formula mein simplify ho jata hai.
Step 3: Us point ka y-coordinate nikalo, x = π/2 value substitute kar ke.
Step 4: Point-slope formula lagao: y − y₀ = m(x − x₀)
Result: Tangent line ka equation mil jata hai jisme slope aur point dono shamil hain.
Yaad rakho: Tangent line hamesha is formula se banti hai: pehle derivative (slope), phir specific point par values substitute karo.
Q2. Horizontal Tangent ka point dhoondo (y = tan⁻¹(x/3) − x/3)
Concept: Horizontal tangent ka matlab hai jahan slope (dy/dx) zero ho.
Step 1: dy/dx nikalo — tan⁻¹(x/3) ki derivative aur x/3 ki derivative subtract karte hain.
Step 2: dy/dx = 0 set karo aur x nikalo. Simplify karne se x² = 0, isliye x = 0.
Step 3: x=0 par y ki value nikalo: y(0) = 0.
Step 4: Chunke slope zero hai, tangent line horizontal hogi (y = 0), aur normal (jo tangent se perpendicular hoti hai) vertical hogi (x = 0).
Result: Point (0,0), Tangent: y=0, Normal: x=0
Q3. Vertical Tangent ka point dhoondo (x + y − eʸ = 0)
Concept: Vertical tangent wahan hoti hai jahan dy/dx undefined ho (matlab denominator zero ho jaye).
Step 1: Implicit differentiation karo — dono taraf x ke sath differentiate karo, y ko x ka function samajh kar.
Step 2: dy/dx = 1/(eʸ−1) milta hai. Ye undefined tab hoga jab eʸ − 1 = 0, matlab y = 0.
Step 3: y = 0 par x nikalo original equation se: x = 1.
Step 4: Vertical tangent: x=1, aur normal (perpendicular to vertical) horizontal hogi: y=0.
Result: Point (1,0), Tangent: x=1, Normal: y=0
Q4. Ellipse par Horizontal Tangent ke points (x² − xy + y² = 1)
Concept: Isme implicit differentiation aur horizontal tangent dono concepts mile hue hain.
Step 1: Implicit differentiation karo — xy term mein Product Rule bhi lagegi.
Step 2: dy/dx formula milta hai: (y−2x)/(2y−x)
Step 3: Horizontal tangent ke liye numerator zero karo: y − 2x = 0, isliye y = 2x.
Step 4: Ye y=2x wapas original ellipse equation mein substitute karo, x² nikal ke x ki values milti hain: x = ±1/√3.
Step 5: Corresponding y values nikalo: y = ±2/√3.
Step 6: Verify karo ke denominator (2y−x) zero na ho un points par (warna dy/dx undefined hoga, invalid point hoga).
Result: Do points milte hain: (1/√3, 2/√3) aur (−1/√3, −2/√3)
Q5. Second Derivative (y₂) nikalna — 7 alag functions
Ye sab mein pehle y₁ (first derivative) nikalte hain, phir y₁ ko dobara differentiate kar ke y₂ (second derivative) milta hai.
(i) y = x⁴−3x²+4x−5 (Simple polynomial):
Direct power rule 2 baar lagao. y₁ = 4x³−6x+4, phir y₂ = 12x²−6
(ii) y = ln(2x+3) (Log function):
y₁ = 2/(2x+3), phir isko dobara differentiate karne se y₂ = −4/(2x+3)²
(iii) y = x³e^(2x) (Product Rule + Chain Rule):
Product rule se y₁ milta hai, phir usay dobara differentiate karte hain (fir se product rule lagti hai) taake y₂ mile.
(iv) x²−y²=4 (Implicit):
Pehle implicit differentiation se y₁ = x/y milta hai. Phir y₁ ko dobara differentiate karne ke liye Quotient Rule lagti hai, aur original equation (x²−y²=4) use kar ke simplify karte hain.
Result: y₂ = −4/y³
(v) x²−xy+y²=7 (Implicit):
Isi tarah pehle y₁ nikalte hain, phir usay differentiate kar ke y₂ nikalte hain.
(vi) x=a cos t, y=b sin t (Parametric):
Parametric mein formula: y₁ = (dy/dt)/(dx/dt). Phir y₂ = [d/dt(y₁)]/(dx/dt) formula se doosri derivative nikalte hain.
Result: y₂ = −b/(a²sin³t)
(vii) x=t², y=2t³ (Parametric):
Isi parametric formula se y₁ = 3t milta hai, phir y₂ nikalte hain.
Result: y₂ = 3/(2t)
Sabaq: Parametric aur Implicit second derivatives mein hamesha pehle y₁ nikalo, phir usay dobara differentiate karo — direct doosri baar differentiate nahi karte original equation ko.
Q6. Higher-Order Derivatives (y₄, y₁₀, y₁₁) — Pattern Dhoondna
(i) y = sin(x/2):
Har baar differentiate karne se pattern repeat hota hai (sin↔cos, sign change hoti hai). 4 baar differentiate karne ke baad y₄ = (1/16)sin(x/2) milta hai.
General Pattern: yₙ = (1/2)ⁿ sin(x/2 + nπ/2) — is formula se hum bina step-by-step kiye seedha y₁₀ aur y₁₁ nikal sakte hain.
(ii) y = ln(x²−a²) = ln(x−a)+ln(x+a):
Trick: Pehle log ko 2 hisso mein split kar do (property of logs), phir har hissa alag differentiate karo. Pattern dikhta hai ke har derivative mein factorial aur negative power aati hai.
General Pattern: yₙ = (−1)ⁿ⁻¹(n−1)! [(x−a)⁻ⁿ + (x+a)⁻ⁿ]
Sabaq: Higher-order derivatives mein pehle 3-4 derivatives nikal kar dekho ke koi pattern ban raha hai, phir us pattern se general formula bana kar seedha 10th ya 11th derivative likh sakte ho — har baar step by step nikalna zaroori nahi.
Q7. Prove karo: y³(d²y/dx²) + 25 = 0 (agar x²+y²=25)
Step 1: Implicit differentiation se y₁ = −x/y milta hai.
Step 2: y₁ ko dobara differentiate karo (Quotient Rule) — y₂ mein (x²+y²) aata hai.
Step 3: Chunke x²+y² = 25 hai (given), isay substitute kar ke y₂ = −25/y³ milta hai.
Step 4: Dono taraf y³ se multiply karo: y³y₂ = −25, jo y³y₂+25=0 ban jata hai — Proved!
Q8. Prove karo: y₂ − 2ay₁ + (a²+b²)y = 0 (agar y = e^(ax)cos bx)
Step 1: Product Rule se y₁ nikalo.
Step 2: y₁ ko dobara Product Rule se differentiate kar ke y₂ nikalo.
Step 3: Teeno (y₂, y₁, y) ko given equation mein substitute karo.
Step 4: cos bx aur sin bx ke terms alag alag group karo — dono groups zero ho jate hain (algebra se check karo), isliye poori expression zero ban jati hai — Proved!
Q9. Prove karo: d²y/dx² + 2 = 0 (agar x=sin t, y=cos²t)
Step 1: dx/dt aur dy/dt nikalo.
Step 2: y₁ = (dy/dt)/(dx/dt) = −2 sin t milta hai (simplify hone ke baad).
Step 3: y₂ nikalne ke liye y₁ ko t ke sath differentiate karo, phir dx/dt se divide karo. Result −2 milta hai.
Step 4: y₂+2 = −2+2 = 0 — Proved!
Q10. Prove karo: d²y/dx² − y = 0 (agar x=ln t, y=1/t)
Step 1: dx/dt = 1/t, dy/dt = −1/t²
Step 2: y₁ = (dy/dt)/(dx/dt) = −1/t
Step 3: y₂ nikalne ke liye isi tarah dobara differentiate karo — result 1/t milta hai.
Step 4: Chunke y=1/t hai, y₂ bhi 1/t ke barabar hai, isliye y₂−y=0 — Proved!
Q11. Prove karo: x²y₂ + xy₁ + y = 0 (agar y = a cos(ln x) + b sin(ln x))
Step 1: y₁ nikalo — Chain Rule use hogi kyunke ln x andar hai.
Step 2: Dono taraf x se multiply karo taake x·y₁ = [−a sin(ln x) + b cos(ln x)] mile.
Step 3: Is naye equation ko x ke sath differentiate karo (LHS mein Product Rule, RHS mein Chain Rule).
Step 4: Dono sides equate kar ke simplify karo — end mein x²y₂+xy₁+y = 0 ban jata hai — Proved!
Is Exercise ke Important Concepts:
- Tangent/Normal Lines: Slope (derivative) nikalo → point ka coordinate nikalo → point-slope formula lagao.
- Horizontal Tangent: dy/dx = 0 rakho aur solve karo.
- Vertical Tangent: dy/dx undefined ho (denominator = 0) — wahan solve karo.
- Second Derivative (Implicit/Parametric): Pehle y₁ nikalo, phir usay dobara differentiate karo — original equation ko dobara differentiate mat karo.
- Higher-order derivative patterns: Pehle 3-4 derivatives nikal kar pattern dhoondo, phir general formula bana lo.
- “Show that” / “Prove” questions: In sab mein hamesha given equation ko differentiate karo, phir required expression mein substitute kar ke simplify karo taake zero ya required result mile.