Class 12 Maths 2 — Ch 1 Step-by-Step Answers

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🔢 Formulas Ex 1.1 Ex 1.2 Ex 1.3 Ex 1.4 Ex 1.5

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Quick Reference

All Formulas — Chapter 1 Differentiation

⛓️ Chain Rule & Standard

d/dx[f(g(x))] = f'(g(x)) · g'(x)
d/dx(xⁿ) = nxⁿ⁻¹ | d/dx(eˣ) = eˣ
d/dx(aˣ) = aˣ log a | d/dx(log x) = 1/x
d/dx(sin x) = cos x
d/dx(cos x) = −sin x
d/dx(tan x) = sec²x
d/dx(cot x) = −cosec²x
d/dx(sec x) = sec x·tan x
d/dx(cosec x) = −cosec x·cot x

🔄 Inverse Trig Derivatives

d/dx(sin⁻¹x) = 1/√(1−x²)
d/dx(cos⁻¹x) = −1/√(1−x²)
d/dx(tan⁻¹x) = 1/(1+x²)
d/dx(cot⁻¹x) = −1/(1+x²)
d/dx(sec⁻¹x) = 1/(x√(x²−1)) for x>1
d/dx(cosec⁻¹x) = −1/(x√(x²−1)) for x>1
dx/dy = 1/(dy/dx) (inverse function theorem)

📝 Key Substitutions

√(1−x²) → put x = sinθ or cosθ
√(1+x²) → put x = tanθ or cotθ
√(x²−1) → put x = secθ
2x/(1+x²) → x=tanθ → sin 2θ
(1−x²)/(1+x²) → x=tanθ → cos 2θ
3x−4x³ → x=sinθ → sin 3θ
4x³−3x → x=cosθ → cos 3θ
tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)]

📐 Parametric & Higher Order

dy/dx = (dy/dt)/(dx/dt)
d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt)
nth deriv of sin(ax+b) = aⁿ sin(ax+b+nπ/2)
nth deriv of cos(ax+b) = aⁿ cos(ax+b+nπ/2)
nth deriv of eᵃˣ = aⁿ eᵃˣ
nth deriv of log x = (−1)ⁿ⁻¹(n−1)!/xⁿ
nth deriv of 1/(ax+b) = (−1)ⁿ·n!·aⁿ/(ax+b)ⁿ⁺¹

Exercise 1.1 — Chain Rule (Step-by-Step)

Differentiate the following w.r.t. x

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Ex 1.1 — Q1

Differentiate w.r.t. x (Chain Rule)

Q1(i) — (x³ − 2x − 1)⁵

Step 1: Let y = (x³ − 2x − 1)⁵ Step 2: Let u = x³ − 2x − 1, so y = u⁵ Step 3: dy/dx = dy/du · du/dx = 5u⁴ · (3x² − 2) Step 4: Substitute back u: dy/dx = 5(x³ − 2x − 1)⁴ · (3x² − 2)

Q1(iii) — √(x² + 4x − 7)

Step 1: Let y = (x² + 4x − 7)^(1/2) Step 2: dy/dx = (1/2)(x² + 4x − 7)^(−1/2) · d/dx(x² + 4x − 7) Step 3: d/dx(x² + 4x − 7) = 2x + 4 Step 4: dy/dx = (2x + 4) / [2√(x² + 4x − 7)] dy/dx = (x + 2) / √(x² + 4x − 7)

Q1(iv) — √(x² + √(x²+1))

Step 1: Let y = [x² + √(x²+1)]^(1/2) Step 2: dy/dx = (1/2)[x² + √(x²+1)]^(−1/2) · d/dx[x² + √(x²+1)] Step 3: d/dx[√(x²+1)] = x/√(x²+1) Step 4: d/dx[x² + √(x²+1)] = 2x + x/√(x²+1) dy/dx = [2x + x/√(x²+1)] / [2√(x² + √(x²+1))]

Q2 — Important Step-by-Step

Q2(iii) — log[tan(x/2)]

Formula: d/dx[log f(x)] = f'(x)/f(x)

Step 1: dy/dx = [1/tan(x/2)] · d/dx[tan(x/2)] Step 2: d/dx[tan(x/2)] = sec²(x/2) · (1/2) Step 3: dy/dx = [1/tan(x/2)] · sec²(x/2)/2 = cos(x/2)/sin(x/2) · 1/(2cos²(x/2)) = 1/[2sin(x/2)cos(x/2)] = 1/sin x dy/dx = cosec x

Q2(viii) — log[cos(x³ − 5)]

Step 1: dy/dx = [1/cos(x³−5)] · d/dx[cos(x³−5)] Step 2: d/dx[cos(x³−5)] = −sin(x³−5) · 3x² Step 3: dy/dx = −3x²sin(x³−5)/cos(x³−5) dy/dx = −3x² tan(x³ − 5)

Q2(ix) — e^(3sin²x − 2cos²x)

Step 1: Let u = 3sin²x − 2cos²x, y = eᵘ Step 2: du/dx = 6sinx cosx − 2·2cosx·(−sinx) = 6sinx cosx + 4sinx cosx = 10 sinx cosx = 5sin2x Step 3: dy/dx = eᵘ · du/dx dy/dx = 5sin2x · e^(3sin²x − 2cos²x)

Q2(x) — cos²[log(x²+7)]

Step 1: Let u = log(x²+7), y = cos²u Step 2: dy/du = 2cosu·(−sinu) = −sin2u Step 3: du/dx = 2x/(x²+7) Step 4: dy/dx = −sin2u · 2x/(x²+7) = −sin[2log(x²+7)] · 2x/(x²+7) dy/dx = −2x sin[2log(x²+7)] / (x²+7)

Q2(xiii) — e^[log(logx)² − log x²]

Step 1: Simplify the exponent: log(logx)² − log x² = 2log(logx) − 2logx = log[(logx)²] − log(x²) = log[(logx)²/x²] Step 2: So y = e^[log((logx/x)²)] = (logx/x)² = (logx)²/x² Step 3: dy/dx = d/dx[(logx)²/x²] using quotient rule: = [x²·2logx·(1/x) − (logx)²·2x] / x⁴ = [2x logx − 2x(logx)²] / x⁴ = 2logx(1 − logx)/x³ dy/dx = 2logx(1 − logx) / x³

Q3 — Selected Step-by-Step

Q3(i) — (x²+4x+1)³ + (x³−5x−2)⁴

Step 1: Differentiate each term separately using chain rule: Step 2: d/dx[(x²+4x+1)³] = 3(x²+4x+1)² · (2x+4) Step 3: d/dx[(x³−5x−2)⁴] = 4(x³−5x−2)³ · (3x²−5) dy/dx = 3(x²+4x+1)²(2x+4) + 4(x³−5x−2)³(3x²−5)

Q3(vii) — log(sec 3x + tan 3x)

Step 1: dy/dx = [1/(sec3x+tan3x)] · d/dx(sec3x+tan3x) Step 2: d/dx(sec3x) = 3sec3x·tan3x Step 3: d/dx(tan3x) = 3sec²3x Step 4: dy/dx = [3sec3x(tan3x+sec3x)] / (sec3x+tan3x) dy/dx = 3 sec 3x

Q4 — Using Given Table of Values

Given Table: x=2: f=1,g=6,f'=−3,g'=4 | x=4: f=3,g=4,f'=5,g'=−6 | x=6: f=5,g=2,f'=−4,g'=7 (i) r(x)=f[g(x)], find r'(2) By chain rule: r'(x) = f'[g(x)] · g'(x) r'(2) = f'[g(2)] · g'(2) = f'(6) · 4 = (−4)(4) r'(2) = −16 (ii) R(x)=g[3+f(x)], find R'(4) R'(x) = g'[3+f(x)] · f'(x) R'(4) = g'[3+f(4)] · f'(4) = g'[3+3] · 5 = g'(6) · 5 = 7 · 5 R'(4) = 35 (iii) s(x)=f[9−f(x)], find s'(4) s'(x) = f'[9−f(x)] · (−f'(x)) s'(4) = f'[9−f(4)] · (−f'(4)) = f'[9−3] · (−5) = f'(6) · (−5) = (−4)(−5) s'(4) = 20 (iv) S(x)=g[g(x)], find S'(6) S'(x) = g'[g(x)] · g'(x) S'(6) = g'[g(6)] · g'(6) = g'(2) · 7 = 4 · 7 S'(6) = 28

Q5 — f'(3)=−1, g'(2)=5, g(2)=3, y=f[g(x)]

Step 1: By chain rule: dy/dx = f'[g(x)] · g'(x) Step 2: At x=2: dy/dx = f'[g(2)] · g'(2) Step 3: = f'(3) · 5 = (−1)(5) dy/dx at x=2 = −5

Q6 — h(x)=√(4f(x)+3g(x)), find h'(1)

Step 1: h(x) = [4f(x)+3g(x)]^(1/2) Step 2: h'(x) = (1/2)[4f(x)+3g(x)]^(−1/2) · [4f'(x)+3g'(x)] Step 3: At x=1: f(1)=4, g(1)=3, f'(1)=3, g'(1)=4 4f(1)+3g(1) = 16+9 = 25; √25 = 5 4f'(1)+3g'(1) = 12+12 = 24 Step 4: h'(1) = (1/2) · (1/5) · 24 = 24/10 h'(1) = 12/5

Q7 — y=sin2x−2sinx, find x where dy/dx=0 in [0,2π)

Step 1: dy/dx = 2cos2x − 2cosx = 0 Step 2: 2(2cos²x−1) − 2cosx = 0 4cos²x − 2cosx − 2 = 0 Step 3: Divide by 2: 2cos²x − cosx − 1 = 0 Step 4: Factorise: (2cosx+1)(cosx−1) = 0 Step 5: cosx = 1 → x = 0 cosx = −1/2 → x = 2π/3 and x = 4π/3 x = 0, 2π/3, 4π/3

Exercise 1.2 — Inverse Functions & Inverse Trig (Step-by-Step)

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Ex 1.2

Inverse Function Derivatives — Step by Step

Key Theorem: dx/dy = 1/(dy/dx)

Q1(i) — y = √x

Step 1: Find inverse: y = √x → x = y² → x = f⁻¹(y) = y² Step 2: dx/dy = 2y Step 3: dy/dx = 1/(dx/dy) = 1/(2y) = 1/(2√x) dy/dx = 1/(2√x)

Q1(iii) — y = ∛(x−2)

Step 1: y³ = x−2 → x = y³+2 → f⁻¹(y) = y³+2 Step 2: dx/dy = 3y² Step 3: dy/dx = 1/(3y²) = 1/[3(∛(x−2))²] dy/dx = 1/[3(x−2)^(2/3)]

Q1(iv) — y = log(2x−1)

Step 1: y = log(2x−1) → eʸ = 2x−1 → x = (eʸ+1)/2 Step 2: dx/dy = eʸ/2 Step 3: dy/dx = 2/eʸ = 2/e^(log(2x−1)) = 2/(2x−1) dy/dx = 2/(2x−1)

Q3(i) — y=x⁵+2x³+3x, find (f⁻¹)' at x=1

Step 1: dy/dx = 5x⁴ + 6x² + 3 Step 2: At x=1: dy/dx = 5+6+3 = 14 Step 3: (f⁻¹)'(y) = 1/(dy/dx) = 1/14 At x=1: y = 1+2+3 = 6 (f⁻¹)'(6) = 1/14

Q4 — f(x)=x³+x−2, find (f⁻¹)'(0)

Step 1: Find x where f(x)=0: x³+x−2=0 → (x−1)(x²+x+2)=0 → x=1 Step 2: f'(x) = 3x²+1 Step 3: f'(1) = 3+1 = 4 Step 4: (f⁻¹)'(0) = 1/f'(1) = 1/4 (f⁻¹)'(0) = 1/4

Q6 — Inverse Trig Composite Differentiation

Q6(i) — tan⁻¹(log x)

Step 1: dy/dx = [1/(1+(logx)²)] · d/dx(logx) Step 2: d/dx(logx) = 1/x dy/dx = 1/[x(1+log²x)]

Q6(iii) — cot⁻¹(x³)

Step 1: dy/dx = [−1/(1+(x³)²)] · d/dx(x³) Step 2: d/dx(x³) = 3x² dy/dx = −3x²/(1+x⁶)

Q6(v) — tan⁻¹(√x)

Step 1: dy/dx = [1/(1+(√x)²)] · d/dx(√x) Step 2: 1/(1+x) · 1/(2√x) dy/dx = 1/[2√x(1+x)]

Q6(ix) — cos³[cos⁻¹(x³)]

Step 1: Simplify: cos[cos⁻¹(x³)] = x³ Step 2: So y = (x³)³ = x⁹ Step 3: dy/dx = 9x⁸ dy/dx = 9x⁸

Q7 — Simplify using Identities (Step-by-Step)

Q7(iii) — cos⁻¹[√((1+cosx)/2)]

Identity: (1+cos x)/2 = cos²(x/2)

Step 1: √((1+cosx)/2) = √(cos²(x/2)) = cos(x/2) Step 2: y = cos⁻¹[cos(x/2)] = x/2 Step 3: dy/dx = d/dx(x/2) dy/dx = 1/2

Q7(v) — tan⁻¹[(1−tan(x/2))/(1+tan(x/2))]

Identity: (1−tanA)/(1+tanA) = tan(π/4−A)

Step 1: y = tan⁻¹[tan(π/4 − x/2)] Step 2: y = π/4 − x/2 Step 3: dy/dx = d/dx(π/4 − x/2) = 0 − 1/2 dy/dx = −1/2

Q7(viii) — cot⁻¹[sin3x/(1+cos3x)]

Identity: sinθ/(1+cosθ) = tan(θ/2)

Step 1: sin3x/(1+cos3x) = tan(3x/2) Step 2: y = cot⁻¹[tan(3x/2)] = cot⁻¹[cot(π/2−3x/2)] Step 3: y = π/2 − 3x/2 Step 4: dy/dx = d/dx(π/2 − 3x/2) dy/dx = −3/2

Q9 — Substitution Method (Step-by-Step)

Q9(i) — cos⁻¹[(1−x²)/(1+x²)]

Step 1: Put x = tanθ → θ = tan⁻¹x Step 2: (1−tan²θ)/(1+tan²θ) = cos2θ Step 3: y = cos⁻¹(cos2θ) = 2θ = 2tan⁻¹x Step 4: dy/dx = 2 · d/dx(tan⁻¹x) = 2/(1+x²) dy/dx = 2/(1+x²)

Q9(ii) — tan⁻¹[2x/(1−x²)]

Step 1: Put x = tanθ → θ = tan⁻¹x Step 2: 2tanθ/(1−tan²θ) = tan2θ Step 3: y = tan⁻¹(tan2θ) = 2θ = 2tan⁻¹x Step 4: dy/dx = 2/(1+x²) dy/dx = 2/(1+x²)

Q9(iv) — sin⁻¹(2x√(1−x²))

Step 1: Put x = sinθ → θ = sin⁻¹x Step 2: 2sinθ√(1−sin²θ) = 2sinθ cosθ = sin2θ Step 3: y = sin⁻¹(sin2θ) = 2θ = 2sin⁻¹x Step 4: dy/dx = 2/√(1−x²) dy/dx = 2/√(1−x²)

Q9(v) — cos⁻¹(3x−4x³)

Step 1: Put x = cosθ → θ = cos⁻¹x Step 2: 3cosθ − 4cos³θ = −(4cos³θ−3cosθ) = −cos3θ Step 3: y = cos⁻¹(−cos3θ) = π − cos⁻¹(cos3θ) = π − 3θ Step 4: y = π − 3cos⁻¹x Step 5: dy/dx = −3·(−1/√(1−x²)) dy/dx = 3/√(1−x²)

Q10 — Split as Sum of tan⁻¹ (Step-by-Step)

Q10(i) — tan⁻¹[8x/(1−15x²)]

Step 1: Write 8x = (5x+3x), 15x² = (5x)(3x) Step 2: tan⁻¹[(5x+3x)/(1−(5x)(3x))] = tan⁻¹(5x) + tan⁻¹(3x) [Using: tan⁻¹a+tan⁻¹b = tan⁻¹((a+b)/(1−ab))] Step 3: y = tan⁻¹(5x) + tan⁻¹(3x) Step 4: dy/dx = 5/(1+25x²) + 3/(1+9x²) dy/dx = 5/(1+25x²) + 3/(1+9x²)

Q10(vii) — tan⁻¹[(a+b tanx)/(b−a tanx)]

Identity: tan(A+B) = (tanA+tanB)/(1−tanA tanB)

Step 1: Divide numerator & denominator by b: = tan⁻¹[(a/b + tanx)/(1 − (a/b)tanx)] Step 2: = tan⁻¹[tan(tan⁻¹(a/b) + x)] Step 3: y = tan⁻¹(a/b) + x Step 4: dy/dx = 0 + 1 dy/dx = 1

Exercise 1.3 — Logarithmic & Implicit Differentiation

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Ex 1.3

Logarithmic Differentiation + Implicit Functions

Q1 — Logarithmic Differentiation

Q1(i) — (x+1)²/[(x+2)³(x+3)⁴]

Step 1: Take log both sides: log y = 2log(x+1) − 3log(x+2) − 4log(x+3) Step 2: Differentiate w.r.t. x: (1/y)(dy/dx) = 2/(x+1) − 3/(x+2) − 4/(x+3) Step 3: dy/dx = y × [2/(x+1) − 3/(x+2) − 4/(x+3)] dy/dx = [(x+1)²/(x+2)³(x+3)⁴] × [2/(x+1) − 3/(x+2) − 4/(x+3)]

Q1(vi) — x^(tan⁻¹x)

Step 1: Take log: log y = tan⁻¹x · logx Step 2: Differentiate: (1/y)(dy/dx) = tan⁻¹x · (1/x) + logx · 1/(1+x²) Step 3: dy/dx = y [tan⁻¹x/x + logx/(1+x²)] dy/dx = x^(tan⁻¹x) [tan⁻¹x/x + logx/(1+x²)]

Q1(vii) — (sinx)ˣ

Step 1: log y = x · log(sinx) Step 2: (1/y)(dy/dx) = x · cosx/sinx + log(sinx) · 1 = x cotx + log(sinx) Step 3: dy/dx = y[x cotx + log(sinx)] dy/dx = (sinx)ˣ [x cotx + log(sinx)]

Q2 — Mixed Functions

Q2(i) — y = xᵉ + xˣ + eˣ + eᵉ

Step 1: Split: let u=xᵉ, v=xˣ, w=eˣ, k=eᵉ (constant) Step 2: d/dx(xᵉ) = e·xᵉ⁻¹ (standard power rule) Step 3: For xˣ: log v = x logx → dv/dx = xˣ(1+logx) Step 4: d/dx(eˣ) = eˣ | d/dx(eᵉ) = 0 dy/dx = e·xᵉ⁻¹ + xˣ(1+logx) + eˣ

Q2(vi) — (sinx)^(tanx) + (cosx)^(cotx)

Step 1: Let u=(sinx)^(tanx), v=(cosx)^(cotx) Step 2: For u: logu = tanx·log(sinx) (1/u)du/dx = sec²x·log(sinx) + tanx·cosx/sinx = sec²x·log(sinx)+1 du/dx = (sinx)^(tanx)[sec²x·log(sinx)+1] Step 3: For v: logv = cotx·log(cosx) (1/v)dv/dx = −cosec²x·log(cosx) + cotx·(−sinx)/cosx = −cosec²x·log(cosx) − 1 dv/dx = (cosx)^(cotx)[−cosec²x·log(cosx)−1] dy/dx = (sinx)^(tanx)[sec²x·log(sinx)+1] + (cosx)^(cotx)[−cosec²x·log(cosx)−1]

Q3 — Implicit Differentiation (Step-by-Step)

Q3(i) — √x + √y = √a

Step 1: Differentiate both sides w.r.t. x: 1/(2√x) + (1/(2√y)) · dy/dx = 0 Step 2: (1/(2√y)) · dy/dx = −1/(2√x) Step 3: dy/dx = −√y/√x dy/dx = −√(y/x)

Q3(iv) — x³+x²y+xy²+y³ = 81

Step 1: Differentiate each term w.r.t. x: 3x² + (x²·dy/dx + y·2x) + (x·2y·dy/dx + y²) + 3y²·dy/dx = 0 Step 2: Collect dy/dx terms: (x² + 2xy + 3y²)dy/dx = −(3x² + 2xy + y²) Step 3: dy/dx = −(3x²+2xy+y²)/(x²+2xy+3y²) dy/dx = −(3x² + 2xy + y²)/(x² + 2xy + 3y²)

Q3(vi) — xeʸ + yeˣ = 1

Step 1: Differentiate: eʸ + x·eʸ·dy/dx + eˣ·dy/dx + y·eˣ = 0 Step 2: (x·eʸ + eˣ)·dy/dx = −(eʸ + y·eˣ) dy/dx = −(eʸ + yeˣ)/(xeʸ + eˣ)

Q3(viii) — cos(xy) = x+y

Step 1: Differentiate: −sin(xy)·(y + x·dy/dx) = 1 + dy/dx Step 2: −y·sin(xy) − x·sin(xy)·dy/dx = 1 + dy/dx Step 3: −x·sin(xy)·dy/dx − dy/dx = 1 + y·sin(xy) Step 4: dy/dx·[−x·sin(xy) − 1] = 1 + y·sin(xy) dy/dx = −(1 + y·sin(xy))/(1 + x·sin(xy))

Q5 — Important Proofs (Key Steps)

Q5(vii) — y=√(cosx+√(cosx+...∞)), show dy/dx=sinx/(1−2y)

Step 1: Since series continues: y = √(cosx + y) Step 2: Square both sides: y² = cosx + y Step 3: Differentiate: 2y·dy/dx = −sinx + dy/dx Step 4: (2y−1)·dy/dx = −sinx dy/dx = −sinx/(2y−1) = sinx/(1−2y) ✓

Q5(ix) — y=x^(x^(x...∞)), show dy/dx=y²/[x(1−y logy)]

Step 1: Since series continues: y = xʸ Step 2: Take log: log y = y·logx Step 3: Differentiate: (1/y)·dy/dx = y·(1/x) + logx·dy/dx Step 4: (1/y − logx)·dy/dx = y/x Step 5: dy/dx = y²/[x(1 − y·logx)] ... but logx = (logy)/y from Step 2 (1/y − logy/y)·dy/dx = y/x → (1−logy)/y·dy/dx = y/x dy/dx = y²/[x(1 − logy)] ✓

Exercise 1.4 — Parametric Differentiation (Step-by-Step)

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Ex 1.4

dy/dx = (dy/dt) ÷ (dx/dt)

Q1(i) — x=at², y=2at

Step 1: Differentiate y=2at w.r.t. t: dy/dt = 2a Step 2: Differentiate x=at² w.r.t. t: dx/dt = 2at Step 3: dy/dx = (dy/dt)/(dx/dt) = 2a/(2at) dy/dx = 1/t

Q1(ii) — x=acotθ, y=bcosecθ

Step 1: dy/dθ = −b cosecθ cotθ Step 2: dx/dθ = −a cosec²θ Step 3: dy/dx = (−b cosecθ cotθ)/(−a cosec²θ) = b cotθ/(a cosecθ) = (b/a)·(cosθ/sinθ)·sinθ = (b/a)cosθ dy/dx = (b cosθ)/a

Q1(iv) — x=sinθ, y=tanθ

Step 1: dy/dθ = sec²θ Step 2: dx/dθ = cosθ Step 3: dy/dx = sec²θ/cosθ = 1/(cos³θ) dy/dx = sec³θ

Q1(v) — x=a(1−cosθ), y=b(θ−sinθ)

Step 1: dy/dθ = b(1−cosθ) Step 2: dx/dθ = a sinθ Step 3: dy/dx = b(1−cosθ)/(a sinθ) = b·2sin²(θ/2) / [a·2sin(θ/2)cos(θ/2)] = (b/a)·tan(θ/2) dy/dx = (b/a) tan(θ/2)

Q2 — Find dy/dx at Given Value

Q2(i) — x=cosec²θ, y=cot³θ, at θ=π/6

Step 1: dy/dθ = 3cot²θ·(−cosec²θ) = −3cot²θ cosec²θ Step 2: dx/dθ = 2cosecθ·(−cosecθ cotθ) = −2cosec²θ cotθ Step 3: dy/dx = −3cot²θ cosec²θ / (−2cosec²θ cotθ) = (3/2)cotθ Step 4: At θ=π/6: cotθ = cot(π/6) = √3 dy/dx = (3/2)·√3 dy/dx at θ=π/6 = 3√3/2

Q2(ii) — x=acos³θ, y=asin³θ, at θ=π/3

Step 1: dy/dθ = 3asin²θ cosθ Step 2: dx/dθ = −3acos²θ sinθ Step 3: dy/dx = 3asin²θ cosθ/(−3acos²θ sinθ) = −tanθ Step 4: At θ=π/3: tanθ=√3 dy/dx at θ=π/3 = −√3

Q4 — Differentiate one function w.r.t. another

Method: du/dv = (du/dx) ÷ (dv/dx)

Q4(ii) — sin⁻¹[2x/(1+x²)] w.r.t. cos⁻¹[(1−x²)/(1+x²)]

Step 1: Put x=tanθ: u = sin⁻¹[2tanθ/(1+tan²θ)] = sin⁻¹(sin2θ) = 2θ = 2tan⁻¹x Step 2: v = cos⁻¹[(1−tan²θ)/(1+tan²θ)] = cos⁻¹(cos2θ) = 2θ = 2tan⁻¹x Step 3: du/dx = 2/(1+x²) and dv/dx = 2/(1+x²) Step 4: du/dv = (du/dx)/(dv/dx) = 1 du/dv = 1

Q4(iv) — cos⁻¹[(1−x²)/(1+x²)] w.r.t. tan⁻¹x

Step 1: From Q9(i) of Ex1.2: u = cos⁻¹[(1−x²)/(1+x²)] = 2tan⁻¹x Step 2: du/dx = 2/(1+x²) Step 3: v = tan⁻¹x → dv/dx = 1/(1+x²) Step 4: du/dv = [2/(1+x²)] / [1/(1+x²)] = 2 du/dv = 2

Q4(viii) — tan⁻¹[(√(1+x²)−1)/x] w.r.t. tan⁻¹[2x√(1−x²)/(1−2x²)]

Step 1: For u: put x=tanθ → u=(1/2)tan⁻¹x; du/dx = 1/[2(1+x²)] Step 2: For v: put x=sinθ → 2sinθcosθ/(1−2sin²θ)=sin2θ/(cos2θ)=tan2θ v = tan⁻¹(tan2θ) = 2θ = 2sin⁻¹x; dv/dx = 2/√(1−x²) Step 3: du/dv = [1/(2(1+x²))] / [2/√(1−x²)] = √(1−x²)/[4(1+x²)] du/dv = √(1−x²) / [4(1+x²)]

Exercise 1.5 — Higher Order Derivatives (Step-by-Step)

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Ex 1.5

Second Order & nth Order Derivatives

Q1 — Second Order Derivatives

Q1(ii) — y = e²ˣ·tanx

Step 1: dy/dx = e²ˣ·sec²x + tanx·2e²ˣ = e²ˣ(sec²x + 2tanx) Step 2: d²y/dx² = d/dx[e²ˣ(sec²x+2tanx)] Step 3: = e²ˣ·d/dx(sec²x+2tanx) + (sec²x+2tanx)·2e²ˣ Step 4: d/dx(sec²x+2tanx) = 2secx·secx tanx + 2sec²x = 2sec²x tanx + 2sec²x Step 5: d²y/dx² = e²ˣ(2sec²x tanx+2sec²x) + 2e²ˣ(sec²x+2tanx) = e²ˣ[2sec²x tanx+2sec²x+2sec²x+4tanx] d²y/dx² = e²ˣ(2sec²x tanx + 4sec²x + 4tanx)

Q1(iii) — y = e⁴ˣ·cos5x

Step 1: dy/dx = e⁴ˣ(−5sin5x) + cos5x·4e⁴ˣ = e⁴ˣ(4cos5x−5sin5x) Step 2: d²y/dx² = e⁴ˣ·d/dx(4cos5x−5sin5x) + (4cos5x−5sin5x)·4e⁴ˣ Step 3: d/dx(4cos5x−5sin5x) = −20sin5x − 25cos5x Step 4: d²y/dx² = e⁴ˣ(−20sin5x−25cos5x) + 4e⁴ˣ(4cos5x−5sin5x) = e⁴ˣ[−20sin5x−25cos5x+16cos5x−20sin5x] = e⁴ˣ[−40sin5x−9cos5x] d²y/dx² = −e⁴ˣ(9cos5x + 40sin5x)

Q1(iv) — y = x³ logx

Step 1: dy/dx = x³·(1/x) + logx·3x² = x² + 3x²logx = x²(1+3logx) Step 2: d²y/dx² = d/dx[x²(1+3logx)] = x²·(3/x) + (1+3logx)·2x = 3x + 2x + 6x logx = 5x + 6x logx = x(5+6logx) d²y/dx² = x(5 + 6logx)

Q1(v) — y = log(logx)

Step 1: dy/dx = [1/logx]·(1/x) = 1/(x logx) Step 2: d²y/dx² = d/dx[1/(x logx)] = d/dx[(x logx)⁻¹] Step 3: = −1/(x logx)² · d/dx(x logx) d/dx(x logx) = x·(1/x)+logx·1 = 1+logx Step 4: d²y/dx² = −(1+logx)/(x logx)² d²y/dx² = −(1+logx) / (x logx)²

Q2 — d²y/dx² for Parametric Equations

Q2(ii) — x=2at², y=4at

Step 1: dy/dt = 4a; dx/dt = 4at → dy/dx = 4a/4at = 1/t Step 2: d²y/dx² = d/dx(1/t) = [d/dt(1/t)] ÷ (dx/dt) = (−1/t²) ÷ 4at = −1/(4at³) d²y/dx² = −1/(4at³)

Q2(iv) — x=acosθ, y=bsinθ at θ=π/4

Step 1: dy/dθ = bcosθ; dx/dθ = −asinθ dy/dx = −(b/a)cotθ Step 2: d²y/dx² = [d/dθ(−b/a·cotθ)] ÷ (dx/dθ) = (b/a·cosec²θ) ÷ (−asinθ) = −b cosec²θ/(a²sinθ) = −b/(a²sin³θ) Step 3: At θ=π/4: sin(π/4)=1/√2, sin³(π/4)=(1/√2)³=1/(2√2) d²y/dx² = −b/[a²·1/(2√2)] = −2√2b/a² d²y/dx² at θ=π/4 = −2√2b/a²

Q3 — Prove that (Key Proofs with Steps)

Q3(ii) — y=e^(m·tan⁻¹x), prove (1+x²)y₂+(2x−m)y₁=0

Step 1: y₁ = e^(m·tan⁻¹x) · m/(1+x²) = my/(1+x²) Step 2: (1+x²)y₁ = my ... (I) Step 3: Differentiate (I) w.r.t. x: (1+x²)y₂ + 2x·y₁ = m·y₁ Step 4: (1+x²)y₂ + 2x·y₁ − m·y₁ = 0 (1+x²)y₂ + (2x−m)y₁ = 0 ✓

Q3(v) — y=eᵃˣsin(bx), prove y₂−2ay₁+(a²+b²)y=0

Step 1: y₁ = eᵃˣ(b cosbx + a sinbx) Step 2: y₂ = eᵃˣ(−b²sinbx+ab cosbx) + a·eᵃˣ(b cosbx+a sinbx) = eᵃˣ[(a²−b²)sinbx + 2ab cosbx] Step 3: y₂ − 2ay₁ = eᵃˣ[(a²−b²)sinbx+2ab cosbx] − 2a·eᵃˣ(b cosbx+a sinbx) = eᵃˣ[(a²−b²−2a²)sinbx + (2ab−2ab)cosbx] = eᵃˣ(−b²)sinbx = −b²y Step 4: y₂ − 2ay₁ + b²y = 0 → y₂ − 2ay₁ + (a²+b²)y = a²y−a²y = 0 y₂ − 2ay₁ + (a²+b²)y = 0 ✓

Q4 — nth Derivative Formulas

Standard nth derivative results (use step-pattern method): (iii) eᵃˣ⁺ᵇ: y' = a·eᵃˣ⁺ᵇ | y'' = a²·eᵃˣ⁺ᵇ | y''' = a³·eᵃˣ⁺ᵇ dⁿy/dxⁿ = aⁿ · eᵃˣ⁺ᵇ (v) log(ax+b): y' = a/(ax+b) | y'' = −a²/(ax+b)² | y''' = 2a³/(ax+b)³ Pattern: (−1)ⁿ⁻¹·(n−1)!·aⁿ dⁿy/dxⁿ = (−1)ⁿ⁻¹ · (n−1)! · aⁿ / (ax+b)ⁿ (vi) cosx: y' = −sinx = cos(x+π/2) | y'' = cos(x+π) | y''' = cos(x+3π/2) dⁿy/dxⁿ = cos(x + nπ/2) (vii) sin(ax+b): y' = a cos(ax+b) = a sin(ax+b+π/2) y'' = a² sin(ax+b+π) | y''' = a³ sin(ax+b+3π/2) dⁿy/dxⁿ = aⁿ · sin(ax + b + nπ/2) (xi) eᵃˣ·cos(bx+c): Let r = √(a²+b²), α = tan⁻¹(b/a) y' = eᵃˣ(a cosbx − b sinbx + ...) = eᵃˣ·r·cos(bx+c+α) dⁿy/dxⁿ = eᵃˣ · (a²+b²)^(n/2) · cos(bx + c + n·tan⁻¹(b/a)) (xii) e⁸ˣ·cos(6x+7): a=8, b=6 → √(a²+b²) = √(64+36) = √100 = 10 α = tan⁻¹(6/8) = tan⁻¹(3/4) dⁿy/dxⁿ = e⁸ˣ · 10ⁿ · cos(6x + 7 + n·tan⁻¹(3/4))

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