Solutions — Differentiating Composite Log Functions and Applications

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1. Show Solution
   f(x) = 3x + 1, f′(x) = 3
   dy/dx = 3/(3x + 1)
2. Show Solution
   Log law: ln((x+1)²) = 2 ln(x+1)
   dy/dx = 2 × 1/(x+1) = 2/(x+1)
   
   Direct chain rule check: f(x) = (x+1)², f′(x) = 2(x+1)
   derivative = 2(x+1)/(x+1)² = 2/(x+1) ✓
3. Show Solution
   y = ln(x²) − ln(x+1) = 2 ln x − ln(x+1)
   dy/dx = 2/x − 1/(x+1)
   
   Combined: [2(x+1) − x] / [x(x+1)] = (x+2) / [x(x+1)]
4. Show Solution
   f(x) = e^x + 1, f′(x) = e^x
   dy/dx = e^x/(e^x + 1)
   
   Note: as x → ∞, dy/dx → 1 (since e^x/(e^x+1) → 1).
5. Show Solution
   At x = 1: y = ln(2 × 1) = ln 2
   dy/dx = 2/(2x) = 1/x; at x = 1: gradient m = 1
   Tangent: y − ln 2 = 1(x − 1)
   y = x − 1 + ln 2
6. Show Solution
   Domain: x > 0
   dy/dx = 2x − 1/x
   Set = 0: 2x = 1/x ⇒ 2x² = 1 ⇒ x² = 1/2 ⇒ x = 1/√2 (positive root only)
   y = (1/√2)² − ln(1/√2) = 1/2 − (−½ ln 2) = 1/2 + ½ ln 2
   d²y/dx² = 2 + 1/x²; at x = 1/√2 this is positive
   Local minimum at (1/√2, ½ + ½ ln 2) ≈ (0.707, 0.847)
7. Show Solution
   N = 500e^−0.1t
   dN/dt = 500 × (−0.1)e^−0.1t = −50e^−0.1t
   At t = 10: dN/dt = −50e⁻¹ = −50/e ≈ −18.4 per unit time
   
   This means the population is decreasing at approximately 18.4 individuals per unit time when t = 10.
8. Show Solution
   y = ln(√(x²+4)) = ln((x²+4)^1/2) = ½ ln(x²+4)
   dy/dx = ½ × 2x/(x²+4) = x/(x²+4)
9. Show Solution
   At x = 1: y = ln(1+1) = ln 2
   dy/dx = 2x/(x²+1); at x = 1: tangent gradient = 2/2 = 1
   Normal gradient = −1/1 = −1
   Normal equation: y − ln 2 = −1(x − 1)
   y = −x + 1 + ln 2
10. Show Solution
    When t = 20, N = 0.6N₀:
    0.6N₀ = N₀e^−20k
    0.6 = e^−20k
    ln(0.6) = −20k
    k = −ln(0.6)/20 = ln(5/3)/20 ≈ 0.0255
    
    Half-life: set N = N₀/2:
    0.5 = e^−kt ⇒ t = ln(2)/k = ln(2)/0.0255 ≈ 27.2 years