Practice Maths

Solutions — The Natural Logarithm and Differentiating ln(x)

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  1. f(x) = 2x, f′(x) = 2
    d/dx[ln(2x)] = f′(x)/f(x) = 2/(2x) = 1/x

    Alternative: ln(2x) = ln 2 + ln x, so d/dx = 0 + 1/x = 1/x.
  2. Use log law: ln(x3) = 3 ln x
    d/dx[3 ln x] = 3 × (1/x) = 3/x

    (Using chain rule directly: f(x) = x3, f′(x) = 3x²; derivative = 3x²/x3 = 3/x — same result.)
  3. f(x) = x² + 4, f′(x) = 2x
    d/dx[ln(x² + 4)] = 2x/(x² + 4)

    Note: x² + 4 > 0 for all real x, so domain is all real numbers.
  4. f(x) = sin x, f′(x) = cos x
    d/dx[ln(sin x)] = cos x / sin x = cot x

    Domain: sin x > 0, so x ∈ (0, π) ∪ (2π, 3π) ∪ ...
  5. dy/dx = 3/(3x − 1)
    At x = 2: dy/dx = 3/(6 − 1) = 3/5 = 0.6
  6. At x = e: y = ln(e) = 1
    dy/dx = 1/x; at x = e: gradient = 1/e
    Tangent: y − 1 = (1/e)(x − e)
    y = x/e − 1 + 1 = y = x/e
  7. Domain: 2x − 1 > 0 ⇒ 2x > 1 ⇒ x > 1/2
    Let f(x) = 2x − 1, f′(x) = 2
    dy/dx = 2/(2x − 1)
  8. Using log laws:
    y = ln(x²) + ln((x+1)3) = 2 ln x + 3 ln(x + 1)
    dy/dx = 2/x + 3/(x + 1)

    This can be written as (2(x+1) + 3x)/(x(x+1)) = (5x+2)/(x(x+1)), but either form is acceptable.
  9. f(x) = x ln x, domain x > 0
    f′(x) = 1 × ln x + x × (1/x) = ln x + 1   (product rule)
    Set f′(x) = 0: ln x = −1 ⇒ x = e−1 = 1/e
    f(1/e) = (1/e) ln(1/e) = (1/e)(−1) = −1/e
    f′′(x) = 1/x; f′′(1/e) = e > 0
    Local minimum at (1/e, −1/e) ≈ (0.368, −0.368)
  10. Key features of y = ln(2x − 1):
    • Domain: x > 1/2 (vertical asymptote at x = 1/2)
    • x-intercept: ln(2x − 1) = 0 ⇒ 2x − 1 = 1 ⇒ x = 1; point (1, 0)
    • dy/dx = 2/(2x − 1); at x = 1, gradient = 2
    • As x → (1/2)+, y → −∞; as x → ∞, y → +∞
    • d²y/dx² = −4/(2x−1)² < 0 always → concave down throughout
    Graph: increasing curve from bottom-left (near asymptote x = 1/2), through (1, 0), concave down; identical shape to y = ln x but shifted right by 1/2 unit and horizontally compressed.