Practice Maths

Solutions — The Quotient Rule and Combined Rules

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  1. Q1 — Basic quotient rule

    (a) f = x + 1, f′ = 1; g = x − 2, g′ = 1. dy/dx = [1(x − 2) − (x + 1)(1)]/(x − 2)² = (x − 2 − x − 1)/(x − 2)² = −3/(x − 2)²

    (b) f = x², f′ = 2x; g = 3x + 1, g′ = 3. dy/dx = [2x(3x + 1) − x²(3)]/(3x + 1)² = (6x² + 2x − 3x²)/(3x + 1)² = (3x² + 2x)/(3x + 1)²

    (c) f = 2x − 3, f′ = 2; g = x², g′ = 2x. dy/dx = [2 × x² − (2x − 3)(2x)] / x4 = (2x² − 4x² + 6x)/x4 = (−2x² + 6x)/x4 = x(−2x + 6)/x4 = 2(3 − x)/x³

    (d) f = 1, f′ = 0; g = x² + 1, g′ = 2x. dy/dx = [0 × (x² + 1) − 1 × 2x]/(x² + 1)² = −2x/(x² + 1)²

  2. Q2 — Rational functions simplified

    (a) f = 3x + 2, f′ = 3; g = x + 1, g′ = 1. dy/dx = [3(x + 1) − (3x + 2)(1)]/(x + 1)² = (3x + 3 − 3x − 2)/(x + 1)² = 1/(x + 1)²

    (b) f = x, f′ = 1; g = x² + 4, g′ = 2x. dy/dx = [x² + 4 − x(2x)]/(x² + 4)² = (x² + 4 − 2x²)/(x² + 4)² = (4 − x²)/(x² + 4)²

    (c) Simplify first: (x² − 1)/(x + 1) = (x − 1)(x + 1)/(x + 1) = x − 1 (for x ≠ −1). d/dx[x − 1] = 1. Quotient rule (verify): f = x² − 1, f′ = 2x; g = x + 1, g′ = 1. dy/dx = [2x(x + 1) − (x² − 1)]/(x + 1)² = (x² + 2x + 1)/(x + 1)² = (x + 1)²/(x + 1)² = 1 ✓

  3. Q3 — Power form equivalents

    (a) y = 4x−2. dy/dx = −8x−3 = −8/x³. (Quotient rule: f = 4, f′ = 0; g = x², g′ = 2x. dy/dx = [0 − 4(2x)]/x4 = −8x/x4 = −8/x³ ✓)

    (b) Rewrite: y = (x + 1)/x³ = x−2 + x−3. dy/dx = −2x−3 − 3x−4 = −2/x³ − 3/x4. Alternatively: (−2x + 3)/x4 after combining. Factor from quotient rule: f = x + 1, f′ = 1; g = x³, g′ = 3x². dy/dx = [x³ − (x + 1)(3x²)]/x6 = (x³ − 3x³ − 3x²)/x6 = (−2x³ − 3x²)/x6 = x²(−2x − 3)/x6 = (−2x − 3)/x4

    (c) y = 3x−1/2. dy/dx = −(3/2)x−3/2 = −3/(2x3/2) = −3/(2x√x)

  4. Q4 — Turning points

    (a) f = x, f′ = 1; g = x² + 1, g′ = 2x. dy/dx = (x² + 1 − 2x²)/(x² + 1)² = (1 − x²)/(x² + 1)². Set 1 − x² = 0: x = ±1

    (b) f = x², f′ = 2x; g = x − 1, g′ = 1. dy/dx = [2x(x − 1) − x²]/(x − 1)² = (2x² − 2x − x²)/(x − 1)² = (x² − 2x)/(x − 1)² = x(x − 2)/(x − 1)². Set x(x − 2) = 0: x = 0 or x = 2

    (c) f = x − 1, f′ = 1; g = x² + 3, g′ = 2x. dy/dx = [x² + 3 − (x − 1)(2x)]/(x² + 3)² = (x² + 3 − 2x² + 2x)/(x² + 3)² = (−x² + 2x + 3)/(x² + 3)². Set −x² + 2x + 3 = 0: x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0. x = 3 or x = −1

  5. Q5 — Tangent equations

    (a) dy/dx = −3/(x − 2)² (from Q1a). At x = 3: y = 4/1 = 4, gradient = −3/1 = −3. Tangent: y − 4 = −3(x − 3) → y = −3x + 13

    (b) f = x², f′ = 2x; g = x + 1, g′ = 1. dy/dx = [2x(x + 1) − x²]/(x + 1)² = (x² + 2x)/(x + 1)². At x = 1: y = 1/2, gradient = (1 + 2)/4 = 3/4. Tangent: y − 1/2 = (3/4)(x − 1) → y = (3/4)x − 1/4

    (c) dy/dx = −2x/(x² + 1)² (from Q1d). At x = 0: y = 1, gradient = 0. Tangent: y = 1 (horizontal tangent)

  6. Q6 — Simplification proofs

    (a) Quotient rule: f = x² + 1, f′ = 2x; g = x, g′ = 1. dy/dx = [2x × x − (x² + 1) × 1]/x² = (2x² − x² − 1)/x² = (x² − 1)/x² ✓
    Rewrite: y = x + x−1. dy/dx = 1 − x−2 = 1 − 1/x² = (x² − 1)/x² ✓

    (b) From Q1c: d/dx[(2x − 3)/x²] = (−2x² + 6x)/x4 = x(−2x + 6)/x4 = (−2x + 6)/x³ = 2(3 − x)/x³ ✓

  7. Q7 — Chain rule within quotient rule

    (a) f = (2x + 1)³, f′ = 3(2x + 1)² × 2 = 6(2x + 1)²; g = x, g′ = 1. dy/dx = [6(2x + 1)² × x − (2x + 1)³ × 1]/x² = (2x + 1)²[6x − (2x + 1)]/x² = (2x + 1)²(4x − 1)/x²

    (b) f = x1/2, f′ = ½x−1/2; g = x² + 1, g′ = 2x. dy/dx = [½x−1/2(x² + 1) − x1/2(2x)] / (x² + 1)². Multiply numerator top and bottom by 2x1/2: [(x² + 1) − 4x²] / (2x1/2(x² + 1)²) = (1 − 3x²) / (2√x(x² + 1)²)

    (c) f = (x² + 3)², f′ = 2(x² + 3)(2x) = 4x(x² + 3); g = x − 1, g′ = 1. dy/dx = [4x(x² + 3)(x − 1) − (x² + 3)²] / (x − 1)² = (x² + 3)[4x(x − 1) − (x² + 3)] / (x − 1)² = (x² + 3)(4x² − 4x − x² − 3) / (x − 1)² = (x² + 3)(3x² − 4x − 3) / (x − 1)²

  8. Q8 — Product and quotient rules combined

    (a) Expand numerator: f = x(x + 1) = x² + x, f′ = 2x + 1; g = x² + 1, g′ = 2x. dy/dx = [(2x + 1)(x² + 1) − (x² + x)(2x)] / (x² + 1)². Numerator: 2x³ + 2x + x² + 1 − 2x³ − 2x² = −x² + 2x + 1. dy/dx = (−x² + 2x + 1)/(x² + 1)²

    (b) Expand numerator: f = x²(x − 1) = x³ − x², f′ = 3x² − 2x; g = 2x + 3, g′ = 2. dy/dx = [(3x² − 2x)(2x + 3) − (x³ − x²)(2)] / (2x + 3)². Numerator: 6x³ + 9x² − 4x² − 6x − 2x³ + 2x² = 4x³ + 7x² − 6x = x(4x² + 7x − 6). dy/dx = x(4x² + 7x − 6) / (2x + 3)²

  9. Q9 — Drug concentration related rates

    (a) f = 5t, f′ = 5; g = t² + 4, g′ = 2t. C′(t) = [5(t² + 4) − 5t(2t)] / (t² + 4)² = (5t² + 20 − 10t²)/(t² + 4)² = (20 − 5t²) / (t² + 4)² = 5(4 − t²) / (t² + 4)²

    (b) Set C′(t) = 0: 5(4 − t²) = 0 → t² = 4 → t = 2 (taking positive value). Maximum concentration at t = 2 hours.

    (c) C(2) = 5(2)/(4 + 4) = 10/8 = 5/4 = 1.25 mg/L

    (d) C′(3) = 5(4 − 9)/(9 + 4)² = 5(−5)/169 = −25/169 < 0. The concentration is decreasing at t = 3 hours.

  10. Q10 — Full combined rules

    (a) Numerator u = x²(x + 1)³. Apply product rule: u′ = 2x(x + 1)³ + x² × 3(x + 1)² = 2x(x + 1)³ + 3x²(x + 1)² = x(x + 1)²[2(x + 1) + 3x] = x(x + 1)²(5x + 2). Denominator v = 2x − 1, v′ = 2.

    (b) f′(x) = [x(x + 1)²(5x + 2)(2x − 1) − x²(x + 1)³(2)] / (2x − 1)²

    (c) Factor x(x + 1)² from numerator: f′(x) = x(x + 1)²[(5x + 2)(2x − 1) − 2x(x + 1)] / (2x − 1)². Expand bracket: 10x² − 5x + 4x − 2 − 2x² − 2x = 8x² − 3x − 2 = (8x + 5)(x − ...). Check: using quadratic formula on 8x² − 3x − 2: discriminant = 9 + 64 = 73 (does not factor nicely). f′(x) = x(x + 1)²(8x² − 3x − 2) / (2x − 1)²

    (d) At x = 1: f′(1) = 1 × (2)² × (8 − 3 − 2) / (1)² = 4 × 3 / 1 = 12