Practice Maths

Solutions — The Chain Rule

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  1. Q1 — Linear inner functions

    (a) u = x + 4, dy/dx = 6(x + 4)5 × 1 = 6(x + 4)5

    (b) u = 3x − 1, du/dx = 3. dy/dx = 4(3x − 1)3 × 3 = 12(3x − 1)3

    (c) u = 2x + 5, du/dx = 2. dy/dx = 3(2x + 5)2 × 2 = 6(2x + 5)2

    (d) u = 5 − x, du/dx = −1. dy/dx = 7(5 − x)6 × (−1) = −7(5 − x)6

  2. Q2 — Quadratic inner functions

    (a) u = x2 + 1, du/dx = 2x. dy/dx = 5(x2 + 1)4 × 2x = 10x(x2 + 1)4

    (b) u = x2 − 4x, du/dx = 2x − 4. dy/dx = 3(x2 − 4x)2(2x − 4) = 6(x − 2)(x2 − 4x)2

    (c) u = 2x2 + x, du/dx = 4x + 1. dy/dx = 4(2x2 + x)3(4x + 1) = 4(4x + 1)(2x2 + x)3

    (d) u = x3 + 1, du/dx = 3x2. dy/dx = 6(x3 + 1)5 × 3x2 = 18x2(x3 + 1)5

  3. Q3 — Square root functions

    (a) y = (2x + 3)1/2. dy/dx = ½(2x + 3)−1/2 × 2 = 1/√(2x + 3)

    (b) y = (x2 + 5)1/2. dy/dx = ½(x2 + 5)−1/2 × 2x = x/√(x2 + 5)

    (c) y = (4 − x)1/2. dy/dx = ½(4 − x)−1/2 × (−1) = −1/(2√(4 − x))

    (d) y = (3x2 − x)1/2. du/dx = 6x − 1. dy/dx = ½(3x2 − x)−1/2(6x − 1) = (6x − 1)/(2√(3x2 − x))

  4. Q4 — Negative power forms

    (a) y = (x + 2)−3. dy/dx = −3(x + 2)−4 × 1 = −3/(x + 2)4

    (b) y = (2x − 1)−5. dy/dx = −5(2x − 1)−6 × 2 = −10/(2x − 1)6

    (c) y = 4(x2 + 3)−2. du/dx = 2x. dy/dx = 4 × (−2)(x2 + 3)−3 × 2x = −16x/(x2 + 3)3

  5. Q5 — Gradient at a point

    (a) dy/dx = 4(2x + 1)3 × 2 = 8(2x + 1)3. At x = 0: 8(1)3 = 8

    (b) dy/dx = 3(x2 + 3)2 × 2x = 6x(x2 + 3)2. At x = 1: 6(1)(4)2 = 6 × 16 = 96

    (c) dy/dx = ½(x + 5)−1/2. At x = 4: ½(9)−1/2 = ½ × 1/3 = 1/6

    (d) dy/dx = 5(3 − x)4 × (−1) = −5(3 − x)4. At x = 2: −5(1)4 = −5

  6. Q6 — Tangent equations

    (a) y = (x + 1)4. dy/dx = 4(x + 1)3. At x = 1: y = 24 = 16, gradient = 4(2)3 = 32. Tangent: y − 16 = 32(x − 1) → y = 32x − 16

    (b) y = (2x − 3)3. dy/dx = 3(2x − 3)2 × 2 = 6(2x − 3)2. At x = 2: y = 13 = 1, gradient = 6(1)2 = 6. Tangent: y − 1 = 6(x − 2) → y = 6x − 11

    (c) y = (2x + 1)1/2. dy/dx = ½(2x + 1)−1/2 × 2 = 1/√(2x + 1). At x = 4: y = √9 = 3, gradient = 1/3. Tangent: y − 3 = (1/3)(x − 4) → y = x/3 + 5/3

  7. Q7 — Complex composite functions

    (a) u = x2 + 2x + 1 = (x + 1)2, du/dx = 2x + 2. dy/dx = 4(x2 + 2x + 1)3(2x + 2) = 8(x + 1)(x2 + 2x + 1)3

    (b) u = x3 − 3x, du/dx = 3x2 − 3. dy/dx = 5(x3 − 3x)4(3x2 − 3) = 15(x2 − 1)(x3 − 3x)4

    (c) u = 1 − 2x2, du/dx = −4x. dy/dx = 3(1 − 2x2)2 × (−4x) = −12x(1 − 2x2)2

    (d) Simplify first: ((x + 1)2)3 = (x + 1)6. dy/dx = 6(x + 1)5 × 1 = 6(x + 1)5

  8. Q8 — Chain rule with power rule

    (a) f′(x) = 12x3 + 2 × 5(x + 1)4 × 1 = 12x3 + 10(x + 1)4

    (b) f′(x) = 3(x2 + 1)2 × 2x − 3x2 = 6x(x2 + 1)2 − 3x2. Factor: 3x[2(x2 + 1)2 − x]

    (c) f′(x) = 5 × 4(2x − 1)3 × 2 + 3 = 40(2x − 1)3 + 3. At any x: f′(x) = 40(2x − 1)3 + 3

  9. Q9 — Related rates: balloon

    (a) r = 3t + 1. V = (4/3)πr3 = (4/3)π(3t + 1)3

    (b) dV/dr = 4πr2 = 4π(3t + 1)2. dr/dt = 3. dV/dt = 4π(3t + 1)2 × 3 = 12π(3t + 1)2

    (c) At t = 1: dV/dt = 12π(3 + 1)2 = 12π × 16 = 192π cm3/s

  10. Q10 — Chain rule and motion

    (a) s = (t2 + 2)3. v = ds/dt = 3(t2 + 2)2 × 2t = 6t(t2 + 2)2 m/s

    (b) At t = 1: v = 6(1)(1 + 2)2 = 6 × 9 = 54 m/s

    (c) a = dv/dt = d/dt[6t(t2 + 2)2]. Using product rule: 6(t2 + 2)2 + 6t × 2(t2 + 2) × 2t = 6(t2 + 2)2 + 24t2(t2 + 2). At t = 1: 6(3)2 + 24(1)(3) = 54 + 72 = 126 m/s2

  11. Q11 — Stationary points

    (a) dy/dx = 4(2x − 4)3 × 2 = 8(2x − 4)3

    (b) Set dy/dx = 0: 8(2x − 4)3 = 0 → 2x − 4 = 0 → x = 2. y = (0)4 − 3 = −3. Stationary point: (2, −3)

    (c) For x < 2: (2x − 4) < 0, so (2x − 4)3 < 0, so dy/dx < 0 (decreasing). For x > 2: dy/dx > 0 (increasing). Gradient changes from negative to positive: minimum turning point at (2, −3).