Solutions — The Product Rule — Grade 11 Maths Methods

Q1 — Basic product rule

(a) u = x, u′ = 1; v = x + 1, v′ = 1. dy/dx = 1 · (x + 1) + x · 1 = x + 1 + x = 2x + 1. (Or expand: x² + x, d/dx = 2x + 1 ✓)

(b) u = x², u′ = 2x; v = 3x − 1, v′ = 3. dy/dx = 2x(3x − 1) + x²(3) = 6x² − 2x + 3x² = 9x² − 2x

(c) u = x + 2, u′ = 1; v = x² − x + 3, v′ = 2x − 1. dy/dx = 1 · (x² − x + 3) + (x + 2)(2x − 1) = x² − x + 3 + 2x² + 3x − 2 = 3x² + 2x + 1

(d) u = x³, u′ = 3x²; v = x^1/2, v′ = ½x^−1/2. dy/dx = 3x² · x^1/2 + x³ · ½x^−1/2 = 3x^5/2 + ½x^5/2 = (7/2)x^5/2. (Verify: x^7/2, d/dx = (7/2)x^5/2 ✓)

Q2 — Simplifying products

(a) u = 2x + 3, u′ = 2; v = x − 4, v′ = 1. dy/dx = 2(x − 4) + (2x + 3)(1) = 2x − 8 + 2x + 3 = 4x − 5

(b) u = x, u′ = 1; v = x² + 5, v′ = 2x. dy/dx = x² + 5 + x(2x) = x² + 5 + 2x² = 3x² + 5

(c) Let p(x) = (x + 1)(x + 2) = x² + 3x + 2, so y = p(x)(x + 3). u = x² + 3x + 2, u′ = 2x + 3; v = x + 3, v′ = 1. dy/dx = (2x + 3)(x + 3) + (x² + 3x + 2)(1) = 2x² + 9x + 9 + x² + 3x + 2 = 3x² + 12x + 11

Q3 — Surd products

(a) y = x² × x^1/2 = x^5/2. dy/dx = (5/2)x^3/2. Or product rule: u = x², u′ = 2x; v = x^1/2, v′ = ½x^−1/2. dy/dx = 2x · x^1/2 + x² · ½x^−1/2 = 2x^3/2 + ½x^3/2 = (5/2)x^3/2

(b) u = 2x + 1, u′ = 2; v = x^1/2, v′ = ½x^−1/2. dy/dx = 2x^1/2 + (2x + 1) · ½x^−1/2 = 2√x + (2x + 1)/(2√x) = [4x + 2x + 1]/(2√x) = (6x + 1)/(2√x)

(c) u = x, u′ = 1; v = (x + 1)^1/2, v′ = ½(x + 1)^−1/2. dy/dx = 1 · (x + 1)^1/2 + x · ½(x + 1)^−1/2 = √(x + 1) + x/(2√(x + 1)) = [2(x + 1) + x]/(2√(x + 1)) = (3x + 2)/(2√(x + 1))

Q4 — Evaluating f′(x) at a point

(a) f(x) = x(x + 4). u = x, u′ = 1; v = x + 4, v′ = 1. f′(x) = x + 4 + x = 2x + 4. f′(2) = 4 + 4 = 8

(b) f(x) = x²(x − 3). u = x², u′ = 2x; v = x − 3, v′ = 1. f′(x) = 2x(x − 3) + x² = 2x² − 6x + x² = 3x² − 6x. f′(1) = 3 − 6 = −3

(c) f(x) = (x + 1)(x² + 2). u = x + 1, u′ = 1; v = x² + 2, v′ = 2x. f′(x) = x² + 2 + (x + 1)(2x) = x² + 2 + 2x² + 2x = 3x² + 2x + 2. f′(0) = 0 + 0 + 2 = 2

Q5 — Tangent equations

(a) y = x(x + 3). dy/dx = 2x + 3. At x = 1: y = 1(4) = 4, gradient = 5. Tangent: y − 4 = 5(x − 1) → y = 5x − 1

(b) y = x²(2x − 1). u = x², u′ = 2x; v = 2x − 1, v′ = 2. dy/dx = 2x(2x − 1) + 2x² = 4x² − 2x + 2x² = 6x² − 2x. At x = 2: y = 4(3) = 12, gradient = 24 − 4 = 20. Tangent: y − 12 = 20(x − 2) → y = 20x − 28

(c) y = (x + 2)(x² + 1). u = x + 2, u′ = 1; v = x² + 1, v′ = 2x. dy/dx = x² + 1 + (x + 2)(2x) = x² + 1 + 2x² + 4x = 3x² + 4x + 1. At x = 0: y = 2(1) = 2, gradient = 1. Tangent: y − 2 = 1(x − 0) → y = x + 2

Q6 — Verification proofs

(a) Product rule: u = x + 1, u′ = 1; v = x + 2, v′ = 1. dy/dx = 1(x + 2) + (x + 1)(1) = x + 2 + x + 1 = 2x + 3 ✓
Expand: y = x² + 3x + 2. dy/dx = 2x + 3 ✓

(b) Product rule: u = x², u′ = 2x; v = x + 1, v′ = 1. dy/dx = 2x(x + 1) + x²(1) = 2x² + 2x + x² = 3x² + 2x ✓
Expand: y = x³ + x². dy/dx = 3x² + 2x ✓

Q7 — Product and chain rule combined

(a) u = x, u′ = 1; v = (2x + 1)³, v′ = 3(2x + 1)² × 2 = 6(2x + 1)². dy/dx = 1 · (2x + 1)³ + x · 6(2x + 1)² = (2x + 1)³ + 6x(2x + 1)² = (2x + 1)²[(2x + 1) + 6x] = (2x + 1)²(8x + 1)

(b) u = x², u′ = 2x; v = (x² + 1)⁴, v′ = 4(x² + 1)³ × 2x = 8x(x² + 1)³. dy/dx = 2x(x² + 1)⁴ + x² × 8x(x² + 1)³ = 2x(x² + 1)³[(x² + 1) + 4x²] = 2x(5x² + 1)(x² + 1)³

Q8 — Stationary points

(a) u = x², u′ = 2x; v = x − 3, v′ = 1. dy/dx = 2x(x − 3) + x² = 2x² − 6x + x² = 3x² − 6x = 3x(x − 2)

(b) Set dy/dx = 0: 3x(x − 2) = 0 → x = 0 or x = 2

(c) Sign diagram for dy/dx = 3x(x − 2): negative for 0 < x < 2, positive for x < 0 and x > 2.
At x = 0: gradient changes from + to − → local maximum.
At x = 2: gradient changes from − to + → local minimum.

(d) y(0) = 0²(0 − 3) = 0. y(2) = 4(2 − 3) = −4. Stationary points: (0, 0) local max and (2, −4) local min

Q9 — Rectangle area rate of change

(a) A(t) = (2t + 1)(t² + 3)

(b) u = 2t + 1, u′ = 2; v = t² + 3, v′ = 2t. dA/dt = 2(t² + 3) + (2t + 1)(2t) = 2t² + 6 + 4t² + 2t = 6t² + 2t + 6

(c) dA/dt at t = 2: 6(4) + 2(2) + 6 = 24 + 4 + 6 = 34 cm²/s

(d) Set 6t² + 2t + 6 = 20 → 6t² + 2t − 14 = 0 → 3t² + t − 7 = 0. t = (−1 ± √(1 + 84))/6 = (−1 + √85)/6 ≈ (−1 + 9.22)/6 ≈ t ≈ 1.37 seconds (taking positive root)

Q10 — Product rule with fractional powers

(a) u = x² − 1, u′ = 2x; v = (x + 2)^1/2, v′ = ½(x + 2)^−1/2

(b) f′(x) = 2x(x + 2)^1/2 + (x² − 1) × ½(x + 2)^−1/2

(c) Common denominator √(x + 2): f′(x) = [2x(x + 2) + ½(x² − 1)] / √(x + 2) = [2x² + 4x + (x² − 1)/2] / √(x + 2) = [(4x² + 8x + x² − 1)/2] / √(x + 2) = (5x² + 8x − 1) / (2√(x + 2))

(d) At x = 2: f′(2) = (5(4) + 8(2) − 1) / (2√4) = (20 + 16 − 1)/(2 × 2) = 35/4 = 8.75