Topic Review — Further Differentiation — Solutions

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1. Fluency

   Q1 — Chain rule: identify and differentiate

   Differentiate each function using the chain rule.

   (a) y = (3x + 1)⁵   (b) y = (x² − 4)³   (c) y = √(2x + 7)   (d) y = 1/(x + 3)²

   ▶ Show Answer

   (a) Inner u = 3x + 1, du/dx = 3. dy/dx = 5(3x + 1)⁴ × 3 = 15(3x + 1)⁴

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

   (c) y = (2x + 7)^1/2. dy/dx = ½(2x + 7)^−1/2 × 2 = 1/√(2x + 7)

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

2. Fluency

   Q2 — Product rule: basic applications

   Differentiate each function using the product rule.

   (a) y = x(x + 5)   (b) y = x²(2x − 3)   (c) y = (x + 1)(x² + 4)   (d) y = x³√x

   ▶ Show Answer

   (a) u = x, v = x + 5. dy/dx = (x + 5) + x = 2x + 5

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

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

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

3. Fluency

   Q3 — Quotient rule: basic applications

   Differentiate each function using the quotient rule.

   (a) y = (x + 3)/(x − 1)   (b) y = x²/(x + 2)   (c) y = (2x + 1)/x²   (d) y = 1/(x² + 4)

   ▶ Show Answer

   (a) dy/dx = [1(x − 1) − (x + 3)(1)]/(x − 1)² = (x − 1 − x − 3)/(x − 1)² = −4/(x − 1)²

   (b) dy/dx = [2x(x + 2) − x²(1)]/(x + 2)² = (2x² + 4x − x²)/(x + 2)² = (x² + 4x)/(x + 2)²

   (c) f = 2x + 1, f′ = 2; g = x², g′ = 2x. dy/dx = (2x² − (2x + 1)(2x))/x⁴ = (2x² − 4x² − 2x)/x⁴ = (−2x² − 2x)/x⁴ = −2(x + 1)/x³

   (d) f = 1, f′ = 0; g = x² + 4, g′ = 2x. dy/dx = −2x/(x² + 4)²

4. Understanding

   Q4 — Chain rule: gradient at a point

   For each curve, find the gradient at the specified point.

   (a) y = (2x − 1)&sup4; at x = 1   (b) y = (x² + 2)³ at x = 1   (c) y = √(3x + 1) at x = 1   (d) y = (1 − 2x)³ at x = 0

   ▶ Show Answer

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

   (b) dy/dx = 3(x² + 2)²(2x) = 6x(x² + 2)². At x = 1: 6(1)(3)² = 6 × 9 = 54

   (c) dy/dx = ½(3x + 1)^−1/2 × 3 = 3/(2√(3x + 1)). At x = 1: 3/(2√4) = 3/4 = 3/4

   (d) dy/dx = 3(1 − 2x)² × (−2) = −6(1 − 2x)². At x = 0: −6(1)² = −6

5. Understanding

   Q5 — Product rule: equation of tangent

   Find the equation of the tangent to each curve at the given point.

   (a) y = x(x + 4) at x = 2   (b) y = x²(x − 1) at x = 2   (c) y = (x + 1)(x² − 2) at x = 1

   ▶ Show Answer

   (a) dy/dx = 2x + 4. At x = 2: y = 2(6) = 12, gradient = 8. Tangent: y − 12 = 8(x − 2) → y = 8x − 4

   (b) Product rule: u = x², v = x − 1. dy/dx = 2x(x − 1) + x² = 3x² − 2x. At x = 2: y = 4(1) = 4, gradient = 12 − 4 = 8. Tangent: y − 4 = 8(x − 2) → y = 8x − 12

   (c) u = x + 1, v = x² − 2. dy/dx = x² − 2 + (x + 1)(2x) = x² − 2 + 2x² + 2x = 3x² + 2x − 2. At x = 1: y = 2(−1) = −2, gradient = 3 + 2 − 2 = 3. Tangent: y + 2 = 3(x − 1) → y = 3x − 5

6. Understanding

   Q6 — Quotient rule: turning points

   Find all x-values where dy/dx = 0 for each function.

   (a) y = x/(x² + 3)   (b) y = x²/(x + 2)   (c) y = (x − 2)/(x² + 1)

   ▶ Show Answer

   (a) dy/dx = (x² + 3 − 2x²)/(x² + 3)² = (3 − x²)/(x² + 3)². Set 3 − x² = 0: x = ±√3

   (b) dy/dx = [2x(x + 2) − x²]/(x + 2)² = (x² + 4x)/(x + 2)² = x(x + 4)/(x + 2)². Set x(x + 4) = 0: x = 0 or x = −4

   (c) f = x − 2, f′ = 1; g = x² + 1, g′ = 2x. dy/dx = [x² + 1 − (x − 2)(2x)]/(x² + 1)² = (x² + 1 − 2x² + 4x)/(x² + 1)² = (−x² + 4x + 1)/(x² + 1)². Set −x² + 4x + 1 = 0: x² − 4x − 1 = 0. x = (4 ± √20)/2 = 2 ± √5. x = 2 + √5 ≈ 4.24 or x = 2 − √5 ≈ −0.24

7. Understanding

   Q7 — Choosing the correct rule

   State which rule (chain, product, quotient, or power rule alone) you would use first, and then differentiate each function.

   (a) y = (5x − 2)³   (b) y = x(x + 1)⁴   (c) y = (x + 1)/(x² + x)   (d) y = 3x⁴ − 7x² + 2

   ▶ Show Answer

   (a) Chain rule. Inner u = 5x − 2. dy/dx = 3(5x − 2)² × 5 = 15(5x − 2)²

   (b) Product rule (with chain rule for second factor). u = x, u′ = 1; v = (x + 1)⁴, v′ = 4(x + 1)³. dy/dx = (x + 1)⁴ + 4x(x + 1)³ = (x + 1)³[(x + 1) + 4x] = (x + 1)³(5x + 1)

   (c) Quotient rule (or simplify: (x + 1)/(x(x + 1)) = 1/x for x ≠ −1, so dy/dx = −1/x²). Using quotient rule: f = x + 1, f′ = 1; g = x² + x = x(x + 1), g′ = 2x + 1. dy/dx = [x² + x − (x + 1)(2x + 1)]/(x² + x)² = [x² + x − 2x² − 3x − 1]/(x²+x)² = (−x² − 2x − 1)/(x²+x)² = −(x+1)²/[x(x+1)]² = −1/x²

   (d) Power rule (term by term). dy/dx = 12x³ − 14x

8. Problem Solving

   Q8 — Chain rule: stationary points and nature

   For the curve y = (x² − 4)⁴:

   (a) Find dy/dx.   (b) Find all stationary points.   (c) Classify each stationary point using a sign diagram.   (d) Find the coordinates of each.

   ▶ Show Answer

   (a) dy/dx = 4(x² − 4)³ × 2x = 8x(x² − 4)³

   (b) Set 8x(x² − 4)³ = 0: x = 0 or x² = 4 → x = 0, x = 2, x = −2

   (c) Sign diagram for dy/dx = 8x(x − 2)³(x + 2)³:
   x < −2: (−)(−)³(−)³ = (−)(+) ... let's test x = −3: 8(−3)(9−4)³ = (−)(+)³ < 0.
   −2 < x < 0: test x = −1: 8(−1)(1−4)³ = (−)(−)³ = (−)(−) > 0.
   0 < x < 2: test x = 1: 8(1)(1−4)³ = (+)(−) < 0.
   x > 2: test x = 3: 8(3)(9−4)³ = (+)(+) > 0.
   x = −2: − to + → local minimum. x = 0: + to − → local maximum. x = 2: − to + → local minimum.

   (d) y(−2) = (4−4)⁴ = 0. y(0) = (−4)⁴ = 256. y(2) = (4−4)⁴ = 0. Stationary points: (−2, 0) min, (0, 256) max, (2, 0) min

9. Problem Solving

   Q9 — Product rule combined with chain rule

   Find dy/dx for each function. Factorise your answers fully.

   (a) y = x²(x + 1)⁵   (b) y = (x − 1)(2x + 3)&sup4;   (c) y = x(x² − 1)³

   ▶ Show Answer

   (a) u = x², u′ = 2x; v = (x + 1)⁵, v′ = 5(x + 1)⁴. dy/dx = 2x(x + 1)⁵ + 5x²(x + 1)⁴ = x(x + 1)⁴[2(x + 1) + 5x] = x(x + 1)⁴(7x + 2)

   (b) u = x − 1, u′ = 1; v = (2x + 3)⁴, v′ = 8(2x + 3)³. dy/dx = (2x + 3)⁴ + 8(x − 1)(2x + 3)³ = (2x + 3)³[(2x + 3) + 8(x − 1)] = (2x + 3)³[10x − 5] = 5(2x − 1)(2x + 3)³

   (c) u = x, u′ = 1; v = (x² − 1)³, v′ = 3(x² − 1)²(2x) = 6x(x² − 1)². dy/dx = (x² − 1)³ + 6x²(x² − 1)² = (x² − 1)²[(x² − 1) + 6x²] = (x² − 1)²(7x² − 1)

10. Problem Solving

    Q10 — Quotient rule: nature of stationary points

    For y = (x² + 1)/(x):

    (a) Find dy/dx using the quotient rule.   (b) Find all stationary points.   (c) Determine the nature of each.   (d) Verify by rewriting y = x + 1/x and differentiating using the power rule.

    ▶ Show Answer

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

    (b) Set (x² − 1)/x² = 0: x² = 1 → x = 1 or x = −1 (x ≠ 0)

    (c) For x < −1: (x² − 1) > 0 → dy/dx > 0. For −1 < x < 0: (x² − 1) < 0 → dy/dx < 0. For 0 < x < 1: dy/dx < 0. For x > 1: dy/dx > 0.
    x = −1: + to − → local maximum. y(−1) = (1 + 1)/(−1) = −2. Point: (−1, −2).
    x = 1: − to + → local minimum. y(1) = 2/1 = 2. Point: (1, 2).

    (d) y = x + x⁻¹. dy/dx = 1 − x⁻² = 1 − 1/x² = (x² − 1)/x² ✓

11. Problem Solving

    Q11 — Motion: chain rule application

    A particle moves along a line. Its displacement at time t ≥ 0 is s(t) = (t² + 1)³ metres.

    (a) Find the velocity v = ds/dt.   (b) Find the acceleration a = dv/dt.   (c) Find the velocity when t = 1.   (d) Is the particle ever stationary (v = 0)? Justify.

    ▶ Show Answer

    (a) v = ds/dt = 3(t² + 1)² × 2t = 6t(t² + 1)² m/s

    (b) a = dv/dt. Apply product rule to 6t(t² + 1)²: u = 6t, u′ = 6; v = (t² + 1)², v′ = 4t(t² + 1). a = 6(t² + 1)² + 6t × 4t(t² + 1) = 6(t² + 1)² + 24t²(t² + 1) = 6(t² + 1)[(t² + 1) + 4t²] = 6(t² + 1)(5t² + 1) m/s²

    (c) v(1) = 6(1)(1 + 1)² = 6 × 4 = 24 m/s

    (d) v = 6t(t² + 1)² = 0 when t = 0 (since (t² + 1)² > 0 for all t). At t = 0, s = 1 and v = 0. The particle is stationary only at t = 0; for all t > 0 the velocity is positive.

12. Problem Solving

    Q12 — Applied optimisation using quotient rule

    A function modelling efficiency is E(x) = x/(x² + 1) for x ≥ 0.

    (a) Find E′(x).   (b) Find the value of x that maximises E(x).   (c) Find the maximum value of E.   (d) Confirm it is a maximum using a sign diagram or second derivative argument.

    ▶ Show Answer

    (a) f = x, f′ = 1; g = x² + 1, g′ = 2x. E′(x) = [x² + 1 − 2x²]/(x² + 1)² = (1 − x²)/(x² + 1)²

    (b) Set 1 − x² = 0: x² = 1, so x = 1 (taking x ≥ 0)

    (c) E(1) = 1/(1 + 1) = 1/2 = 0.5

    (d) Sign diagram: for 0 < x < 1, 1 − x² > 0 so E′ > 0 (increasing). For x > 1, 1 − x² < 0 so E′ < 0 (decreasing). Gradient changes from positive to negative → local maximum at x = 1.

13. Problem Solving

    Q13 — Full mixed: identify rule, differentiate, apply

    For each function, state the rule(s) required and find dy/dx. Then find the gradient at x = 1.

    (a) y = x²(3x − 1)³   (b) y = (x + 1)² / (x² + 1)   (c) y = (x² − 1)(x² + 1)⁴

    ▶ Show Answer

    (a) Product & chain rule. u = x², u′ = 2x; v = (3x − 1)³, v′ = 9(3x − 1)². dy/dx = 2x(3x − 1)³ + 9x²(3x − 1)² = x(3x − 1)²[2(3x − 1) + 9x] = x(3x − 1)²(15x − 2). At x = 1: 1(2)²(13) = 52

    (b) Quotient & chain rule. f = (x + 1)², f′ = 2(x + 1); g = x² + 1, g′ = 2x. dy/dx = [2(x + 1)(x² + 1) − (x + 1)²(2x)]/(x² + 1)² = 2(x + 1)[(x² + 1) − x(x + 1)]/(x² + 1)² = 2(x + 1)(x² + 1 − x² − x)/(x² + 1)² = 2(x + 1)(1 − x)/(x² + 1)² = 2(1 − x²)/(x² + 1)². At x = 1: 0. Gradient = 0 (stationary point).

    (c) Product & chain rule. u = x² − 1, u′ = 2x; v = (x² + 1)⁴, v′ = 8x(x² + 1)³. dy/dx = 2x(x² + 1)⁴ + 8x(x² − 1)(x² + 1)³ = 2x(x² + 1)³[(x² + 1) + 4(x² − 1)] = 2x(x² + 1)³(5x² − 3). At x = 1: 2(1)(2)³(5 − 3) = 2 × 8 × 2 = 32

14. Problem Solving

    Q14 — Rate of change application

    A population of bacteria at time t hours is modelled by P(t) = 100t / (t² + 9) (hundreds of bacteria).

    (a) Find P′(t).   (b) Find the time at which the population peaks.   (c) What is the maximum population?   (d) Find the rate of change of population at t = 6. Is the population increasing or decreasing?

    ▶ Show Answer

    (a) f = 100t, f′ = 100; g = t² + 9, g′ = 2t. P′(t) = [100(t² + 9) − 100t(2t)]/(t² + 9)² = 100(t² + 9 − 2t²)/(t² + 9)² = 100(9 − t²)/(t² + 9)²

    (b) Set 9 − t² = 0: t² = 9 → t = 3 (t ≥ 0). Peak at t = 3 hours.

    (c) P(3) = 300/(9 + 9) = 300/18 = 50/3 ≈ 16.7 hundred bacteria

    (d) P′(6) = 100(9 − 36)/(36 + 9)² = 100(−27)/(45)² = −2700/2025 = −4/3 hundred bacteria per hour. Negative, so the population is decreasing at t = 6.

15. Problem Solving

    Q15 — Extended: all three rules

    Consider the function f(x) = x(x + 1)² / (x² + 1).

    (a) Find the numerator's derivative using the product and chain rule.   (b) Apply the quotient rule to find f′(x).   (c) Evaluate f′(0).   (d) Determine whether f has a stationary point at x = 0.

    ▶ Show Answer

    (a) Numerator: u(x) = x(x + 1)². Product rule: u′ = 1 × (x + 1)² + x × 2(x + 1) = (x + 1)² + 2x(x + 1) = (x + 1)[(x + 1) + 2x] = (x + 1)(3x + 1)

    (b) f = x(x + 1)², f′ = (x + 1)(3x + 1); g = x² + 1, g′ = 2x.
    f′(x) = [(x + 1)(3x + 1)(x² + 1) − x(x + 1)²(2x)] / (x² + 1)²
    Factor (x + 1): f′(x) = (x + 1)[(3x + 1)(x² + 1) − 2x²(x + 1)] / (x² + 1)²
    Expand: (3x + 1)(x² + 1) = 3x³ + 3x + x² + 1. 2x²(x + 1) = 2x³ + 2x².
    Bracket: 3x³ + x² + 3x + 1 − 2x³ − 2x² = x³ − x² + 3x + 1.
    f′(x) = (x + 1)(x³ − x² + 3x + 1) / (x² + 1)²

    (c) f′(0) = (0 + 1)(0 − 0 + 0 + 1)/(0 + 1)² = (1)(1)/1 = 1

    (d) f′(0) = 1 ≠ 0. The function does not have a stationary point at x = 0.