0606/11 May/Jun 2025, Question 6: The Product and Quotient Rules

Part (a) · 5 marks

Differentiate the curve and show dy/dx equals Ax(x² − 1)³/(x² + 1)⁵, stating the positive integer A.

1. 1d/dx[(x² − 1)/(x² + 1)] = ((x² + 1)(2x) − (x² − 1)(2x))/(x² + 1)²
2. 2= 4x/(x² + 1)²
3. 3dy/dx = 4((x² − 1)/(x² + 1))³ × ((x² + 1)(2x) − (x² − 1)(2x))/(x² + 1)²
4. 4= 4 × (x² − 1)³/(x² + 1)³ × 4x/(x² + 1)²
5. 5= 16x(x² − 1)³/(x² + 1)⁵, so A = 16

Answerdy/dx = 16x(x² − 1)³/(x² + 1)⁵ shown, A = 16

Part (bi) · 1 mark

Show the stationary points occur at x = −1, x = 0 and x = 1.

1. 1Stationary points where dy/dx = 0
2. 216x(x² − 1)³ = 0
3. 3x = 0 or x² − 1 = 0, so x = ±1

Answer16x(x² − 1)³ = 0 gives x = 0 or x² = 1, so x = −1, 0, 1 (shown)

Part (bii) · 2 marks

Apply the first derivative test to identify which two stationary points share the same nature and whether they are maxima or minima.

1. 1x < −1: dy/dx negative; −1 < x < 0: dy/dx positive, so x = −1 is a minimum
2. 2−1 < x < 0: positive; 0 < x < 1: negative, so x = 0 is a maximum
3. 30 < x < 1: negative; x > 1: positive, so x = 1 is a minimum
4. 4x = 1 and x = −1 have the same nature: both minimum points

Answerx = 1 and x = −1 are minimum points

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