Mathematics Advanced • Year 11 • Module 3 • Lesson 6
Product and Quotient Rules
Practise HSC-style writing on product/quotient differentiation, including a structured tangent + stationary-point response.
1. Short-answer questions
1.1 Differentiate y = x³ sin x with respect to x. Show your u, v, u′, v′ explicitly. 3 marks Band 3-4
1.2 Differentiate y = (2x + 1) / (x − 3). Give your answer in fully simplified form. 3 marks Band 3-4
1.3 Find dy/dx for y = x² sin(3x), naming each differentiation rule used. 4 marks Band 4
Stuck on 1.3? You will need product rule (for the multiplication) and chain rule (for sin(3x)).2. Extended response
2.1 The curve C is defined by f(x) = x / (x² + 1).
(a) Show that f′(x) = (1 − x²) / (x² + 1)².
(b) Hence find the coordinates of the stationary points of C and classify each using the sign of f′(x) on either side.
(c) Write down the equation of the tangent to C at the origin in the form y = mx, and explain in one sentence why this tangent meets C at exactly one point. 7 marks Band 5-6
Explicit marking criteria
Part (a), 2 marks
• 1 mark applies quotient rule correctly with u = x, v = x² + 1, u′ = 1, v′ = 2x.
• 1 mark simplifies the numerator (x² + 1) − 2x² to obtain 1 − x², and writes the final form (1 − x²)/(x² + 1)².
Part (b), 3 marks
• 1 mark solves 1 − x² = 0 to give x = ±1, then computes y-coordinates (1, 1/2) and (−1, −1/2).
• 1 mark uses a sign table (or second derivative) to classify (1, 1/2) as local maximum.
• 1 mark classifies (−1, −1/2) as local minimum.
Part (c), 2 marks
• 1 mark finds f′(0) = 1 and f(0) = 0, then writes the tangent y = x.
• 1 mark explains that solving x/(x² + 1) = x gives x · (x²)/(x² + 1) = 0, i.e. only x = 0, so the tangent meets C at the origin only.
Your response:
Stuck on (c)? Substitute the tangent equation y = x into the curve equation and solve.How did this worksheet feel?
What I'll revisit before next class:
1.1, y = x³ sin x (3 marks)
Sample response. Let u = x³ and v = sin x. Then u′ = 3x² and v′ = cos x. Using the product rule y′ = u′v + uv′,
y′ = (3x²)(sin x) + (x³)(cos x) = 3x² sin x + x³ cos x.
Marking notes. 0.5, names u, v. 0.5, finds u′, v′ correctly. 2, substitutes into the product rule and writes the final answer in either expanded or factored x²(3 sin x + x cos x) form. A response that "just writes" the answer without u, v, u′, v′ scores 2/3 only.
1.2, y = (2x + 1)/(x − 3) (3 marks)
Sample response. Let u = 2x + 1 and v = x − 3. Then u′ = 2 and v′ = 1. By the quotient rule,
y′ = [u′v − uv′] / v² = [2(x − 3) − (2x + 1)(1)] / (x − 3)² = (2x − 6 − 2x − 1) / (x − 3)² = −7 / (x − 3)².
Marking notes. 0.5, names u, v. 0.5, finds u′, v′. 1, substitutes into quotient rule with the minus sign correctly in the numerator. 1, simplifies the numerator. Sign errors in the numerator (using +7 instead of −7) lose 1 mark.
1.3, y = x² sin(3x) (4 marks)
Sample response. Rules used: product rule (for x² · sin(3x)) and chain rule (inside sin(3x)).
Let u = x² and v = sin(3x). Then u′ = 2x. For v′, by the chain rule with inner function w = 3x and outer sin(w), v′ = cos(3x) · 3 = 3 cos(3x).
By the product rule: y′ = (2x)(sin(3x)) + (x²)(3 cos(3x)) = 2x sin(3x) + 3x² cos(3x).
Marking notes. 1, names both rules and identifies u, v. 1, correct u′ and v′ (with the factor of 3 from the chain rule). 2, correct application of product rule and final simplified answer. Omitting the chain-rule factor of 3 in v′ is the most common error (loses 1 mark for v′ and 1 mark for the final answer).
2.1, f(x) = x / (x² + 1) (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Let u = x and v = x² + 1, so u′ = 1 and v′ = 2x.
f′(x) = (u′v − uv′) / v² = [(1)(x² + 1) − (x)(2x)] / (x² + 1)² [1 mark, quotient rule with correct u, v, u′, v′.]
= (x² + 1 − 2x²) / (x² + 1)² = (1 − x²) / (x² + 1)². [1 mark, numerator simplified.]
Part (b). Stationary points occur where f′(x) = 0. Since (x² + 1)² > 0 for all x, this reduces to 1 − x² = 0, giving x = ±1.
At x = 1: f(1) = 1/2, so a stationary point at (1, 1/2). At x = −1: f(−1) = −1/2, so a stationary point at (−1, −1/2). [1 mark, both stationary points found with y-coordinates.]
Sign of f′(x) (driven by sign of 1 − x²):
x: −2 −1 0 1 2
1 − x²: − 0 + 0 −
f′(x): − 0 + 0 −
At x = −1: sign changes − → +, so (−1, −1/2) is a local minimum. [1 mark, minimum.]
At x = +1: sign changes + → −, so (1, 1/2) is a local maximum. [1 mark, maximum.]
Part (c). f(0) = 0 and f′(0) = (1 − 0)/(0 + 1)² = 1. Tangent at the origin: y = x. [1 mark, tangent equation.]
To find intersections of the tangent with C, set x/(x² + 1) = x. Multiplying through (x² + 1 > 0): x = x(x² + 1), so x · x² = 0, i.e. x³ = 0, giving x = 0 only. Therefore the tangent meets C only at the origin (it is tangent everywhere else either above or below the curve). [1 mark, unique intersection justified.]
Total: 7/7.
Band descriptors for marker.
Band 3: Sets up quotient rule but has algebra slip in numerator; finds one stationary point only; no classification or tangent. ≈ 2-3 marks.
Band 4: Completes part (a) cleanly. Finds both stationary points but classifies only one, or classifies by inspection without justification. ≈ 4-5 marks.
Band 5: Both stationary points classified with a sign table or second derivative. Tangent equation correct but no uniqueness reasoning. ≈ 5-6 marks.
Band 6: All three parts complete; uses (x² + 1)² > 0 to reduce f′(x) = 0 to numerator only; tangent equation supported by both f(0) and f′(0); uniqueness shown algebraically. 7/7.