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.