Mathematics Advanced • Year 11 • Module 3 • Lesson 6
Product and Quotient Rules
Apply the product and quotient rules to multi-step problems set in business, motion and modelling contexts.
Problem 1 — Revenue from price × quantity (financial)
A small cafe sells x cups of coffee per day at a price of p(x) = 8 − 0.05x dollars per cup (price falls as quantity grows). Revenue is
R(x) = x · p(x) = x · (8 − 0.05x), 0 ≤ x ≤ 160
Set up: What are we solving for?
(i) Differentiate R(x) using the product rule, treating x and (8 − 0.05x) as u and v. Show u, v, u′, v′ explicitly. 3 marks
(ii) Expand R(x) first, then differentiate using the power rule. Confirm the two methods give the same R′(x). 2 marks
(iii) Find the value of x at which R′(x) = 0 and state the maximum daily revenue. 2 marks
Stuck? Revisit lesson § Concept (why product rule has two cross-terms).Problem 2 — Population with growth and decay (product rule)
A pest population in a Sydney greenhouse, t weeks after introducing a biological control, is modelled by
N(t) = 50 t² · e−0.5t, t ≥ 0
Set up: What are we solving for?
(i) Find N′(t) using the product rule with u = 50t² and v = e−0.5t. Recall d/dt[ekt] = k ekt. 3 marks
(ii) Factorise N′(t) and find the value(s) of t for which N′(t) = 0 (with t ≥ 0). 2 marks
(iii) Interpret your answer to (ii) in the context: at what week does the pest population peak, and what happens to N(t) for very large t? 2 marks
Problem 3 — Average cost per item (quotient rule)
The total cost of producing x widgets, in dollars, is C(x) = x² + 100x + 900. The average cost per item is
A(x) = C(x) / x = (x² + 100x + 900) / x, x > 0
Set up: What are we solving for?
(i) Find A′(x) using the quotient rule. Show u, v, u′, v′. 3 marks
(ii) A classmate solves the same problem by writing A(x) = x + 100 + 900/x = x + 100 + 900 x−1, then differentiating directly. Find A′(x) by this method and confirm it equals your answer from (i). 2 marks
(iii) Set A′(x) = 0 to find the production level x that minimises average cost. Justify briefly that this is a minimum. 2 marks
Stuck on (ii)? "Simplify first" can be just as valid as a rule, and often quicker.Problem 4 — Velocity of a damped spring (combined rules)
The position of a mass on a damped spring, t seconds after release, is
s(t) = e−t · cos(2t), t ≥ 0 (metres)
Set up: What are we solving for?
(i) Find v(t) = s′(t) using the product rule. Recall d/dt[e−t] = −e−t and d/dt[cos(2t)] = −2 sin(2t). 3 marks
(ii) Factor e−t out of your answer to (i). 1 mark
(iii) Find v(0) and interpret its sign physically (is the mass moving in the positive or negative s-direction at the instant of release?). 2 marks
Problem 5 — Tangent to a quotient curve (geometric)
Consider the curve y = x² / (x + 2), defined for x ≠ −2.
Set up: What are we solving for?
(i) Find dy/dx using the quotient rule. Simplify the numerator. 3 marks
(ii) Find the gradient and y-coordinate of the curve at x = 2, hence write the equation of the tangent there in the form y = mx + b. 3 marks
(iii) The curve has a horizontal tangent where dy/dx = 0. Find the x-values where this occurs. 2 marks
Stuck on (iii)? A fraction is zero when its numerator is zero (and denominator is non-zero).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Revenue
Set up. We are finding the rate at which revenue changes with quantity sold, then the quantity that maximises revenue.
(i) u = x, v = 8 − 0.05x. u′ = 1, v′ = −0.05. R′(x) = (1)(8 − 0.05x) + (x)(−0.05) = 8 − 0.05x − 0.05x = 8 − 0.1x.
(ii) Expand: R(x) = 8x − 0.05x². Differentiate: R′(x) = 8 − 0.1x. ✓ Same answer.
(iii) R′(x) = 0 ⇒ 8 = 0.1x ⇒ x = 80 cups. Maximum daily revenue R(80) = 80 · (8 − 4) = 80 · 4 = $320.
Problem 2 — Pest population
Set up. We are finding when the population stops growing and starts shrinking.
(i) u = 50t², v = e−0.5t; u′ = 100t, v′ = −0.5 e−0.5t. N′(t) = (100t)(e−0.5t) + (50t²)(−0.5 e−0.5t) = 100t e−0.5t − 25t² e−0.5t.
(ii) Factor: N′(t) = 25 t · e−0.5t (4 − t). e−0.5t > 0 always, so N′(t) = 0 when t = 0 or t = 4.
(iii) t = 0 is the starting point (N = 0). t = 4 weeks is the peak: for 0 < t < 4 we have (4 − t) > 0 so N′(t) > 0 (population growing), and for t > 4 we have N′(t) < 0 (population shrinking). For large t the exponential decay dominates and N(t) → 0: the biological control eventually wins.
Problem 3 — Average cost
Set up. We are differentiating an average cost in two ways (quotient rule and simplify-first) and using it to find the most cost-efficient production level.
(i) u = x² + 100x + 900, v = x; u′ = 2x + 100, v′ = 1. A′(x) = [(2x + 100)(x) − (x² + 100x + 900)(1)] / x² = (2x² + 100x − x² − 100x − 900) / x² = (x² − 900) / x².
(ii) A(x) = x + 100 + 900 x−1 ⇒ A′(x) = 1 − 900 x−2 = 1 − 900/x² = (x² − 900)/x². ✓ Same as (i).
(iii) A′(x) = 0 ⇒ x² = 900 ⇒ x = 30 (rejecting x = −30 since x > 0). Minimum confirmed because A′(x) < 0 for x < 30 and A′(x) > 0 for x > 30 (sign change from − to +). Minimum average cost A(30) = 30 + 100 + 30 = $160 per widget.
Problem 4 — Damped spring
Set up. We are computing the velocity of an oscillating mass whose amplitude shrinks exponentially.
(i) u = e−t, v = cos(2t); u′ = −e−t, v′ = −2 sin(2t). v(t) = (−e−t)(cos(2t)) + (e−t)(−2 sin(2t)) = −e−t cos(2t) − 2 e−t sin(2t).
(ii) v(t) = −e−t [cos(2t) + 2 sin(2t)].
(iii) v(0) = −e⁰ [cos 0 + 2 sin 0] = −1 · [1 + 0] = −1 m/s. The sign is negative, so at the instant of release the mass is moving in the negative s-direction (back toward equilibrium from its initial positive displacement s(0) = e⁰ · cos 0 = 1 m).
Problem 5 — Tangent to y = x²/(x + 2)
Set up. We are differentiating a rational function with the quotient rule, then evaluating the derivative to build a tangent line and to find where the tangent is horizontal.
(i) u = x², v = x + 2; u′ = 2x, v′ = 1. dy/dx = [2x(x + 2) − x²(1)] / (x + 2)² = (2x² + 4x − x²) / (x + 2)² = (x² + 4x) / (x + 2)² = x(x + 4) / (x + 2)².
(ii) At x = 2: y = 4/4 = 1, so the point is (2, 1). Gradient m = (4 + 8) / 16 = 12/16 = 3/4. Tangent: y − 1 = (3/4)(x − 2) ⇒ y = (3/4)x − 1/2.
(iii) dy/dx = 0 ⇒ numerator x(x + 4) = 0 ⇒ x = 0 or x = −4. Both lie in the domain x ≠ −2, so the curve has horizontal tangents at x = 0 and x = −4.