Mathematics Advanced • Year 11 • Module 3 • Lesson 6
Product and Quotient Rules
Build procedural fluency in choosing and applying the product rule, the quotient rule, and "simplify first" for differentiation.
1. Quick recall
Answer each in the space provided. 1 mark each
Q1.1 Complete the formulas. Use brackets exactly as written.
Product rule: if y = u·v, then y′ = ____________ + ____________
Quotient rule: if y = u / v, then y′ = ( ____________ − ____________ ) / ____________
Q1.2 For each expression, decide which rule is the most efficient first move: product, quotient, or simplify first.
(a) y = x² · x⁵ → ______________________
(b) y = x³ · sin x → ______________________
(c) y = (x² + 3x) / x → ______________________
Q1.3 The quotient rule has the denominator ______________ . The numerator is u′v ____________ uv′ (state the sign).
2. Worked example, y = x² sin x (product rule)
Follow the four-step pattern: name u and v, find u′ and v′, substitute, simplify.
Problem. Differentiate y = x² · sin x.
Step 1, Name the two factors.
Let u = x² and v = sin x
Reason: a product splits into two pieces; we must differentiate each separately.
Step 2, Differentiate each factor.
u′ = 2x and v′ = cos x
Reason: power rule for x²; derivative of sin is cos.
Step 3, Substitute into y′ = u′v + uv′.
y′ = (2x)(sin x) + (x²)(cos x)
Reason: the product rule has two cross-terms, never just one.
Step 4, Tidy.
y′ = 2x sin x + x² cos x
Conclusion. y′ = 2x sin x + x² cos x.
3. Faded example, fill in the missing steps
Differentiate y = x³ / (x + 1) using the quotient rule. Fill in each blank. 4 marks
Step 1, Name u and v:
u = ______________ v = ______________
Step 2, Differentiate each:
u′ = ______________ v′ = ______________
Step 3, Substitute into y′ = (u′v − uv′) / v²:
y′ = [ ( ______ )( ______ ) − ( ______ )( ______ ) ] / ( ______ )²
Step 4, Expand the numerator and simplify:
y′ = ____________________________ / (x + 1)²
Conclusion. y′ = ____________________________
4. Graduated practice, differentiate each function
State which rule you used. Show your u, v, u′, v′ work for every Standard and Extension question.
Foundation, clean numbers, no surprises (4 questions)
| Q | Function | Rule chosen | y′ |
|---|---|---|---|
| 4.1 1 | y = x · x⁴ (simplify first if you can) | ||
| 4.2 1 | y = x² · ex | ||
| 4.3 1 | y = (x² + 3x) / x | ||
| 4.4 1 | y = x · cos x |
Standard, typical HSC difficulty (6 questions)
Show u, v, u′, v′ then substitute. Leave answers in unfactored or factored form as you prefer.
4.5 y = x³ sin x 2 marks
4.6 y = (2x + 1) / (x − 3) 2 marks
4.7 y = (x + 1)(x² − 2) (try both methods: product rule, then expand-then-differentiate) 2 marks
4.8 y = (cos x) / x 2 marks
4.9 y = x² · ln x 2 marks
4.10 y = sin x / (x² + 1) 2 marks
Extension, combine rules (2 questions)
4.11 Differentiate y = x² · sin(3x). Identify which rules you used at each step. 3 marks
4.12 Find the gradient of y = x² · ex at x = 0. Show your derivative first, then substitute. 3 marks
5. Self-check the easy 3
Tick the first three once you have checked your method.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, Formulas
Product: y′ = u′v + uv′. Quotient: y′ = (u′v − uv′) / v².
Q1.2, Rule choice
(a) x² · x⁵ → simplify first to x⁷, then power rule. (b) x³ sin x → product rule. (c) (x² + 3x)/x → simplify first to x + 3, then power rule.
Q1.3, Quotient rule shape
Denominator is v² (the original denominator, squared). Numerator sign is − (minus): u′v − uv′. The order matters, swapping gives the wrong sign.
Q3, Faded example y = x³ / (x + 1)
Step 1: u = x³, v = x + 1.
Step 2: u′ = 3x², v′ = 1.
Step 3: y′ = [(3x²)(x + 1) − (x³)(1)] / (x + 1)².
Step 4: numerator = 3x³ + 3x² − x³ = 2x³ + 3x².
Conclusion: y′ = (2x³ + 3x²) / (x + 1)², or equivalently x²(2x + 3) / (x + 1)².
Q4.1, y = x · x⁴
Simplify first: y = x⁵, so y′ = 5x⁴. (Using the product rule gives the same answer but is slower.)
Q4.2, y = x² · ex
Product rule. u = x², v = ex; u′ = 2x, v′ = ex. y′ = 2x · ex + x² · ex = ex(2x + x²) = x ex(2 + x).
Q4.3, y = (x² + 3x)/x
Simplify first: y = x + 3, so y′ = 1. (The quotient rule would give the same answer through more algebra.)
Q4.4, y = x cos x
Product rule. u = x, v = cos x; u′ = 1, v′ = −sin x. y′ = (1)(cos x) + (x)(−sin x) = cos x − x sin x.
Q4.5, y = x³ sin x
Product rule. u = x³, v = sin x; u′ = 3x², v′ = cos x. y′ = 3x² sin x + x³ cos x.
Q4.6, y = (2x + 1)/(x − 3)
Quotient rule. u = 2x + 1, v = x − 3; u′ = 2, v′ = 1. y′ = [2(x − 3) − (2x + 1)(1)] / (x − 3)² = (2x − 6 − 2x − 1) / (x − 3)² = −7 / (x − 3)².
Q4.7, y = (x + 1)(x² − 2)
Method A (product rule): u = x + 1, v = x² − 2; u′ = 1, v′ = 2x. y′ = (1)(x² − 2) + (x + 1)(2x) = x² − 2 + 2x² + 2x = 3x² + 2x − 2.
Method B (expand first): y = x³ + x² − 2x − 2, so y′ = 3x² + 2x − 2. ✓ Same answer.
Q4.8, y = cos x / x
Quotient rule. u = cos x, v = x; u′ = −sin x, v′ = 1. y′ = [(−sin x)(x) − (cos x)(1)] / x² = (−x sin x − cos x) / x².
Q4.9, y = x² ln x
Product rule. u = x², v = ln x; u′ = 2x, v′ = 1/x. y′ = (2x)(ln x) + (x²)(1/x) = 2x ln x + x = x(2 ln x + 1).
Q4.10, y = sin x / (x² + 1)
Quotient rule. u = sin x, v = x² + 1; u′ = cos x, v′ = 2x. y′ = [cos x · (x² + 1) − sin x · (2x)] / (x² + 1)².
Q4.11, y = x² sin(3x)
Rules used: product rule (because x² and sin(3x) are multiplied), and inside it chain rule (for sin(3x)).
u = x², v = sin(3x). u′ = 2x. For v′, chain rule: d/dx[sin(3x)] = 3 cos(3x). y′ = (2x)(sin(3x)) + (x²)(3 cos(3x)) = 2x sin(3x) + 3x² cos(3x).
Q4.12, gradient of y = x² ex at x = 0
From Q4.2: y′ = ex(2x + x²). At x = 0: y′(0) = e⁰(0 + 0) = 1 · 0 = 0. The curve has a horizontal tangent at the origin.