Study Guide
Differentiation Basics: The Power Rule
Introduces the derivative as a gradient function and sets out the power rule for differentiating polynomials, including negative and…
- Difficulty
- Level 3 · Challenge
- Time
- 20 min
Study GuideYear 11Level 3 · ChallengeHSCVCE
Free Year 11 Differentiation: Product, Quotient and Chain Rules
An HSC/VCE-style study guide to the three rules that let you differentiate products, quotients and composite functions. Each worked example names u and v (or the inside and outside function) before differentiating, then simplifies the result by factorising, and one example uses the derivative to find a gradient at a point. Includes a six-question self-check with answers.
Instant download · No sign-up · Free for personal, classroom and homeschool use
- Year
- Year 11
- Subject
- Mathematics
- Topic
- Calculus
- Difficulty
- Level 3 · Challenge
- Estimated time
- 30 minutes
- Curriculum
- Australian Curriculum
- Answers
- Not applicable
- Format
- PDF (A4) + print
Students will practise
- differentiating a product of two functions using the product rule
- differentiating a quotient using the quotient rule
- differentiating a composite function using the chain rule, including square roots
- choosing and combining rules, then simplifying by factorising
Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.
What's next?
Completed: Differentiation: Product, Quotient and Chain Rules
How to use this study guide
- Read it together first, pausing at each worked example to try the step before reading the answer.
- Attempt the "Check yourself" questions at the end without looking back.
- Then practise with a worksheet from the pathway above and finish with the topic test.
Common questions
Who is this study guide for?
Year 11 students (typically ages 16–17) working on calculus. It is pitched at level 3 · challenge.
Are the answers included?
This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.
How long does it take?
About 30 minutes. Short, regular sessions work best: two or three a week beats one long one.
Do I need to sign up to download?
No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.
What should we do next?
Try Year 11 Differentiation (Power Rule) Worksheet — Level 1.
Preview
Year 11 · Mathematics · Calculus
Differentiation: Product, Quotient and Chain Rules
Success Tutoring
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What you need to know
The power rule handles sums of powers of x, but most functions you meet in senior maths are products, quotients or functions inside other functions. Three rules cover all of these. The method is always the same: name the pieces, differentiate each piece on its own, then assemble the result with the rule.
| Rule | If y = … | Then = … |
|---|---|---|
| Product rule | u × v | u′v + uv′ |
| Quotient rule | ||
| Chain rule | f(g(x)), i.e. an inside function g inside an outside function f | f′(g(x)) × g′(x) — differentiate the outside, keep the inside, multiply by the derivative of the inside |
| Chain rule shortcut | (ax + b)n | n(ax + b)n − 1 × a |
Product rule: y = (2x + 1)(x2 − 3)
- u = 2x + 1, so u′ = 2. v = x2 − 3, so v′ = 2x.
- = u′v + uv′ = 2(x2 − 3) + (2x + 1)(2x)
- = 2x2 − 6 + 4x2 + 2x
- = 6x[sup 2] + 2x − 6
- Check by expanding first: y = 2x3 + x2 − 6x − 3, so = 6x2 + 2x − 6. Same answer.
Chain rule: y = (3x2 − 5)4
- Inside: g(x) = 3x2 − 5, so g′(x) = 6x. Outside: ( )4, whose derivative is 4( )3.
- = 4(3x2 − 5)3 × 6x
- = 24x(3x[sup 2] − 5)[sup 3]
Chain rule with a square root: y = √(x2 + 9)
- Rewrite: y = (x2 + 9)1/2. Inside derivative: 2x.
- = (x2 + 9)−1/2 × 2x
- = = [frac x/√(x[sup 2] + 9)]
Quotient rule: y =
- u = x2 + 1, so u′ = 2x. v = 2x − 3, so v′ = 2.
- = =
- Numerator: 4x2 − 6x − 2x2 − 2 = 2x2 − 6x − 2
- = [frac 2x[sup 2] − 6x − 2/(2x − 3)[sup 2]] (leave the denominator as a power — do not expand it)
Product and chain together: y = x(x + 1)5
- u = x, so u′ = 1. v = (x + 1)5, so v′ = 5(x + 1)4 (chain rule, inside derivative 1).
- = 1 × (x + 1)5 + x × 5(x + 1)4
- Factorise out the common factor (x + 1)4: = (x + 1)4[(x + 1) + 5x]
- = (x + 1)[sup 4](6x + 1)
Find the gradient of y = (x2 − 1)3 at x = 2
- = 3(x2 − 1)2 × 2x = 6x(x2 − 1)2
- At x = 2: 6(2)(4 − 1)2 = 12 × 9 = 108
Common mistake: forgetting the inside derivative in the chain rule. The derivative of (3x2 − 5)4 is not 4(3x2 − 5)3; you must multiply by 6x. If the inside is just x, the inside derivative is 1 and nothing changes — that is why the power rule is a special case.
Common mistake: swapping the order in the quotient rule. It is u′v − uv′ ("derivative of the top times the bottom, minus the top times the derivative of the bottom"), all over v2. The order matters because of the minus sign. In the product rule the order does not matter.
Check yourself
- Differentiate y = (5x − 2)3.
- Differentiate y = (x + 4)(x2 − 1) using the product rule.
- Differentiate y = .
- Differentiate y = √(4x + 1).
- Differentiate y = (x2 + 3)−2.
- Find the gradient of y = (2x − 1)4 at x = 1.
Answers: 1. 15(5x − 2)2 2. (x2 − 1) + (x + 4)(2x) = 3x2 + 8x − 1 3. = 4. (4x + 1)−1/2 × 4 = 5. −2(x2 + 3)−3 × 2x = 6. = 8(2x − 1)3, so the gradient is 8
Preview — the PDF contains the full resource.
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- Last reviewed
- 1 October 2026
- How it was made
- Written by our team
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