Study Guide
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…
- Difficulty
- Level 3 · Challenge
- Time
- 30 min
Study GuideYear 11Level 3 · Challenge
Free Year 11 Differentiation Basics: The Power Rule
Introduces the derivative as a gradient function and sets out the power rule for differentiating polynomials, including negative and fractional powers. Worked examples cover rewriting terms before differentiating, finding the gradient at a point, tangent lines and stationary points.
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
- 20 minutes
- Curriculum
- Australian Curriculum
- Answers
- Not applicable
- Format
- PDF (A4) + print
Students will practise
- understanding the derivative as a gradient function
- applying the power rule to polynomial terms
- rewriting roots and reciprocals as powers before differentiating
- finding gradients, tangents and stationary points
Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.
What's next?
Completed: Differentiation Basics: The Power Rule
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 20 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 Basics: The Power Rule
Success Tutoring
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What you need to know
The derivative of a function tells you the gradient (steepness) of its graph at any point — in other words, how fast y is changing as x changes. It is written as or f′(x). For a straight line the gradient is the same everywhere, but for a curve the gradient changes from point to point, so the derivative is itself a function of x.
The power rule. If y = xn, then [frac dy/dx] = nx[sup n − 1]. Bring the power down to the front as a multiplier, then reduce the power by 1. A constant multiplier stays where it is, and you differentiate a sum term by term.
| Function | Derivative | Why |
|---|---|---|
| x2 | 2x | power 2 comes down, power becomes 1 |
| x3 | 3x2 | power 3 comes down, power becomes 2 |
| 4x5 | 20x4 | 4 × 5 = 20, power becomes 4 |
| x | 1 | x = x1, so 1 × x0 = 1 |
| 7 (a constant) | 0 | a constant does not change |
| x−1 | −x−2 | −1 comes down, power becomes −2 |
| x1/2 | x−1/2 | comes down, power becomes − |
Differentiate y = 3x4 − 2x2 + 7x − 5
- 3x4 → 3 × 4x3 = 12x3
- −2x2 → −2 × 2x = −4x
- 7x → 7
- −5 → 0
- = 12x[sup 3] − 4x + 7
Gradient at a point: f(x) = x3 − 6x at x = 2
- f′(x) = 3x2 − 6
- f′(2) = 3 × 4 − 6 = 12 − 6 = 6
- So the curve has gradient 6 at the point where x = 2.
Rewrite first: y = √x +
- Write each term as a power of x: y = x1/2 + x−2
- Differentiate: = x−1/2 − 2x−3
- Rewrite if required: = −
Expand first: y = (x + 2)(x − 3)
- Expand: y = x2 − 3x + 2x − 6 = x2 − x − 6
- = 2x − 1
Common mistake: differentiating a constant to itself instead of 0, and forgetting to multiply the existing coefficient by the power (4x5 becomes 20x4, not 4x4). Another common slip: the power rule cannot be applied to a product or quotient directly — expand or rewrite first.
Equation of the tangent to y = x2 at x = 3
- Point: when x = 3, y = 9, so the point is (3, 9).
- Gradient: = 2x, so at x = 3 the gradient is 6.
- Line through (3, 9) with gradient 6: y − 9 = 6(x − 3)
- y = 6x − 9
A stationary point is where the gradient is zero. Set f′(x) = 0 and solve for x, then substitute back into the original function to find y. For y = x2 − 4x: = 2x − 4 = 0 gives x = 2, and y = 4 − 8 = −4, so the stationary point is (2, −4).
Check yourself
- Differentiate y = x6.
- Differentiate y = 5x3 − 2x.
- If f(x) = 2x2 + 3x, find f′(1).
- Differentiate y = .
- Differentiate y = (x + 1)2.
- Find the stationary point of y = x2 − 4x.
Answers: 1. 6x5 2. 15x2 − 2 3. f′(x) = 4x + 3, so f′(1) = 7 4. y = x−1, so = −x−2 = − 5. Expand to x2 + 2x + 1, so = 2x + 2 6. (2, −4)
Preview — the PDF contains the full resource.
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- Last reviewed
- 1 October 2026
- How it was made
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