60-minute lesson
Year 13 Mathematics: Differentiation rules and applications
Lesson objective
- Differentiate polynomials using the power rule.
- Use the chain rule and product rule for composite functions and products.
- Find and classify stationary points.
- Solve a simple optimisation problem.
Success criteria:
- I can differentiate expressions such as 2x^3 − 5x^2 + 4 and (3x + 1)^4.
- I can use the product rule on an expression such as x^2(x + 1).
- I can find a stationary point and decide whether it is a maximum or minimum.
- I can set up and solve a maximum-area problem.
Materials
- Paper and pencil
- A calculator
- A length of string (for the area investigation)
- Grid paper
Introduction
6 minIntroduce today's words:
- Derivative: The gradient function of a curve, written f'(x) or dy/dx.
- Chain rule: A rule for differentiating a function of a function: dy/dx = dy/du × du/dx.
- Product rule: A rule for differentiating a product: if y = uv, then dy/dx = u'v + uv'.
- Stationary point: A point on a curve where the gradient is zero, so dy/dx = 0.
- Optimisation: Finding the largest or smallest possible value of a quantity, such as maximum area or minimum cost.
Ask your child what they already know about differentiation rules and applications.
Explanation
12 minRecall the power rule: the derivative of x^n is n x^(n − 1). Constants multiplying a term stay as multipliers, the derivatives of separate terms are found one at a time, and the derivative of a constant is 0. So the derivative of 2x^3 − 5x^2 + 4 is 6x^2 − 10x.
The chain rule handles a function inside another function, such as (3x + 1)^4. Treat the bracket as a single quantity u = 3x + 1, so y = u^4. Then dy/dx = (derivative of the outside) × (derivative of the inside) = 4u^3 × 3 = 12(3x + 1)^3.
The product rule handles two expressions multiplied together. If y = uv, then dy/dx = u'v + uv'. For y = x^2(x + 1): u = x^2 and v = x + 1, so u' = 2x and v' = 1, giving dy/dx = 2x(x + 1) + x^2 × 1 = 3x^2 + 2x. You can check by expanding first: x^3 + x^2 differentiates to 3x^2 + 2x.
A stationary point is where the tangent is horizontal, so dy/dx = 0. To find one, differentiate, set the derivative equal to zero, solve for x, then substitute back into the original function to find y.
To classify a stationary point, check the gradient just either side of it. If the gradient changes from negative to positive, it is a minimum turning point; from positive to negative, a maximum. For a parabola, the sign of the x^2 coefficient also tells you: positive opens upwards (minimum), negative opens downwards (maximum).
Optimisation problems use the same idea. Write the quantity to be maximised or minimised as a function of one variable, differentiate, set the derivative equal to zero and solve. Always check that the answer makes sense in the context.
Worked examples
9 minUsing the chain rule
Differentiate y = (3x + 1)^4.
Step 1: Let u = 3x + 1, so y = u^4.
Step 2: Derivative of the outside: 4u^3. Derivative of the inside: du/dx = 3.
Step 3: Multiply: dy/dx = 4(3x + 1)^3 × 3 = 12(3x + 1)^3.
Answer: dy/dx = 12(3x + 1)^3
Finding and classifying a stationary point
Find the stationary point of y = x^2 − 6x + 5 and state its nature.
Step 1: dy/dx = 2x − 6.
Step 2: Set dy/dx = 0: 2x − 6 = 0, so x = 3.
Step 3: Find y: 3^2 − 6 × 3 + 5 = 9 − 18 + 5 = −4.
Step 4: At x = 2 the gradient is −2 and at x = 4 it is +2, so the gradient changes from negative to positive.
Answer: A minimum turning point at (3, −4).
Maximum area
A rectangular garden bed has a perimeter of 20 m. Find the largest possible area.
Step 1: Let the width be x m. Since 2 × (length + width) = 20, the length is 10 − x m.
Step 2: Area A = x(10 − x) = 10x − x^2.
Step 3: dA/dx = 10 − 2x. Setting this to 0 gives x = 5.
Step 4: The gradient changes from positive to negative at x = 5, so this is a maximum. Area = 5 × 5 = 25.
Answer: 25 m^2 (a 5 m by 5 m square)
Guided practice (do together)
12 min1. Differentiate 4x^3.
(a) 12x^2 (b) 4x^2 (c) 12x^3 (d) 3x^4
2. Differentiate y = (2x + 5)^3.
(a) 3(2x + 5)^2 (b) 6(2x + 5)^2 (c) 6(2x + 5)^3 (d) 2(2x + 5)^2
3. Stationary points of a curve occur where:
(a) y = 0 (b) x = 0 (c) dy/dx = 0 (d) dy/dx = 1
4. Use the product rule to differentiate y = x^2(x + 1).
(a) 2x (b) 3x^2 + 1 (c) 2x^2 + 2x (d) 3x^2 + 2x
Independent practice
15 min5. Which describes the stationary point of y = −x^2 + 4x?
(a) A minimum at (2, 4) (b) A maximum at (2, 4) (c) A maximum at (4, 0) (d) A minimum at (0, 0)
6. If f(x) = x^4 − 2x, find f'(1).
7. Find the x-coordinate of the stationary point of y = x^2 + 8x + 1.
8. A rectangle has a perimeter of 36 cm. What is its largest possible area, in cm^2?
Questions to check understanding
- Can you differentiate expressions such as 2x^3 − 5x^2 + 4 and (3x + 1)^4?
- Can you use the product rule on an expression such as x^2(x + 1)?
- Can you find a stationary point and decide whether it is a maximum or minimum?
- Can you set up and solve a maximum-area problem?
- What was the trickiest part today?
Answer guide
1. 12x^2 — Bring down the power and reduce it by 1: 4 × 3x^2 = 12x^2.
2. 6(2x + 5)^2 — Chain rule: 3(2x + 5)^2 × 2 = 6(2x + 5)^2.
3. dy/dx = 0 — At a stationary point the tangent is horizontal, so the gradient dy/dx equals 0.
4. 3x^2 + 2x — u'v + uv' = 2x(x + 1) + x^2 × 1 = 2x^2 + 2x + x^2 = 3x^2 + 2x.
5. A maximum at (2, 4) — dy/dx = −2x + 4 = 0 gives x = 2, and y = −4 + 8 = 4. The x^2 coefficient is negative, so the parabola opens downwards and the point is a maximum.
6. 2 — f'(x) = 4x^3 − 2, so f'(1) = 4 − 2 = 2.
7. x = −4 — dy/dx = 2x + 8 = 0 gives x = −4.
8. 81 cm^2 — With width x, the length is 18 − x and A = 18x − x^2. dA/dx = 18 − 2x = 0 gives x = 9, so the maximum area is 9 × 9 = 81 cm^2.
Review
6 minRecap the success criteria together. Watch for these common misconceptions:
- Forgetting to multiply by the derivative of the inside when using the chain rule, such as differentiating (2x + 5)^3 as 3(2x + 5)^2.
- Thinking the derivative of a product is the product of the derivatives. The derivative of x^2(x + 1) is not 2x × 1.
- Stopping after finding x at a stationary point. A point needs both coordinates, and its nature should be checked.
- Assuming every stationary point is a maximum or minimum. Some curves, such as y = x^3 at x = 0, have a stationary point where the curve levels off and keeps rising.
Extension activities
- Use a 40 cm length of string to form different rectangles. Measure and record their areas, then compare your best result with the calculus answer.
- A box with an open top is made by cutting equal squares from the corners of a 30 cm by 30 cm sheet of card and folding up the sides. Find the size of square that gives the largest volume.
- Find the stationary points of y = x^3 − 3x and sketch the curve.
Suggested follow-up
- Revisit differentiation rules and applications tomorrow with two or three quick questions from memory.
- Try the online practice check for this topic and look at any questions that need another go.
Tips for parents
- Encourage your learner to check chain and product rule answers by expanding and differentiating where that is practical.
- Ask your learner to sketch a quick graph to see whether a stationary point should be a maximum or a minimum.
- Optimisation problems are easier when the learner draws and labels a diagram first. Prompt for this before any algebra.