Year 12 · Mathematics · Algebra
Differentiation rules and applications
Use the power, product and chain rules to differentiate functions, and apply derivatives to find stationary points and solve optimisation problems.
Awaiting educator review. This material was drafted by our content team with AI assistance and is waiting for review by a qualified educator. Please check it suits your child before using it. Created 10 October 2026.
Learning goals
Learners will
- 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 looks like
- 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.
The big idea
Recall 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
Using the chain rule
Differentiate y = (3x + 1)^4.
- Let u = 3x + 1, so y = u^4.
- Derivative of the outside: 4u^3. Derivative of the inside: du/dx = 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.
- dy/dx = 2x − 6.
- Set dy/dx = 0: 2x − 6 = 0, so x = 3.
- Find y: 3^2 − 6 × 3 + 5 = 9 − 18 + 5 = −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.
- Let the width be x m. Since 2 × (length + width) = 20, the length is 10 − x m.
- Area A = x(10 − x) = 10x − x^2.
- dA/dx = 10 − 2x. Setting this to 0 gives x = 5.
- 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)
Practice check
Have a go, then check your answers. Each answer comes with an explanation.
Answer guide for parents
Differentiate 4x^3.
12x^2 — Bring down the power and reduce it by 1: 4 × 3x^2 = 12x^2.
Differentiate y = (2x + 5)^3.
6(2x + 5)^2 — Chain rule: 3(2x + 5)^2 × 2 = 6(2x + 5)^2.
Stationary points of a curve occur where:
dy/dx = 0 — At a stationary point the tangent is horizontal, so the gradient dy/dx equals 0.
Use the product rule to differentiate y = x^2(x + 1).
3x^2 + 2x — u'v + uv' = 2x(x + 1) + x^2 × 1 = 2x^2 + 2x + x^2 = 3x^2 + 2x.
Which describes the stationary point of y = −x^2 + 4x?
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.
If f(x) = x^4 − 2x, find f'(1).
2 — f'(x) = 4x^3 − 2, so f'(1) = 4 − 2 = 2.
Find the x-coordinate of the stationary point of y = x^2 + 8x + 1.
x = −4 — dy/dx = 2x + 8 = 0 gives x = −4.
A rectangle has a perimeter of 36 cm. What is its largest possible area, in cm^2?
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.
Watch out for
- 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.
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.
Go further
- 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.