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Year 12 Mathematics · Differentiation rules and applications

Name: ______________________

Date: ____________

1. 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
5. 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?

Answer sheet

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.