These questions practise the skills in differentiation techniques: powers, products, the chain rule, ln and ex forms, and checking a result. They run from easy to harder, and each answer is fully worked. Check which functions are in scope on the current 0606 syllabus page.
How to use this set: write the rule you plan to use in your first line, work the question on paper, then open the answer. Mark the method, not only the final expression. Use the mistake log and retest queue to record any question you missed, and retry it after a day.
Questions
1. Differentiate y = 5x⁴ − 3x² + 7.
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Each term: 5x⁴ gives 20x³, −3x² gives −6x, and the constant 7 gives 0.
dy/dx = 20x³ − 6x
2. Differentiate y = 6√x − 4/x.
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Rewrite: y = 6x1/2 − 4x−1.
6x1/2 gives 3x−1/2. Then −4x−1 gives +4x−2.
dy/dx = 3/√x + 4/x²
3. Differentiate y = (x + 4)(2x − 3), by the product rule and by expanding.
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Product rule: u′ = 1 and v′ = 2, so dy/dx = (2x − 3) + 2(x + 4) = 4x + 5.
Expanding: y = 2x² + 5x − 12, so dy/dx = 4x + 5. ✓
dy/dx = 4x + 5
4. Differentiate y = (3x − 1)⁵, and find the gradient at x = 1.
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dy/dx = 5(3x − 1)⁴ × 3 = 15(3x − 1)⁴.
At x = 1: 15 × 2⁴ = 15 × 16 = 240.
5. Find the gradient of y = x³ + 2√x − 1/x at x = 4, given as an exact fraction.
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Rewrite: y = x³ + 2x1/2 − x−1. Then dy/dx = 3x² + x−1/2 + x−2.
At x = 4: 3 × 16 = 48, then 1/√4 = 1/2, then 1/16.
48 + 1/2 + 1/16 = 48 + 8/16 + 1/16 = 48 9/16 = 777/16.
6. Differentiate y = √(x² + 9) and find the gradient at x = 4.
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y = (x² + 9)1/2, so dy/dx = ½(x² + 9)−1/2 × 2x = x/√(x² + 9).
At x = 4: x² + 9 = 25, √25 = 5, so the gradient is 4/5.
7. Differentiate y = x²(x − 2)³ and factorise your answer. Find the gradient at x = 3.
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Let u = x² and v = (x − 2)³. Then u′ = 2x and v′ = 3(x − 2)².
dy/dx = 2x(x − 2)³ + 3x²(x − 2)² = x(x − 2)²[2(x − 2) + 3x] = x(x − 2)²(5x − 4).
At x = 3: 3 × 1 × 11 = 33. Direct check: 6 × 1 + 27 × 1 = 33. ✓
8. Differentiate y = e4x − 1 and find the gradient at x = ¼.
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The inside is 4x − 1 with derivative 4, so dy/dx = 4e4x − 1.
At x = ¼: 4x − 1 = 0, and e⁰ = 1, so the gradient is 4.
9. Differentiate y = ln(5x² + 1) and find the gradient at x = 1.
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The inside has derivative 10x, so dy/dx = 10x/(5x² + 1).
At x = 1: 10/6 = 5/3.
10. Differentiate y = x ln x and find the gradient at x = e.
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Let u = x and v = ln x, so u′ = 1 and v′ = 1/x.
dy/dx = ln x + x × (1/x) = ln x + 1.
At x = e: ln e = 1, so the gradient is 2.
11. Differentiate y = xe−2x. Find the value of x where the gradient is zero.
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Let u = x and v = e−2x, so u′ = 1 and v′ = −2e−2x.
dy/dx = e−2x − 2xe−2x = (1 − 2x)e−2x.
The factor e−2x is never zero, so set 1 − 2x = 0, giving x = ½.
12. A student says the gradient of y = (x² + 1)³ at x = 1 is 12, using dy/dx = 3(x² + 1)². Check with the values f(1.01) ≈ 8.24363 and f(0.99) ≈ 7.76357. Is the student right?
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Estimate: (8.24363 − 7.76357) / 0.02 = 0.48006 / 0.02 ≈ 24.0.
The student is wrong, because 12 does not match 24. The inner derivative 2x was missed: dy/dx = 3(x² + 1)² × 2x = 6x(x² + 1)².
At x = 1: 6 × 4 = 24, which matches the estimate.
If you got these wrong
| What went wrong | Questions | Go back to |
|---|---|---|
| Power rule, constants or index arithmetic | 1, 2, 5 | Differentiate a polynomial with fractional powers |
| Multiplied u′ and v′, or mixed up the product structure | 3, 7, 10, 11 | Differentiate a product where applicable |
| Missed the inner derivative | 4, 6, 7, 8, 9 | Apply a chain rule to a composite expression |
| Wrote the derivative of ln or ex incorrectly | 8, 9, 10, 11 | Differentiate logarithmic and exponential forms |
| Could not tell whether an answer was right | 12 | Check a derivative by comparing local gradients |
The calculus shape and rate explorer and the non-calculator working trainer both support this set.
When a pattern repeats across several questions, the cause is usually one habit in your first line, and it is quick to fix with a teacher watching. Our online one-to-one Additional Mathematics tuition is built for exactly that kind of review.