IGCSE Additional Mathematics: Differentiation Practice Questions
Differentiating x to the power n gives n times x to the power n minus 1. The derivative is the gradient function, so setting it to zero locates stationary points, and the second derivative classifies them.
Topic 15 of Cambridge IGCSE Additional Mathematics 0606 carries a large share of the marks on both papers. Classifying stationary points is where candidates lose marks by stating the answer without the evidence. The questions below insist on the second derivative test.
What you need to know for Differentiation
- Power ruleThe derivative of xn is n xn-1. Constants differentiate to zero.
- Chain ruleFor y = (ax + b)n, the derivative is n a (ax + b)n-1. Differentiate the outside, then multiply by the derivative of the inside.
- Stationary pointsWhere the derivative equals zero. Solve dy over dx = 0 to find the x coordinates, then substitute back for the y coordinates.
- Second derivative testIf the second derivative is positive the point is a minimum, and if negative a maximum. State the value, not just the conclusion.
- Tangents and normalsThe gradient of the tangent at a point is the derivative evaluated there. The normal gradient is its negative reciprocal.
IGCSE Additional Mathematics Differentiation questions and answers
4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.
Differentiate y = 3x4 minus 2x2 + 7 with respect to x.
Show the worked answer
- Differentiate each term separately using the power rule.
- For 3x4: multiply by the index and reduce it by 1, giving 12x3.
- For minus 2x2: this gives minus 4x. The constant 7 differentiates to zero.
- So dy over dx = 12x3 minus 4x.
Differentiate y = (2x + 1)5 with respect to x.
Show the worked answer
- Use the chain rule. Differentiate the outside function first, treating the bracket as a single item.
- That gives 5(2x + 1)4.
- Multiply by the derivative of the inside, which is 2.
- dy over dx = 5 x 2 x (2x + 1)4 = 10(2x + 1)4.
Find the coordinates of the stationary points of y = x3 minus 3x2 minus 9x + 5 and determine their nature.
Show the worked answer
- Differentiate: dy over dx = 3x2 minus 6x minus 9.
- Set to zero: 3x2 minus 6x minus 9 = 0. Divide by 3 to get x2 minus 2x minus 3 = 0.
- Factorise: (x minus 3)(x + 1) = 0, so x = 3 or x = minus 1.
- Substitute for the y coordinates: at x = 3, y = 27 minus 27 minus 27 + 5 = minus 22. At x = minus 1, y = minus 1 minus 3 + 9 + 5 = 10.
- Second derivative: 6x minus 6. At x = 3 it equals 12, which is positive, so (3, minus 22) is a minimum.
- At x = minus 1 it equals minus 12, which is negative, so (minus 1, 10) is a maximum.
Find the equation of the tangent to the curve y = x2 minus 4x + 1 at the point where x = 3.
Show the worked answer
- Find the y coordinate: y = 9 minus 12 + 1 = minus 2, so the point is (3, minus 2).
- Differentiate: dy over dx = 2x minus 4.
- Evaluate at x = 3: gradient = 6 minus 4 = 2.
- Use the point and gradient: y minus (minus 2) = 2(x minus 3), so y + 2 = 2x minus 6.
- Rearrange: y = 2x minus 8.
Common mistakes in this topic
- Leaving a constant term in a derivative.
- Omitting the inside derivative when using the chain rule.
- Classifying a stationary point without the second derivative value.
- Forgetting to find the y coordinates of stationary points.
- Confusing the tangent gradient with the normal gradient.
Exam tips
- Differentiate term by term and write each step. Method marks are awarded per term.
- For a bracket raised to a power, use the chain rule rather than expanding.
- Always quote the second derivative value, then state the conclusion.
- Stationary point questions ask for coordinates, so find y as well as x.
- Normal gradient is the negative reciprocal of the tangent gradient. Check the product is minus 1.
Practise 20 more questions like this, free
vStudyWise marks every answer instantly, tracks the topics you keep dropping marks on and turns them into a weekly study plan.
Differentiation FAQs
How do I differentiate a bracket raised to a power?
Use the chain rule. Differentiate the outside, reducing the index by one and multiplying by the original index, then multiply by the derivative of the expression inside the bracket. For (2x plus 1) to the power 5 this gives 10 times (2x plus 1) to the power 4.
How do I find and classify stationary points?
Set the first derivative equal to zero and solve for x, then substitute back into the original equation for the y coordinates. Find the second derivative and evaluate it at each point: positive means a minimum and negative means a maximum. Quote the value as evidence.
How do I find the equation of a tangent?
Find the y coordinate of the point, differentiate the curve, and evaluate the derivative at that x value to get the gradient. Then use the point and the gradient in the equation of a straight line and rearrange into the required form.
Why does a constant disappear when differentiating?
The derivative measures the rate of change. A constant term has the same value everywhere, so it contributes no change and its gradient is zero. Adding a constant shifts a curve vertically without altering its gradient at any point.
Continue through the IGCSE Additional Mathematics syllabus
Related IGCSE Additional Mathematics topics
See all 17 IGCSE Additional Mathematics practice topics ›
Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.