OxfordAQA International GCSE Mathematics: Differentiation, tangents and stationary points (9260) – Revision Notes

Quick recall for OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), section 3.2.2, statements A13 to A15. All of it is Extension only. For full explanations, start with the study guide, then test yourself on the practice questions. Course pages: the OxfordAQA IGCSE Mathematics hub and the checklist. Not sure what to revise first? Take a free 10-minute diagnostic.

Key results

Ref	Result	Tier
A13	dy/dx is the gradient function: the gradient of the curve at any x	Extension only
A13	kxⁿ → k × n × xⁿ⁻¹ (multiply by the power, then reduce the power by 1)	Extension only
A13	A term kx → k. A constant → 0. Differentiate sums term by term	Extension only
A14	Gradient of the curve at a point = gradient of the tangent there	Extension only
A14	Tangent: y − y₁ = m(x − x₁), with m = dy/dx at that point	Extension only
A15	Stationary point: dy/dx = 0	Extension only
A15	Maximum, minimum, point of inflection; sketch with known stationary points	Extension only

Reading the gradient

• Positive dy/dx: the curve is rising as x increases.
• Negative dy/dx: the curve is falling as x increases.
• Zero dy/dx: the tangent is horizontal, so the point is stationary.
• A larger gradient value, positive or negative, means a steeper curve at that point.

You can use this to check an answer. A maximum must have the curve rising to its left and falling to its right, so the gradient values you tested should be positive then negative.

A13: differentiating

Method in steps:

1. Multiply out brackets, or divide each term by the denominator, until every term has the form kxⁿ.
2. Differentiate each term: multiply by the power, then subtract 1 from the power.
3. Keep the signs between terms. Constants vanish.

Worked reminders:

y = 4x³ − x² + 6          →  dy/dx = 12x² − 2x
y = (x + 5)(x − 1)        →  y = x² + 4x − 5   →  dy/dx = 2x + 4
y = (10x⁴ + 5x²)/(5x)     →  y = 2x³ + x       →  dy/dx = 6x² + 1
y = 8 − x⁵                →  dy/dx = −5x⁴

The last one shows that the minus sign stays attached to the term: −x⁵ gives −5x⁴.

A14: tangents

Method in steps:

1. Put the x-value into y = f(x) to get the y-coordinate of the point.
2. Put the x-value into dy/dx to get the gradient m.
3. Write y − y₁ = m(x − x₁) and tidy.

Worked reminder: tangent to y = x² − 4x + 7 at x = 3.

y = 9 − 12 + 7 = 4        point (3, 4)
dy/dx = 2x − 4  →  m = 2
y − 4 = 2(x − 3)  →  y = 2x − 2

If the question gives the gradient instead of the x-value, solve dy/dx = that gradient first, then find y.

A15: stationary points

Method in steps:

1. Solve dy/dx = 0 for x. Factorise where possible.
2. Substitute into y = f(x) for the y-coordinate.
3. Find the sign of dy/dx just before and just after each x-value.
4. Classify using the table.

Signs of dy/dx (before, after)	Type
+, −	Maximum
−, +	Minimum
+, + or −, −	Point of inflection

Worked reminder: y = 2x³ − 3x² − 36x + 10.

dy/dx = 6x² − 6x − 36 = 6(x − 3)(x + 2)
x = 3:   y = 54 − 27 − 108 + 10 = −71
x = −2:  y = −16 − 12 + 72 + 10 = 54

Test x = −3, 0, 4: dy/dx = 36, −36, 36. So (−2, 54) is a maximum and (3, −71) is a minimum.

Inflection reminder: y = x³ has dy/dx = 3x², which is zero at x = 0 but positive either side. The origin is a stationary point of inflection.

Sketching

1. Plot the stationary points with their coordinates.
2. Find the y-intercept (put x = 0).
3. Use the highest power to decide what happens at each end.
4. Draw a smooth curve with a horizontal tangent at each stationary point.

For a quartic with a positive x⁴ term, both ends go up. For a cubic with a positive x³ term, the left end goes down and the right end goes up.

Worked reminders for harder questions

Tangent with a given gradient

Find the equations of the tangents to y = x³ − 3x that have gradient 9.

dy/dx = 3x² − 3 = 9   →   x² = 4   →   x = 2 or x = −2
x = 2:   y = 8 − 6 = 2       y − 2 = 9(x − 2)   →  y = 9x − 16
x = −2:  y = −8 + 6 = −2     y + 2 = 9(x + 2)   →  y = 9x + 16

Two points, two tangents. Do not stop after the positive root.

Finding an unknown constant

The curve y = x³ + ax + 5 has a stationary point where x = 2. Find a, then the point and its type.

dy/dx = 3x² + a
x = 2 gives dy/dx = 0:  12 + a = 0   →   a = −12
y = 8 − 24 + 5 = −11    point (2, −11)

Test x = 1: dy/dx = 3 − 12 = −9. Test x = 3: dy/dx = 27 − 12 = 15. The gradient goes − then +, so (2, −11) is a minimum.

The key idea is that “stationary at x = 2” means “put x = 2 into dy/dx and set it equal to 0”.

Sign-table template

For fast, tidy working, copy this layout and fill it in for each stationary x-value:

x        | a (just below) | stationary x | b (just above)
dy/dx    | value          | 0            | value
sign     | + or −         |              | + or −

Choose a and b so that no other stationary x-value lies between them and the one you are testing.

Must-know distinctions

Idea	Compare
y-value and gradient	y = f(x) gives height. dy/dx gives steepness. Stationary points use dy/dx = 0 to find x and y = f(x) to find the height
Stationary point and turning point	Every maximum and minimum is stationary. A stationary point of inflection is stationary but the curve does not turn round
Gradient of curve and tangent	Same number at the point of contact. The tangent is a line, the gradient function belongs to the curve
Expanding and differentiating	Expand first. The product of the derivatives of two brackets is not the derivative of the product

Quick self-test

1. Differentiate y = 9x² − 5x + 11.
2. Differentiate y = x⁴/x.
3. What is the derivative of the constant 7?
4. Find the gradient of y = x³ − 2x² at x = 3.
5. Find the x-values of the stationary points of y = x³ − x² − x.
6. Find the equation of the tangent to y = x² + 5 at x = 2.
7. Find and classify the stationary points of y = x³ − 3x.
8. For y = x⁴ − 2x², state the coordinates of the stationary points and their types.
9. The gradient of a curve at the point P is 0. What can you say about the tangent at P?
10. For y = x² − 6x, find the x-value where the gradient is 0 and state whether it is a maximum, minimum or inflection.

Answers

1. dy/dx = 18x − 5.
2. y = x³ after simplifying, so dy/dx = 3x².
4. dy/dx = 3x² − 4x = 27 − 12 = 15.
5. dy/dx = 3x² − 2x − 1 = (3x + 1)(x − 1) = 0, so x = −1/3 or x = 1.
6. Point (2, 9). dy/dx = 2x = 4. y − 9 = 4(x − 2), so y = 4x + 1.
7. dy/dx = 3x² − 3 = 3(x − 1)(x + 1). At x = −1, y = 2. At x = 1, y = −2. Testing x = −2, 0, 2 gives gradients 9, −3, 9. So (−1, 2) is a maximum and (1, −2) is a minimum.
8. dy/dx = 4x³ − 4x = 4x(x − 1)(x + 1). Points: (−1, −1), (0, 0), (1, −1). Testing x = −2, −0.5, 0.5, 2 gives gradients −24, 1.5, −1.5, 24. So (−1, −1) and (1, −1) are minima, and (0, 0) is a maximum.
9. The tangent at P is horizontal (parallel to the x-axis).
10. dy/dx = 2x − 6 = 0 gives x = 3. The gradient goes from negative to positive (try x = 2 and x = 4), so it is a minimum, at (3, −9).

Where marks are usually lost

• Not expanding brackets before differentiating.
• Forgetting to subtract 1 from the power, or forgetting to multiply by the old power.
• Leaving off the y-coordinate when the question asks for the coordinates of a point.
• Putting stationary x-values into dy/dx when finding y.
• Naming a maximum or minimum with no sign test or no working shown. Show the gradient values you tested.
• Choosing test values that jump past a second stationary point.
• Writing a tangent equation with the wrong sign, for example y − 2 = 6(x + 3) when the point is x = 3.
• Leaving a tangent equation as y − y₁ = m(x − x₁) with numbers in it when asked for the form y = mx + c.
• Sketching a stationary point as a sharp corner instead of a smooth curve with a flat tangent.

Official syllabus

These notes follow the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards), published by OxfordAQA: section 3.2.2, statements A13, A14 and A15.