OxfordAQA International AS and A-level Mathematics: Differentiation: gradients, tangents, normals and stationary points (9660) – Revision Notes

These notes condense specification section P1.3 (Differentiation) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. It is part of International AS Unit P1, which also counts towards the International A-level. For full explanations and longer worked examples, read the study guide. Test yourself with the practice questions, then check the course hub, the printable checklist and the free diagnostics. For the whole unit, see the Unit P1 revision notes.

Definitions

Term	Meaning
Derivative f’(x) or dy/dx	The gradient of the tangent to y = f(x) at the point with that x-value.
Limit idea	The gradient of a chord to a nearby point tends to the tangent gradient as the gap h tends to 0.
Rate of change	dy/dx is how fast y changes per unit of x. Its unit is (unit of y) per (unit of x).
Second derivative	f’‘(x) or d²y/dx²: the derivative of f’(x). It measures how the gradient is changing.
Stationary point	A point where dy/dx = 0.
Increasing / decreasing	f’(x) > 0 / f’(x) < 0 on an interval.
Normal	The line perpendicular to the tangent at the same point.

The spec says differentiation from first principles is not tested, and points of inflection are not required.

Rules in one table

Task	Rule or result
Power rule	axⁿ → anxⁿ⁻¹, for rational n
Constant	k → 0
Sum or difference	Differentiate term by term
Square root	√x = x^(1/2), derivative (1/2)x^(−1/2)
Reciprocal	1/xⁿ = x⁻ⁿ, derivative −nx^(−n−1)
Parallel gradients	m₁ = m₂
Perpendicular gradients	m₁ × m₂ = −1
Line through (x₁, y₁)	y − y₁ = m(x − x₁)

Quick reminder: x^(5/2) differentiates to (5/2)x^(3/2). The expression 3x² − √x differentiates to 6x − (1/2)x^(−1/2).

Method boxes

Rewrite before differentiating

1. Split any fraction with a sum on top: (x³ + 4)/x becomes x² + 4x⁻¹.
2. Turn roots and reciprocals into powers.
3. Combine products of powers into one power: x√x is x^(3/2).
4. Differentiate term by term.

Tangent and normal at x = a

1. y-coordinate: substitute a into the curve.
2. Gradient m = f’(a).
3. Tangent: y − y₁ = m(x − a).
4. Normal: gradient −1/m, same point.
5. Rearrange to the requested form.

Stationary points

1. Solve dy/dx = 0.
2. Substitute each x into y to get the full coordinates.
3. Find d²y/dx² and substitute: negative means maximum, positive means minimum.
4. If it is 0, test the sign of dy/dx either side instead.

Increasing or decreasing intervals

1. Solve f’(x) = 0 to find the boundaries.
2. Test the sign of f’(x) in each interval (factorise first).
3. State intervals using inequalities and respect the domain.

Optimisation

1. Define variables and write the fixed constraint.
2. Eliminate to one variable and write the quantity as a function of it.
3. Differentiate, set to 0, solve.
4. Prove maximum or minimum.
5. Answer the question asked, with units.

Worked reminders

Tangent and normal. Curve y = x² − 6x at x = 5. The y-value is 25 − 30 = −5 and dy/dx = 2x − 6 = 4.

Tangent: y + 5 = 4(x − 5)           → y = 4x − 25
Normal:  y + 5 = −(1/4)(x − 5)      → x + 4y + 15 = 0

Stationary point. For y = 3x² − 12x + 7 we have dy/dx = 6x − 12 = 0 at x = 2, and y = 12 − 24 + 7 = −5. Then d²y/dx² = 6 > 0, so (2, −5) is a minimum.

Optimisation shape. If a quantity is S = x² + k/x, then dS/dx = 2x − k/x². Setting it to 0 gives x³ = k/2. The second derivative 2 + 2k/x³ is positive for x > 0, which confirms a minimum.

Common question patterns

Points with a given gradient. For y = x³ − 3x², find the points where the gradient is 9. Set 3x² − 6x = 9, so x² − 2x − 3 = 0 and (x − 3)(x + 1) = 0. Then x = 3 or x = −1, and substituting into the curve gives the points (3, 0) and (−1, −4). A tangent parallel to a given line has the same gradient as that line.

Unknown constants. If y = kx² − 5x has gradient 7 at x = 3, then dy/dx = 2kx − 5 gives 6k − 5 = 7, so k = 2. Differentiate first, substitute second, then solve.

Interpreting a rate. If V litres is the volume in a tank after t minutes and dV/dt = 18√t, then at t = 16 the volume is increasing at 72 litres per minute. A negative value means the quantity is falling. Only a general appreciation of this idea is required.

Increasing or decreasing everywhere. To show a function is always increasing, show f’(x) > 0 for every x. A quadratic derivative can be written in completed-square form: a positive square plus a positive constant is always positive.

Sketching what you have found. A cubic with a positive x³ coefficient rises, turns at a maximum, falls to a minimum and rises again. If you know its two stationary points, the shape follows. A table of x, f’(x) sign and behaviour is a tidy way to present the argument:

Type of point	before the point	at the point	after the point
Maximum	f’ > 0	f’ = 0	f’ < 0
Minimum	f’ < 0	f’ = 0	f’ > 0
Neither	same sign	f’ = 0	same sign

Must-know distinctions

Often confused	The difference
Tangent and normal	The tangent has gradient f’(a). The normal has gradient −1/f’(a).
Stationary point and maximum	Stationary means dy/dx = 0. It may be a maximum, a minimum, or neither.
f’(x) = 0 and f’’(x) = 0	The first locates stationary points. The second only makes the second-derivative test inconclusive.
Local and overall maximum	A local maximum is the highest value nearby, not necessarily the largest value of the function.
dy/dx and a gradient number	dy/dx is a function of x. Substitute an x-value to get a gradient.

Quick self-test

Try these without looking below. Use the working space on paper.

1. Differentiate 9x^(2/3).
2. Differentiate 3/(4x²).
3. Find the gradient of y = x³ − 5x where x = −2.
4. A tangent has gradient −3/5. State the gradient of the normal.
5. Find the equation of the tangent to y = x² + 3x at the point where x = 1.
6. Find and classify the stationary point of y = x² − 10x + 3.
7. For what values of x is f(x) = x³ − 12x increasing?
8. Find d²y/dx² when y = x⁴ − 2x², and evaluate it at x = 1.
9. What sign does f’’(x) have at a maximum point?
10. A particle’s distance is s = 4t^(3/2) metres after t seconds. Find ds/dt at t = 9 and give the unit.
11. For y = 2x^(3/2) − 12x with x > 0, find the coordinates of the stationary point and say whether it is a maximum or a minimum.

Answers

1. 6x^(−1/3), which is 6/∛x.
2. Write as (3/4)x⁻², giving −(3/2)x⁻³.
3. 3x² − 5 = 7 at x = −2.
4. 5/3.
5. Gradient 2x + 3 = 5 and y = 4 at x = 1, so y − 4 = 5(x − 1), giving y = 5x − 1.
6. dy/dx = 2x − 10 = 0 at x = 5, y = −22, and d²y/dx² = 2 > 0, so (5, −22) is a minimum.
7. f’(x) = 3x² − 12 > 0, so x < −2 or x > 2.
8. d²y/dx² = 12x² − 4, which is 8 at x = 1.
9. Negative (when the test is conclusive).
10. ds/dt = 6t^(1/2), which is 18 metres per second at t = 9.
11. dy/dx = 3x^(1/2) − 12 = 0 gives x = 16, and y = 2(64) − 192 = −64. The second derivative (3/2)x^(−1/2) is 3/8 > 0 at x = 16, so (16, −64) is a minimum.

Where marks are usually lost

• Differentiating before rewriting, especially terms like 5/x² or x√x.
• Dropping the minus sign on a negative power: x⁻² differentiates to −2x⁻³.
• Writing the normal with the tangent’s gradient, or flipping the gradient without the sign change.
• Not showing the y-coordinate calculation, so the line has no justified point.
• Stating the nature of a stationary point with no second-derivative value or sign check to support it.
• Giving only the x-value of a stationary point when the coordinates were requested.
• In a justification, writing “f’’ is positive so it is a maximum” (the wrong way round).
• In optimisation, not proving that the value is a maximum or minimum, or not giving the final quantity with units.
• Rounding too early, which spoils later answers; keep exact values until the end.
• Losing a fractional power when simplifying: 2x^(3/2) differentiates to 3x^(1/2), and the coefficient 3 is easy to miscopy.
• Using the second-derivative test on a point that is not stationary; always solve dy/dx = 0 first.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: section 3.1, International AS Unit P1 (Pure Maths), content P1.3 Differentiation.