Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 Differentiation – Revision Notes

For full explanations and longer worked examples, use the study guide.

These notes cover section 2.4, Differentiation, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). Section 2.4 is part of Pure Mathematics 2 and is examined on Paper 2 (1 hour 15 minutes, 50 marks), the AS Level Pure Mathematics route. Paper 1 knowledge is assumed, so tangents, normals and stationary points from section 1.7 are used freely. The same outcomes are section 3.4 of Pure Mathematics 3. A scientific calculator is allowed, but unsupported calculator answers earn no marks.

Course links: A Level Mathematics hub, printable 9709 checklist, practice questions for this unit, the whole-paper Pure Mathematics 2 revision notes, and the Paper 1 differentiation revision notes for the chain rule and stationary points.

2.4 at a glance

Outcome	In one line
Standard derivatives	eˣ, ln x, sin x, cos x, tan x, with multiples, sums, differences and composites
Products and quotients	Product rule and quotient rule, often combined with the chain rule
Parametric and implicit	First derivative only; used for gradients, tangents and normals

Paper 1 tools still apply on top of these rules. For an explicit curve y = f(x), classify a stationary point with d²y/dx² or with the sign of dy/dx either side, and use whichever method the question names. For parametric and implicit curves you only ever need dy/dx.

Formulas

Result	In MF19?
d/dx eˣ = eˣ	Yes
d/dx ln x = 1/x	Yes
d/dx sin x = cos x; d/dx cos x = −sin x	Yes
d/dx tan x = sec²x	Yes
Product: d/dx (uv) = v du/dx + u dv/dx	Yes
Quotient: d/dx (u/v) = (v du/dx − u dv/dx)/v²	Yes
Parametric: dy/dx = (dy/dt) ÷ (dx/dt)	Yes
Chain rule: dy/dx = (dy/du) × (du/dx)	No (learn)
d/dx e^(f(x)) = f′(x)e^(f(x)); d/dx ln f(x) = f′(x)/f(x)	No (learn as chain-rule patterns)

Trig derivatives hold only with x in radians.

Composite patterns to know on sight

e^(kx)        →  k e^(kx)
ln(ax + b)    →  a/(ax + b)
ln(kx)        →  1/x              (ln k is a constant)
sin(kx)       →  k cos(kx)
cos(kx)       →  -k sin(kx)
tan(kx)       →  k sec²(kx)
sinⁿx         →  n sinⁿ⁻¹x cos x

Small reminders:

• d/dx (ln x)² = 2 ln x × (1/x) = (2 ln x)/x. This is not the same as ln(x²), whose derivative is 2/x.
• d/dx e^(x²) = 2x e^(x²). The power stays; only a factor appears in front.
• d/dx sin²(3x) = 2 sin(3x) × 3 cos(3x) = 3 sin(6x) after the double-angle identity.

Method in steps

Product rule

1. Name u and v.
2. Write du/dx and dv/dx separately (chain rule inside each if needed).
3. Substitute into u dv/dx + v du/dx.
4. Take out common factors, such as a power of x or an exponential.

Example: y = x e^(−x). dy/dx = x(−e^(−x)) + e^(−x) = e^(−x)(1 − x). Stationary at x = 1, because e^(−x) is never 0.

Quotient rule

1. u is the top, v is the bottom.
2. Top of the answer: v du/dx minus u dv/dx.
3. Bottom of the answer: v².
4. Simplify the numerator only; leave v² factorised.

Parametric

1. Find dx/dt and dy/dt.
2. dy/dx = (dy/dt) ÷ (dx/dt); simplify, leaving it in t.
3. For a point: find t first, then x, y and the gradient from that t.

Example: x = ln t, y = t³. dx/dt = 1/t, dy/dt = 3t², so dy/dx = 3t² ÷ (1/t) = 3t³.

Implicit

1. Differentiate every term with respect to x, both sides.
2. Each y-term gets × dy/dx; each xy-type term needs the product rule.
3. Collect the dy/dx terms on one side, factorise, divide.
4. Substitute the point’s x and y (both are needed).

Example: y eˣ + y³ = 10. Product rule on y eˣ: y eˣ + eˣ dy/dx. Then 3y² dy/dx. So dy/dx (eˣ + 3y²) = −y eˣ, giving dy/dx = −y eˣ/(eˣ + 3y²).

Tangent and normal at a point

1. Gradient m from dy/dx at the point.
2. Tangent: y − y₁ = m(x − x₁). Normal: gradient −1/m.
3. Give the form asked for, such as ax + by + c = 0 with integers.

Two more worked reminders

Quotient to stationary point. y = eˣ/(x − 2), for x ≠ 2.

dy/dx = ((x - 2)eˣ - eˣ × 1) / (x - 2)²
      = eˣ(x - 3) / (x - 2)²
eˣ > 0, so dy/dx = 0 only when x = 3;  y = e³

The stationary point is (3, e³). Take out the common eˣ before setting the top equal to zero.

Parametric special points. x = t² + 1, y = t³ − 3t.

dx/dt = 2t,   dy/dt = 3t² - 3
Horizontal tangent: 3t² - 3 = 0  →  t = ±1  →  points (2, -2) and (2, 2)
Vertical tangent:   2t = 0       →  t = 0   →  point (1, 0)   (dy/dt = -3 ≠ 0)

Find t first, then turn each t into a pair of coordinates. Two values of t can give the same x.

Must-know distinctions

• ln(x²) vs (ln x)². The first is 2 ln x, derivative 2/x. The second needs the chain rule: (2 ln x)/x.
• sin x² vs sin²x. sin(x²) differentiates to 2x cos(x²). sin²x = (sin x)² differentiates to 2 sin x cos x.
• Product vs constant multiple. 4 tan x is a constant multiple: derivative 4 sec²x. No product rule is needed.
• Parametric vs implicit. Parametric gives x and y in terms of t: divide dy/dt by dx/dt. Implicit links x and y directly: differentiate term by term with respect to x.
• Horizontal vs vertical tangent (parametric). Horizontal where dy/dt = 0; vertical where dx/dt = 0.
• Tangent vs normal gradient. Normal gradient is −1/m, not −m and not 1/m.
• Exact vs 3 s.f. “Exact” means leave e, ln, π or surds in the answer. Otherwise give 3 significant figures.

Quick self-test

Differentiate with respect to x (questions 1 to 9).

1. e^(4x + 1)
2. ln(7x)
3. ln √(x + 2)
4. sin 5x
5. cos²x
6. x eˣ
7. x² sin x
8. (cos x)/x
9. ln(cos x)
10. Find the gradient of y = tan 3x at x = 0.
11. A curve has x = t³, y = t². Find dy/dx in terms of t.
12. The curve x² + y³ = 9 passes through (1, 2). Find the gradient there.

Answers

1. 4e^(4x + 1)
2. ln 7 + ln x, so 1/x
3. ½ ln(x + 2), so 1/(2(x + 2))
4. 5 cos 5x
5. 2 cos x × (−sin x) = −2 sin x cos x, or −sin 2x
6. x eˣ + eˣ = eˣ(x + 1)
7. 2x sin x + x² cos x
8. (x(−sin x) − cos x × 1)/x² = −(x sin x + cos x)/x²
9. (−sin x)/(cos x) = −tan x
10. 3 sec²(0) = 3
11. 2t ÷ 3t² = 2/(3t)
12. 2x + 3y² dy/dx = 0, so dy/dx = −2x/(3y²) = −1/6

Where marks are usually lost

• The inside derivative is missed: d/dx e^(3 − 2x) written as e^(3 − 2x) instead of −2e^(3 − 2x).
• The quotient rule is written with the top reversed, giving the negative of the correct answer.
• dy/dx left off a y-term in implicit work, such as d/dx (y²) written as 2y.
• The product rule skipped on an xy term, so only x dy/dx appears.
• A minus sign in front of a product term not applied to both parts of the product-rule result.
• In parametric questions, x-values substituted into a gradient that is written in terms of t.
• Trig derivatives used with degrees: dy/dx = cos x is only true when x is in radians.
• Solutions lost when dy/dx = 0 is divided by a factor such as cos x without checking whether that factor could be zero.
• Exact answers turned into decimals, or an exponential such as e^(ln 3) not simplified to 3.
• A “show that” result reached with a step missing, such as the unsimplified quotient-rule line.

Next steps

Try the practice questions, then check your AS readiness with the free AS diagnostic or the 9709 self-check bank. For eˣ and ln x themselves, see the logarithms and exponentials revision notes. A Level candidates meet this content again, with tan⁻¹ x added, in the Pure Mathematics 3 revision notes.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 2, Pure Mathematics 2 (for Paper 2): section 2.4 Differentiation.