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06062025–2027 syllabus

ADDITIONAL MATHEMATICS · CHAPTER 14

Calculus

Use rates of change and accumulation to analyse curves, optimise quantities, measure areas and model motion.

Additional Mathematics6 connected sectionsSyllabus-aligned guide

LEARNING OBJECTIVES

What you will be able to do

  • differentiate standard, composite, product and quotient functions
  • use derivatives for gradients, rates and extrema
  • integrate standard forms and evaluate areas
  • connect displacement, velocity and acceleration

AT A GLANCE

Syllabus0606Coverage2025–2027Sections6LevelAdditional Mathematics

INTRODUCTION · THE BIG IDEA

Use rates of change and accumulation to analyse curves, optimise quantities, measure areas and model motion.

Differentiation measures instantaneous rate of change; integration reverses differentiation and accumulates change. The two processes connect algebraic formulas to the shape and motion they describe.

No calculus formulas are supplied in the 0606 formula list. Accurate recall, clear chain-rule factors and constants of integration are therefore essential.

01

SECTION 01

Derivative rules and notation

Core concept

Differentiate term by term. For a composite y = f(g(x)), multiply the derivative of the outer function by the derivative of the inner function. Trigonometric differentiation uses radians.

RULE 1
d/dx(xⁿ) = nxⁿ⁻¹
RULE 2
d/dx(sin x) = cos x
RULE 3
d/dx(cos x) = −sin x
RULE 4
d/dx(tan x) = sec²x
RULE 5
d/dx(eˣ) = eˣ
RULE 6
d/dx(ln x) = 1/x
RULE 7
d/dx[f(g(x))] = f′(g(x))g′(x)
Original worked example

Differentiate a composite power

  1. Let y = (3x² + 1)⁴.
  2. Differentiate the outer fourth power: 4(3x² + 1)³.
  3. Multiply by the inner derivative 6x.
  4. Simplify the constant factor.

Answer: dy/dx = 24x(3x² + 1)³.

02

SECTION 02

Products, quotients, tangents and normals

Core concept

Use the product rule when both factors vary and the quotient rule when a variable expression is in the denominator. Evaluate the derivative at a point to obtain the tangent gradient; the normal has negative reciprocal gradient.

DETAILED EXPLANATION

  • Find the point coordinate as well as the gradient.
  • A horizontal tangent has a vertical normal.
  • Simplify only after the derivative structure is correct.
RULE 1
(uv)′ = u′v + uv′
RULE 2
(u/v)′ = (vu′ − uv′)/v²
RULE 3
m_normal = −1/m_tangent
03

SECTION 03

Stationary points and optimisation

Core concept

Stationary points satisfy f′(x) = 0. Classify them by a sign change in f′ or by f″: positive second derivative gives a local minimum and negative gives a local maximum. Points of inflexion are not included in this syllabus.

RULE 1
f′(x) = 0 at a stationary point
RULE 2
f″(x) > 0 ⇒ local minimum
RULE 3
f″(x) < 0 ⇒ local maximum
ORIGINAL STUDY DIAGRAMOptimise a model
1Define the target quantity
2Use constraints to get one variable
3Differentiate
4Solve f′(x)=0
5Classify and interpret with units
Original worked example

Maximise a rectangle under a constraint

  1. A rectangle has perimeter 40, so y = 20 − x.
  2. Area A = x(20 − x) = 20x − x².
  3. A′ = 20 − 2x = 0 gives x = 10.
  4. A″ = −2 < 0, so this is a maximum; then y = 10.

Answer: Maximum area 100 square units, for a 10 by 10 square.

04

SECTION 04

Connected rates and small changes

Core concept

When quantities depend on time through another variable, link their rates with the chain rule. For a small change δx, the corresponding change δy is approximately f′(x)δx.

DETAILED EXPLANATION

  • Use signed rates: decreasing quantities have negative derivatives.
  • Keep units through the chain.
  • Small-increment results are approximations, not exact identities.
  • State the point at which the derivative is evaluated.
RULE 1
dy/dt = (dy/dx)(dx/dt)
RULE 2
δy ≈ (dy/dx)δx
RULE 3
f(x+δx) ≈ f(x) + f′(x)δx
05

SECTION 05

Integration and plane areas

Core concept

An indefinite integral is a family of antiderivatives, so include + C. Definite integrals give signed area; geometric area may require splitting where the upper curve changes or where a graph crosses the axis.

RULE 1
∫xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ −1
RULE 2
∫1/x dx = ln|x| + C
RULE 3
∫(ax+b)ⁿdx = (ax+b)ⁿ⁺¹/[a(n+1)] + C
RULE 4
∫sin(ax+b)dx = −cos(ax+b)/a + C
RULE 5
∫cos(ax+b)dx = sin(ax+b)/a + C
RULE 6
∫sec²(ax+b)dx = tan(ax+b)/a + C
RULE 7
∫e^(ax+b)dx = e^(ax+b)/a + C
Area setup
RegionIntegral
curve above x-axis∫f(x) dx
curve below x-axis−∫f(x) dx
between upper f and lower g∫[f(x)−g(x)] dx
boundary changessplit into separate integrals
06

SECTION 06

Kinematics and motion graphs

Core concept

For straight-line motion, velocity is the derivative of displacement and acceleration is the derivative of velocity. Integrating reverses these relationships, with constants found from initial conditions.

DETAILED EXPLANATION

  • A change of direction occurs when v changes sign.
  • Gradient links displacement–time to velocity and velocity–time to acceleration.
  • Area under velocity–time gives displacement; area under speed–time gives distance.
RULE 1
v = ds/dt
RULE 2
a = dv/dt = d²s/dt²
RULE 3
displacement change = ∫v dt
RULE 4
velocity change = ∫a dt
RULE 5
distance travelled = ∫|v| dt
ORIGINAL STUDY DIAGRAMAnalyse a motion model
1Differentiate or integrate as required
2Use initial conditions
3Find times when v = 0
4Split at direction changes
5Interpret signs and units

QUICK CHAPTER SUMMARY

The ideas to carry forward

  • Differentiation measures instantaneous change and integration reverses it.
  • Chain, product and quotient rules handle composite expressions.
  • Stationary points require both solving and classification.
  • Indefinite integrals include an arbitrary constant.
  • Area may require signed-integral splitting.
  • Kinematics links s, v and a by differentiation and integration.

QUICK REVISION CHECKLIST

Can you do each of these without your notes?

  • differentiate standard, composite, product and quotient functions
  • use derivatives for gradients, rates and extrema
  • integrate standard forms and evaluate areas
  • connect displacement, velocity and acceleration
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