Differential Calculus for Matric Maths

First principles, rules of differentiation, cubic graphs, optimisation and rates of change for NSC Paper 1. AI step-by-step explanations.

Calculus appears in Paper 1 and is usually worth 30–35 marks. CAPS Grade 12 covers first principles, rules of differentiation, cubic graph sketching, optimisation and rates of change. Calculus connects directly to the cubic function: the first derivative gives the stationary points, the second derivative tells you their nature, and the point of inflection lies where f''(x) = 0. In optimisation and rate-of-change questions, the real skill is translating the words into a function you can differentiate.

Key concepts

  • Differentiation from first principles
  • Rules: power, sum, constant multiple
  • Cubic graphs: stationary points, points of inflection
  • Optimisation (maxima and minima)
  • Rates of change
  • Notation: f'(x), dy/dx and Dₓ[…]
  • Equation of a tangent using the gradient f'(x) at the point of contact
  • Concavity from the sign of f''(x)
  • Interpreting a graph of f'(x) to describe f(x)

Worked examples

Find f'(x) from first principles for f(x) = 3x² − 2.

f'(x) = lim(h→0) [f(x+h) − f(x)] / h = lim(h→0) [3(x+h)² − 2 − (3x² − 2)] / h = lim(h→0) (6xh + 3h²) / h = 6x.

Find dy/dx if y = 4x³ − 2/x + √x.

Rewrite as powers: y = 4x³ − 2x⁻¹ + x^(1/2). Then dy/dx = 12x² + 2x⁻² + ½x^(−1/2) = 12x² + 2/x² + 1/(2√x).

Find the equation of the tangent to f(x) = x³ − 3x at x = 2.

f(2) = 8 − 6 = 2, so the point is (2, 2). f'(x) = 3x² − 3, so the gradient is f'(2) = 9. y − 2 = 9(x − 2) gives y = 9x − 16.

Find and classify the stationary points of f(x) = x³ − 6x² + 9x + 1.

f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0, so x = 1 or x = 3. f(1) = 5 and f(3) = 1. f''(x) = 6x − 12: f''(1) < 0 gives a local maximum at (1, 5) and f''(3) > 0 a local minimum at (3, 1). The point of inflection is at x = 2: (2, 3).

A rectangle has a perimeter of 40 cm. Find its maximum area.

Let the length be x, so the width is 20 − x and A = x(20 − x) = 20x − x². A'(x) = 20 − 2x = 0 gives x = 10, and A''(x) = −2 < 0 confirms a maximum. Maximum area = 10 × 10 = 100 cm².

Common mistakes

  • Dropping the limit notation when working from first principles (mark loss).
  • Mixing up local maximum and local minimum from f''(x).
  • Forgetting to test whether a stationary point is in the valid domain of an optimisation problem.
  • Differentiating before rewriting surds and fractions as powers of x.
  • Substituting into f'(x) instead of f(x) to find the y-coordinate of a stationary point.

Exam tips

  • Always set up an equation for the quantity to be optimised AND a constraint — then substitute.
  • Show f'(x) = 0 explicitly when finding stationary points.
  • Rewrite every term as a power of x before you differentiate.
  • In optimisation, state what the variable represents and confirm the maximum or minimum with f'' or a sign test.

In past papers

  • NSC Nov 2024 Paper 1 Q8–Q9 — cubic graph and optimisation

Frequently asked questions

Is integration in the Matric Maths CAPS syllabus?

No — Grade 12 Mathematics covers only differential calculus. Integration is not examined.

What is the difference between the derivative and the gradient?

f'(x) is a function that gives the gradient of the tangent at any x-value. Substituting a particular x-value gives the gradient of the tangent at that point.

How do I find the point of inflection of a cubic?

Solve f''(x) = 0. For a cubic, the x-coordinate is also exactly halfway between the x-coordinates of the two turning points.

Can I lose marks for notation?

Yes. Write f'(x) = or dy/dx = correctly, and keep 'lim h→0' on every line of a first-principles answer until you actually take the limit.

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