Differential Calculus — Complete Notes

What you'll learn

• What a limit is and how it leads to the derivative.
• How to find the derivative from first principles.
• The power rule and how to differentiate sums, constants and roots.
• How to find the gradient at a point and the equation of a tangent line.
• How to locate and classify stationary points on a cubic function.
• How to sketch cubic graphs and solve practical optimisation problems.

The big picture

Calculus is the mathematics of change. In Grade 12 you study differential calculus, which answers one central question: how fast is something changing at a particular instant?

Think of a car's speedometer. It does not tell you the average speed for the whole trip — it tells you the speed at that exact moment. The number on the speedometer is a derivative. Everything in this chapter is built to find that instantaneous rate of change.

CAPS examiners almost always combine three skills in one question: differentiate correctly, use the derivative to find a gradient or turning point, and then interpret the answer in context. So do not just memorise the rules — practise using them.

Limits — the foundation

A limit describes the value a function approaches as the input gets closer and closer to some number — without necessarily reaching it.

We write:

lim (h → 0) f(x + h) = L

This means "as h gets very small, f(x + h) approaches L". We never actually set h = 0, because that would create division by zero. We simplify first, then let h shrink.

Key idea: if you substitute and get 0/0, the expression can usually be factorised or simplified to remove the problem.

The derivative from first principles

The derivative of a function f(x) is the limit of the average gradient as the interval shrinks to zero:

f'(x) = lim (h → 0) [ f(x + h) − f(x) ] / h

This is called differentiation from first principles. The formula comes directly from the gradient formula m = (y₂ − y₁)/(x₂ − x₁), where the two points are (x, f(x)) and (x + h, f(x + h)).

Gradient of secant through   (x, f(x)) and (x+h, f(x+h))

        rise      f(x+h) - f(x)
  m  = -------  = --------------
        run            h

As h -> 0 the secant becomes the TANGENT,
and its gradient becomes f'(x).

Notation

There are several ways to write the derivative. They all mean the same thing:

Rules for differentiation

Once you know the rules, you rarely use first principles again — but you must understand it because it is often examined directly.

The power rule

If f(x) = xⁿ, then f'(x) = n·xⁿ⁻¹. Multiply by the old power, then reduce the power by one.

The constant rule

The derivative of a constant is 0. A constant does not change, so its rate of change is zero.

Constant multiple rule

d/dx [ k·f(x) ] = k·f'(x). A constant multiplier stays outside and is carried through.

Sum and difference rule

Differentiate each term separately: d/dx [ f(x) ± g(x) ] = f'(x) ± g'(x).

Roots and fractions

Rewrite them as powers first:

• √x = x^(1/2), so its derivative is (1/2)·x^(−1/2) = 1/(2√x)
• 1/x = x⁻¹, so its derivative is −x⁻² = −1/x²

Gradient at a point and the tangent line

The derivative gives the gradient of the tangent at any point. To find the equation of a tangent at x = a:

1. Find the point: calculate f(a) to get the y-coordinate.
2. Find the gradient: calculate f'(a).
3. Use the straight-line formula: y − y₁ = m(x − x₁).

Stationary points and the shape of a cubic

A stationary point is where the gradient is zero: f'(x) = 0. These are the turning points (or flat points) of the graph.

To classify each stationary point, use the second derivative f''(x):

A cubic function f(x) = ax³ + bx² + cx + d can have at most two turning points and always has exactly one point of inflection.

Sketching a cubic function

Follow this checklist every time:

1. Find the y-intercept by setting x = 0.
2. Find the x-intercepts by factorising f(x) = 0.
3. Find the stationary points from f'(x) = 0.
4. Classify them using f''(x).
5. Plot the points and join with a smooth curve. A positive cubic rises on the right; a negative cubic falls on the right.

   local max
     (1,5)          Cubic  y = x^3 - 6x^2 + 9x + 1
       *             rises from the bottom-left, peaks
      / \            at (1,5), dips to a valley at (3,1),
     /   \           then rises steeply again.
    /     \  local min
   /       * (3,1)
  /

Worked example 1 — first principles

Find the derivative of f(x) = 3x² − 2x from first principles.

Step 1 — write f(x + h). f(x + h) = 3(x + h)² − 2(x + h) = 3(x² + 2xh + h²) − 2x − 2h = 3x² + 6xh + 3h² − 2x − 2h

Step 2 — subtract f(x). f(x + h) − f(x) = (3x² + 6xh + 3h² − 2x − 2h) − (3x² − 2x) = 6xh + 3h² − 2h

Step 3 — divide by h. [ f(x + h) − f(x) ] / h = 6x + 3h − 2

Step 4 — let h → 0. f'(x) = 6x − 2

Worked example 2 — different rules

Differentiate f(x) = 5x⁴ − 3√x + 2/x.

Step 1 — rewrite with powers. f(x) = 5x⁴ − 3x^(1/2) + 2x⁻¹

Step 2 — apply the power rule to each term. f'(x) = 20x³ − 3·(1/2)x^(−1/2) + 2·(−1)x⁻²

Step 3 — simplify. f'(x) = 20x³ − 3/(2√x) − 2/x²

Worked example 3 — sketch a cubic

Sketch f(x) = x³ − 6x² + 9x + 1.

y-intercept: f(0) = 1, so the point is (0, 1).

Stationary points: f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). So f'(x) = 0 when x = 1 or x = 3. f(1) = 1 − 6 + 9 + 1 = 5 → (1, 5) f(3) = 27 − 54 + 27 + 1 = 1 → (3, 1)

Classify: f''(x) = 6x − 12. f''(1) = −6 < 0 → local maximum at (1, 5). f''(3) = 6 > 0 → local minimum at (3, 1).

Sketch: the curve rises from the bottom-left, reaches a peak at (1, 5), dips to a valley at (3, 1), then rises steeply. It crosses the y-axis at y = 1.

Common mistakes

• Setting h = 0 too early. You must simplify the fraction before letting h approach zero, otherwise you get 0/0.
• Dropping the sign when differentiating 1/x. The derivative of x⁻¹ is −x⁻², not x⁻².
• Confusing f(x) with f'(x). f(a) gives a y-value; f'(a) gives a gradient. They are different numbers.
• Forgetting to find the y-coordinate. When asked for a turning point, give both coordinates, not just x.
• Misclassifying turning points. Always test with f''(x); do not assume the first one is a maximum.
• Not rewriting roots and fractions as powers before differentiating — the power rule only works on powers.

Quick summary / cheat sheet

Quick memory hook: multiply by the power, subtract one from the power.

Practice questions

1. Differentiate f(x) = x² + 3x from first principles.
2. Differentiate f(x) = 2x⁵ − 8x² + 7.
3. Find the gradient of the tangent to f(x) = x³ − 6x at x = −1.
4. Find and classify the stationary points of f(x) = x³ − 3x² − 9x + 5.
5. Find the equation of the tangent to f(x) = x² − 4x + 1 at x = 3.
6. A rectangular enclosure is fenced on three sides using 40 m of fencing (the fourth side is a wall). Find the dimensions that give the maximum area.

Answers

1. f(x+h) − f(x) = 2xh + h² + 3h; divide by h → 2x + h + 3; h → 0 gives f'(x) = 2x + 3.
2. f'(x) = 10x⁴ − 16x.
3. f'(x) = 3x² − 6, so f'(−1) = 3 − 6 = −3.
4. f'(x) = 3(x − 3)(x + 1): stationary at x = 3 and x = −1. f(3) = −22 and f(−1) = 10. f''(x) = 6x − 6, so f''(3) = 12 > 0 → local minimum at (3, −22) and f''(−1) = −12 < 0 → local maximum at (−1, 10).
5. f(3) = −2 and f'(3) = 2, so y + 2 = 2(x − 3), giving y = 2x − 8.
6. Let the two equal sides be x, so the third side is 40 − 2x. Area A = 40x − 2x². A'(x) = 40 − 4x = 0 → x = 10. The sides are 10 m, 10 m and 20 m, giving a maximum area of 200 m² (A'' = −4 < 0 confirms a maximum).

Official EduRank Team notes. Always cross-check with your textbook and teacher.