Differentiation: The Rules and How Exams Use Them

The power rule plus every way an exam applies it — differentiating polynomials, rewriting surds and fractions first, first principles, gradients and tangents, and turning points — each worked in full, with a worksheet.

Differentiation finds the gradient of a curve at any point. The result, the derivative, is itself a new function you can evaluate at any x. In Grade 12 it unlocks tangents, rates of change and optimisation — a large slice of Paper 1. This guide covers the rule and every way exams use it.

The derivative gives the gradient of the tangent line that just touches the curve at a point P.

1The power rule

For a term axⁿ, multiply by the power, then reduce the power by one:

if y = axⁿ then dydx = n·a·xⁿ⁻¹

The derivative of a constant is 0 (a flat line has zero gradient).
The derivative of 5x is 5 (x¹ becomes x⁰ = 1).
Differentiate a sum term by term.

2Type 1: Differentiate a polynomial

Worked ExampleExample 1: differentiate y = 3x³ − 5x² + 4x − 7

1
3x³ → 9x².
×3, reduce power to 2.
2
−5x² → −10x.
×2, reduce power to 1.
3
4x → 4; −7 → 0.
x¹→x⁰=1; the derivative of a constant is 0.
4
dy/dx = 9x² − 10x + 4.
Combine the terms.

3Type 2: Rewrite surds and fractions first

The power rule only works on powers, so rewrite 1x as x⁻¹ and x as x¹² before differentiating.

Worked ExampleExample 2: differentiate y = x + 2⁄x

1
Rewrite: y = x^12 + 2x⁻¹.
Turn the surd and fraction into powers.
2
x^12 → 12x^(−12); 2x⁻¹ → −2x⁻².
Apply the power rule to each.
3
dy/dx = 12x^(−12) − 2x⁻² = 1⁄(2x) − 2⁄x².
Rewrite back into surd/fraction form if asked.

4Type 3: First principles (from the definition)

The definition of the derivative: f′(x) = lim_h→0 f(x+h) − f(x)h.

Worked ExampleExample 3: differentiate f(x) = x² from first principles

1
f(x + h) = (x + h)² = x² + 2xh + h².
Expand.
2
f(x+h) − f(x) = 2xh + h². Divide by h: 2x + h.
The x² terms cancel; factor out h.
3
As h → 0: f′(x) = 2x.
Let h go to zero.

5Type 4: Gradient and tangent at a point

Worked ExampleExample 4: find the gradient of y = x² − 4x at x = 3

1
Differentiate: dy/dx = 2x − 4.
The derivative gives the gradient function.
2
At x = 3: 2(3) − 4 = 2.
Substitute x = 3.
3
The gradient of the tangent there is 2.
The derivative's value = the tangent's gradient.

6Type 5: Turning points (gradient = 0)

Worked ExampleExample 5: find the turning points of y = x³ − 3x

1
dy/dx = 3x² − 3. Set = 0: 3x² − 3 = 0.
Turning points occur where the gradient is zero.
2
x² = 1 → x = 1 or x = −1.
Solve.
3
y-values: at x=1, y = 1−3 = −2; at x=−1, y = −1+3 = 2. Points (1,−2) and (−1,2).
Substitute back for the y-coordinates.

💡

Rewrite awkward terms as powers first (fractions → negative powers, surds → fractional powers). Then the mechanical power rule applies to everything.

⚠️

Reduce the power, don't drop it. The derivative of x³ is 3x², not 3x or 3. Only the power decreases by exactly one.

Differentiating is quick marks; the challenge is what comes after (tangents, turning points, optimisation). Drill the worksheet so the differentiation step is fast and flawless.

7First principles: the four forms you must know

f′(x) = lim_h→0 f(x + h) − f(x)h

The ATP prescribes exactly four forms. In every one the h in the denominator must cancel before you let h → 0 — if it does not, you have made an algebra slip.

Worked ExampleDetermine f′(x) from first principles if f(x) = ax² + bx + c

1
f(x + h) = a(x + h)² + b(x + h) + c = ax² + 2axh + ah² + bx + bh + c.
Expand fully.
2
f(x + h) − f(x) = 2axh + ah² + bh.
Every term without an h cancels.
3
h(2ax + ah + b)h = 2ax + ah + b.
Factor out h and cancel.
4
Let h → 0: f′(x) = 2ax + b.

Worked ExampleDetermine f′(x) from first principles if f(x) = ax³

1
f(x + h) = a(x³ + 3x²h + 3xh² + h³).
Cube the bracket.
2
f(x + h) − f(x) = a(3x²h + 3xh² + h³).
3
ah(3x² + 3xh + h²)h = a(3x² + 3xh + h²).
4
Let h → 0: f′(x) = 3ax².

Worked ExampleDetermine f′(x) from first principles if f(x) = ax

1
f(x + h) − f(x) = ax + h − ax = ax − a(x + h)x(x + h).
Common denominator.
2
= −ahx(x + h).
The ax terms cancel.
3
Divide by h: −ax(x + h).
4
Let h → 0: f′(x) = −ax².

Worked ExampleDetermine f′(x) from first principles if f(x) = c

1
f(x + h) − f(x) = c − c = 0.
A constant never changes.
2
0h = 0, so f′(x) = 0.
A horizontal line has zero gradient.

⚠️

You may not substitute h = 0 before cancelling — that gives 00, which is undefined. Cancel the h first, then take the limit.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Differentiate y = x⁴. (power rule)
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4x³
2
Differentiate y = 5x³ − 2x² + 7. (polynomial)
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15x² − 4x
3
Differentiate y = 2x² − 3x + 9. (polynomial)
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4x − 3
4
Differentiate y = 3⁄x (write as 3x⁻¹). (rewrite)
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−3x⁻² = −3⁄x²
5
Differentiate y = x (write as x^12). (rewrite)
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12x^(−12) = 1⁄(2x)
6
From first principles, differentiate f(x) = 3x. (first principles)
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[3(x+h)−3x]/h = 3h/h = 3
7
Gradient of y = x² + 2x at x = 1. (gradient)
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2x+2 at x=1 = 4
8
Gradient of y = x³ at x = 2. (gradient)
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3x² at x=2 = 12
9
Find x where y = x² − 6x has a turning point. (turning point)
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2x−6 = 0 → x = 3
10
Find the turning points of y = x³ − 12x. (turning point)
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3x²−12=0 → x = ±2 → (2,−16) and (−2,16)

Now practise it

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