Differentiation — Grade 12 Maths (CAPS)

The power rule and the ways Grade 12 exams apply 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 the ways 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 dy/dx = n·a·x^{n − 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 1/x as x⁻¹ and √x as x^1/2 before differentiating.

Worked ExampleExample 2: differentiate y = √x + 2/x

1
Rewrite: y = x^1/2 + 2x⁻¹.
Turn the surd and fraction into powers.
2
x^1/2 → 1/2x^(−1/2); 2x⁻¹ → −2x⁻².
Apply the power rule to each.
3
dy/dx = 1/2x^(−1/2) − 2x⁻² = 1/(2√x) − 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) = lim_h→0 (f(x + h) − f(x))/h
Always start from the definition.
2
= lim_h→0 ((x + h)² − x²)/h = lim_h→0 (2xh + h²)/h
Expand; the x² terms cancel.
3
= lim_h→0 (2x + h)
Divide by h (h ≠ 0 until the limit is taken).
4
= 2x
Only now let h go to zero — the lim sign stays on every line until this one.

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, and a first-principles question is worth five or six marks. The method never changes — only the algebra in the middle does. 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. Write the limit on every line until you take it, exactly as below.

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 before subtracting.
2
f(x + h) − f(x) = 2axh + ah² + bh.
The ax², bx and c all cancel.
3
f′(x) = lim_h→0 (2axh + ah² + bh)/h = lim_h→0 (h(2ax + ah + b))/h
Every remaining term has a factor of h — that is what makes it work. Keep the limit on this line.
4
= lim_h→0 (2ax + ah + b)
Cancel the h — only now is it safe to let h → 0.
5
= 2ax + b
The constant c has vanished, which is why a vertical shift never changes the gradient.

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

1
f(x + h) = a(x + h)³ = a(x³ + 3x²h + 3xh² + h³).
Cube the bracket.
2
f(x + h) − f(x) = a(3x²h + 3xh² + h³).
The ax³ terms cancel.
3
f′(x) = lim_h→0 (ah(3x² + 3xh + h²))/h = lim_h→0 a(3x² + 3xh + h²)
Factor out h and cancel it.
4
= 3ax²
Let h → 0: the 3xh and h² terms vanish.

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

1
f(x + h) − f(x) = a/(x + h) − a/x = (ax − a(x + h))/(x(x + h)) = −ah/(x(x + h)).
One common denominator — this is the step that decides the question. The ax terms cancel.
2
f′(x) = lim_h→0 −ah/(hx(x + h)) = lim_h→0 −a/(x(x + h))
Dividing by h cancels the h in the numerator.
3
= −a/x²
Let h → 0: x(x + h) becomes x².

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

1
f(x + h) − f(x) = c − c = 0.
A constant function has the same value everywhere.
2
f′(x) = lim_h→0 0/h = 0
A horizontal line has gradient zero — as it must.

⚠️

Writing lim_h→0 only on the final line. The limit notation must appear on every line until you actually let h go to zero — markers deduct for dropping it, and it is free to keep.

💡

Never substitute h = 0 before cancelling. You get 0/0, which is undefined. The whole point of the algebra is to cancel the h in the denominator 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^1/2). (rewrite)
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1/2x^(−1/2) = 1/(2√x)
6
From first principles, differentiate f(x) = 3x. (first principles)
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f′(x) = lim_h→0 (3(x + h) − 3x)/h = lim_h→0 3h/h = lim_h→0 3 = 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)
11
Determine f′(x) from first principles if f(x) = 2x² − 3x.
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f(x + h) = 2(x + h)² − 3(x + h) = 2x² + 4xh + 2h² − 3x − 3h ✓; f(x + h) − f(x) = 4xh + 2h² − 3h ✓; ÷ h: 4x + 2h − 3 ✓; lim as h→0 ✓; f′(x) = 4x − 3 ✓
12
Determine f′(x) from first principles if f(x) = 3/x.
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f′(x) = lim_h→0 (3/(x + h) − 3/x)/h ✓
= lim_h→0 (3x − 3(x + h))/(hx(x + h)) = lim_h→0 −3h/(hx(x + h)) ✓✓
= lim_h→0 −3/(x(x + h)) ✓
= −3/x² ✓
13
Use first principles to show that the derivative of f(x) = 7 is 0.
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f(x + h) − f(x) = 7 − 7 = 0 ✓; 0 ÷ h = 0, so f′(x) = 0 ✓
14
Explain why you may NOT substitute h = 0 at the start of a first-principles question.
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It gives 0 ÷ 0, which is undefined ✓; the h in the denominator must be cancelled by simplifying first ✓