Key Idea: Differentiating a polynomial finds its gradient function f'(x) — the rate of change at every point. It's a pure Paper 1 by-hand skill and the foundation for tangents, maxima and minima later in the course.
📉 The power rule
- the power — bring it down to multiply
- the new power — one less than before
Multiply by the power, then subtract 1 from it. A constant multiple stays: d/dx(a·xⁿ) = a·n·xⁿ⁻¹. A standalone constant → 0 (its graph is flat). And x means x¹, so it differentiates to 1.
🔁 Rewrite BEFORE you differentiate
| Term | Rewrite as a power | Then differentiate |
|---|---|---|
| Polynomial term, e.g. 2x³ | already a power | Each term separately: 6x² |
| A fraction, $\frac{1}{xⁿ}$ | x⁻ⁿ | Power rule → more-negative power (−2 → −3) |
| A root, √x | x¹/² | ½x⁻¹⁄² = 1/(2√x), since ½ − 1 = −½ |
| A constant, e.g. −9 | — | 0 |
✏️ IB-style worked examples
IB-style question — differentiate a polynomial
Differentiate f(x) = 4x³ − 6x² + 2x − 11.
Step by step:
Apply the power rule to each term, keeping the signs.
The constant −11 differentiates to 0.
f'(x) = 12x² − 12x + 2
IB-style question — rewrite a fraction and a root first
Differentiate g(x) = 5/x² + √x.
Step by step:
Rewrite each term as a power.
Power rule: 5(−2)x⁻³ and ½x⁻¹⁄².
Tidy back into fraction / root form.
g'(x) = −10/x³ + 1/(2√x)
IB-style question — gradient at a point
Find the gradient of y = x³ − 5x at the point where x = 2.
Step by step:
Differentiate first.
Now substitute x = 2.
The gradient at x = 2 is 7.
IB-style question — find x for a given gradient
The curve y = x² − 6x has gradient 4 at one point. Find x there.
Step by step:
Differentiate and set equal to the gradient.
Solve for x.
x = 5.
Important: You cannot apply the power rule to √x or 1/x² as written. Rewrite them as x¹/² and x⁻² before differentiating — forgetting this is the single biggest error in this topic. (And differentiate before you substitute a value, never after.)
Tap each card to reveal the answer.
Differentiate 7x⁴ 28x³ — multiply by 4, drop the power to 3.
Differentiate 8x − 3 8 — the x¹ gives 8, the constant −3 gives 0.
Differentiate 1/x³ (write as a power first) −3x⁻⁴ = −3/x⁴ — rewrite as x⁻³, then −3 down, power → −4.
Differentiate √x ½x⁻¹⁄² = 1/(2√x) — write √x = x¹/², then ½ − 1 = −½.
Gradient of y = x² + 3x at x = 1 5 — f'(x) = 2x + 3, then sub x = 1: 2 + 3.
Where does y = x² have gradient equal to tan 45°? x = 0.5 — tan 45° = 1, so 2x = 1.
Exam tips
- Power rule: multiply by the power, then subtract 1 from it.
- Differentiate a polynomial term by term; standalone constants become 0.
- Rewrite fractions (1/xⁿ = x⁻ⁿ) and roots (√x = x¹/²) as powers BEFORE differentiating.
- Gradient at a point = f'(a): differentiate first, then substitute.
- For 'where is the gradient m?', solve f'(x) = m — a quadratic f' may give two answers. A tangent's gradient is tan(angle with the x-axis).