Key Idea: The second derivative tells you how a curve bends — and gives the fastest way to classify a stationary point as a max or min. It's a pure non-calculator skill (Paper 1).
🔁 Differentiate twice
- first derivative — the gradient
- second derivative — how the gradient itself changes
There's nothing new to learn — just apply the power rule a second time to f′(x). e.g. f(x) = x⁴ → f′(x) = 4x³ → f″(x) = 12x².
🥣 Concavity & the second-derivative test
| Sign of f″(x) | Concavity | At a stationary point (f′ = 0) |
|---|---|---|
| f″(x) > 0 | Concave up (∪, holds water) | Minimum |
| f″(x) < 0 | Concave down (∩, spills water) | Maximum |
| f″(x) = 0 | Possible point of inflexion | Test inconclusive — check sign of f′ |
Tip: Concave up is a cup ∪ (gradient increasing); concave down is a cap ∩ (gradient decreasing).
✏️ IB-style worked examples
IB-style question — find the second derivative
Given f(x) = 2x³ − 5x², find f′(x) and f″(x).
Step by step:
Differentiate once with the power rule.
Differentiate f′(x) again.
f′(x) = 6x² − 10x; f″(x) = 12x − 10.
IB-style question — where is the curve concave up?
For f(x) = 2x³ − 5x², find the values of x for which the curve is concave up.
Step by step:
Concave up where f″(x) > 0.
Solve the inequality.
Concave up for x > 5/6 (and concave down for x < 5/6).
IB-style question — classify stationary points with f″
For f(x) = x³ − 12x, find the stationary points and classify each using the second-derivative test.
Step by step:
Stationary where f′(x) = 0.
Find the second derivative.
Test each x-value.
Minimum at x = 2, maximum at x = −2.
Important: If f″(x) = 0 at a stationary point, the second-derivative test is inconclusive — it is not automatically a point of inflexion. Fall back on the sign of f′ just left and right of the point.
Tap each card to reveal the answer.
f(x) = x⁵, find f″(x) f′ = 5x⁴, so f″(x) = 20x³ — just power-rule twice.
What does f″(x) > 0 tell you about the curve? The curve is concave up (∪) — the gradient is increasing.
At a stationary point, f″ < 0. Max or min? Maximum — concave down (∩).
f(x) = x³ − 3x has f′ = 0 at x = 1. f″(1) = ?, and is it max or min? f″(x) = 6x, so f″(1) = 6 > 0 → minimum.
At a stationary point f″ = 0. What's the verdict? Inconclusive — check the sign of f′ on each side instead.
Exam Tips
- f″(x) = differentiate f′(x) again (= d²y/dx²) — no new rule, just the power rule twice.
- f″ > 0 → concave up (∪); f″ < 0 → concave down (∩).
- Second-derivative test at a stationary point: f″ > 0 → min, f″ < 0 → max.
- If f″ = 0 at the stationary point the test fails — use the sign of f′ either side.
- This is a Paper 1 by-hand skill — show f′(x) and f″(x) clearly to earn the method marks.