The derivative of the derivative: The second derivative is what you get by differentiating f'(x) again. It is written f''(x) or d²y/dx², and it measures how the gradient itself is changing.
IB-style question — find f''
Given f(x) = x³ − 4x², find f'(x) and f''(x).
Step by step
- Differentiate once.
- Differentiate again.
Final answer
f'(x) = 3x² − 8x; f''(x) = 6x − 8.
Just differentiate twice: There's no new rule — apply the power rule a second time to f'(x).
f'' > 0 cups up, f'' < 0 cups down: The sign of f'' gives the concavity: f''(x) > 0 → concave up (cup shape, ∪); f''(x) < 0 → concave down (cap shape, ∩). It tells you how the curve bends.
IB-style question — where concave up
For f(x) = x³ − 4x², find where the curve is concave up.
Step by step
- f''(x) = 6x − 8; concave up where f''(x) > 0.
- Solve.
Final answer
Concave up for x > 4/3 (and concave down for x < 4/3).
Up = ∪, down = ∩: Picture the shape: concave up holds water (∪); concave down spills it (∩).
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At a stationary point, check the sign of f'': At a stationary point (f'(x) = 0): if f''(x) > 0 it's a minimum (concave up); if f''(x) < 0 it's a maximum (concave down). This is the second-derivative test.
IB-style question — classify the stationary points
For f(x) = x³ − 3x, find and classify the stationary points using f''.
Step by step
- Stationary: f'(x) = 3x² − 3 = 0 ⇒ x = ±1. Second derivative f''(x) = 6x.
- Test each.
Final answer
Minimum at x = 1, maximum at x = −1.
If f'' = 0, the test fails: If f''(x) = 0 at the stationary point, the test is inconclusive — fall back on the sign of f' on each side.