Derivatives of derivatives
The first derivative f′(x) tells how f changes. Differentiating again gives the second derivative f″(x): how the rate of change itself changes.
Example: f(x) = x⁴ − 2x³ + 5
- f′(x) = 4x³ − 6x²
- f″(x) = 12x² − 12x
Notation
| First derivative | Second derivative |
|---|---|
| f′(x) | f″(x) |
| dy/dx | d²y/dx² |
Higher-order derivatives continue: f‴(x), f⁽⁴⁾(x), …
What the first derivative tells you
| f′(x) | f(x) |
|---|---|
| > 0 | increasing |
| < 0 | decreasing |
| = 0 | horizontal tangent (possible max or min) |
What the second derivative tells you
| f″(x) | Shape of f |
|---|---|
| > 0 | concave up (holds water, like ∪) |
| < 0 | concave down (spills water, like ∩) |
| = 0 and changes sign | point of inflection |
Motion: position, velocity, acceleration
| Quantity | Relation |
|---|---|
| position s(t) | |
| velocity v(t) | s′(t) |
| acceleration a(t) | v′(t) = s″(t) |
Example: s(t) = t³ − 6t² + 9t (m, s)
- v(t) = 3t² − 12t + 9
- a(t) = 6t − 12
- At t = 1: v = 0, a = −6 m/s² (momentarily stopped, about to move backwards)
The second derivative test
At a point where f′(a) = 0:
- f″(a) > 0 → local minimum (concave up)
- f″(a) < 0 → local maximum (concave down)
- f″(a) = 0 → test is inconclusive
Example: f(x) = x³ − 3x
- f′(x) = 3x² − 3 = 0 → x = ±1
- f″(x) = 6x
- f″(1) = 6 > 0 → local min at x = 1
- f″(−1) = −6 < 0 → local max at x = −1
Interpreting in context
- If a company’s profit P(t) has P′ > 0 but P″ < 0, profit is increasing, but more slowly (growth is levelling off).
- Speeding up vs slowing down: an object speeds up when velocity and acceleration have the same sign.
Worked examples
Example 1: f(x) = 5x³. f″(x)? f′ = 15x², f″ = 30x.
Example 2: f(x) = eˣ. f″(x)? eˣ.
Example 3: s(t) = 4t² + 2t. Acceleration? 8 (constant).
Common mistakes
- Mixing up f′ and f″ meanings.
- Assuming f″ = 0 always means an inflection point (it must change sign).
- Forgetting units: acceleration is m/s².
Try it yourself
- f(x) = x⁵ − x². Find f″(x).
- Is y = x² concave up or down?
- s(t) = −5t² + 30t. Find v(3) and a(3).
- Use the second derivative test on f(x) = x² − 4x at its critical point.
Answers: 1) 20x³ − 2 2) concave up (f″ = 2) 3) v = 0, a = −10 4) x = 2; f″ = 2 > 0 → local minimum
Quick recap
- The first derivative shows increasing/decreasing; the second derivative shows concavity.
- Concave up: f″ > 0; concave down: f″ < 0.
- In motion, the second derivative of position is acceleration.