Key Idea: The derivative f′(x) gives the steepness of a curve at any x. In context it is a rate of change, so always give units.
| Notation | Means |
|---|---|
| f′(x) or dy/dx | The gradient function |
| f′(a) | The gradient at x = a |
| f′(x) = 0 | Flat here: a stationary point |
The power rule
- axⁿ → naxⁿ⁻¹: multiply by n, then reduce the power by 1.
- Differentiate a polynomial one term at a time.
- A constant gives 0; a linear term ax gives a.
- For the gradient at one point, find f′(x), then substitute. Expand brackets first.
Common mix-ups
- f(x) is the height; f′(x) is the steepness.
- A high f(x) does not mean a high f′(x).
- f′(x) = 0 means flat, not zero height.
Bring the power down to the front, then take one off the power.
T(t) = 80 − 2t². Find T′(3) and explain it. T′(t) = −4t, so T′(3) = −12: the drink cools at 12 °C per minute.
Find f′(x) for f(x) = 5x³ − 2x + 9. 15x² − 2.
Where does y = x³ − 6x² + 5 have gradient 0? 3x(x − 4) = 0, so x = 0 or x = 4.
Profit P (thousands of dollars) has P′(t) = −3. What does this mean? Profit is falling by 3 thousand dollars per year at that time.