Rate of change of quantities
If y depends on x, then dy/dx is the rate of change of y with respect to x. At x = a we write (dy/dx) at a, or f′(a).
If both x and y change with time t, use the chain rule: dy/dt = (dy/dx) × (dx/dt).
Example: a circle's radius grows at 0.5 cm/s. Area A = πr², so dA/dt = 2πr × dr/dt. When r = 4 cm, dA/dt = 2π(4)(0.5) = 4π cm²/s.
Marginal cost and marginal revenue in economics are also derivatives: the change in cost or revenue for one more item.
Increasing and decreasing functions
On an interval where f is differentiable:
- f′(x) > 0 for all x → f is strictly increasing (graph climbs left to right).
- f′(x) < 0 for all x → f is strictly decreasing.
- f′(x) = 0 for all x → f is constant.
How to find the intervals
- Find f′(x) and solve f′(x) = 0.
- These points split the number line into intervals.
- Test the sign of f′ in each interval.
For f(x) = x³ − 3x: f′ = 3(x − 1)(x + 1). Increasing on (−∞, −1) and (1, ∞), decreasing on (−1, 1).
Maxima and minima
A local maximum is a point higher than all nearby points; a local minimum is lower than all nearby points. At such a point (if the function is smooth there) f′(c) = 0. Points with f′(c) = 0 or f′ not defined are called critical points.
First derivative test
- f′ changes from + to − at c → local maximum.
- f′ changes from − to + at c → local minimum.
- No sign change → neither (a point of inflection, like x³ at 0).
Second derivative test
- f′(c) = 0 and f″(c) < 0 → local maximum.
- f′(c) = 0 and f″(c) > 0 → local minimum.
- f″(c) = 0 → the test fails; go back to the first derivative test.
Absolute (global) max and min on [a, b]
Find all critical points inside [a, b], work out f at those points and at the ends a and b. The biggest value is the absolute maximum, the smallest is the absolute minimum.
Real-life problems
- Name the quantity to make largest/smallest (volume, area, cost).
- Write it in terms of one variable using the given condition.
- Differentiate, set the derivative to 0, solve.
- Use the second derivative (or first) test to confirm max or min.
- Answer the question with units.
Box example: V = x(12 − 2x)², dV/dx = 12(x − 2)(x − 6). x = 6 gives no box, so x = 2; V″(2) = −48 < 0, a maximum of 128 cm³.
Key formulas and definitions
- Rate of change: dy/dx; related rates dy/dt = (dy/dx)(dx/dt)
- f′(x) > 0 → increasing, f′(x) < 0 → decreasing
- Critical point: f′(c) = 0 or f′(c) not defined
- First derivative test: + → − max, − → + min
- Second derivative test: f″(c) < 0 max, f″(c) > 0 min
Worked examples
1. The side of a square grows at 3 cm/s. How fast is its area growing when the side is 10 cm?
A = x², dA/dt = 2x·dx/dt = 2(10)(3) = 60 cm²/s.
2. A balloon's radius increases at 2 cm/s. Find the rate of change of volume when r = 5 cm.
V = (4/3)πr³, dV/dt = 4πr²·dr/dt = 4π(25)(2) = 200π cm³/s.
3. The cost of making x items is C(x) = 0.5x² + 10x + 100. Find the marginal cost when x = 20.
C′(x) = x + 10. C′(20) = 30. Making the 21st item costs about ₹30 more.
4. Find the intervals where f(x) = x² − 4x + 1 is increasing or decreasing.
f′(x) = 2x − 4 = 0 at x = 2. For x < 2, f′ < 0 (decreasing); for x > 2, f′ > 0 (increasing).
5. Find local maxima and minima of f(x) = 2x³ − 9x² + 12x + 1.
f′ = 6x² − 18x + 12 = 6(x − 1)(x − 2) = 0 → x = 1, 2. f″ = 12x − 18. f″(1) = −6 < 0 → local max f(1) = 6. f″(2) = 6 > 0 → local min f(2) = 5.
6. Find the absolute max and min of f(x) = x³ − 3x on [0, 3].
Critical point inside: x = 1 (x = −1 is outside). f(0) = 0, f(1) = −2, f(3) = 18. Absolute max 18 at x = 3, absolute min −2 at x = 1.
7. Two positive numbers add up to 20. Find them so their product is largest.
Numbers x and 20 − x. P = 20x − x². P′ = 20 − 2x = 0 → x = 10. P″ = −2 < 0 → max. Numbers 10 and 10, product 100.
8. A rectangle has perimeter 40 m. Show that its area is largest when it is a square.
Sides x and 20 − x, A = 20x − x², A′ = 20 − 2x = 0 → x = 10, A″ = −2 < 0. Both sides 10 m: a square of area 100 m².
Common mistakes
- Thinking f′(c) = 0 always means a max or min. For x³ at 0 the slope is 0 but the curve keeps rising.
- Forgetting to check the end points when finding the absolute max or min on a closed interval.
- In word problems, differentiating before writing the quantity in one variable.
- Mixing units in rates: if r is in cm and time in s, the answer is in cm²/s or cm³/s.