Rate of change
If y = f(x), then dy/dx is the rate at which y changes with respect to x. When both depend on time t, the chain rule links their rates: dy/dt = (dy/dx)(dx/dt).
Example. The radius of a circular ripple grows at 3 cm/s. How fast is its area growing when r = 10 cm?
A = πr², so dA/dt = 2πr (dr/dt) = 2π × 10 × 3 = 60π cm²/s.
Example: marginal cost. If the cost of making x items is C(x) = 0.005x³ − 0.02x² + 30x + 5000, the marginal cost C′(x) = 0.015x² − 0.04x + 30 is the approximate cost of making one more item. At x = 3, it is 0.135 − 0.12 + 30 = ₹30.015.
Increasing and decreasing functions
On an interval:
if f′(x) > 0, f is strictly increasing;
if f′(x) < 0, f is strictly decreasing;
if f′(x) = 0 throughout, f is constant.
Method: find f′(x), find where f′(x) = 0 (the critical points), split the number line at those points, and check the sign of f′(x) in each part.
Example. f(x) = x³ − 3x² − 9x + 5. f′(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1), which is zero at x = −1 and x = 3.
Interval | Sign of f′(x) | f is |
|---|---|---|
(−∞, −1) | + | increasing |
(−1, 3) | − | decreasing |
(3, ∞) | + | increasing |
Local maxima and minima
A point c is a local maximum if f(c) is greater than or equal to every nearby value, and a local minimum if it is less than or equal to every nearby value. At such a point (where f is differentiable), f′(c) = 0.
But f′(c) = 0 doesn't guarantee an extremum. For f(x) = x³, f′(0) = 0, yet 0 is neither a maximum nor a minimum; it is a point of inflection.
First derivative test. At a critical point c:
f′ changes from + to −: local maximum.
f′ changes from − to +: local minimum.
f′ doesn't change sign: neither.
Second derivative test. If f′(c) = 0:
f″(c) < 0: local maximum.
f″(c) > 0: local minimum.
f″(c) = 0: the test fails; use the first derivative test.
Example (continued). f″(x) = 6x − 6. At x = −1, f″ = −12 < 0: local maximum, f(−1) = −1 − 3 + 9 + 5 = 10. At x = 3, f″ = 12 > 0: local minimum, f(3) = 27 − 27 − 27 + 5 = −22.
Absolute maximum and minimum on a closed interval
A function continuous on a closed interval [a, b] always has an absolute maximum and minimum there. To find them:
Find all critical points in (a, b).
Evaluate f at those points and at the end points a and b.
The largest value is the absolute maximum; the smallest is the absolute minimum.
Example. f(x) = x³ − 3x² − 9x + 5 on [−2, 4]. Values: f(−2) = −8 − 12 + 18 + 5 = 3, f(−1) = 10, f(3) = −22, f(4) = 64 − 48 − 36 + 5 = −15. Absolute maximum 10 at x = −1; absolute minimum −22 at x = 3.
Optimisation problems
Identify the quantity to maximise or minimise and write it in terms of one variable, using the given condition.
Differentiate, set the derivative to zero and solve.
Use the second derivative test (or the context) to confirm it is a maximum or minimum.
Answer the question that was asked.
Example: largest rectangle. Of all rectangles with perimeter 40 cm, which has the greatest area? With sides x and 20 − x, A = x(20 − x) = 20x − x². A′ = 20 − 2x = 0 gives x = 10, and A″ = −2 < 0, a maximum. The best rectangle is a 10 cm square, with area 100 cm².
Example: open box. Equal squares of side x are cut from the corners of an 18 cm square sheet, and the sides are folded up to make an open box. V = x(18 − 2x)². V′ = (18 − 2x)² − 4x(18 − 2x) = (18 − 2x)(18 − 6x), which is zero at x = 3 (x = 9 gives no box). V″ at x = 3 is negative, so the volume is greatest when x = 3 cm: V = 3 × 12² = 432 cm³.
Example: two numbers. Two positive numbers have sum 16. Their product P = x(16 − x) is greatest when x = 8, giving 64. The sum of their cubes, S = x³ + (16 − x)³, is least when S′ = 3x² − 3(16 − x)² = 0, also at x = 8.
Summary
dy/dx is a rate of change; link time rates with the chain rule.
f′ > 0: increasing; f′ < 0: decreasing.
Local max where f′ goes + to − (or f″ < 0); local min where f′ goes − to + (or f″ > 0).
On [a, b], compare the values at critical points and end points.
In optimisation, reduce to one variable before differentiating.