1. Chapter at a Glance
How fast is the area of an oil spill growing? Where is a function rising and where is it falling? What dimensions give a box the largest volume for a fixed amount of cardboard? Derivatives answer all these. This chapter applies differentiation to rates of change, increasing and decreasing functions, and maxima and minima (optimisation) — the most practical part of school calculus, and a regular source of 5-mark board questions.
- Derivative as a rate of change; related rates.
- Increasing and decreasing functions; the sign of f′(x).
- Critical points; local maxima and minima.
- First derivative test and second derivative test.
- Absolute maximum and minimum on a closed interval.
- Optimisation word problems (areas, volumes, costs).
2. Rate of Change
dy/dx is the rate of change of y with respect to x. If both depend on time t, the chain rule gives dy/dt = (dy/dx)(dx/dt) — the basis of "related rates" problems.
The radius of a circular ripple increases at 5 cm/s. How fast is the area increasing when the radius is 8 cm?
Solution: A = πr² → dA/dt = 2πr dr/dt = 2π(8)(5) = 80π cm²/s.
The total cost of producing x units is C(x) = 0.007x³ − 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.
Solution: MC = C′(x) = 0.021x² − 0.006x + 15. At x = 17: 0.021 × 289 − 0.102 + 15 = 6.069 − 0.102 + 15 = ₹20.967.
3. Increasing and Decreasing Functions
If f is continuous on [a, b] and differentiable on (a, b):
- f′(x) > 0 on (a, b) ⇒ f is strictly increasing.
- f′(x) < 0 on (a, b) ⇒ f is strictly decreasing.
- f′(x) = 0 on (a, b) ⇒ f is constant.
Find the intervals in which f(x) = 2x³ − 3x² − 36x + 7 is increasing or decreasing.
Solution: f′(x) = 6x² − 6x − 36 = 6(x − 3)(x + 2). Zeros at −2 and 3.
| Interval | Sign of f′ | f is |
|---|---|---|
| (−∞, −2) | + | Increasing |
| (−2, 3) | − | Decreasing |
| (3, ∞) | + | Increasing |
4. Maxima and Minima
A point c is a critical point of f if f′(c) = 0 or f′(c) does not exist. Local extrema can occur only at critical points.
- f′ changes from + to − at c ⇒ local maximum.
- f′ changes from − to + at c ⇒ local minimum.
- No sign change ⇒ point of inflection (neither), e.g. x³ at 0.
- f″(c) < 0 ⇒ local maximum.
- f″(c) > 0 ⇒ local minimum.
- f″(c) = 0 ⇒ test fails; use the first derivative test.
Find the local maximum and minimum values of f(x) = 2x³ − 3x² − 36x + 7.
Solution: Critical points −2 and 3. f″(x) = 12x − 6. f″(−2) = −30 < 0 → local max f(−2) = −16 − 12 + 72 + 7 = 51. f″(3) = 30 > 0 → local min f(3) = 54 − 27 − 108 + 7 = −74.
A continuous function on a closed interval has an absolute maximum and minimum. Find them by evaluating f at the critical points inside (a, b) and at the endpoints a and b, then comparing.
Find the absolute maximum and minimum of f(x) = 2x³ − 15x² + 36x + 1 on [1, 5].
Solution: f′(x) = 6(x − 2)(x − 3). Values: f(1) = 24, f(2) = 29, f(3) = 28, f(5) = 56. Absolute max 56 at x = 5; absolute min 24 at x = 1.
5. Optimisation Problems
- Name the quantity to maximise/minimise and write it as a function of one variable (use the given constraint to eliminate others).
- Find the critical points.
- Use the second derivative (or endpoints) to confirm max/min.
- Answer the question asked, with units.
Equal squares of side x are cut from the corners of an 18 cm × 18 cm sheet, and the sides are folded up to make an open box. Find x for maximum volume.
Solution: V = x(18 − 2x)², 0 < x < 9. V′ = (18 − 2x)² − 4x(18 − 2x) = (18 − 2x)(18 − 6x) = 0 → x = 3. V″(3) < 0 → max. x = 3 cm, V = 3 × 144 = 432 cm³.
Show that among all rectangles of given perimeter, the square has the maximum area.
Solution: Perimeter 2p: sides x and p − x. A = x(p − x); A′ = p − 2x = 0 → x = p/2; A″ = −2 < 0 → max. Sides p/2 and p/2 — a square. ∎
6. Solved Problems (Board Pattern)
A balloon remains spherical while air is pumped in at 900 cm³/s. Find the rate at which the radius increases when r = 15 cm.
Solution: V = (4/3)πr³ → dV/dt = 4πr² dr/dt → 900 = 4π(225) dr/dt → dr/dt = 1/π cm/s.
Show that the right circular cylinder of maximum volume that can be inscribed in a sphere of radius R has height 2R/√3.
Solution: Height h, radius r with r² = R² − h²/4. V = πh(R² − h²/4); V′ = π(R² − 3h²/4) = 0 → h = 2R/√3; V″ = −(3π/2)h < 0 → max. ∎
Show that f(x) = x³ − 3x² + 3x − 100 is increasing on ℝ.
Solution: f′(x) = 3(x − 1)² ≥ 0, zero only at x = 1 → f is increasing on ℝ.
7. Master Formula Table
| Idea | Rule |
|---|---|
| Rate of change | dy/dt = (dy/dx)(dx/dt) |
| Increasing / decreasing | f′ > 0 / f′ < 0 |
| Critical point | f′(c) = 0 or undefined |
| Second derivative test | f″ < 0 max; f″ > 0 min |
| Absolute extrema | Compare f at critical points and endpoints |
| Marginal cost / revenue | C′(x), R′(x) |
| Useful areas/volumes | A = πr², V = (4/3)πr³, V = πr²h, S = 4πr² |
8. Master Summary Table
| Question type | Key step | Watch out |
|---|---|---|
| Related rates | Differentiate the formula w.r.t. t | Substitute values after differentiating |
| Monotonicity | Sign chart of f′ | Factorise f′ fully |
| Local extrema | f′ = 0 then test | f″ = 0 is inconclusive |
| Absolute extrema | Include endpoints | Closed interval needed |
| Optimisation | One-variable function from constraint | Check domain |
9. Mind Map
- dy/dt chain
- Marginal cost
- Sign of f′
- Intervals
- First/second test
- Endpoints
- Boxes, cylinders
- Fixed perimeter
10. Tips and Tricks
- Maximising √(something) or something² is the same as maximising "something" (if positive) — simplify first.
- For a sign chart, test one value in each interval between the zeros of f′.
- Draw and label a diagram for every optimisation question.
- Substituting the given value (e.g. r = 8) before differentiating.
- Forgetting endpoints when finding absolute extrema.
- Reporting x instead of the maximum value (or vice versa) — read the question.
11. Important One-liners
- f′ > 0 means f is increasing.
- Local extrema occur only at critical points.
- f′(c) = 0 and f″(c) < 0 give a local maximum.
- Among rectangles of fixed perimeter, the square has the largest area.
- A continuous function on a closed interval attains an absolute max and min.
12. Exam-pattern Questions with Detailed Answers
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is:
- 10π
- 12π
- 8π
- 11π
Show answer
dA/dr = 2πr = 12π.
Answer: (B)
Show that f(x) = 3x + 17 is increasing on ℝ.
Show answer
f′(x) = 3 > 0.
Answer: Proved
Find the minimum value of f(x) = (2x − 1)² + 3.
Show answer
The square is ≥ 0, zero at x = ½ → minimum 3.
Answer: 3
The volume of a cube increases at 8 cm³/s. How fast is its surface area increasing when an edge is 12 cm?
Show answer
V = a³ → 8 = 3a² da/dt → da/dt = 8/432 = 1/54. S = 6a² → dS/dt = 12a da/dt = 144/54 = 8/3 cm²/s.
Answer: 8/3 cm²/s
Find the intervals where f(x) = x² − 4x + 6 is increasing and decreasing.
Show answer
f′ = 2x − 4. Decreasing on (−∞, 2), increasing on (2, ∞).
Answer: As above
Find the local extrema of f(x) = x³ − 6x² + 9x + 15.
Show answer
f′ = 3(x − 1)(x − 3). f″ = 6x − 12: f″(1) = −6 → max f(1) = 19; f″(3) = 6 → min f(3) = 15.
Answer: Local max 19 at x = 1; local min 15 at x = 3
Find two positive numbers whose sum is 15 and the sum of whose squares is minimum.
Show answer
S = x² + (15 − x)²; S′ = 4x − 30 = 0 → x = 7.5; S″ = 4 > 0. Numbers 7.5 and 7.5.
Answer: 15/2 and 15/2
Find the absolute maximum and minimum of f(x) = sin x + cos x on [0, π].
Show answer
f′ = cos x − sin x = 0 → x = π/4. f(0) = 1, f(π/4) = √2, f(π) = −1. Max √2, min −1.
Answer: Max √2, min −1
A window is a rectangle surmounted by a semicircle. Its perimeter is 10 m. Find the dimensions for maximum light (maximum area).
Show answer
Width 2r, height h: 2h + 2r + πr = 10 → h = (10 − 2r − πr)/2. A = 2rh + πr²/2 = r(10 − 2r − πr) + πr²/2 = 10r − 2r² − πr²/2. A′ = 10 − 4r − πr = 0 → r = 10/(π + 4). Then h = 10/(π + 4) = r. Width 20/(π + 4) m, height of rectangle 10/(π + 4) m.
Answer: r = h = 10/(π + 4) m
An open tank with a rectangular base must hold 8 m³ and be 2 m deep. Building costs ₹70 per m² for the base and ₹45 per m² for the sides. Find the least cost.
Show answer
Base x × y with 2xy = 8 → y = 4/x. Cost C = 70(4) + 45 × 2 × 2(x + y) = 280 + 180(x + 4/x). C′ = 180(1 − 4/x²) = 0 → x = 2 (C″ > 0, minimum), y = 2. C = 280 + 180 × 4 = ₹1000.
Answer: ₹1000 (square base 2 m × 2 m)
Show that the right circular cone of least curved surface area for a given volume has height √2 times the base radius.
Show answer
V = ⅓πr²h fixed → h = 3V/(πr²). S² = π²r²(r² + h²) = π²r⁴ + 9V²/r². d(S²)/dr = 4π²r³ − 18V²/r³ = 0 → 2π²r⁶ = 9V² = π²r⁴h² → h² = 2r² → h = √2 r; second derivative positive → minimum. ∎
Answer: Proved
A man 2 m tall walks at 5 km/h away from a lamp 6 m high. How fast is the tip of his shadow moving? How fast is the shadow lengthening?
Show answer
Let x = man's distance from lamp, s = shadow length. Similar triangles: 6/(x + s) = 2/s → s = x/2. ds/dt = ½ × 5 = 2.5 km/h (shadow lengthening). Tip moves at d(x + s)/dt = 7.5 km/h.
Answer: Tip 7.5 km/h; shadow grows at 2.5 km/h
13. Quick Revision Summary
- Rates: differentiate the relation w.r.t. t, then substitute.
- f′ > 0 increasing, f′ < 0 decreasing; use a sign chart.
- Critical points → first or second derivative test for local extrema.
- Absolute extrema on [a, b]: compare critical values and endpoints.
- Optimisation: one-variable function from the constraint, then test.