Applications of Derivatives
'The derivative is the instantaneous rate of change — it tells you how fast something is changing at a single moment.'
1. Chapter Overview
This chapter applies DERIVATIVES to real-world and geometric problems. Topics include: RATE OF CHANGE of quantities, EQUATIONS OF TANGENTS AND NORMALS, INCREASING AND DECREASING functions, APPROXIMATIONS using differentials, and MAXIMA AND MINIMA (including absolute and local extrema). These are the practical tools of calculus — used everywhere from physics to economics.
2. Rate of Change
- If y = f(x), then dy/dx represents the RATE OF CHANGE of y with respect to x.
- 'If a quantity changes with time, its derivative gives the speed of that change.'
Worked Example 1
Problem: The radius of a circle is increasing at the rate of 0.7 cm/s. Find the rate of increase of its area when r = 5 cm. Solution: A = πr². dA/dt = (dA/dr) × (dr/dt) = 2πr × (dr/dt) = 2π(5)(0.7) = 7π cm²/s.
3. Tangents and Normals
Tangent
- Slope of tangent to curve y = f(x) at (x₀, y₀): m = f'(x₀).
- Equation of tangent: y − y₀ = f'(x₀)(x − x₀).
Normal
- Slope of normal = −1/f'(x₀) (provided f'(x₀) ≠ 0).
- Equation of normal: y − y₀ = (−1/f'(x₀))(x − x₀).
Worked Example 2
Problem: Find the equations of the tangent and normal to the curve y = x³ − 2x² + 1 at x = 2. Solution: y(2) = 8 − 8 + 1 = 1. Point: (2, 1). dy/dx = 3x² − 4x. At x = 2: m = 12 − 8 = 4. Tangent: y − 1 = 4(x − 2) ⇒ y = 4x − 7. Normal: y − 1 = (−1/4)(x − 2) ⇒ 4y − 4 = −x + 2 ⇒ x + 4y = 6.
4. Increasing and Decreasing Functions
| Condition | Function Behaviour | Example |
|---|---|---|
| f'(x) > 0 for all x in (a, b) | STRICTLY INCREASING | f(x) = x³ on R |
| f'(x) < 0 for all x in (a, b) | STRICTLY DECREASING | f(x) = −x on R |
| f'(x) ≥ 0 for all x in (a, b) | NON-DECREASING | f(x) = x³ (non-decreasing) |
| f'(x) ≤ 0 for all x in (a, b) | NON-INCREASING | — |
'If the derivative is positive, the function climbs. If negative, it falls. Zero? It could be a turning point.'
5. Tangents and Normals — Approximation Using Differentials
- ∆y ≈ dy = f'(x) · dx
- f(x + ∆x) ≈ f(x) + f'(x) · ∆x
Worked Example 3
Problem: Approximate √(25.3) using differentials. Solution: Let f(x) = √x. f'(x) = 1/(2√x). Take x = 25, ∆x = 0.3. √(25.3) ≈ √25 + (1/(2√25))(0.3) = 5 + (1/10)(0.3) = 5 + 0.03 = 5.03. Actual: √(25.3) ≈ 5.02998. Close match!
6. Maxima and Minima
6.1 Local Maxima and Minima
| Condition | Type of Point |
|---|---|
| f'(x) = 0 and f''(x) < 0 | Local MAXIMUM |
| f'(x) = 0 and f''(x) > 0 | Local MINIMUM |
| f'(x) = 0 and f''(x) = 0 | Test FAILS. Use first derivative test |
6.2 First Derivative Test
- If f'(x) changes from + to − at c → Local MAXIMUM at x = c.
- If f'(x) changes from − to + at c → Local MINIMUM at x = c.
- If f'(x) does NOT change sign → NEITHER (point of inflection).
6.3 Absolute Maxima and Minima on [a, b]
- Find all critical points (f'(x) = 0 or undefined) in (a, b).
- Evaluate f at critical points AND at endpoints a and b.
- The largest value is the ABSOLUTE MAXIMUM. The smallest is the ABSOLUTE MINIMUM.
Worked Example 4
Problem: Find the local maxima and minima of f(x) = x³ − 6x² + 9x + 1. Solution: f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). Critical points: x = 1, x = 3. f''(x) = 6x − 12. At x = 1: f''(1) = −6 < 0 ⇒ Local MAXIMUM. f(1) = 1 − 6 + 9 + 1 = 5. At x = 3: f''(3) = 6 > 0 ⇒ Local MINIMUM. f(3) = 27 − 54 + 27 + 1 = 1.
7. Optimisation Problems
Optimisation uses the same concepts to solve real-world problems. 'Find the derivative, set it to zero, check whether it is a maximum or minimum.'
Worked Example 5
Problem: A farmer wants to fence a rectangular field adjacent to a river. No fence is needed along the river. If 100 m of fencing is available, find the dimensions that maximise the area. Solution: Let width = x (perpendicular to river), length = y (parallel to river). Fencing: 2x + y = 100 ⇒ y = 100 − 2x. Area A = xy = x(100 − 2x) = 100x − 2x². dA/dx = 100 − 4x = 0 ⇒ x = 25 m. y = 100 − 50 = 50 m. d²A/dx² = −4 < 0 ⇒ Maximum. Max area = 25 × 50 = 1250 m².
8. Common Mistakes
- Confusing increasing with positive derivative: A function can be increasing even where derivative is zero at isolated points (e.g., f(x) = x³ at x = 0).
- Second derivative test confusion: f''(x) < 0 means MAXIMUM (concave down), f''(x) > 0 means MINIMUM (concave up). Many students get this BACKWARDS.
- Forgetting to check endpoints: For absolute extrema on a closed interval, ALWAYS evaluate endpoints.
- Rate of change chain rule: dA/dt = (dA/dr)(dr/dt) — forgetting the chain rule leads to wrong answers.
9. CBSE Exam Focus
- Rate of change problems — especially areas, volumes, and related rates
- Equations of tangents and normals
- Finding intervals of increase/decrease
- Local maxima and minima using first and second derivative tests
- Absolute maxima and minima on closed intervals
- Word problems — optimisation (maximising area, minimising cost, etc.)
10. Self-Test
Q1: Find the intervals in which f(x) = 2x³ − 15x² + 36x + 7 is strictly increasing or decreasing. A1: f'(x) = 6x² − 30x + 36 = 6(x² − 5x + 6) = 6(x − 2)(x − 3). f'(x) > 0 when x < 2 or x > 3. f'(x) < 0 when 2 < x < 3. Increasing on (−∞, 2) ∪ (3, ∞). Decreasing on (2, 3).
Q2: Find the equations of tangent and normal to the circle x² + y² = 25 at (3, 4). A2: 2x + 2y(dy/dx) = 0 ⇒ dy/dx = −x/y. At (3, 4): m = −3/4. Tangent: y − 4 = (−3/4)(x − 3) ⇒ 3x + 4y = 25. Normal: y − 4 = (4/3)(x − 3) ⇒ 4x − 3y = 0.
Q3: Find two positive numbers whose sum is 16 and whose product is maximum. A3: Let numbers be x and (16 − x). P = x(16 − x) = 16x − x². dP/dx = 16 − 2x = 0 ⇒ x = 8. d²P/dx² = −2 < 0 ⇒ Max. Numbers: 8, 8. Product = 64.
Q4: A circular disc of radius 7 cm is being heated. Its radius increases at 0.02 cm/s. Find the rate of increase of area when r = 7 cm. A4: A = πr². dA/dt = 2πr(dr/dt) = 2π(7)(0.02) = 0.28π cm²/s.
Q5: Find the absolute maximum and minimum of f(x) = 2x³ − 3x² − 12x + 5 on [−2, 3]. A5: f'(x) = 6x² − 6x − 12 = 6(x² − x − 2) = 6(x − 2)(x + 1). Critical: x = 2, x = −1. f(−2) = −16 − 12 + 24 + 5 = 1. f(−1) = −2 − 3 + 12 + 5 = 12. f(2) = 16 − 12 − 24 + 5 = −15. f(3) = 54 − 27 − 36 + 5 = −4. Absolute max = 12 at x = −1. Absolute min = −15 at x = 2.
11. Conclusion
Applications of derivatives show why calculus MATTERS:
- RATE OF CHANGE: 'How fast is it changing right now?'
- TANGENTS: 'What is the slope at this exact point?'
- OPTIMISATION: 'What is the best possible outcome given constraints?'
- APPROXIMATION: 'What is a good estimate without the exact value?'
'Derivatives are not just abstract mathematics — they are the language of change in physics, engineering, economics, and biology.'
