Applications of Derivatives
Learn Applications of Derivatives with notes, examples, and practice questions.
Chapter Overview
Derivatives measure instantaneous rate of change. This chapter applies them to curve sketching, optimization, and approximation — the most practical part of calculus for board exams.
Rate of Change
Velocity
v = ds/dt
Acceleration
a = dv/dt = d²s/dt²
Related rates
Differentiate both sides w.r.t. time t
Increasing / Decreasing & Critical Points
- f′(x) > 0 → f increasing; f′(x) < 0 → f decreasing
- Critical points: where f′(x) = 0 or f′ undefined
- First derivative test: sign change of f′ at critical point → max or min
Second Derivative & Curvature
Concavity
f″ > 0 → concave up; f″ < 0 → concave down
Second derivative test
f′(c)=0, f″(c)>0 → local min; f″(c)<0 → local max
Point of inflection
f″ changes sign
Maxima and Minima (Optimization)
Closed interval method: evaluate f at critical points and endpoints; pick largest/smallest.
Word problems: define variable, write function to optimize, find critical points, verify max/min.
Approximation
Linear approximation
f(x+h) ≈ f(x) + h·f′(x)
Marginal concepts
Marginal cost ≈ dC/dx
Solved Examples
Step-by-step solutions — read each step before checking the final answer.
Example 1: Find Local Max
Find local extrema of f(x) = x³ − 3x² + 2.
- 1f′ = 3x² − 6x = 3x(x−2)
- 2Critical: x = 0, 2
- 3f″ = 6x−6; f″(0)<0 max; f″(2)>0 min
- 4f(0)=2 max, f(2)=−2 min
Answer
Local max 2 at x=0; local min −2 at x=2
Example 2: Optimization
Rectangle perimeter 40. Maximize area.
- 1Let sides x and 20−x
- 2A = x(20−x) = 20x − x²
- 3A′ = 20−2x = 0 → x = 10
- 4Square gives max area 100
Answer
Max area 100 when x = 10 (square)
Example 3: Related Rates
Circle radius increases 2 cm/s. Rate of area change when r=5?
- 1A = πr²
- 2dA/dt = 2πr·dr/dt
- 3dA/dt = 2π(5)(2) = 20π cm²/s
Answer
20π cm²/s
Key Points to Remember
- ✓ Set f′(x)=0 for critical points — don't forget endpoints on closed intervals
- ✓ Second derivative test fails when f″(c)=0 — use first derivative test
- ✓ Optimization: one variable, one objective function
Exam Tips
- • Draw sign chart for f′ in max/min problems
- • State units in applied problems
- • Verify answer makes physical sense (positive length, etc.)
Formula Cheat Sheet
Quick reference — NCERT board exam formulas
Applications of Derivatives
- Product:(uv)′ = u′v + uv′
- Quotient:(u/v)′ = (u′v − uv′)/v²
- Chain:[f(g(x))]′ = f′(g(x))·g′(x)
- f″(x) = d²y/dx²
- Tangent:slope = f′(a); y − y₁ = f′(a)(x − x₁)
- Normal:slope = −1/f′(a)
- Max/min:f′(a)=0; f″(a)<0 → max, f″(a)>0 → min
Practice MCQs — Applications of Derivatives
Test your understanding with topic-wise multiple choice questions. Explanations appear after each answer.
If f′(x)>0 on (a,b), f is: