Master Applications of Derivatives – Class 12 Maths Chapter 6 | Tangents, Max‑Min, Approximation, NCERT & PYQs

 Here's the complete guide to Chapter 6: Applications of Derivatives (Class 12 CBSE Maths, 2025–26 syllabus):


🧠 A. Key Concepts & Applications

  1. Rate of Change

    • The derivative dy/dxdy/dx represents the rate at which yy changes w.r.t. xx.

    • Applicable in physics (velocity, acceleration), economics (marginal cost/revenue), biology, etc.

  2. Tangents and Normals

    • Slope of tangent at x=ax = a: m=f(a)m = f'(a).

    • Equation of tangent: yf(a)=f(a)(xa)y - f(a) = f'(a)(x - a).

    • Normal line slope: 1/f(a)-1/f'(a).

  3. Increasing/Decreasing Functions & Stationary Points

    • If f(x)>0f'(x) > 0 → increasing; f(x)<0f'(x) < 0 → decreasing.

    • Stationary points where f(x)=0f'(x) = 0.

    • Test using sign of f(x)f'(x) or second derivative f(x)f''(x) (if >0>0: min, if <0<0: max).

  4. Maxima and Minima

    • Locally highest/lowest values.

    • Applications include geometry (maximizing area/volume) and optimization problems (cost, revenue).

  5. Approximations via Derivatives

    • Use linearization: Δyf(a)Δx\Delta y ≈ f'(a) \cdot \Delta x.

    • Practical uses: estimating small errors, measuring slope near a point, etc.


📘 B. NCERT Exercise-Wise Detailed Solutions

  1. Ex 6.1: Rate-of-change and tangent-line problems.

  2. Ex 6.2: Find derivatives of complex functions using chain rule and apply in tangent equations.

  3. Ex 6.3: Analyze increasing/decreasing intervals, locate stationary points.

  4. Ex 6.4: Solve maxima/minima word problems (e.g., rectangular fencing, revenue).

  5. Ex 6.5: Show differentiation-based approximations.

Each solution includes step-by-step:

  1. Statement of the problem

  2. Expression of f(x)f'(x)

  3. Analysis and final result


🏆 C. CBSE Previous Year Questions (with Best Answers)

  • 2023: Tangent to curve y2=4x+3y^2 = 4x + 3 at point (1,3)(1,3).
    Solution: Differentiate implicitly → equation of tangent.

  • 2022: Find maxima of function f(x)=x1/3(1x)2/3f(x) = x^{1/3}(1 - x)^{2/3}.
    Solution: Compute f(x)f'(x), locate critical points, confirm using sign chart or f(x)f''(x).

  • 2021: A 20 m fencing problem to enclose a rectangular area with one side by a wall.
    Solution: Let width xx, length yy; express area A(x)A(x); find A(x)=0A'(x)=0 → max.

#Tags:
#Class12Maths, #Calculus, #Derivatives, #MaximaMinima, #Tangents, #Optimization, #CBSEPYQs, #NCERTSolutions, #Chapter6Maths


Comments

Popular posts from this blog

Rise of Nationalism in Europe – Class 10 CBSE Detailed Notes | Chapter 1 History

Nationalism in India – Class 10 CBSE History Notes | Chapter 2 | Key Events, Movements & Leaders

Resources and Development – Class 10 Geography Notes, Concepts, and Map Work