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MathematicsClass 12Application of Derivatives

Application of Derivatives | Class 12 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 3 min read

Application of Derivatives – this guide gives you a concise, exam-ready overview of Application of Derivatives from Class 12 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

Introduction

The chapter 'Application of Derivatives' in Class 12 Mathematics builds upon the foundational concept of derivatives introduced in earlier classes. Derivatives represent the rate of change of a function with respect to its variable and provide a powerful tool to analyze the behavior of functions. This chapter explores various practical and theoretical applications of derivatives, including how they help in understanding the rate at which quantities change, determining intervals where functions increase or decrease, finding tangents and normals to curves, approximating function values near points, and identifying maxima and minima of functions. These applications are essential in solving real-world problems across physics, economics, engineering, and other fields. The chapter emphasizes the use of derivatives not just as a mathematical concept but as a means to model and solve problems involving change and optimization.

📊 Diagram: No specific diagrams in the introduction section; serves as an overview.

🧪 Activity: No activity in the introduction section.

🔗 Connection: Leads to the next section on 'Rate of Change of Quantities' which formalizes the concept of derivatives as rates of change.

Frequently asked questions

1. Find the rate of change of the area of a circle with respect to its radius when the radius is 5 cm.

Let the area of the circle be A = πr^2. The rate of change of area with respect to radius is dA/dr = d(πr^2)/dr = 2πr. When r = 5 cm, dA/dr = 2π × 5 = 10π cm^2/cm.

2. The volume of a cube is increasing at the rate of 8 cm^3/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Let the edge of the cube be x cm. Volume V = x^3, Surface area S = 6x^2. Given dV/dt = 8 cm^3/s. dV/dt = 3x^2 dx/dt ⇒ dx/dt = dV/dt / (3x^2) = 8 / (3 × 12^2) = 8 / 432 = 1/54 cm/s. Now, dS/dt = 12x dx/dt = 12 × 12 × (1/54) = 144 / 54 = 8/3 cm^2/s.

3. Find the intervals in which the function f(x) = x^2 - 4x + 6 is increasing or decreasing.

First, find f'(x): f'(x) = 2x - 4. Set f'(x) = 0 ⇒ 2x - 4 = 0 ⇒ x = 2. For x < 2, f'(x) < 0 ⇒ f(x) is decreasing on (-∞, 2). For x > 2, f'(x) > 0 ⇒ f(x) is increasing on (2, ∞).

4. Find the equation of the tangent and normal to the curve y = x^2 + 2x at the point (1, 3).

First, find dy/dx: dy/dx = 2x + 2. At x = 1, dy/dx = 2 × 1 + 2 = 4. Equation of tangent: y - 3 = 4(x - 1) ⇒ y = 4x - 1. Slope of normal = -1/4. Equation of normal: y - 3 = -1/4(x - 1) ⇒ y = -1/4x + 3.25.

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