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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.

Tangents and Normals

This section focuses on the geometric application of derivatives in finding the equations of tangents and normals to curves at given points. A tangent to a curve at a point is a straight line that touches the curve at that point without crossing it, having the same slope as the curve there. The slope of the tangent line to the curve y = f(x) at point (x1, y1) is given by the derivative f'(x1). Using the point-slope form of a line, the equation of the tangent is y - y1 = f'(x1)(x - x1). The normal to the curve at the same point is the line perpendicular to the tangent, and its slope is the negative reciprocal of the tangent's slope, i.e., -1/f'(x1). The equation of the normal is y - y1 = -1/f'(x1) (x - x1). The section includes examples of finding tangent and normal equations for various curves such as polynomials, circles, and implicit functions. This application is fundamental in geometry, physics, and engineering for analyzing curve behavior and constructing related lines.

📊 Diagram: Diagram of a curve with a point marked, showing the tangent line touching the curve at that point and the normal line perpendicular to the tangent.

🧪 Activity: Activity includes finding tangent and normal lines to given curves at specified points.

🔗 Connection: Prepares for 'Approximations' section by understanding linearization via tangents.

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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