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.

Increasing and Decreasing Functions

This section explains how the derivative of a function helps determine whether the function is increasing or decreasing over an interval. A function f(x) is said to be increasing on an interval if for any two points x1 and x2 in the interval, with x1 < x2, we have f(x1) ≤ f(x2). Similarly, it is decreasing if f(x1) ≥ f(x2). The derivative f'(x) provides a test for this behavior: if f'(x) > 0 for all x in an interval, then f is increasing there; if f'(x) < 0, then f is decreasing. The section also discusses the concept of stationary points where f'(x) = 0 and how these points can be candidates for local maxima or minima. The section includes examples showing how to determine intervals of increase and decrease by analyzing the sign of the derivative and interpreting the results graphically. This understanding is crucial for sketching graphs and analyzing function behavior in calculus.

📊 Diagram: Graph showing a curve with intervals marked where derivative is positive (increasing) and negative (decreasing), and points where derivative is zero (stationary points).

🧪 Activity: Activity includes finding intervals of increase and decrease for given functions and verifying with graphs.

🔗 Connection: Leads to the next section on 'Tangents and Normals' by applying derivatives to geometric problems.

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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