# NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

## NCERT Solutions for Class 12 Maths Chapter-6 Application of Derivatives

In chapter 5, Continuity and Differentiability, we already learnt how to find a derivative of functions like composite functions, implicit functions, inverse trigonometric functions, logarithmic functions and exponential functions. In this chapter, we will learn about applications of derivatives in various fields, e.g., in social science, science, engineering etc. In this chapter, there are five exercises with 102 questions.  The NCERT Solutions for Class 12 Maths Chapter-6 Application of Derivatives are prepared by subject experts.

The derivative can be used to find equations of tangent and normal to curves, to determine the rate of change in quantities, to find turning points on the graph of the function, use the derivative to find approximate value of certain quantities and also use derivative to find intervals on which a function is increasing or decreasing. The five exercises of NCERT Class 12 Maths Chapter-6 Application of Derivatives covers the rate of change of quantities, increasing and decreasing functions, approximations, tangents and normals and maxima and minima.

## What is the derivative?

The derivative  $\dpi{100} dS/dt$ is the rate of change of distance(S) with respect to the time(t). In a similar manner, whenever one quantity (y) varies with another quantity (x), and also satisfy $\dpi{100} y=f(x)$,then $\dpi{100} \frac{dy}{dx}$ or $\dpi{100} f^{'}(x)$ represents the rate of change of y with respect to x  and $\dpi{100} \dpi{100} \frac{dy}{dx} ]_{x=x_{o}}$ or $\dpi{100} f^{'}(x_{o})$ represents the rate of change of y with respect to x at $\dpi{100} x=x_{o}$.  Let's take an example of a derivative

 Example- Find the rate of change of the area of a circle per second with respect to its radius r when r = 5 cm. Solution-  The area A of a circle with radius r is given by $\dpi{100} A=\pi r^{2}$. Therefore, the rate of change of the area(A) with respect to its radius(r) is given by -                             $\dpi{100} \frac{dA}{dr}=\frac{d}{dr}(\pi r^{2})=2\pi r$                When   $\dpi{100} r=5cm,\:\:\: \frac{dA}{dr}=10\pi$ Thus, the area of the circle is changing at the rate of $\dpi{100} 10\pi \:\:cm^2/s$

## Topics of NCERT Grade 12 Maths Chapter-6 Application of Derivatives

6.1 Introduction

6.2 Rate of Change of Quantities

6.3 Increasing and Decreasing Functions

6.4 Tangents and Normals

6.5 Approximations

6.6 Maxima and Minima

6.6.1 Maximum and Minimum Values of a Function in a Closed Interval

## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives- Solved Exercise Questions

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.1

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.3

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.4

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.5

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous

## NCERT Solutions for class 12- Maths

 Chapter 1 Relations and Functions Chapter 2 Inverse Trigonometric Functions Chapter 3 Matrices Chapter 4 Determinants Chapter 5 Continuity and Differentiability Chapter 7 Integrals Chapter 8 Application of Integrals Chapter 9 Differential Equations Chapter 10 Vector Algebra Chapter 11 Three Dimensional Geometry Chapter 12 Linear Programming Chapter 13 Probability