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

Last updated date: 14th Aug 2024
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## NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 - FREE PDF Download

NCERT Chapter 6 Exercise 6.1 of Class 12 Maths, "Application of Derivatives," is an important chapter that deals with real life uses of derivatives in various fields. Exercise 6.1 focuses on the fundamental applications such as rate of change of quantities. This concept is crucial for understanding the real-world applications of calculus. It's important to focus on understanding how derivatives are used to analyze the behavior of functions.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 6 Exercise 6.1 Class 12 | Vedantu
3. Access NCERT Solutions for Maths Class 12 Chapter 6- Application of Derivatives Exercise 6.1
4. Class 12 Maths Chapter 6: Exercises Breakdown
5. CBSE Class 12 Maths Chapter 6 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

By understanding the application of derivatives class 12 exercise 6.1, students will be well-prepared for advanced problems and applications in calculus. Vedantu's detailed NCERT solutions for class 12 provide clear, step-by-step guidance to help you grasp these essential concepts and excel in your exams.

## Glance on NCERT Solutions Maths Chapter 6 Exercise 6.1 Class 12 | Vedantu

• Class 12 Maths Chapter 6 Exercise 6.1, "Application of Derivatives," explores the practical uses of derivatives in various fields

• The derivative of a function represents the rate at which the function's value changes as its input changes.

• The rate of change of a quantity refers to how one quantity changes in relation to another. In mathematical terms, it is the derivative of one variable with respect to another.

• Ensure you are comfortable with different techniques of differentiation, including the power rule, product rule, quotient rule, and chain rule before solving ex 6.1 class 12.

• If Y=f(x), then the derivative  $f^{-1}(x)$ represents that rate of change of y with respect to x.

• Maths Class 12 Application of Derivatives article contains exercise notes, important questions, exemplar solutions, exercises and video links for exercise 6.1 -  Application of Derivatives, which you can download as PDFs.

• Class 12 Maths Exercise 6.1 NCERT solutions has over all 18 questions.

Competitive Exams after 12th Science

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

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

a) $r=3cm$

b) $r=4cm$

Ans: The area of a circle (A) with radius $(r)$ is $A=\pi {{r}^{2}}$.

Now, for a given radius, the change in the area of the circle is given by,

$\frac{dA}{dr}=\frac{d}{dr}\left( \pi {{r}^{2}} \right)=2\pi r$

1. When $r=3cm$

$\frac{dA}{dr}=2\pi (3)$

$=6\pi$

As a result, when the radius of the circle is 3cm, the area of the circle changes at a rate of $6\pi$ cm.

2. When $r=4cm$

$\frac{dA}{dr}=2\pi (4)$

$=8\pi$

As a result, when the radius of the circle is 4cm, the area of the circle changes at a rate of $8\pi$cm.

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

Ans:

Let $x$ be the length of a side, $v$ be the volume, and $s$ be the surface area of the cube.

When $x$ is a function of time $t$, we have

$V={{x}^{3}}$ and $S=6{{x}^{2}}$

It is given that $\frac{dV}{dt}=8c{{m}^{3}}/s$

Then, by chain rule we have

$\therefore 8=\frac{dv}{dt}=\left( {{x}^{3}} \right)$

$=\frac{d}{x}\left( x^3 \right).\frac{dx}{dt}=3{{x}^{2}}.\frac{dx}{dt}$

$\Rightarrow \frac{dx}{dt}=\frac{8}{3{{x}^{2}}}$                                                                                    …… (1)

Now, $\Rightarrow \frac{ds}{dt}=\frac{d}{dt}\left( 6{{x}^{2}} \right)=\frac{d}{dx}\left( 6{{x}^{2}} \right).\frac{dx}{dt}$                                 (By chain rule)

$=12x.\frac{dx}{dt}=\frac{32}{x}$

Thus, when $x=12cm$,

$\frac{dS}{dt}=\frac{32}{12}c{{m}^{2}}/s$

$=\frac{8}{3}c{{m}^{2}}/s.$

As a result, if the cube's edge length is $12cm$, the surface area is rising at a rate of $\frac{8}{3}c{{m}^{2}}/s.$

3. The radius of a circle is increasing uniformly at the rate of $3$cm/s. Find the rate at which the area of the circle is increasing when the radius is $10$ cm.

Ans: For the radius $(r)$, the area of a circle (A) is $A=\pi {{r}^{2}}$.

Now, for time ($t$), the rate of change of area (A) is given by,

$\frac{dA}{dt}=\frac{d}{d}\left( \pi {{r}^{2}} \right)$

$=2\pi r\frac{dr}{dt}$                                                          (By chain rule)

It is given that the increase in radius of circle is,

$\frac{dr}{dt}=3cm/s$

$\frac{dA}{dt}=2\pi r\left( 3 \right)=6\pi r$

Thus, when $r=10cm$

$\frac{dA}{dt}=6\pi \left( 10 \right)=60\pi c{{m}^{2}}/s$

As a result, when the radius of the circle is $10cm$, the rate at which the area of the circle increases is $60\pi c{{m}^{2}}/s$.

4. An edge of a variable cube is increasing at the rate of $3$cm/s. How fast is the volume of the cube increasing when the edge is $10$cm long?

Ans: Let $x$ be the length of a side and $v$ be the volume of the cube. Then,        $V={{x}^{3}}$

$\therefore \frac{dV}{dt}=3{{x}^{2}}.\frac{dx}{dt}$  (By chain rule)

It is given that,

$\frac{dx}{dt}=3cm/s$

$\therefore \frac{dV}{dt}=3{{x}^{2}}(3)=9{{x}^{2}}$

Then, when $x=10cm$,

$\frac{dV}{dt}=9{{\left( 10 \right)}^{2}}=900c{{m}^{3}}/s$

As a result, when the edge is $10cm$ long, the volume of the cube increases at the rate of $900c{{m}^{3}}/s$.

5. A stone is dropped into a quiet lake and waves move in circles at the speed of $5$ cm/s. At the instant when the radius of the circular wave is $8$ cm, how fast is the enclosed area increasing?

Ans: The area of a circle (A) with radius $(r)$ is given by,

$A=\pi {{r}^{2}}$

Now, for time ($t$), the rate of change of area (A) is given by,

$\therefore \frac{dA}{dt}=\frac{d}{dt}\left( \pi {{r}^{2}} \right)$

$=\frac{d}{dr}\left( \pi {{r}^{2}} \right)\frac{dr}{dt}=2\pi r\frac{dr}{dt}$ (By chain rule)

It is given that $\frac{dr}{dt}=5cm/s$.

Thus, when $r=8cm$

$\frac{dA}{dt}=80\pi$

As a result, when the circular wave's radius is $8$cm, the enclosed area grows at a pace of $80\pi c{{m}^{2}}/s$.

6. The radius of a circle is increasing at the rate of $0.7$cm/s. What is the rate of increase of its circumference?

Ans. The circumference of a circle (C) with the radius $\left( r \right)$ is given by

$C=2\pi r$.

Now, the rate of change of circumference (C) for time ($t$) is given by,

$\frac{dC}{dt}=\frac{dC}{dr}.\frac{dr}{dt}$                                                            (By chain rule)

$=\frac{d}{dr}\left( 2\pi r \right)\frac{dr}{dt}$

$=2\pi .\frac{dr}{dt}$

It is given that $\frac{dr}{dt}=0.7cm/s$.

Hence, the rate of increase of the circumference is $1.4\pi cm/s$.

7. The length $x$ of a rectangle is decreasing at the rate of $5$ cm/minute and the width $y$ is increasing at the rate of $4$ cm/minute. When  $x=8cm$ and $y=6cm$, find the rates of change of $\left( a \right)$ the perimeter, and $\left( b \right)$ the area of the rectangle.

Ans: It is given that the length $x$ is decreasing at the rate of $5cm/\min$ and the width $y$ is increasing at the rate of $4cm/\min$i.e,

$\frac{dx}{dt}=-5cm/\min ,$

$\frac{dy}{dt}=4cm/\min$

(a) The perimeter (P) of a rectangle is given by

$P=2\left( x+y \right)$

$\therefore \frac{dP}{dt}=2\left( \frac{dx}{dt}+\frac{dy}{dt} \right)$

$=-2cm/\min$

Hence, the rate of decrease in the perimeter of rectangle is $2cm/\min$.

(b) The area (A) of a rectangle is given by,

$A=x\times y$

$\frac{dA}{dt}=\frac{dx}{dt}.y+x.\frac{dy}{dt}$

$=-5y+4x$

When $x=8cm$ and $y=6cm$,

$\frac{dA}{dt}=\left( -5\times 6+4\times 8 \right)c{{m}^{2}}/\min$

$=2c{{m}^{2}}/\min$

Hence, the rate of increase in the area of the rectangle is $2c{{m}^{2}}/\min$.

8. A balloon, which always remains spherical on inflation, is being inflated by pumping in $900$ cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is $15$ cm.

Ans: The volume of a sphere (V) with radius ($r$) is given by,

$V=\frac{4}{3}\pi {{r}^{3}}$

For time ($t$), the rate of change of volume (V) is given by,

$\frac{dV}{dt}=\frac{dV}{dr}.\frac{dr}{dt}$                                                   (By chain rule)

$=\frac{d}{dr}\left( \frac{4}{3}\pi {{r}^{3}} \right).\frac{dr}{dt}$

$=4\pi {{r}^{2}}.\frac{dr}{dt}$

It is given that

$\frac{dv}{dt}=900c{{m}^{3}}/s$

$\therefore 900=4\pi {{r}^{2}}.\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{900}{4\pi {{r}^{2}}}$

$=\frac{225}{\pi {{r}^{2}}}$

Therefore, when the radius is $15$cm,

$\frac{dr}{dt}=\frac{225}{\pi {{\left( 15 \right)}^{2}}}=\frac{1}{\pi }$

As a result, the rate at which the balloon's radius increases when the radius is $15cm$ is $\frac{1}{\pi }cm/\sec$.

9. A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the latter is $10$ cm.

Ans: The volume of a sphere (V) with radius ($r$) is given by,

$V=\frac{4}{3}\pi {{r}^{3}}$

Rate of change of volume (V) for time ($t$) is given by,

$\frac{dV}{dr}=\frac{d}{dr}\left( \frac{4}{3}\pi {{r}^{3}} \right)$

$=4\pi {{r}^{2}}$

Therefore, when the radius is $10cm$,

$\frac{dV}{dr}=4\pi {{\left( 10 \right)}^{2}}$

$=400\pi$

Hence, the rate of increase in the volume of the balloon is $400\pi c{{m}^{3}}/\sec$.

10. A ladder $5$ m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2$ cm/s. How fast is its height on the wall decreasing when the foot of the ladder is $4$  m away from the wall?

Ans: Let the height of the wall be $y$m and the foot of the ladder be $x$m away from the wall.

Then, by Pythagoras Theorem, we have:

${{x}^{2}}+{{y}^{2}}=25$  (Length of the ladder is $5m$)

$\Rightarrow y=\sqrt{25-{{x}^{2}}}$

Then, the rate of change of height $y$for time $t$ is given by,

$\frac{dy}{dt}=\frac{-x}{\sqrt{25-{{x}^{2}}}}.\frac{dx}{dt}$

It is given that $\frac{dx}{dt}=2cm/s$

$\therefore \frac{dy}{dt}=\frac{-2x}{\sqrt{25-{{x}^{2}}}}$

Now, when $x=4m$ we have:

$\frac{dy}{dt}=\frac{8}{3}$

Hence, the rate of decrease in the height of the ladder on the wall is    $\frac{8}{2}cm/\sec$.

11. A particle moving along the curve $6y={{x}^{3}}+2$. Find the points on the curve at which the $y$coordinate is changing $8$ times as fast as the x-coordinate.

Ans: Given the equation of the curve as $6y={{x}^{3}}+2$.

The rate of change of the position of the particle for time

(t)  is given by,

$6\frac{dy}{dt}=3{{x}^{2}}\frac{dx}{dt}+0$

$\Rightarrow 2\frac{dy}{dt}={{x}^{2}}\frac{dx}{dt}$

When the particle's y-coordinate changes $8$ times as fast as its x-coordinate i.e., $\left( \frac{dy}{dt}=8\frac{dx}{dt} \right)$ , we have:

$2\left( 8\frac{dx}{dt} \right)={{x}^{2}}\frac{dx}{dt}$

$\Rightarrow 16\frac{dx}{dt}={{x}^{2}}\frac{dx}{dt}$

$\Rightarrow \left( {{x}^{2}}-16 \right)\frac{dx}{dt}=0$

$\Rightarrow {{x}^{2}}=16$

$\Rightarrow x=\pm 4$

When $x=-4,$

$y=\frac{-31}{3}$

and when $x=4,$

$y=11$

Hence, the points required on the curve are $\left( 4,11 \right)$ and $\left( -4,\frac{-31}{3} \right)$.

12. The radius of an air bubble is increasing at the rate of $1/2$ cm/s. At what rate is the volume of the bubbles of the bubble increasing when the radius is $1$cm?

Ans: The air bubble is in the shape of a sphere.

The volume of an air bubble (V) with radius (r) is given by,

$V=\frac{4}{3}\pi {{r}^{3}}$

Rate of change of volume (V) for time ($t$) is given by,

$\frac{dV}{dt}=\frac{dV}{dr}.\frac{dr}{dt}$                                                   (By chain rule)

$=\frac{d}{dr}\left( \frac{4}{3}\pi {{r}^{3}} \right).\frac{dr}{dt}$

$=4\pi {{r}^{2}}.\frac{dr}{dt}$

It is given that $\frac{dr}{dt}=\frac{1}{2}cm/s$

Therefore, when r is $1$cm,

$\frac{dV}{dt}=4\pi {{\left( 1 \right)}^{2}}\left( \frac{1}{2} \right)$

$=2\pi c{{m}^{3}}/s$

Hence, the volume of the bubble increases at the rate of $2\pi$cm3/s.

13. A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}\left( 2x+1 \right)$ . Find the rate of change of its volume with respect to $x$.

Ans: The volume of a sphere $\left( V \right)$ with radius $\left( r \right)$ is given by,

$V=\frac{4}{3}\pi {{r}^{2}}$.

Diameter is $\frac{3}{2}\left( 2x+1 \right)$

$r=\frac{3}{4}\left( 2x+1 \right)$

$V=\frac{4}{3}\pi {{r}^{3}}$

$=\frac{9}{16}\pi {{\left( 2x+1 \right)}^{3}}$

Hence, the rate of change of volume with respect to $x$ is as

$\frac{dV}{dt}=\frac{9}{16}\pi \frac{d}{dt}{{\left( 2x+1 \right)}^{3}}$

$=\frac{27}{8}\pi {{\left( 2x+1 \right)}^{3}}$.

14. Sand is pouring from a pipe at the rate of $12c{{m}^{3}}/\sec$. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is $4$ cm?

Ans: The volume of a cone ($V$) with radius and height ($h$) is given by.

$V=\frac{1}{3}\pi {{r}^{2}}h$

It is provided that,

$h=\frac{1}{6}r$

$\Rightarrow r=6h$

$\therefore V=\frac{1}{3}\pi {{\left( 6h \right)}^{2}}h$

$=12\pi {{h}^{3}}$

The rate of change of volume for time ($t$) is given by

$\frac{dV}{dt}=12\pi \frac{d}{dh}\left( {{h}^{3}} \right).\frac{dh}{dt}$      (by chain rule)

$=12\pi \left( 3{{h}^{2}} \right)\frac{dh}{dt}$

$=36\pi {{h}^{2}}\frac{dh}{dt}$

It is provided that $\frac{dV}{dt}=12c{{m}^{2}}/s$.

Therefore, when $h=4cm$ we have:

$12=36\pi {{\left( 4 \right)}^{2}}\frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt}=\frac{12}{36\pi \left( 16 \right)}$

$\frac{dh}{dt}=\frac{1}{48\pi }$

Hence, when the height of the sand cone is $4$ cm, its height is increasing at the rate of $\frac{1}{48}cm/\sec$.

15. The total cost $C\left( x \right)$ in Rupees associated with the production of $x$ units of an item is given by $C\left( x \right)=0.007{{x}^{3}}-0.003{{x}^{2}}+15x+4000$.

Find the marginal cost when $17$ units are produced.

Ans: The rate of change in total cost for output is known as marginal cost.

$M C=\frac{d C}{d x}=0.007\left(3 x^{2}\right)-0.003(2 x)+15$

$=0.021 x^{2}-0.006 x+15$

When

$x=17,MC=0.021{{\left( 17 \right)}^{2}}-0.006\left( 17 \right)+15$

$=0.21\left( 289 \right)-0.006\left( 17 \right)+15$

$=6.069-0.102+15$

$=20.967$

Hence, the marginal cost when $17$ units are produced is $Rs.20.967$.

16. The total revenue in Rupees received from the sale of $x$ units of a product is given by $R\left( x \right)=13{{x}^{2}}+26x+15$. Find the marginal revenue when $x=7$.

Ans: The rate of change in total cost for the number of units sold is known as marginal cost. Let the number of units sold be x.

$MR=\frac{dR}{dx}=13\left( 2x \right)+26$

$=26x+26$

When $x=7$,

$MR=208$

Hence, the required marginal revenue is $Rs.208$.

17. The rate of change of the area of a circle with respect to its radius $r$ at $r=6cm$ is $\left( A \right)10\pi$ $\left( B \right)12\pi$ $\left( C \right)=8\pi$ $\left( D \right)11\pi$.

Ans: The area of a circle $\left( A \right)$ with radius $\left( r \right)$ is given by, $A=\pi {{r}^{2}}$

As a result, the area's rate of change in relation to its radius r is  $\frac{dA}{dt}=\frac{d}{dr}\left( \pi {{r}^{2}} \right)=2\pi r$

Therefore, when $r=6cm$,

$\frac{dA}{dr}=2\pi \times 6$

$=12\pi c{{m}^{2}}/s$

Hence, the required rate of change of the area of a circle is $12\pi c{{m}^{2}}/s$.

Therefore, option B is correct.

18. The total revenue in Rupees received from the sale of $x$units of a product is given by $R\left( x \right)=3{{x}^{2}}+36x+5$. The marginal revenue, when $x=15$ is $\left( A \right)116$ $\left( B \right)96$ $\left( C \right)90$ $\left( D \right)126$.

Ans: The rate of change in total cost for the number of units sold is known as marginal cost.

$\therefore MR=\frac{dR}{dx}=6x+36$

Therefore, when $x=15$,

$MR=90+36=126$

Hence, the required marginal revenue is $Rs.126$.

The correct option is D.

## Conclusion

NCERT Exercise 6.1 Class 12 Maths covers essential applications of derivatives, including the rate of change of quantities, and the behavior of functions as increasing or decreasing. These topics are foundational for understanding how calculus applies to real-world situations and for solving practical problems involving rates and slopes. Practicing these problems will enhance your problem-solving skills and conceptual understanding.  Vedantu's solutions offer clear, step-by-step explanations, making it easier to understand these concepts. By thoroughly studying class 12 maths chapter 6 exercise 6.1, you will be well-prepared to tackle similar questions in your exams and apply these principles effectively.

## Class 12 Maths Chapter 6: Exercises Breakdown

 Exercise Number of Questions Exercise 6.2 19 Questions and Solutions Exercise 6.3 29 Questions and Solutions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 6 - Application of Derivatives Exercise 6.1

1. What is the Weightage Mark of Chapter 6 in Class 12 Maths?

Out of the total 100 marks for maths, 20 marks are for internal assessments which include midterm marks as well as project work, and 80 marks for the final written paper. There are 35 marks dedicated to the topic calculus in the final written examination. Calculus is spread out over several chapters in both the NCERT maths textbooks of class 12 and chapter 6 is one of these calculus chapters. A part of those 35 marks for calculus includes a few important questions from chapter 6, called “Applications of Derivatives.” From these 35 marks, some marks are awarded for chapter 6.

2. What are the Topics Included in Class 12 Maths Chapter 6?

The topics included in class 12 maths chapter 6 Applications of Derivatives are the various ways in which differentiation and derivatives can be used in real-life ways - this is what 12th maths chapter 6 exercise 6.1 pertains to. The following is the exhaustive list of the topics which come under class 12 maths chapter 6:

• A brief introduction and recap of chapter 5.

• Rate of change of quantities (applying what we’ve learnt about derivatives to practical examples).

• Increasing and decreasing functions.

• Tangents and normals.

• Approximations.

• Maxima and minima concepts.

• Several miscellaneous examples of the topics mentioned above for a clearer understanding of the same.

3. What is the Chapter 6 Application of Derivatives of Class 12 Maths?

The application of derivatives is an important chapter of Class 12. In this students learn about the rate of change of quantities, increasing and decreasing functions using graphical methods, Tangents and Normals, Approximations, Maxima and Minima concepts. These topics to be understood in detail to get the proper grip of the subject. This can be best understood with Vedantu’s NCERT solutions where the practice and preparation can be done thoroughly.

4. How many questions are there in Exercise 6.1 of Chapter 6 Application of Derivatives of Class 12 Maths?

There are a total number of 18 questions in Exercise 6.1 of  Chapter 6 Application Of Derivatives. Each numericals in the exercise is of a different kind and has to be practised with focus and understanding. These solutions are given systematically in NCERT solutions from Vedantu which has been curated by expert professionals having the depth knowledge of the subject. The solutions are easy to understand and even the complex numericals are solved and shown in a simple manner.  Solving all these one can be sure of being thorough with the chapter.

5. An edge of a variable cube is increasing at the rate of 3cm/s.  When the edge is 10cm how fast will the volume of the cube increase?

Let x be the edge  of the cube and v the volume respectively. We know that the edge of the cube is increasing at 3 cm/sec.

We have to calculate that how fast it would increase when the edge is 10 cm

Let us find dv/dt when x =10 cm

Now, Volume =(Edge)3

V=X3

Now dv/dt=d(x)3/dt

dv/dt=d(x)3/dt x dx/dx

dv/dt=d(x)3/dt x dx/dt

dv/dt=3x2x . dx/dt

dv/dt=3x2x3……………………….(1)

dv/dt=9x2

Now when x=10

Then dv/dtx=10=9(10)2

dv/dtx=10=900

Since volume has to be in cm3/sec

dv/dt=900 cm3/sec

6. What is the weightage of the marks of different questions?

Relation and functions will have a weightage of 8 marks.

Algebra has a weightage of 10 marks.

Calculus has a weightage 35

Vectors and Three-Dimensional Geometry has the weightage14

Linear programming has a weightage of 5 marks.

Probability has a weightage of 8 marks.

The total will be 80.

If the preparation is correct then it's not difficult to score full marks. Practising the numerical regularly is the best method to get the proper hold of the subject.

7. What do you understand about Maxima And Minima concepts?

The maxima and minima is the concept of differential calculus. The maxima and the minima are dealt with by the branch of Mathematics called the calculus of variations. The words maxima and minima mean the maximum and the minimum value respectively. The chapter has many numericals for which conceptual understanding is required. Students can refer to Vedantu’s NCERT solutions for the details of the chapters and the solutions from the exercises. This also contains the extra questions at the end of each chapter for better practice.

8. Why are applications of derivatives important ex 6.1 class 12?

They help in understanding real-world problems involving rates of change and the behavior of functions.

9. How do you find the rate of change using derivatives in application of derivatives class 12 exercise 6.1 ?

The derivative of a function gives the rate of change of the dependent variable with respect to the independent variable.

10. What should I focus on in class 12 ex 6.1?

Focus on understanding how to apply derivatives to find rates of change, and analyze functions' behavior, and practice problems on tangents and normals.

11. How can Vedantu’s solutions help me to solve class 12 maths ch 6 ex 6.1?

Vedantu provides detailed, step-by-step solutions that make complex concepts easier to understand and apply.

12. What are common mistakes to avoid in application of derivatives class 12 exercise 6.1?

In ex 6.1 class 12 it ensure accurate calculation of derivatives and careful application of concepts like increasing/decreasing functions and tangents/normals.