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NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5 - Continuity and Differentiability

NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5 - Continuity and Differentiability

Edited By Ramraj Saini | Updated on Dec 03, 2023 05:13 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.7

NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5 Continuity and Differentiability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In the previous exercises, you have learned about finding the first-order differentiation of different types of functions. In NCERT solutions for Class 12 Maths chapter 5 exercise 5.7, you will learn about finding the derivatives of second order. If a function y=f(x) is given, you can find its first-order derivative by differentiation of y y w.r.t x i.e. f'(x) = dy/dx and if you again differentiate f'(x) w.r.t the x, you will get the second-order f''(x).

If you have command on the first-order differentiation then exercise 5.7 Class 12 Maths will be very easy for you as you just need to differentiate the given function two times. The first and second-order derivatives are useful in finding minimum and maximum values of the functions, finding the domain and range of the function, other subjects also. 12th class Maths exercise 5.7 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Continuity and Differentiability Exercise: 5.7

Question:1 Find the second order derivatives of the functions given in Exercises 1 to 10.

x^2 + 3x+ 2

Answer:

Given function is
y=x^2 + 3x+ 2
Now, differentiation w.r.t. x
\frac{dy}{dx}= 2x+3
Now, second order derivative
\frac{d^2y}{dx^2}= 2
Therefore, the second order derivative is \frac{d^2y}{dx^2}= 2

Question:2 Find the second order derivatives of the functions given in Exercises 1 to 10.

x ^{20}

Answer:

Given function is
y=x ^{20}
Now, differentiation w.r.t. x
\frac{dy}{dx}= 20x^{19}
Now, the second-order derivative is
\frac{d^2y}{dx^2}= 20.19x^{18}= 380x^{18}
Therefore, second-order derivative is \frac{d^2y}{dx^2}= 380x^{18}

Question:3 Find the second order derivatives of the functions given in Exercises 1 to 10.

x \cos x

Answer:

Given function is
y = x \cos x
Now, differentiation w.r.t. x
\frac{dy}{dx}= \cos x + x(-\sin x ) = \cos x-x\sin x
Now, the second-order derivative is
\frac{d^2y}{dx^2}= -\sin x-(\sin x+x\cos x) = -2\sin x - x\sin x
Therefore, the second-order derivative is \frac{d^2y}{dx^2}= -2\sin x - x\sin x

Question:4 Find the second order derivatives of the functions given in Exercises 1 to 10.

\log x

Answer:

Given function is
y=\log x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{1}{x}
Now, second order derivative is
\frac{d^2y}{dx^2}= \frac{-1}{x^2}
Therefore, second order derivative is \frac{d^2y}{dx^2}= \frac{-1}{x^2}

Question:5 Find the second order derivatives of the functions given in Exercises 1 to 10.

x ^3 \log x

Answer:

Given function is
y=x^3\log x
Now, differentiation w.r.t. x
\frac{dy}{dx}=3x^2.\log x+x^3.\frac{1}{x}= 3x^2.\log x+ x^2
Now, the second-order derivative is
\frac{d^2y}{dx^2}= 6x.\log x+3x^2.\frac{1}{x}+2x=6x.\log x+3x+2x = x(6.\log x+5)
Therefore, the second-order derivative is \frac{d^2y}{dx^2} = x(6.\log x+5)

Question:6 Find the second order derivatives of the functions given in Exercises 1 to 10.

e ^x \sin5 x

Answer:

Given function is
y= e^x\sin 5x
Now, differentiation w.r.t. x
\frac{dy}{dx}=e^x.\sin 5x +e^x.5\cos 5x = e^x(\sin5x+5\cos5x)
Now, second order derivative is
\frac{d^2y}{dx^2}= e^x(\sin5x+5\cos5x)+e^x(5\cos5x+5.(-5\sin5x))
= e^x(\sin5x+5\cos5x)+e^x(5\cos5x-25\sin5x)=e^x(10\cos5x-24\sin5x)
=2e^x(5\cos5x-12\sin5x)
Therefore, second order derivative is \frac{dy}{dx}=2e^x(5\cos5x-12\sin5x)

Question:7 Find the second order derivatives of the functions given in Exercises 1 to 10.

e ^{6x}\cos 3x

Answer:

Given function is
y= e^{6x}\cos 3x
Now, differentiation w.r.t. x
\frac{dy}{dx}=6e^{6x}.\cos 3x +e^{6x}.(-3\sin 3x)= e^{6x}(6\cos 3x-3\sin 3x)
Now, second order derivative is
\frac{d^2y}{dx^2}= 6e^{6x}(6\cos3x-3\sin3x)+e^{6x}(6.(-3\sin3x)-3.3\cos3x)
= 6e^{6x}(6\cos3x-3\sin3x)-e^{6x}(18\sin3x+9\cos3x)
e^{6x}(27\cos3x-36\sin3x) = 9e^{6x}(3\cos3x-4\sin3x)
Therefore, second order derivative is \frac{dy}{dx} = 9e^{6x}(3\cos3x-4\sin3x)

Question:8 Find the second order derivatives of the functions given in Exercises 1 to 10.

\tan ^{-1} x

Answer:

Given function is
y = \tan^{-1}x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(\tan^{-1}x)}{dx}=\frac{1}{1+x^2}
Now, second order derivative is
\frac{d^2y}{dx^2}= \frac{-1}{(1+x^2)^2}.2x = \frac{-2x}{(1+x^2)^2}
Therefore, second order derivative is \frac{d^2y}{dx^2} = \frac{-2x}{(1+x^2)^2}

Question:9 Find the second order derivatives of the functions given in Exercises 1 to 10.

\log (\log x )

Answer:

Given function is
y = \log(\log x)
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(\log(\log x))}{dx}=\frac{1}{\log x}.\frac{1}{x}= \frac{1}{x\log x}
Now, second order derivative is
\frac{d^2y}{dx^2}= \frac{-1}{(x\log x)^2}.(1.\log x+x.\frac{1}{x}) = \frac{-(\log x+1)}{(x\log x)^2}
Therefore, second order derivative is \frac{d^2y}{dx^2} = \frac{-(\log x+1)}{(x\log x)^2}

Question:10 Find the second order derivatives of the functions given in Exercises 1 to 10.

\sin (\log x )

Answer:

Given function is
y = \sin(\log x)
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(\sin(\log x))}{dx}=\cos (\log x).\frac{1}{x}= \frac{\cos (\log x)}{x}
Now, second order derivative is
Using Quotient rule
\frac{d^2y}{dx^2}=\frac{-\sin(\log x)\frac{1}{x}.x-\cos(\log x).1}{x^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}
Therefore, second order derivative is \frac{d^2y}{dx^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}

Question:11 If y = 5 \cos x - 3 \sin x prove that \frac{d^2y}{dx^2}+y = 0

Answer:

Given function is
y = 5 \cos x - 3 \sin x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(5\cos x-3\sin x)}{dx}=-5\sin x-3\cos x
Now, the second-order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(-5\sin x-3\cos x)}{dx^2}=-5\cos x+3\sin x
Now,
\frac{d^2y}{dx^2}+y=-5\cos x+3\sin x+5\cos x-3\sin x = 0
Hence proved

Question:12 If y = \cos ^{-1} x Find \frac{d ^2 y }{dx^2 } in terms of y alone.

Answer:

Given function is
y = \cos ^{-1} x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d( \cos ^{-1} x)}{dx}=\frac{-1}{\sqrt{1-x^2}}
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{\sqrt{1-x^2}})}{dx^2}=\frac{-(-1)}{(\sqrt{1-x^2})^2}.(-2x) = \frac{-2x}{1-x^2} -(i)
Now, we want \frac{d^2y}{dx^2} in terms of y
y = \cos ^{-1} x
x = \cos y
Now, put the value of x in (i)
\frac{d^2y}{dx^2} = \frac{-2\cos y }{1-\cos^2 y } = \frac{-2\cos y}{\sin ^2 y}= -2\cot y cosec y
(\because 1-\cos^2x =\sin^2 x\ and \ \frac{\cos x}{\sin x} = \cot x \ and \ \frac{1}{\sin x}= cosec x)
Therefore, answer is \frac{d^2y}{dx^2} = -2\cot y cosec y

Question:13 If y = 3 \cos (\log x) + 4 \sin (\log x), show that x^2 y_2 + xy_1 + y = 0

Answer:

Given function is
y = 3 \cos (\log x) + 4 \sin (\log x)
Now, differentiation w.r.t. x
y_1=\frac{dy}{dx}=\frac{d( 3 \cos (\log x) + 4 \sin (\log x))}{dx}=-3\sin(\log x).\frac{1}{x}+4\cos (\log x).\frac{1}{x}
=\frac{4\cos (\log x)-3\sin(\log x)}{x} -(i)
Now, second order derivative is
By using the Quotient rule
y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{4\cos (\log x)-3\sin(\log x)}{x})}{dx^2}= \frac{(-4\sin(\log x).\frac{1}{x}-3\cos(\log x).\frac{1}{x}).x-1.(4\cos (\log x)-3\sin(\log x))}{x^2}
=\frac{-\sin(\log x)+7\cos (\log x)}{x^2} -(ii)
Now, from equation (i) and (ii) we will get y_1 \ and \ y_2
Now, we need to show
x^2 y_2 + xy_1 + y = 0
Put the value of y_1 \ and \ y_2 from equation (i) and (ii)
x^2\left ( \frac{-\sin(\log x)+7\cos (\log x)}{x^2} \right )+x\left ( \frac{4\cos (\log x)-3\sin(\log x)}{x} \right )+ 3 \cos (\log x) +4\sin(\log x)
-\sin(\log x)-7\cos(\log x)+4\cos(\log x)-3\sin(\log x)+3\ cos (\log x) +4\sin(\log x)
=0
Hence proved

Question:14 If y = A e ^{mx} + Be ^{nx} , show that \frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0

Answer:

Given function is
y = A e ^{mx} + Be ^{nx}
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(A e ^{mx} + Be ^{nx})}{dx}=mAe^{mx}+nBe^{nx} -(i)
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(mAe^{mx}+nBe^{nx})}{dx^2}= m^2Ae^{mx}+n^2Be^{nx} -(ii)
Now, we need to show
\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0
Put the value of \frac{d^2y}{dx^2} \ and \ \frac{dy}{dx} from equation (i) and (ii)
m^2Ae^{mx}+n^2Be^{nx}-(m+n)(mAe^{mx}+nBx^{nx}) +mn(Ae^{mx}+Be^{nx})
m^2Ae^{mx}+n^2Be^{nx}-m^2Ae^{mx}-mnBx^{nx}-mnAe^{mx} -n^2Be^{nx}+mnAe^{mx}+mnBe^{nx}
=0
Hence proved

Question:15 If y = 500 e ^{7x} + 600 e ^{- 7x } , show that \frac{d^2 y}{dx ^2} = 49 y
Answer:

Given function is
y = 500 e ^{7x} + 600 e ^{- 7x }
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(500 e ^{7x} + 600 e ^{- 7x })}{dx}=7.500e^{7x}-7.600e^{-7x} =3500e^{7x}-4200e^{-7x} -(i)
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(3500e^{7x}-4200e^{-7x})}{dx^2}
= 7.3500e^{7x}-(-7).4200e^{-7x}= 24500e^{7x}+29400e^{-7x} -(ii)
Now, we need to show
\frac{d^2 y}{dx ^2} = 49 y
Put the value of \frac{d^2y}{dx^2} from equation (ii)
24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})
= 24500e^{7x}+29400e^{-7x}
Hence, L.H.S. = R.H.S.
Hence proved

Question:16 If e ^y (x+1) = 1 show that \frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2

Answer:

Given function is
e ^y (x+1) = 1
We can rewrite it as
e^y = \frac{1}{x+1}
Now, differentiation w.r.t. x
\frac{d(e^y)}{dx}=\frac{d(\frac{1}{x+1})}{dx}\\ e^y.\frac{dy}{dx}= \frac{-1}{(x+1)^2}\\ \frac{1}{x+1}.\frac{dy}{dx}= \frac{-1}{(x+1)^2} \ \ \ \ \ \ \ \ \ (\because e^y = \frac{1}{x+1})\\ \frac{dy}{dx}= \frac{-1}{x+1} -(i)
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{x+1})}{dx^2}=\frac{-(-1)}{(x+1)^2} = \frac{1}{(x+1)^2} -(ii)
Now, we need to show
\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2
Put value of \frac{d^2y}{dx^2} \ and \ \frac{dy}{dx} from equation (i) and (ii)
\frac{1}{(x+1)^2}=\left ( \frac{-1}{x+1} \right )^2
=\frac{1}{(x+1)^2}
Hence, L.H.S. = R.H.S.
Hence proved

Question:17 If y = (\tan^{-1} x)^2 show that (x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2

Answer:

Given function is
y = (\tan^{-1} x)^2
Now, differentiation w.r.t. x
y_1=\frac{dy}{dx}=\frac{d((\tan^{-1}x)^2)}{dx}= 2.\tan^{-1}x.\frac{1}{1+x^2}= \frac{2\tan^{-1}x}{1+x^2} -(i)
Now, the second-order derivative is
By using the quotient rule
y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{2\tan^{-1}x}{1+x^2})}{dx^2}=\frac{2.\frac{1}{1+x^2}.(1+x^2)-2\tan^{-1}x(2x)}{(1+x^2)^2}=\frac{2-4x\tan^{-1}x}{(1+x^2)^2} -(ii)
Now, we need to show
(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2
Put the value from equation (i) and (ii)
(x^2+1)^2.\frac{2-4x\tan^{-1}x}{(1+x^2)^2}+2x(x^2+1).\frac{2\tan^{-1}x}{x^2+1}\\ \Rightarrow 2-4x\tan^{-1}x+4x\tan^{-1}x = 2
Hence, L.H.S. = R.H.S.
Hence proved

More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7

In Class 12th Maths chapter 5 exercise 5.7, you will get 17 questions related to finding the second-order derivatives. There are four examples given in the NCERT book prior to the ex 5.7 which you can solve to get more familiar with second derivatives before solving the exercise question. After solving examples, you can try to solve Class 12th Maths chapter 5 exercise 5.7 questions. You may not be to solve these exercise 5.7 Class 12 Maths problems by yourself at first. You can go through Class 12 Maths chapter 5 exercise 5.7 solutions to get conceptual clarity.

Also Read| Continuity and Differentiability Class 12th Chapter 5 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7

  • Class 12 Maths chapter 5 exercise 5.7 solutions are designed in a descriptive manner that you can understand easily.
  • Don't go through NCERT solutions for Class 12 Maths chapter 5 exercise 5.7 without trying to solve NCERT problems by yourself.
  • You can use exercise 5.7 Class 12 Maths solutions for reference.
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Key Features Of NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 5.7 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 5.7, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 5.7 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 5.7 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 5.7 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 5.7 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Question (FAQs)

1. If y = c is a function where c is a constant then find dy/dx ?

y = c

dy/dx = 0

2. If y = c is a function where c is a constant then find the second order derivative of y ?

y = c

dy/dx = 0

d(dy/dx)/dx = 0

3. Find the first derivative of y = x ?

Given y = x

dy/dx = 1

4. Find the second order derivative of y = x ?

Given y = x

dy/dx = 1

d(dy/dx)/dx = 0

5. What is the second order derivative of y = e^x ?

y = e^x

dy/dx = e^x

d(dy/dx)/dx = e^x

d^(2)y/dx^2 = e^x

6. Can i get CBSE Class 10 exam pattern ?

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7. Can I get detailed syllabus for CBSE Class 10 ?

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8. Can I get detailed syllabus for CBSE Class 10 Maths ?

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Have a question related to CBSE Class 12th ?
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

You can use them people also used problem

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Divyan 06 August,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

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Yuvan 01 September,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

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kumardurgesh1802 10 July,2024
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be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
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