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NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5

NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability Exercise 5.5

Mathematics is a subject that students might find challenging. The concepts and theorems of the subject require continual practice and revision. Since Mathematics cannot be learned by rote learning and is a conceptual subject, students need a deep understanding of its concepts. Chapter 5 of Class 12 Mathematics is Continuity and Differentiability. Students might find the concepts and the calculations of this chapter a bit difficult at the beginning. The Extramarks website provides the solutions to the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 to help students practice efficiently. Once students get acquainted with the paper pattern by revising the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5, they can eventually manage to frame better answers. With practice and proper guidance, the concepts, and calculations prove to become easy for them.

Class 12 Maths NCERT Textbooks do not contain solutions to the questions mentioned in the respective exercises of the listed chapters. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 provided by Extramarks helps students in understanding the questions given in the NCERT textbooks while having an accurate solution and approach to every question. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 are indispensable as NCERT is the building block for the foundation of the concepts of any chapter of Mathematics. So it is important to practice well the questions mentioned in the NCERT textbook.

Chapter-5 Class 12 Mathematics: Continuity and Differentiability, includes the following topics-Continuity, Differentiability, Exponential and Logarithmic Functions, Logarithmic Differentiation, Derivatives of Functions in Parametric Forms, Second Order Derivatives and Mean Value Theorem. So, the chapter covers a lot of topics with very essential concepts. Extramarks provides students with all the essential means to make the learning experience systematic and effortless for them so that they can score maximum marks in the board examinations.

Click here to download the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5.

NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability Exercise 5.5

Students can download the NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability Exercise 5.5 in PDF format. These solutions are easily accessible and also can be available offline so that students can go through them whenever and wherever they want.

Access NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

Class 12 Continuity and Differentiability Exercise 5.5

NCERT Solutions are used by schools as the first step for teaching students. Many students rely on them for a better understanding of the basic concepts. The NCERT textbooks are the primary source of teaching in many schools, as the question paper of the board examinations is mostly based on the fundamentals of the NCERT. This makes it more important for students to thoroughly go through the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5. For senior classes like Class 11 and 12, these solutions are an essential means of preparation for board examinations. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 help students to have enough practice and understanding of the concepts so that they can also solve any complicated problem that comes in front of them during the board examination. But practising just the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 is not enough because students of Class 12 should practice multiple questions before sitting in the board examinations. The NCERT textbook just contains a few questions that explain the concepts, but it is significant to thoroughly practice those conceptual questions for a deeper understanding. Therefore, Extramarks provides students with learning tools like K12 study material for boards, sample papers, past years’ papers and much more which could help them excel in their studies and score good grades.

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Topics Covered in NCERT Class 12 Maths Chapter 5 Exercise 5.5

Class 12 Maths Ch 5 Ex 5.5 is based on Logarithm Differentiation. It is a topic which is also used in subjects like Physics, and sometimes in Chemistry. Logarithmic Differentiation is a topic that forms the basis of some important topics which are to be studied further in the fields of Engineering, Research, etc. Some students might find it difficult to understand the concepts and calculations of the topic in the beginning. But once they practice and understand the concepts of the topic, they will be able to solve those problems very easily.

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 provided by Extramarks help students with a deep understanding of the concepts of the topic. They will be able to find authentic solutions to their problems without having to look anywhere else. Apart from the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5, Extramarks also provides students with learning resources that will improve their understanding of concepts and also save their time as they will not have to continually search for the answers.

The Class 12 Maths Chapter 5 Exercise 5.5 contains eighteen questions. Students need to practice the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 thoroughly as it is a challenging exercise which requires a lot of effort and determination.

Click on the given link to download the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5.

What are the Steps to Solve Logarithmic Differentiation?

Logarithmic Differentiation is a method that helps students find derivatives of some difficult functions, using logarithms. There are some cases wherein differentiating the logarithm of a given function is simpler as compared to differentiating the function itself. By properly using the properties of logarithms and the chain rule, the process of finding the derivatives becomes easy. This method can be applied to almost all the non-zero functions that are differentiable. Hence, the differentiation of some complicated functions is done by taking logarithms and then the logarithmic derivative is used to solve the certain function. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 provided by the Extramarks’ website helps students to get a better hold of the concept of Logarithmic Differentiation.

Formula for Logarithmic differentiation-

The equations that are in the form of y = f(x) = [u(x)]{v(x)} can be solved effortlessly using the concept of logarithmic differentiation. The formula of this concept is given by:

d/dx(xx) = xx(1+ln x)

For differentiating functions of this type, students take on both sides of the given equation. Therefore, taking log on both sides they get, log y = log[u(x)]{v(x)}

log y = v(x)log u(x)

Then students have to apply the chain rule for solving this type of problem. For a more detailed explanation, students can have a look at the Class 12 Maths NCERT Solutions Chapter 5 Exercise 5.5 provided by the Extramarks website.

Important steps to solve logarithmic differentiation problems are:

The first step is to solve and find the natural log of the function which has to be differentiated.
Then, by using the properties of log functions, allocate the expressions that were initially assembled at the beginning in the original function and were challenging to differentiate.
Then differentiate the equation that was formed in the above step.
Then, multiply the solved equation by the function itself and the required derivative will be solved.

For a detailed explanation of these steps to solve logarithmic equations, students can subscribe to the Extramarks website and can download the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5.

Click on the given link to download the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5.

Summary of NCERT Solution of Maths Class 12 Chapter 5 Exercise 5.5

Mathematics is primarily about critical thinking. It is a conceptual subject that enhances the problem-solving and analytical skills of the students. Mathematics is a subject that needs ample amount of practice. It is essential for them to understand the basic concepts of the subject to score well. Mathematics is not just an academic subject, it is fundamental for many scientific concepts.

Class 12 Mathematics has two volumes of the curriculum, which have a wide range of chapters. Since that is a huge syllabus, students are required to work very hard. The first step for students is to learn the NCERT content readily. One of the many chapters of Mathematics Class 12 is Continuity and Differentiability. It can be an intimidating chapter for students as it includes many new concepts which can be challenging for students. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 provided by the Extramarks website are explained in detail which enhances students’ conceptual clarity. These answers are cross-checked by the subject experts of Extramarks and are curated according to the latest exam pattern. Students can download the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 provided by the Extramarks website and practice these questions for obtaining maximum marks in their board examinations.

Click on the given link to download the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5.

Benefits of Using Class 12 Maths NCERT Solutions Chapter 5 Exercise 5.5

The availability of a question bank before the examinations are not enough for students; they need detailed and step-by-step solutions. The NCERT textbook questions are the primary means of information for most of the question banks. Having access to the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 helps students prepare better for their board examinations. Having a means of learning makes students confident for their board examinations. Understanding the concepts of the NCERT solutions can help students move ahead with their preparation. The concepts of the chapter can be confusing and challenging at times, but having the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 provided by Extramarks can be a good learning module. Using these learning modules can make the learning process of Mathematics easy and enjoyable.

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 provided by Extramarks are easily accessible online and offline, which helps students revise anywhere and anytime they want. The steps and the formulas of the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 can be confusing sometimes. Students need to remember the steps and the calculations involved with the solutions of this exercise. Some portals might not be reliable to access the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5. Some concepts and functions of the chapter are complicated and confusing. Having a credible source to access the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 can authenticate the learning and preparation process. Students can access the NCERT Solutions for the Class 12 Maths Chapter 5 Continuity and Differentiability on the Extramarks website. These solutions are compiled by academic experts, making them a reliable source of knowledge. Students can rely on the Extramarks website if they want to access the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5. Apart from NCERT solutions, the website also has elaborate study material to help students have proper clarification of the concepts before sitting for their exams. Apart from getting access to NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5, students can also go through the entire exercise of all units and chapters to make sure they do not miss anything essential for the examinations.

Properly Organized Study Material

Students need properly organized study material so that their learning process can be systematic. Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 that are properly organized answers for Exercise 5.5. This will speed up their learning process and also save them time as they will not have to search for solutions elsewhere. It will also help the students to make a proper timetable to keep a track record of their syllabus and revisions. The students of Class 12 need to retain the topics that they have already gone through. Extramarks provides students with all the means necessary to obtain high scores in the board examinations. Extramarks also makes available various NCERT Solutions for students to practice. Some of these include

NCERT Solutions Class 12

NCERT Solutions Class 11

NCERT Solutions Class 10

NCERT Solutions Class 9

NCERT Solutions Class 8

NCERT Solutions Class 7

NCERT Solutions Class 6

NCERT Solutions Class 5

NCERT Solutions Class 4

NCERT Solutions Class 3

NCERT Solutions Class 2

NCERT Solutions Class 1

Understanding the Concept

Learning new concepts can be difficult and challenging for students. These new concepts create various doubts. Mathematics is a conceptual subject so understanding all concepts of the respective chapters is very significant. Remembering all these different concepts is difficult for the students. Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 for a deeper understanding of the concepts of the topic. Apart from that, Extramarks provides them with learning modules like K12 study material, live doubt-solving sessions, assessment centres and much more. All these tools help students maintain a proper track of the topics they have covered so that they can keep up with all the topics they have gone through and also those which require more practice. This way, students can achieve their goals and score well in their board examinations.

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The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 provided by the Extramarks website, are curated by expert professionals. They are step-by-step solutions that are properly explained and are very easily accessible. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 provided by Extramarks are authentic answers that students can find easily on the web. Practising these solutions helps increase the academic efficiency of students.

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NCERT Solutions for Class 12 Maths

Solving NCERT questions is good practice for students who want to achieve a steady speed while solving other complicated papers or questions. Learning solutions like the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 target a specific section of a chapter. There are multiple learning modules and study materials for students on Extramarks, like Class 12 NCERT solutions for students, K12 Study material for board examinations, media-rich engaging content, curriculum mapped learning and much more so that students can enjoy learning. It is a one-stop solution for learning, practising and clearing doubts. Another reason to follow up with Extramarks’ NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 is that they are designed and curated by in-house experts, ensuring a smooth learning experience for all students.

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NCERT solutions are easily available online. Multiple portals provide students with easily available NCERT solutions from professional and experienced experts. However, students need to verify the credibility of these sources. One such credible platform is Extramarks. Apart from the NCERT solutions from professional and experienced experts, it has practice modules, sample papers, past years’ papers, and question banks for students. The curriculum and study material on the platform is brought together by the in-house experts of the industry. The information provided to the students is fact-checked. This makes Extramarks a reliable source for NCERT solutions from professional and experienced teachers.

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NCERT Mathematics Class 12 Volume-1

Chapter-1- Relations and Functions	Page- 1-32
Chapter-2- Inverse Trigonometric Functions	Page- 33-55
Chapter-3- Matrices	Page- 56-102
Chapter-4- Determinants	Page- 103-146
Chapter-5- Continuity and Differentiability	Page- 147-193
Chapter-6- Applications of Derivatives	Page-194-246

NCERT Mathematics Class 12 Volume-2

Chapter-7- Integrals	Page- 287-358
Chapter-8- Applications of Integrals	Page- 359-378
Chapter-9- Differential Equations	Page- 379-423
Chapter-10- Vector Algebra	Page- 424-462
Chapter-11- Three Dimensional Geometry	Page- 463-503
Chapter-12- Linear Programming	Page- 504-530
Chapter-13- Probability	Page- 531-587

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NCERT Solutions for Class 12 Maths Chapter 5 Other Exercises

Chapter 5 – Continuity and Differentiability Exercises
Exercise 5.1	10 Short Questions and 24 Long Questions
Exercise 5.2	2 Short Questions 8 Long Questions
Exercise 5.3	9 short Questions and 6 long Questions
Exercise 5.4	5 Short Questions and 5 Long Questions
Exercise 5.6	1 Short Question and 1 Long Question
Exercise 5.7	10 Short Questions and 7 Long Questions
Exercise 5.8	Questions with Solutions

Q.1 Differentiate the function given with respect to x.
cosx.cos2x.cos3x

Ans

$\begin{array}{l} Let y = cosx . \cos 2 x . \cos 3 x \\ Taking logarithm on both the sides, we get \\ logy = \log (cosx . \cos 2 x . \cos 3 x) \\ = logcosx + logcos 2 x + logcos 3 x \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} logy = \frac{d}{dx} logcosx + \frac{d}{dx} logcos 2 x + \frac{d}{dx} logcos 3 x \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{cosx} \frac{d}{dx} cosx + \frac{1}{\cos 2 x} \frac{d}{dx} \cos 2 x + \frac{1}{\cos 3 x} \frac{d}{dx} \cos 3 x \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{cosx} \times - sinx + \frac{1}{\cos 2 x} \times - \sin 2 x \frac{d}{dx} 2 x \\ + \frac{1}{\cos 3 x} \times - \sin 3 x \times \frac{d}{dx} 3 x \\ \frac{1}{y} \frac{dy}{dx} = - \frac{sinx}{cosx} - \frac{\sin 2 x}{\cos 2 x} \times 2 - \frac{\sin 3 x}{\cos 3 x} \times 3 \\ \frac{dy}{dx} = - y (tanx + 2 \tan 2 x + 3 \tan 3 x) \\ \frac{dy}{dx} = - cosx . \cos 2 x . \cos 3 x (tanx + 2 \tan 2 x + 3 \tan 3 x) \end{array}$

Q.2 Differentiate the function given with respect to x.

$\sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}}$

Ans

$\begin{array}{l} Let y = \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} \\ Taking logarithm on both the sides, we get \\ logy = \log \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} \\ \Rightarrow logy = \frac{1}{2} \log {\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} \\ \Rightarrow logy = \frac{1}{2} {\log (x - 1) + \log (x - 2) - \log (x - 3) - \log (x - 4) - \log (x - 5)} \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} logy = \frac{1}{2} \frac{d}{dx} {\log (x - 1) + \log (x - 2) - \log (x - 3) - \log (x - 4) - \log (x - 5)} \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} (\begin{array}{l} \frac{1}{x - 1} \frac{d}{dx} (x - 1) + \frac{1}{x - 2} \frac{d}{dx} (x - 2) - \frac{1}{x - 3} \frac{d}{dx} (x - 3) \\ - \frac{1}{x - 4} \frac{d}{dx} (x - 4) - \frac{1}{x - 5} \frac{d}{dx} (x - 5) \end{array}) \\ [By chain rule] \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} (\frac{1}{x - 1} \times 1 + \frac{1}{x - 2} \times 1 - \frac{1}{x - 3} \times 1 - \frac{1}{x - 4} \times 1 - \frac{1}{x - 5} \times 1) \\ \frac{dy}{dx} = \frac{1}{2} y (\frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5}) \\ \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} (\frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5}) \end{array}$

Q.3 Differentiate the function given with respect to x.

${(logx)}^{cosx}$

Ans

$\begin{array}{l} Let y = {(logx)}^{cosx} \\ Taking logarithm on both the sides, we get \\ logy = cosx . \log (logx) \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} logy = \frac{d}{dx} {cosx . \log (logx)} \\ \frac{1}{y} \frac{dy}{dx} = cosx \frac{d}{dx} \log (logx) + \log (logx) \frac{d}{dx} cosx [By product rule] \\ \frac{1}{y} \frac{dy}{dx} = cosx . \frac{1}{logx} \frac{d}{dx} logx + \log (logx) . (- sinx) [By Chain rule] \\ \frac{dy}{dx} = y {\frac{cosx}{logx} \times \frac{1}{x} - sinx . \log (logx)} \\ ∴ \frac{dy}{dx} = {(logx)}^{cosx} {\frac{cosx}{xlogx} - sinx . \log (logx)} \end{array}$

Q.4 Differentiate the function given with respect to x.

$x^{x} - 2^{sinx}$

Ans

$\begin{array}{l} Let y = x^{x} - 2^{sinx} \\ = y_{1} + y_{2} \\ So, y_{1} = x^{x} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = \log (x^{x}) \\ = xlogx \\ Differentiating both sides with respect to x, we get \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x \frac{d}{dx} logx + logx \frac{d}{dx} x \\ = x . \frac{1}{x} + logx . 1 \\ = 1 + logx \\ \frac{{dy}_{1}}{dx} = y_{1} (1 + logx) \\ = x^{x} (1 + logx) \\ Now, y_{2} = 2^{sinx} \\ Taking logarithm on both the sides, we get \\ {logy}_{2} = \log (2^{sinx}) \\ = sinx . \log 2 \\ Differentiating both sides with respect to x, we get \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = \log 2 \frac{d}{dx} sinx \\ = \log 2 . cosx \\ \frac{{dy}_{1}}{dx} = y_{2} (\log 2 . cosx) \\ = 2^{sinx} cosx . \log 2 \\ Thus, \frac{dy}{dx} = \frac{{dy}_{1}}{dx} - \frac{{dy}_{2}}{dx} \\ = x^{x} (1 + logx) - 2^{sinx} cosx . \log 2 \end{array}$

Q.5 Differentiate the function given with respect to x.

${(x+3)}^{2} . {(x + 4)}^{3} . {(x + 5)}^{4}$

Ans

$\begin{array}{l} Let y = {(x +3)}^{2} . {(x +4)}^{3} . {(x +5)}^{4} \\ Taking logarithm on both the sides, we get \\ logy = \log {{(x +3)}^{2} . {(x +4)}^{3} . {(x +5)}^{4}} \\ = 2 \log (x + 3) + 3 \log (x + 4) + 4 \log (x + 5) \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} logy = 2 \frac{d}{dx} \log (x + 3) + 3 \frac{d}{dx} \log (x + 4) + 4 \frac{d}{dx} \log (x + 5) \\ \frac{1}{y} \frac{dy}{dx} = 2 \times \frac{1}{x + 3} \frac{d}{dx} (x + 3) + 3 \times \frac{1}{x + 4} \frac{d}{dx} (x + 4) + 4 \times \frac{1}{x + 5} \frac{d}{dx} (x + 5) \\ = \frac{2}{x + 3} \times 1 + \frac{3}{x + 4} \times 1 + \frac{4}{x + 5} \times 1 \\ \frac{dy}{dx} = y (\frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}) \\ = {(x +3)}^{2} . {(x +4)}^{3} . {(x +5)}^{4} (\frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}) \\ = {(x +3)}^{2} . {(x +4)}^{3} . {(x +5)}^{4} (\frac{2 (x + 4) (x + 5) + 3 (x + 3) (x + 5) + 4 (x + 3) (x + 4)}{(x + 3) (x + 4) (x + 5)}) \\ = (x +3) . {(x +4)}^{2} . {(x +5)}^{3} {\begin{cases} 2 (x^{2} + 9 x + 20) + 3 (x^{2} + 8 x + 15) \\ + 4 (x^{2} + 7 x + 12) \end{cases}} \\ = (x +3) . {(x +4)}^{2} . {(x +5)}^{3} {\begin{cases} 2 x^{2} + 18 x + 40 + 3 x^{2} + 24 x \\ + 45 + 4 x^{2} + 28 x + 48 \end{cases}} \\ ∴ \frac{dy}{dx} = (x +3) . {(x +4)}^{2} . {(x +5)}^{3} (9 x^{2} + 70 x + 133) \end{array}$

Q.6 Differentiate the function given with respect to x.

${(x+ \frac{1}{x})}^{x} + x^{(1 + \frac{1}{x})}$

Ans

$\begin{array}{l} Let y = {(x + \frac{1}{x})}^{x} + x^{(1 + \frac{1}{x})} \\ = y_{1} + y_{2} \\ So, y_{1} = {(x + \frac{1}{x})}^{x} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = xlog (x + \frac{1}{x}) \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} {xlog (x + \frac{1}{x})} \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x \frac{d}{dx} \log (x + \frac{1}{x}) + \log (x + \frac{1}{x}) \frac{d}{dx} x \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x . \frac{1}{(x + \frac{1}{x})} \frac{d}{dx} (x + \frac{1}{x}) + \log (x + \frac{1}{x}) \times 1 \\ \frac{{dy}_{1}}{dx} = y_{1} {\frac{x^{2}}{x^{2} + 1} \times (1 - \frac{1}{x^{2}}) + \log (x + \frac{1}{x})} \\ \frac{{dy}_{1}}{dx} = {(x + \frac{1}{x})}^{x} {\frac{x^{2}}{x^{2} + 1} \times (\frac{x^{2} - 1}{x^{2}}) + \log (x + \frac{1}{x})} \\ \frac{{dy}_{1}}{dx} = {(x + \frac{1}{x})}^{x} {\frac{x^{2} - 1}{x^{2} + 1} + \log (x + \frac{1}{x})} \\ Now, y_{2} = x^{(1 + \frac{1}{x})} \\ Taking logarithm on both the sides, we get \\ {logy}_{2} = (1 + \frac{1}{x}) logx \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{2} = \frac{d}{dx} {(1 + \frac{1}{x}) logx} \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = (1 + \frac{1}{x}) \frac{d}{dx} logx + logx \frac{d}{dx} (1 + \frac{1}{x}) [By product rule] \\ \frac{{dy}_{2}}{dx} = y_{2} {(1 + \frac{1}{x}) \frac{1}{x} + logx . (0 - \frac{1}{x^{2}})} \\ \frac{{dy}_{2}}{dx} = x^{(1 + \frac{1}{x})} {\frac{1}{x} + \frac{1}{x^{2}} - \frac{1}{x^{2}} logx} \\ = x^{(1 + \frac{1}{x})} {\frac{x + 1 - logx}{x^{2}}} \\ Thus, \\ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ = {(x + \frac{1}{x})}^{x} {\frac{x^{2} - 1}{x^{2} + 1} + \log (x + \frac{1}{x})} + x^{(1 + \frac{1}{x})} {\frac{x + 1 - logx}{x^{2}}} \end{array}$

Q.7 Differentiate the function given with respect to x.

${(logx)}^{x} + x^{logx}$

Ans

$\begin{array}{l} Let y = {(logx)}^{x} + x^{logx} \\ = y_{1} + y_{2} \\ So, y_{1} = {(logx)}^{x} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = x . \log (logx) \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} {x . \log (logx)} \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x \frac{d}{dx} \log (logx) + \log (logx) \frac{d}{dx} x [By Product Rule] \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x . \frac{1}{logx} \frac{d}{dx} logx + \log (logx) \times 1 \\ \frac{{dy}_{1}}{dx} = y_{1} {x . \frac{1}{logx} \frac{1}{x} + \log (logx)} \\ \frac{{dy}_{1}}{dx} = {(logx)}^{x} {\frac{1}{logx} + \log (logx)} \\ \frac{{dy}_{1}}{dx} = {(logx)}^{x -1} {1 + logx . \log (logx)} \\ Now, y_{2} = x^{logx} \\ Taking logarithm on both the sides, we get \\ {logy}_{2} = logx . logx \\ = {(logx)}^{2} \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{2} = \frac{d}{dx} {(logx)}^{2} \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = 2 logx \frac{d}{dx} logx [By Chain Rule] \\ = 2 logx \times \frac{1}{x} \\ \frac{{dy}_{2}}{dx} = \frac{2 y_{2}}{x} logx \\ = \frac{2 x^{logx}}{x} logx \\ = 2 x^{logx -1} logx \\ Hence, \\ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ = {(logx)}^{x -1} {1 + logx . \log (logx)} + 2 x^{logx -1} logx \end{array}$

Q.8 Differentiate the function given with respect to x.

${(sinx)}^{x} + \sin^{- 1} \sqrt{x}$

Ans

$\begin{array}{l} Let y = {(sinx)}^{x} + \sin^{- 1} \sqrt{x} \\ = y_{1} + y_{2} \\ So, y_{1} = {(sinx)}^{x} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = x . \log (sinx) \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} {x . \log (sinx)} \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x \frac{d}{dx} \log (sinx) + \log (sinx) \frac{d}{dx} x [By Product Rule] \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x . \frac{1}{sinx} \frac{d}{dx} sinx + \log (sinx) \times 1 \\ \frac{{dy}_{1}}{dx} = y_{1} {x . \frac{1}{sinx} cosx + \log (sinx)} \\ \frac{{dy}_{1}}{dx} = {(sinx)}^{x} {x \frac{cosx}{sinx} + \log (sinx)} \\ \frac{{dy}_{1}}{dx} = {(sinx)}^{x} {xcotx + \log (sinx)} \\ Now, y_{2} = \sin^{- 1} \sqrt{x} \\ Differentiating both sides with respect to x, we get \\ \frac{{dy}_{2}}{dx} = \frac{d}{dx} \sin^{- 1} \sqrt{x} \\ \frac{{dy}_{2}}{dx} = \frac{1}{\sqrt{1 - {(\sqrt{x})}^{2}}} \frac{d}{dx} \sqrt{x} [By Chain Rule] \\ = \frac{1}{\sqrt{1 - x}} \times \frac{1}{2 \sqrt{x}} \\ \frac{{dy}_{2}}{dx} = \frac{1}{2 \sqrt{x - x^{2}}} \\ Hence, \\ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ = {(sinx)}^{x} {xcotx + \log (sinx)} + \frac{1}{2 \sqrt{x - x^{2}}} \end{array}$

Q.9 Differentiate the function given with respect to x.

$x^{sinx} + {(sinx)}^{cosx}$

Ans

$\begin{array}{l} Let y = x^{sinx} + {(sinx)}^{cosx} \\ = y_{1} + y_{2} \\ So, y_{1} = x^{sinx} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = sinx . logx \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} {sinx . logx} \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = sinx \frac{d}{dx} logx + logx \frac{d}{dx} sinx [By Product Rule] \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = sinx . \frac{1}{x} + logx \times cosx \\ \frac{{dy}_{1}}{dx} = y_{1} {sinx . \frac{1}{x} + logx \times cosx} \\ \frac{{dy}_{1}}{dx} = x^{sinx} {\frac{sinx}{x} + cosx . logx} \\ Now, y_{2} = {(sinx)}^{cosx} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = cosx . logsinx \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} (cosx . logsinx) \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = cosx \frac{d}{dx} logsinx + logsinx \frac{d}{dx} cosx [By Product Rule] \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = cosx . \frac{1}{sinx} \frac{d}{dx} sinx + logsinx \times - sinx \\ \frac{{dy}_{2}}{dx} = {(sinx)}^{cosx} (cosx . \frac{1}{sinx} . cosx - sinx . logsinx) \\ = {(sinx)}^{cosx} (cotx . cosx - sinx . logsinx) \\ ∴ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ = x^{sinx} (\frac{sinx}{x} + cosx . logx) + {(sinx)}^{cosx} (cosx . cotx - sinx . logsinx) \end{array}$

Q.10 Differentiate the function given with respect to x.

$x^{xcosx} + \frac{x^{2} + 1}{x^{2} - 1}$

Ans

$\begin{array}{l} Let y = x^{xcosx} + \frac{x^{2} + 1}{x^{2} - 1} \\ = y_{1} + y_{2} \\ So, y_{1} = x^{xcosx} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = xcosx . logx \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} {xcosx . logx} \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = xcosx \frac{d}{dx} logx + logx \frac{d}{dx} (xcosx) \\ [By Product Rule] \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = xcosx . \frac{1}{x} + logx (x \frac{d}{dx} cosx + cosx \frac{d}{dx} x) \\ [By Product Rule] \\ \frac{{dy}_{1}}{dx} = y_{1} {cosx + logx (- xsinx + cosx)} \\ \frac{{dy}_{1}}{dx} = x^{xcosx} {cosx - xsinxlogx + logx . cosx} \\ \frac{{dy}_{1}}{dx} = x^{xcosx} {cosx (1 + logx) - xsinxlogx} \\ Now, \\ y_{2} = \frac{x^{2} + 1}{x^{2} - 1} \\ Taking logarithm on both the sides, we get \\ {logy}_{2} = \log (x^{2} + 1) - \log (x^{2} - 1) \\ Differentiating w . r . t . x, we get \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = \frac{1}{x^{2} + 1} \frac{d}{dx} (x^{2} + 1) - \frac{1}{x^{2} - 1} \frac{d}{dx} (x^{2} - 1) [By chain rule] \\ \frac{{dy}_{2}}{dx} = y_{2} (\frac{1}{x^{2} + 1} \times 2 x - \frac{1}{x^{2} - 1} \times 2 x) \\ = 2 x (\frac{x^{2} + 1}{x^{2} - 1}) (\frac{1}{x^{2} + 1} - \frac{1}{x^{2} - 1}) \\ = 2 x (\frac{x^{2} + 1}{x^{2} - 1}) (\frac{(x^{2} - 1) - (x^{2} + 1)}{(x^{2} + 1) ((x^{2} - 1))}) \\ = 2 x (\frac{x^{2} + 1}{x^{2} - 1}) {\frac{x^{2} - 1 - x^{2} - 1}{(x^{2} + 1) (x^{2} - 1)}} \\ = 2 x (\frac{1}{x^{2} - 1}) {\frac{- 2}{(x^{2} - 1)}} \\ \frac{{dy}_{2}}{dx} = \frac{- 4 x}{{(x^{2} - 1)}^{2}} \\ Thus, \\ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ = x^{xcosx} {cosx (1 + logx) - xsinxlogx} - \frac{4 x}{{(x^{2} - 1)}^{2}} \end{array}$

Q.11 $Differentiate the function given with respect to x. {(xcosx)}^{x} + {(xsinx)}^{\frac{1}{x}}$

Ans

$\begin{array}{l} Let y = {(xcosx)}^{x} + {(xsinx)}^{\frac{1}{x}} \\ = y_{1} + y_{2} \\ So, y_{1} = {(xcosx)}^{x} \\ Taking logarithm on both the sides, we get \\ {logy}_{1} = xlog (xcosx) \\ = xlogx + xlogcosx \\ Differentiating both sides with respect to x, we get \\ \frac{d}{dx} {logy}_{1} = \frac{d}{dx} (x . logx + x . logcosx) \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = \frac{d}{dx} (x . logx) + \frac{d}{dx} (x . logcosx) \\ \frac{1}{y_{1}} \frac{{dy}_{1}}{dx} = x \frac{d}{dx} logx + logx \frac{d}{dx} x + x \frac{d}{dx} logcosx + logcosx \frac{d}{dx} x \\ [By Product Rule] \\ \frac{{dy}_{1}}{dx} = y_{1} (x . \frac{1}{x} + logx + x \frac{1}{cosx} . \frac{d}{dx} cosx + logcosx) \\ \frac{{dy}_{1}}{dx} = {(xcosx)}^{x} (1 + logx + x \frac{1}{cosx} \times - sinx + logcosx) \\ = {(xcosx)}^{x} (1 + logx - x \frac{sinx}{cosx} + logcosx) \\ \frac{{dy}_{1}}{dx} = {(xcosx)}^{x} (1 + logx - xtanx + logcosx) \\ Now, \\ y_{2} = {(xsinx)}^{\frac{1}{x}} \\ Taking logarithm on both the sides, we get \\ {logy}_{2} = \frac{1}{x} \log (xsinx) \\ Differentiating w . r . t . x, we get \\ \frac{1}{y_{2}} \frac{{dy}_{2}}{dx} = \frac{1}{x} \frac{d}{dx} \log (xsinx) + \log (xsinx) \frac{d}{dx} . \frac{1}{x} [By product rule] \\ \frac{{dy}_{2}}{dx} = y_{2} {\frac{1}{x} (\frac{1}{xsinx}) \frac{d}{dx} (xsinx) - \log (xsinx) . \frac{1}{x^{2}}} [By chain rule] \\ \frac{{dy}_{2}}{dx} = {(xsinx)}^{\frac{1}{x}} {\frac{1}{x} (\frac{1}{xsinx}) (x \frac{d}{dx} sinx + sinx \frac{d}{dx} x) - \log (xsinx) . \frac{1}{x^{2}}} \\ \frac{{dy}_{2}}{dx} = {(xsinx)}^{\frac{1}{x}} {\frac{1}{x} (\frac{1}{xsinx}) (x . cosx + sinx) - \log (xsinx) . \frac{1}{x^{2}}} \\ \frac{{dy}_{2}}{dx} = {(xsinx)}^{\frac{1}{x}} {(\frac{x . cosx + sinx}{x^{2} sinx}) - \log (xsinx) . \frac{1}{x^{2}}} \\ ∴ \frac{dy}{dx} = \frac{{dy}_{1}}{dx} + \frac{{dy}_{2}}{dx} \\ \frac{dy}{dx} = {(xcosx)}^{x} (1 + logx - xtanx + logcosx) \\ + {(xsinx)}^{\frac{1}{x}} {(\frac{x . cosx + sinx}{x^{2} sinx}) - \log (xsinx) . \frac{1}{x^{2}}} \end{array}$

Q.12 $Find \frac{dy}{dx} of the function x^{y} + y^{x} =$

Ans

$\begin{array}{l} The given function is \\ x^{y} + y^{x} = 1 \\ Let u = x^{y} and v = y^{x} \\ Then, function becomes \\ u + v = 1 \\ Differentiating w . r . t . x, we get \\ \frac{du}{dx} + \frac{dv}{dx} = 0 . .. (i) \\ Since, u = x^{y} \\ Taking \log both sides, we get \\ logu = {logx}^{y} \\ logu = ylogx \\ Differentiating both sides w . r . t . x, we get \\ \frac{d}{dx} logu = \frac{d}{dx} ylogx \\ \frac{1}{u} \frac{du}{dx} = y \frac{d}{dx} logx + logx \frac{dy}{dx} \\ = y \times \frac{1}{x} + logx \frac{dy}{dx} \\ \frac{du}{dx} = u (\frac{y}{x} + logx \frac{dy}{dx}) \\ \frac{du}{dx} = x^{y} (\frac{y}{x} + logx \frac{dy}{dx}) . .. (ii) \\ And v = y^{x} \\ Taking \log both sides, we get \\ logv = {logy}^{x} \\ logv = xlogy \\ Differentiating both sides w . r . t . x, we get \\ \frac{d}{dx} logv = \frac{d}{dx} xlogy \\ \frac{1}{v} \frac{dv}{dx} = x \frac{d}{dx} logy + logy \frac{d}{dx} x \\ = x \times \frac{1}{y} \frac{dy}{dx} + logy (1) \\ \frac{dv}{dx} = v (\frac{x}{y} \frac{dy}{dx} + logy) \\ \frac{dv}{dx} = y^{x} (\frac{x}{y} \frac{dy}{dx} + logy) . .. (iii) \\ From equatin (i), (ii) and (iii), we have \\ x^{y} (\frac{y}{x} + logx \frac{dy}{dx}) + y^{x} (\frac{x}{y} \frac{dy}{dx} + logy) = 0 \\ x^{y - 1} y + x^{y} logx \frac{dy}{dx} + {yx}^{- 1} x \frac{dy}{dx} + y^{x} logy = 0 \\ (x^{y} logx + {yx}^{- 1} x) \frac{dy}{dx} = - x^{y - 1} y - y^{x} logy \\ \frac{dy}{dx} = - \frac{{yx}^{y - 1} + y^{x} logy}{x^{y} logx + {xy}^{x}^{- 1}} \end{array}$

Q.13 $Find \frac{dy}{dx} of the function : y^{x} = x^{y}$

Ans

$\begin{array}{l} The given function is y^{x} = x^{y} \\ Taking logarithm both sides, we get \\ {logy}^{x} = {logx}^{y} \\ xlogy = ylogx \\ Differentiating both sides, w . r . t . x, we get \\ x \frac{d}{dx} logy + logy \frac{d}{dx} x = y \frac{d}{dx} logx + logx \frac{d}{dx} y [By Product Rule] \\ \frac{x}{y} \frac{dy}{dx} + logy \times 1 = \frac{y}{x} + logx \frac{dy}{dx} \\ \frac{x}{y} \frac{dy}{dx} - logx \frac{dy}{dx} = \frac{y}{x} - logy \\ (\frac{x}{y} - logx) \frac{dy}{dx} = \frac{y}{x} - logy \\ \frac{dy}{dx} = \frac{(\frac{y}{x} - logy)}{(\frac{x}{y} - logx)} \\ = \frac{y (y - xlogy)}{x (x - ylogx)} \end{array}$

Q.14 $Find \frac{dy}{dx} of the function : {(cosx)}^{y} = {(cosy)}^{x}$

Ans

$\begin{array}{l} The given function is {(cosx)}^{y} = {(cosy)}^{x} \\ Taking logarithm both sides, we get \\ \log {(cosx)}^{y} = \log {(cosy)}^{x} \\ ylogcosx = xlogcosy \\ Differentiating both sides, w . r . t . x, we get \\ y \frac{d}{dx} logcosx + logcosx \frac{d}{dx} y = x \frac{d}{dx} logcosy + logcosy \frac{d}{dx} x [By Product Rule] \\ y \times \frac{1}{cosx} \frac{d}{dx} cosx + logcosx \frac{dy}{dx} = x \times \frac{1}{cosy} \frac{d}{dx} cosy + logcosy \\ \frac{y}{cosx} \times - sinx + logcosx \frac{dy}{dx} = \frac{x}{cosy} \times - siny \frac{dy}{dx} + logcosy \\ \frac{- ysinx}{cosx} + logcosx \frac{dy}{dx} = \frac{- xsiny}{cosy} \frac{dy}{dx} + logcosy \\ logcosx \frac{dy}{dx} + \frac{xsiny}{cosy} \frac{dy}{dx} = logcosy + \frac{ysinx}{cosx} \\ (logcosx + \frac{xsiny}{cosy}) \frac{dy}{dx} = logcosy + \frac{ysinx}{cosx} \\ \frac{dy}{dx} = \frac{(logcosy + \frac{ysinx}{cosx})}{(logcosx + \frac{xsiny}{cosy})} \\ = \frac{(logcosy + ytanx)}{(logcosx + xtany)} \end{array}$

Q.15 $Find \frac{dy}{dx} of the function : xy = e^{(x - y)}$

Ans

$\begin{array}{l} The given function is xy = e^{(x - y)} \\ Taking logarithm both sides, we get \\ logxy = {loge}^{(x - y)} \\ \Rightarrow logx + logy = (x - y) loge \\ \Rightarrow logx + logy = (x - y) \times 1 ∵ [loge = 1] \\ \Rightarrow logx + logy = (x - y) \\ Differentiating both sides, w . r . t . x, we get \\ \frac{d}{dx} logx + \frac{d}{dx} logy = \frac{d}{dx} x - \frac{dy}{dx} [By Product Rule] \\ \Rightarrow \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 - \frac{dy}{dx} \\ \Rightarrow (1 + \frac{1}{y}) \frac{dy}{dx} = 1 - \frac{1}{x} \\ \Rightarrow \frac{dy}{dx} = \frac{1 - \frac{1}{x}}{(1 + \frac{1}{y})} \\ = \frac{(\frac{x - 1}{x})}{(\frac{y + 1}{y})} \\ \Rightarrow \frac{dy}{dx} = \frac{y (x - 1)}{x (y + 1)} \end{array}$

Q.16

$\begin{array}{l} Find the derivative of the function given by \\ f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) \\ and hence find f ‘ (1) . \end{array}$

Ans

$\begin{array}{l} The given relationship is f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) \\ Taking logarithm on both the sides, we get \\ logf (x) = \log (1 + x) + \log (1 + x^{2}) + \log (1 + x^{4}) + \log (1 + x^{8}) \\ Differentiating both sides with respect to x, we get \\ \frac{1}{f (x)} \frac{d}{dx} f (x) = \frac{1}{1 + x} \frac{d}{dx} (1 + x) + \frac{1}{1 + x^{2}} \frac{d}{dx} (1 + x^{2}) + \frac{1}{1 + x^{4}} \frac{d}{dx} (1 + x^{4}) \\ + \frac{1}{1 + x^{8}} \frac{d}{dx} (1 + x^{8}) \\ = \frac{1}{1 + x} (1) + \frac{1}{1 + x^{2}} (0 + 2 x) + \frac{1}{1 + x^{4}} (0 + 4 x^{3}) \\ + \frac{1}{1 + x^{8}} (0 + 8 x^{7}) \\ \frac{d}{dx} f (x) = f (x) (\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}) \\ ∴ f ‘ (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}) \\ Putting x = 1, we get \\ f ‘ (1) = (1 + 1) (1 + 1^{2}) (1 + 1^{4}) (1 + 1^{8}) (\frac{1}{1 + 1} + \frac{2 (1)}{1 + 1^{2}} + \frac{4 {(1)}^{3}}{1 + 1^{4}} + \frac{8 {(1)}^{7}}{1 + 1^{8}}) \\ = 2 \times 2 \times 2 \times 2 (\frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2}) \\ = 16 (\frac{15}{2}) \\ ∴ f ‘ (1) = 120 \end{array}$

Q.17 Differentiate (x² – 5x + 8)(x³ + 7x + 9) in three ways mentioned below:

(i) By using product rule.

(ii) By expanding the product to obtain a single polynomial.

(iii) By logarithmic differentiation. Do they all give the same answer?

Ans

$\begin{array}{l} Let y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) \\ (i) Differentiating w . r . t . x, by using product rule, we get \\ \frac{dy}{dx} = \frac{d}{dx} {(x^{2} - 5 x + 8) (x^{3} + 7 x + 9)} \\ = (x^{2} - 5 x + 8) \frac{d}{dx} (x^{3} + 7 x + 9) + (x^{3} + 7 x + 9) \frac{d}{dx} (x^{2} - 5 x + 8) \\ = (x^{2} - 5 x + 8) (3 x^{2} + 7) + (x^{3} + 7 x + 9) (2 x - 5) \\ = 3 x^{4} + 7 x^{2} - 15 x^{3} - 35 x + 24 x^{2} + 56 + 2 x^{4} - 5 x^{3} + 14 x^{2} - 35 x + 18 x - 45 \\ ∴ \frac{dy}{dx} = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11 \\ (ii) y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) \\ = x^{5} + 7 x^{3} + 9 x^{2} - 5 x^{4} - 35 x^{2} - 45 x + 8 x^{3} + 56 x + 72 \\ = x^{5} - 5 x^{4} + 15 x^{3} - 26 x^{2} + 11 x + 72 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (x^{5} - 5 x^{4} + 15 x^{3} - 26 x^{2} + 11 x + 72) \\ = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11 \\ (iii) y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) \\ Taking logarithm on both the sides, we get \\ logy = \log {(x^{2} - 5 x + 8) (x^{3} + 7 x + 9)} \\ = \log (x^{2} - 5 x + 8) + \log (x^{3} + 7 x + 9) \\ Differentiating w . r . t . x, we get \\ \frac{d}{dx} logy = \frac{d}{dx} \log (x^{2} - 5 x + 8) + \frac{d}{dx} \log (x^{3} + 7 x + 9) \\ = \frac{1}{(x^{2} - 5 x + 8)} \frac{d}{dx} (x^{2} - 5 x + 8) + \frac{1}{(x^{3} + 7 x + 9)} \frac{d}{dx} (x^{3} + 7 x + 9) \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{(x^{2} - 5 x + 8)} (2 x - 5) + \frac{1}{(x^{3} + 7 x + 9)} (3 x^{2} + 7) \\ \frac{dy}{dx} = y (\frac{(2 x - 5) (x^{3} + 7 x + 9) + (3 x^{2} + 7) (x^{2} - 5 x + 8)}{(x^{2} - 5 x + 8) (x^{3} + 7 x + 9)}) \\ \frac{dy}{dx} = y (\frac{(2 x - 5) (x^{3} + 7 x + 9) + (3 x^{2} + 7) (x^{2} - 5 x + 8)}{y}) \\ [∵ Let y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) \\ \frac{dy}{dx} = 2 x^{4} + 14 x^{2} + 18 x - 5 x^{3} - 35 x - 45 + 3 x^{4} - 15 x^{3} \\ + 24 x^{2} + 7 x^{2} - 35 x + 56 \\ \frac{dy}{dx} = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11 \\ From the above three observations, it can be say that all the \\ results are same . \end{array}$

Q.18

$\begin{array}{l} If u, v and w are functions of x, then show that \\ \frac{d}{dx} (u, v, w) = \frac{du}{dx} v . w + u \frac{dv}{dx} . w + u . v . \frac{dw}{dx} \\ in two ways - first by repeated application of product rule, \\ second by logarithmic differentiation . \end{array}$

Ans

$\begin{array}{l} Let y = u . v . w = u . (v . w) \\ By applying product rule, we get \\ \frac{dy}{dx} = \frac{d}{dx} {u . (v . w)} \\ = u \frac{d}{dx} (v . w) + (v . w) \frac{d}{dx} u \\ = u {v \frac{d}{dx} w + w \frac{d}{dx} v} + (v . w) \frac{d}{dx} u \\ = u . v \frac{d}{dx} w + u . w \frac{d}{dx} v + (v . w) \frac{d}{dx} u \\ \frac{dy}{dx} = \frac{du}{dx} . v . w + \frac{dv}{dx} . u . w + \frac{dw}{dx} . u . v \\ Thus, this is the differentiation by product rule . \\ By taking logarithm on both sides of the equation y = u . v . w, we get \\ logy = logu + logv + logw \\ Differentiating both sides with respect to x, we get \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \\ \frac{dy}{dx} = y (\frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}) \\ = u . v . w (\frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}) \\ \frac{dy}{dx} = \frac{du}{dx} . v . w + \frac{dv}{dx} . u . w + \frac{dw}{dx} . u . v \\ Thus, this is the differentiation by logarithmic differentiation . \end{array}$

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NCERT Solutions for Class 12 Maths

NCERT Solutions for Class 12 Maths Chapter 5 Other Exercises

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FAQs (Frequently Asked Questions)

1. How many books are there in the curriculum of Class 12 Mathematics?

2. How many chapters are there in the curriculum of Class 12 Mathematics?

3. Is it essential to do all the questions of the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5?

4. Are the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 available on Extramarks?

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