# 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.

### 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.

### 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.

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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

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### NCERT Solutions for Class 12 Maths

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

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

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

### 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

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

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

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

Ans

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

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

${\text{(logx)}}^{\mathrm{cosx}}$

Ans

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

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

${\text{x}}^{\mathrm{x}}–{2}^{\mathrm{sinx}}$

Ans

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

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

${\text{(x+3)}}^{2}.{\left(\mathrm{x}+4\right)}^{3}.{\left(\mathrm{x}+5\right)}^{4}$

Ans

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

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

${\text{(x+}\frac{1}{\mathrm{x}}\right)}^{\mathrm{x}}+{\mathrm{x}}^{\left(1+\frac{1}{\mathrm{x}}\right)}$

Ans

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

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

${\text{(logx)}}^{\mathrm{x}}+{\mathrm{x}}^{\mathrm{logx}}$

Ans

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

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

${\text{(sinx)}}^{\mathrm{x}}+{\mathrm{sin}}^{–1}\sqrt{\mathrm{x}}$

Ans

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

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

${\text{x}}^{\mathrm{sinx}}+{\left(\mathrm{sinx}\right)}^{\mathrm{cosx}}$

Ans

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

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

${\mathbf{\text{x}}}^{\mathrm{xcosx}}+\frac{{\mathrm{x}}^{2}+1}{{\mathrm{x}}^{2}–1}$

Ans

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

Q.11

$\text{Differentiate the function given with respect to x.}\phantom{\rule{0ex}{0ex}}{\text{(xcosx)}}^{\mathrm{x}}+{\left(\mathrm{xsinx}\right)}^{\frac{1}{\mathrm{x}}}$

Ans

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

Q.12

$\text{Find }\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{function}\mathrm{ }{\mathrm{x}}^{\mathrm{y}}+{\mathrm{y}}^{\mathrm{x}}=$

Ans

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

Q.13

$\text{Find }\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{function}\mathrm{ }:\mathrm{ }{\mathrm{y}}^{\mathrm{x}}={\mathrm{x}}^{\mathrm{y}}$

Ans

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

Q.14

$\mathrm{Find}\mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{function}\mathrm{ }:\mathrm{ }{\left(\mathrm{cosx}\right)}^{\mathrm{y}}\mathrm{ }=\mathrm{ }{\left(\mathrm{cosy}\right)}^{\mathrm{x}}$

Ans

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

Q.15

$\text{Find }\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{function}\mathrm{ }:\mathrm{xy}\mathrm{ }=\mathrm{ }{\mathrm{e}}^{\left(\mathrm{x}-\mathrm{y}\right)}$

Ans

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

Q.16

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

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{relationship}\mathrm{is} \mathrm{f}\left(\mathrm{x}\right)=\left(1+\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)\left(1+{\mathrm{x}}^{4}\right)\left(1+{\mathrm{x}}^{8}\right)\\ \mathrm{Taking}\mathrm{logarithm}\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \mathrm{logf}\left(\mathrm{x}\right)=\mathrm{log}\left(1+\mathrm{x}\right)+\mathrm{log}\left(1+{\mathrm{x}}^{2}\right)+\mathrm{log}\left(1+{\mathrm{x}}^{4}\right)+\mathrm{log}\left(1+{\mathrm{x}}^{8}\right)\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{1}{\mathrm{f}\left(\mathrm{x}\right)}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{1+\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+\mathrm{x}\right)+\frac{1}{1+{\mathrm{x}}^{2}}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+{\mathrm{x}}^{2}\right)+\frac{1}{1+{\mathrm{x}}^{4}}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+{\mathrm{x}}^{4}\right)\\ +\frac{1}{1+{\mathrm{x}}^{8}}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+{\mathrm{x}}^{8}\right)\\ \mathrm{ }=\frac{1}{1+\mathrm{x}}\left(1\right)+\frac{1}{1+{\mathrm{x}}^{2}}\left(0+2\mathrm{x}\right)+\frac{1}{1+{\mathrm{x}}^{4}}\left(0+4{\mathrm{x}}^{3}\right)\\ +\frac{1}{1+{\mathrm{x}}^{8}}\left(0+8{\mathrm{x}}^{7}\right)\\ \frac{\mathrm{d}}{\mathrm{dx}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\left(\frac{1}{1+\mathrm{x}}+\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}+\frac{4{\mathrm{x}}^{3}}{1+{\mathrm{x}}^{4}}+\frac{8{\mathrm{x}}^{7}}{1+{\mathrm{x}}^{8}}\right)\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=\left(1+\mathrm{x}\right)\left(1+{\mathrm{x}}^{2}\right)\left(1+{\mathrm{x}}^{4}\right)\left(1+{\mathrm{x}}^{8}\right)\left(\frac{1}{1+\mathrm{x}}+\frac{2\mathrm{x}}{1+{\mathrm{x}}^{2}}+\frac{4{\mathrm{x}}^{3}}{1+{\mathrm{x}}^{4}}+\frac{8{\mathrm{x}}^{7}}{1+{\mathrm{x}}^{8}}\right)\\ \mathrm{Putting}\mathrm{x}= 1,\mathrm{we}\mathrm{get}\\ \mathrm{ }\mathrm{f}‘\left(1\right)=\left(1+1\right)\left(1+{1}^{2}\right)\left(1+{1}^{4}\right)\left(1+{1}^{8}\right)\left(\frac{1}{1+1}+\frac{2\left(1\right)}{1+{1}^{2}}+\frac{4{\left(1\right)}^{3}}{1+{1}^{4}}+\frac{8{\left(1\right)}^{7}}{1+{1}^{8}}\right)\\ \mathrm{ }=2×2×2×2\left(\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right)\\ =16\left(\frac{15}{2}\right)\\ \therefore \mathrm{ }\mathrm{f}‘\left(1\right)=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}\mathrm{Let} \mathrm{y}=\left({\mathrm{x}}^{\mathrm{2}}–5\mathrm{x}+ 8\right)\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)\\ \left(\mathrm{i}\right)\mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{by}\mathrm{using}\mathrm{product}\mathrm{rule},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left({\mathrm{x}}^{\mathrm{2}}–5\mathrm{x}+ 8\right)\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)\right\}\\ =\left({\mathrm{x}}^{\mathrm{2}}–5\mathrm{x}+ 8\right)\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)+\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{2}}–5\mathrm{x}+ 8\right)\\ =\left({\mathrm{x}}^{\mathrm{2}}–5\mathrm{x}+ 8\right)\left(3{\mathrm{x}}^{\mathrm{2}}+7\right)+\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+\mathrm{9}\right)\left(2\mathrm{x}-\mathrm{5}\right)\\ =3{\mathrm{x}}^{4}+7{\mathrm{x}}^{2}-15{\mathrm{x}}^{3}-35\mathrm{x}+24{\mathrm{x}}^{2}+56+2{\mathrm{x}}^{4}-5{\mathrm{x}}^{3}+14{\mathrm{x}}^{2}-35\mathrm{x}+18\mathrm{x}-45\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=5{\mathrm{x}}^{4}-20{\mathrm{x}}^{3}+45{\mathrm{x}}^{2}-52\mathrm{x}+11\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{y}=\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)\\ ={\mathrm{x}}^{5}+7{\mathrm{x}}^{3}+9{\mathrm{x}}^{2}-5{\mathrm{x}}^{4}-35{\mathrm{x}}^{2}-45\mathrm{x}+8{\mathrm{x}}^{3}+56\mathrm{x}+72\\ ={\mathrm{x}}^{5}-5{\mathrm{x}}^{4}+15{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+11\mathrm{x}+72\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{ }\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{5}-5{\mathrm{x}}^{4}+15{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+11\mathrm{x}+72\right)\\ =5{\mathrm{x}}^{4}-20{\mathrm{x}}^{3}+45{\mathrm{x}}^{2}-52\mathrm{x}+11\\ \left(\mathrm{iii}\right)\mathrm{ }\mathrm{y}=\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)\\ \mathrm{Taking}\mathrm{logarithm}\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \mathrm{logy}=\mathrm{log}\left\{\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)\right\}\\ =\mathrm{log}\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)+\mathrm{log}\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{ }\mathrm{we}\mathrm{get}\\ \frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logy}=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{log}\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{log}\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)\\ =\frac{1}{\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)+\frac{1}{\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)\\ \frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)}\left(2\mathrm{x}-5\right)+\frac{1}{\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)}\left(3{\mathrm{x}}^{2}+7\right)\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}\left(\frac{\left(2\mathrm{x}-5\right)\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)+\left(3{\mathrm{x}}^{2}+7\right)\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)}{\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)}\right)\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}\left(\frac{\left(2\mathrm{x}-5\right)\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)+\left(3{\mathrm{x}}^{2}+7\right)\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)}{\mathrm{y}}\right)\\ \left[\because \mathrm{Let}\mathrm{ }\mathrm{y}=\left({\mathrm{x}}^{\mathrm{2}}-5\mathrm{x}+ 8\right)\left({\mathrm{x}}^{\mathrm{3}}+7\mathrm{x}+ 9\right)\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=2{\mathrm{x}}^{4}+14{\mathrm{x}}^{2}+18\mathrm{x}-5{\mathrm{x}}^{3}-35\mathrm{x}-45+3{\mathrm{x}}^{4}-15{\mathrm{x}}^{3}\\ +24{\mathrm{x}}^{2}+7{\mathrm{x}}^{2}-35\mathrm{x}+56\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=5{\mathrm{x}}^{4}-20{\mathrm{x}}^{3}+45{\mathrm{x}}^{2}-52\mathrm{x}+11\\ \mathrm{From}\mathrm{the}\mathrm{above}\mathrm{three}\mathrm{observations},\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{say}\mathrm{that}\mathrm{all}\mathrm{the}\\ \mathrm{results}\mathrm{are}\mathrm{same}\mathrm{.}\end{array}$

Q.18

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

Ans

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

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