# NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7

The Central Board of Secondary Education CBSE is one of the biggest and most reputed boards in the country. CBSE is a national-level board with schools affiliated with it all over the country. According to CBSE, education has an objective approach, and its curriculum requires students to have an in-depth understanding of the subject matter. CBSE is the ideal board for students who wish to take a variety of competitive exams and succeed in them. CBSE’s higher secondary curriculum is well known throughout the country for its level of difficulty. It is found that students who switch from another board to CBSE for their eleventh standard face many obstacles. This might seem challenging but there are effective ways that can be followed by the student to get over these struggles.

NCERT stands for National Council of Educational Research and Training and is a resource-based institution founded by the Indian Government. NCERT assists the Central Government and the State Government on matters related to education and academia. CBSE follows all the NCERT suggested textbooks. CBSE emphasises improving the cognitive development of its students. Regular revision is necessary for CBSE students, as well as hard work, excellent reference work, and a solid understanding of all subjects.

Mathematics is one of the core subjects for all the students who have opted for the subject after their Class 10 board examinations. The change from Class 10 to a 10 + 2 level can be challenging. Many students suffer from great anxiety when it comes to Mathematics, and the 10+2 level makes the subject even more difficult. The CBSE Class 12 Mathematics has a massive syllabus and finishing the whole syllabus can be challenging.

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.7) Exercise 5.7

Chapter 5 of Class 12 CBSE is Continuity and Differentiation which introduces the students to concepts from Physics. This exercise primarily focuses on finding out and calculating the second order derivative of functions, calculating the value of d²y/dx² regarding finding its expression in question which asks the students to ‘proof’ something.

Mathematics is a core subject that has interrelations with every science subject. Exercise 5.7 Class 12th Maths NCERT Solutions teaches the students to calculate second-order derivatives and these techniques and concepts are extremely crucial for understanding Coulomb’s Law when a student is preparing for Electric Charges and Fields. The mathematical proof of the law itself implements the concepts of second-order derivatives and therefore the numerical-based questions of the same chapter use the same concepts. These areas of the syllabus which merge different subjects are extremely important, suggesting teachers and students must take these chapters seriously. It is suggested that these chapters should be given a lot of priority not only because they carry a lot of weightage but also because it makes sense in the context of a good studying strategy. When a student is learning Exercise 5.7 they are not just preparing for Mathematics but in turn, they are also making progress in their Physics Syllabus.

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Teachers all over the country have repeatedly emphasised the importance of inculcating a habit of regularly practising mathematical problems to master them. For a student to combat both the vast Mathematics syllabus and the advanced level of difficulty that comes with being in CBSE Class 12. Teachers and Students unanimously have observed that these reasons hinder the smooth progression of the syllabus. Students often complain about how the doubts that arise for them while solving mathematical problems sometimes follow the same trend consecutively for many questions.

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

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Class 12 students can often get intimidated by the vast syllabus of Class 12 Mathematics. If a student has a thorough understanding of all the concepts and topics of the Mathematics syllabus, it can be one of the most scoring subjects in Class 12. Students should have a thorough understanding of all the important concepts and topics of Mathematics. Understanding all the formulas will help the students solve the exercise questions with ease.

### What Do You Learn from NCERT Maths 5.7 Exercise Class 12?

Chapter 5 of the CBSE is Continuity and Differentiability and the 5.7 Exercise is one of the most important exercises of the chapter. Concepts like Differentiation and Integration are introduced to students in Class 11 and Students who have taken up science as their stream know the importance of these concepts. These ideas constantly reiterate themselves in subjects like Chemistry and Physics all the time. As observed in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 the implications of this exercise are huge. One just does not learn the mathematical concepts but also the interdisciplinary context of these mathematical; concepts, and their practical applications.

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### Extramarks Class 12 Physics Chapter 1 NCERT Solutions

Physics is one of the core subjects and therefore students from the Science stream have to give  a lot of importance to it. Physics is a subject which is completely analytical and students take a considerable amount of time understanding the concepts because these topics require the fundamental concepts of the chapter to be very clear. Physics and Mathematics as they are two sides of the same coin. To be good at Physics students have to be good at Mathematics and vice-versa.

Chapter 12 of Class 12 CBSE Physics explicates more on the concept of Electric Charges and Electric Fields. Students have learned about these concepts in their earlier classes but in Class 12 they learn about these concepts in full depth. Having a good hold over the Class 12 Maths NCERT Solutions Chapter 5 Exercise 5.7 is advised by teachers.

Chapter 12 discusses new concepts about conductors, insulators, and charging through induction. The chapter includes diagrams delineating the charge flow and it also helps visualise the force field around a source charge. The chapter discusses electric dipoles, while the theory of these concepts is important, the derivations of the formulas in these chapters are based on everything discussed in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7.

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Q.1

${\text{Find the second order derivative of​ the function : x}}^{20}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{y}={\mathrm{x}}^{20}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{20}\\ =20{\mathrm{x}}^{19}\\ \mathrm{Differentiating}\mathrm{ }\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \therefore \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(20{\mathrm{x}}^{19}\right)\\ \mathrm{ }=20×19\mathrm{ }{\mathrm{x}}^{18}\\ \mathrm{ }=380\mathrm{ }{\mathrm{x}}^{18}\\ \mathrm{Thus},\mathrm{second}\mathrm{derivative}\mathrm{of}\mathrm{given}\mathrm{function}\mathrm{is}380\mathrm{ }{\mathrm{x}}^{18}\mathrm{.}\end{array}$

Q.2

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{second}\mathrm{ }\mathrm{order}\mathrm{ }\mathrm{derivative}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{function}:\\ \mathrm{x}.\mathrm{cosx}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{y}=\mathrm{xcosx}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{xcosx}\right)\\ =\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}+\mathrm{cosx}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\left[\mathrm{By}\mathrm{product}\mathrm{rule}\right]\\ =\mathrm{x}\left(-\mathrm{sinx}\right)+\mathrm{cosx}×1\\ =-\mathrm{xsinx}+\mathrm{cosx}\\ \mathrm{Differentiating}\mathrm{ }\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \therefore \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(-\mathrm{xsinx}+\mathrm{cosx}\right)\\ \mathrm{ }=-\left(\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}+\mathrm{sinx}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}\\ \mathrm{ }=-\left(\mathrm{xcosx}+\mathrm{sinx}×1\right)-\mathrm{sinx}\\ \mathrm{ }=-\mathrm{xcosx}-\mathrm{sinx}-\mathrm{sinx}\\ \mathrm{ }=-\mathrm{xcosx}-2\mathrm{sinx}\\ \mathrm{Thus},\mathrm{second}\mathrm{derivative}\mathrm{of}\mathrm{given}\mathrm{function}\mathrm{is}-\left(\mathrm{xcosx}+2\mathrm{sinx}\right)\mathrm{.}\end{array}$

Q.3

$\text{Find the second order derivative​ of ​the function: logx}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{y}=\mathrm{logx}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\\ =\frac{1}{\mathrm{x}}={\mathrm{x}}^{-1}\\ \mathrm{Differentiating}\mathrm{ }\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \therefore \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{-1}\right)\\ \mathrm{ }=-1\mathrm{ }{\mathrm{x}}^{-2}\\ \mathrm{ }=-{\mathrm{x}}^{-2}\\ \mathrm{Thus},\mathrm{second}\mathrm{derivative}\mathrm{of}\mathrm{given}\mathrm{function}\mathrm{is}-{\mathrm{x}}^{-2}\mathrm{.}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{second}\mathrm{ }\mathrm{order}\mathrm{ }\mathrm{derivative}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{function}:\\ {\mathrm{x}}^{3}\mathrm{ }\mathrm{logx}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{y}={\mathrm{x}}^{3}\mathrm{logx}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{3}\mathrm{logx}\right)\\ ={\mathrm{x}}^{3}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}+\mathrm{logx}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{3}\left[\mathrm{By}\mathrm{product}\mathrm{rule}\right]\\ ={\mathrm{x}}^{3}×\frac{1}{\mathrm{x}}+\mathrm{logx}×3{\mathrm{x}}^{2}\\ ={\mathrm{x}}^{2}+3{\mathrm{x}}^{2}\mathrm{logx}\\ \mathrm{Differentiating}\mathrm{ }\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \therefore \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}+3{\mathrm{x}}^{2}\mathrm{logx}\right)\\ \mathrm{ }=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}+3\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}\mathrm{logx}\right)\\ \mathrm{ }=2\mathrm{x}+3\left({\mathrm{x}}^{2}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}+\mathrm{logx}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}\right)\\ \mathrm{ }=2\mathrm{x}+3\left({\mathrm{x}}^{2}×\frac{1}{\mathrm{x}}+\mathrm{logx}.2\mathrm{x}\right)\\ \mathrm{ }=2\mathrm{x}+3\mathrm{x}+6\mathrm{xlogx}\\ \mathrm{ }=5\mathrm{x}+6\mathrm{xlogx}\\ \mathrm{Thus},\mathrm{second}\mathrm{derivative}\mathrm{of}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{x}\left(5+6\mathrm{logx}\right)\mathrm{.}\end{array}$

Q.5

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{second}\mathrm{ }\mathrm{order}\mathrm{ }\mathrm{derivative}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{function}:\\ {\mathrm{e}}^{\mathrm{x}}\mathrm{ }\mathrm{sin}5\mathrm{x}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{y}={\mathrm{e}}^{\mathrm{x}}\mathrm{sin}5\mathrm{x}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{x}}\mathrm{sin}5\mathrm{x}\right)\\ ={\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sin}5\mathrm{x}+\mathrm{sin}5\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{\mathrm{x}}\left[\mathrm{By}\mathrm{product}\mathrm{rule}\right]\\ ={\mathrm{e}}^{\mathrm{x}}\left(5\mathrm{cos}5\mathrm{x}\right)+\mathrm{sin}5\mathrm{x}.{\mathrm{e}}^{\mathrm{x}}\\ ={\mathrm{e}}^{\mathrm{x}}\left(5\mathrm{cos}5\mathrm{x}+\mathrm{sin}5\mathrm{x}\right)\\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \therefore \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{{\mathrm{e}}^{\mathrm{x}}\left(5\mathrm{cos}5\mathrm{x}+\mathrm{sin}5\mathrm{x}\right)\right\}\\ \mathrm{ }={\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(5\mathrm{cos}5\mathrm{x}+\mathrm{sin}5\mathrm{x}\right)+\left(5\mathrm{cos}5\mathrm{x}+\mathrm{sin}5\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{\mathrm{x}}\\ \left[\mathrm{By}\mathrm{product}\mathrm{rule}\right]\\ \mathrm{ }={\mathrm{e}}^{\mathrm{x}}\left(-25\mathrm{sin}5\mathrm{x}+5\mathrm{cos}5\mathrm{x}\right)+\left(5\mathrm{cos}5\mathrm{x}+\mathrm{sin}5\mathrm{x}\right){\mathrm{e}}^{\mathrm{x}}\\ \mathrm{ }={\mathrm{e}}^{\mathrm{x}}\left(-25\mathrm{sin}5\mathrm{x}+5\mathrm{cos}5\mathrm{x}+5\mathrm{cos}5\mathrm{x}+\mathrm{sin}5\mathrm{x}\right)\\ \mathrm{ }=2{\mathrm{e}}^{\mathrm{x}}\left(-12\mathrm{sin}5\mathrm{x}+5\mathrm{cos}5\mathrm{x}\right)\end{array}$

Q.6

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{second}\mathrm{ }\mathrm{order}\mathrm{ }\mathrm{derivative}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{function}:\\ {\mathrm{e}}^{6\mathrm{x}}\mathrm{ }\mathrm{cos}3\mathrm{x}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{y}={\mathrm{e}}^{6\mathrm{x}}\mathrm{cos}3\mathrm{x}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{6\mathrm{x}}\mathrm{cos}3\mathrm{x}\right)\\ ={\mathrm{e}}^{6\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cos}3\mathrm{x}+\mathrm{cos}3\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{6\mathrm{x}}\left[\mathrm{By}\mathrm{product}\mathrm{rule}\right]\\ ={\mathrm{e}}^{6\mathrm{x}}\left(-3\mathrm{sin}3\mathrm{x}\right)+\mathrm{cos}3\mathrm{x}.6{\mathrm{e}}^{6\mathrm{x}}\\ ={\mathrm{e}}^{6\mathrm{x}}\left(6\mathrm{cos}3\mathrm{x}-3\mathrm{sin}3\mathrm{x}\right)\\ \mathrm{Differentiating}\mathrm{ }\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \therefore \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{{\mathrm{e}}^{6\mathrm{x}}\left(6\mathrm{cos}3\mathrm{x}-3\mathrm{sin}3\mathrm{x}\right)\right\}\\ \mathrm{ }={\mathrm{e}}^{6\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(6\mathrm{cos}3\mathrm{x}-3\mathrm{sin}3\mathrm{x}\right)+\left(6\mathrm{cos}3\mathrm{x}-3\mathrm{sin}3\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{6\mathrm{x}}\\ \left[\mathrm{By}\mathrm{product}\mathrm{rule}\right]\\ \mathrm{ }={\mathrm{e}}^{\mathrm{x}}\left(-18\mathrm{sin}3\mathrm{x}-9\mathrm{cos}3\mathrm{x}\right)+\left(6\mathrm{cos}3\mathrm{x}-3\mathrm{sin}3\mathrm{x}\right)6{\mathrm{e}}^{6\mathrm{x}}\\ \mathrm{ }={\mathrm{e}}^{6\mathrm{x}}\left(-18\mathrm{sin}3\mathrm{x}-9\mathrm{cos}3\mathrm{x}+36\mathrm{cos}3\mathrm{x}-18\mathrm{sin}3\mathrm{x}\right)\\ \mathrm{ }={\mathrm{e}}^{6\mathrm{x}}\left(-36\mathrm{sin}3\mathrm{x}+27\mathrm{cos}3\mathrm{x}\right)\\ \mathrm{ }=9{\mathrm{e}}^{6\mathrm{x}}\left(-4\mathrm{sin}3\mathrm{x}+3\mathrm{cos}3\mathrm{x}\right)\\ \mathrm{Thus},\mathrm{second}\mathrm{derivative}\mathrm{of}\mathrm{given}\mathrm{function}\mathrm{is}\\ 9{\mathrm{e}}^{6\mathrm{x}}\left(3\mathrm{cos}3\mathrm{x}-4\mathrm{sin}3\mathrm{x}\right)\mathrm{.}\end{array}$

Q.7

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{second}\mathrm{ }\mathrm{order}\mathrm{ }\mathrm{derivative}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{function}:\\ {\mathrm{tan}}^{-1}\mathrm{x}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{y}={\mathrm{tan}}^{-1}\mathrm{x}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}\mathrm{x}\\ =\frac{1}{1+{\mathrm{x}}^{2}}\\ \mathrm{Differentiating}\mathrm{ }\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \therefore \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{1+{\mathrm{x}}^{2}}\right)\\ \mathrm{ }=\frac{\mathrm{d}}{\mathrm{dx}}{\left(1+{\mathrm{x}}^{2}\right)}^{-1}\\ \mathrm{ }=-{\left(1+{\mathrm{x}}^{2}\right)}^{-2}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+{\mathrm{x}}^{2}\right)\\ =-{\left(1+{\mathrm{x}}^{2}\right)}^{-2}\left(0+2\mathrm{x}\right)\\ =-\frac{2\mathrm{x}}{{\left(1+{\mathrm{x}}^{2}\right)}^{2}}\\ \mathrm{Thus},\mathrm{second}\mathrm{derivative}\mathrm{of}\mathrm{given}\mathrm{function}\mathrm{is}-\frac{2\mathrm{x}}{{\left(1+{\mathrm{x}}^{2}\right)}^{2}}\mathrm{.}\end{array}$

Q.8

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{scond}\mathrm{orderderivative}\mathrm{of}\mathrm{the}\mathrm{function}:\\ \mathrm{log}\left(\mathrm{logx}\right)\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{y}=\mathrm{log}\left(\mathrm{logx}\right)\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{log}\left(\mathrm{logx}\right)\\ =\frac{1}{\mathrm{logx}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\left[\mathrm{By} \mathrm{chain} \mathrm{rule}\right]\\ =\frac{1}{\mathrm{logx}}\left(\frac{1}{\mathrm{x}}\right)\\ =\frac{1}{\mathrm{xlogx}}\\ \mathrm{Differentiating}\mathrm{ }\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \therefore \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{xlogx}}\right)\\ \mathrm{ }=\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{xlogx}\right)}^{-1}\\ \mathrm{ }=-{\left(\mathrm{xlogx}\right)}^{-2}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{xlogx}\right)\left[\mathrm{By} \mathrm{chain} \mathrm{rule}\right]\\ \mathrm{ }=-\frac{1}{{\left(\mathrm{xlogx}\right)}^{2}}\left(\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}+\mathrm{logx}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\right)\\ \mathrm{ }=-\frac{1}{{\left(\mathrm{xlogx}\right)}^{2}}\left(\mathrm{x}×\frac{1}{\mathrm{x}}+\mathrm{logx}×1\right)\\ \mathrm{ }=-\frac{1}{{\left(\mathrm{xlogx}\right)}^{2}}\left(1+\mathrm{logx}\right)\\ \mathrm{ }=-\frac{\left(1+\mathrm{logx}\right)}{{\left(\mathrm{xlogx}\right)}^{2}}\\ \mathrm{Thus},\mathrm{second}\mathrm{derivative}\mathrm{of}\mathrm{given}\mathrm{function}\mathrm{is}-\frac{\left(1+\mathrm{logx}\right)}{{\left(\mathrm{xlogx}\right)}^{2}}\mathrm{.}\end{array}$

Q.9 Find the second order derivative of the function:
sin (log x)

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{y}=\mathrm{sin}\left(\mathrm{logx}\right)\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sin}\left(\mathrm{logx}\right)\\ =\mathrm{cos}\left(\mathrm{logx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\left[\mathrm{By} \mathrm{chain} \mathrm{rule}\right]\\ =\mathrm{cos}\left(\mathrm{logx}\right)\left(\frac{1}{\mathrm{x}}\right)\\ =\frac{\mathrm{cos}\left(\mathrm{logx}\right)}{\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{ }\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \therefore \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{cos}\left(\mathrm{logx}\right)}{\mathrm{x}}\right)\\ \mathrm{ }=\frac{\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cos}\left(\mathrm{logx}\right)-\mathrm{cos}\left(\mathrm{logx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}}{{\mathrm{x}}^{2}}\\ \mathrm{ }=\frac{\mathrm{x}×-\mathrm{sin}\left(\mathrm{logx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}-\mathrm{cos}\left(\mathrm{logx}\right)×1}{{\mathrm{x}}^{2}}\left[\mathrm{By} \mathrm{chain} \mathrm{rule}\right]\\ \mathrm{ }=\frac{-\mathrm{xsin}\left(\mathrm{logx}\right)×\frac{1}{\mathrm{x}}-\mathrm{cos}\left(\mathrm{logx}\right)}{{\mathrm{x}}^{2}}\\ \mathrm{ }=\frac{-\left\{\mathrm{sin}\left(\mathrm{logx}\right)+\mathrm{cos}\left(\mathrm{logx}\right)\right\}}{{\mathrm{x}}^{2}}\\ \mathrm{Thus},\mathrm{second}\mathrm{derivative}\mathrm{of}\mathrm{given}\mathrm{function}\mathrm{is}\\ \frac{-\left\{\mathrm{sin}\left(\mathrm{logx}\right)+\mathrm{cos}\left(\mathrm{logx}\right)\right\}}{{\mathrm{x}}^{2}}\mathrm{.}\end{array}$

Q.10

$\text{If}\phantom{\rule{0ex}{0ex}}\mathrm{y}=5\mathrm{cosx}-3\phantom{\rule{0ex}{0ex}}\mathrm{sinx},\phantom{\rule{0ex}{0ex}}\mathrm{prove}\phantom{\rule{0ex}{0ex}}\mathrm{that}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{2}}+\mathrm{y}=0$

Ans

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{y}=5\mathrm{cosx}-3\mathrm{sinx}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(5\mathrm{cosx}-3\mathrm{sinx}\right)\\ =5\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}-3\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\\ =-5\mathrm{sinx}-3\mathrm{cosx}\\ \mathrm{Again}\mathrm{differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=-\frac{\mathrm{d}}{\mathrm{dx}}\left(5\mathrm{sinx}+3\mathrm{cosx}\right)\\ =-\left(5\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}+3\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}\right)\\ =-\left(5\mathrm{cosx}-3\mathrm{sinx}\right)\\ =-\mathrm{y}\\ \therefore \mathrm{ }\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}+\mathrm{y}=0\\ \mathrm{Hence}\mathrm{Proved}\mathrm{.}\end{array}$

Q.11

$\text{If}\phantom{\rule{0ex}{0ex}}\mathrm{y}=\mathrm{co}{\mathrm{s}}^{-1}\mathrm{x}.\phantom{\rule{0ex}{0ex}}\mathrm{Find}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{in}\phantom{\rule{0ex}{0ex}}\mathrm{term}\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{o}\mathrm{f}\phantom{\rule{0ex}{0ex}}\mathrm{y}\phantom{\rule{0ex}{0ex}}\mathrm{alone}.$

Ans

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{y}={\mathrm{cos}}^{-1}\mathrm{x}\\ \mathrm{cosy}=\mathrm{x}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{y},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{cosy}=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{x}\\ ⇒ \mathrm{ }\frac{\mathrm{dx}}{\mathrm{dy}}=-\mathrm{siny}\\ ⇒ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{\mathrm{siny}}\\ ⇒ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{cosecy} ...\left(\mathrm{i}\right)\\ \mathrm{Again}\mathrm{differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(-\mathrm{cosecy}\right)\\ =\mathrm{cosecycoty}\frac{\mathrm{dy}}{\mathrm{dx}}\\ =\mathrm{cosecycoty}×-\mathrm{cosecy}\\ =-{\mathrm{cosec}}^{2}\mathrm{ycoty}\left[\mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\right]\end{array}$

Q.12

$\text{If}\phantom{\rule{0ex}{0ex}}\mathrm{y}=3\mathrm{cos}\left(\mathrm{logx}\right)+4\mathrm{sin}\left(\mathrm{logx}\right),\mathrm{show}\phantom{\rule{0ex}{0ex}}\mathrm{that}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}^{2}{\mathrm{y}}_{2}+\mathrm{x}{\mathrm{y}}_{1}+\mathrm{y}=0$

Ans

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{y}=3\mathrm{cos}\left(\mathrm{logx}\right)+4\mathrm{sin}\left(\mathrm{logx}\right)\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\left(={\mathrm{y}}_{1}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{3\mathrm{cos}\left(\mathrm{logx}\right)+4\mathrm{sin}\left(\mathrm{logx}\right)\right\}\\ =\mathrm{3}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cos}\left(\mathrm{logx}\right)+ 4\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sin}\left(\mathrm{logx}\right)\\ =-3\mathrm{sin}\left(\mathrm{logx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}+4\mathrm{cos}\left(\mathrm{logx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\left[\mathrm{By}\mathrm{chain}\mathrm{rule}\right]\\ =-3\mathrm{sin}\left(\mathrm{logx}\right)\frac{1}{\mathrm{x}}+4\mathrm{cos}\left(\mathrm{logx}\right)\frac{1}{\mathrm{x}}\\ =\frac{-3\mathrm{sin}\left(\mathrm{logx}\right)+4\mathrm{cos}\left(\mathrm{logx}\right)}{\mathrm{x}}\\ \mathrm{Again}\mathrm{differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}\left({\mathrm{y}}_{2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{4\mathrm{cos}\left(\mathrm{logx}\right)-3\mathrm{sin}\left(\mathrm{logx}\right)}{\mathrm{x}}\right)\\ =\frac{\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\left\{4\mathrm{cos}\left(\mathrm{logx}\right)-3\mathrm{sin}\left(\mathrm{logx}\right)\right\}-\left\{4\mathrm{cos}\left(\mathrm{logx}\right)-3\mathrm{sin}\left(\mathrm{logx}\right)\right\}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}}{{\mathrm{x}}^{2}}\\ =\frac{\mathrm{x}\left\{-4\mathrm{sin}\left(\mathrm{logx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}-3\mathrm{cos}\left(\mathrm{logx}\right)\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\right\}-\left\{4\mathrm{cos}\left(\mathrm{logx}\right)-3\mathrm{sin}\left(\mathrm{logx}\right)\right\}×1}{{\mathrm{x}}^{2}}\\ =\frac{\mathrm{x}\left\{-4\mathrm{sin}\left(\mathrm{logx}\right)\frac{1}{\mathrm{x}}-3\mathrm{cos}\left(\mathrm{logx}\right)\frac{1}{\mathrm{x}}\right\}-\left\{4\mathrm{cos}\left(\mathrm{logx}\right)-3\mathrm{sin}\left(\mathrm{logx}\right)\right\}}{{\mathrm{x}}^{2}}\\ =\frac{-4\mathrm{sin}\left(\mathrm{logx}\right)-3\mathrm{cos}\left(\mathrm{logx}\right)-4\mathrm{cos}\left(\mathrm{logx}\right)+3\mathrm{sin}\left(\mathrm{logx}\right)}{{\mathrm{x}}^{2}}\\ =\frac{-\mathrm{sin}\left(\mathrm{logx}\right)-7\mathrm{cos}\left(\mathrm{logx}\right)}{{\mathrm{x}}^{2}}\\ \mathrm{Thus},\\ \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}={\mathrm{y}}_{2}=\frac{-\mathrm{sin}\left(\mathrm{logx}\right)-7\mathrm{cos}\left(\mathrm{logx}\right)}{{\mathrm{x}}^{2}}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.={\mathrm{x}}^{2}{\mathrm{y}}_{2}+{\mathrm{xy}}_{1}+\mathrm{y}\\ ={\mathrm{x}}^{2}\left(\frac{-\mathrm{sin}\left(\mathrm{logx}\right)-7\mathrm{cos}\left(\mathrm{logx}\right)}{{\mathrm{x}}^{2}}\right)+\mathrm{x}\left(\frac{-3\mathrm{sin}\left(\mathrm{logx}\right)+4\mathrm{cos}\left(\mathrm{logx}\right)}{\mathrm{x}}\right)\\ +3\mathrm{cos}\left(\mathrm{logx}\right)+4\mathrm{sin}\left(\mathrm{logx}\right)\\ =-\mathrm{sin}\left(\mathrm{logx}\right)-7\mathrm{cos}\left(\mathrm{logx}\right)-3\mathrm{sin}\left(\mathrm{logx}\right)+4\mathrm{cos}\left(\mathrm{logx}\right)+3\mathrm{cos}\left(\mathrm{logx}\right)\\ +\mathrm{ }4\mathrm{sin}\left(\mathrm{logx}\right)\\ =-4\mathrm{sin}\left(\mathrm{logx}\right)+\mathrm{ }4\mathrm{sin}\left(\mathrm{logx}\right)-7\mathrm{cos}\left(\mathrm{logx}\right)+7\mathrm{cos}\left(\mathrm{logx}\right)\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence}\mathrm{proved}\mathrm{.}\end{array}$

Q.13

$\text{If}\phantom{\rule{0ex}{0ex}}\mathrm{y}=\mathrm{A}{\mathrm{e}}^{\mathrm{mx}}+\mathrm{B}{\mathrm{e}}^{\mathrm{nx}},\mathrm{show}\phantom{\rule{0ex}{0ex}}\mathrm{that}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{2}}-\left(\mathrm{m}+\mathrm{n}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{mny}=0$

Ans

$\begin{array}{l}Giventhat\\ y=A{e}^{mx}+B {e}^{nx}\\ Differentiatingw.r.t.x,weget\\ \frac{dy}{dx}=A\frac{d}{dx}{e}^{mx}+B\frac{d}{dx} {e}^{nx}\\ =Am{e}^{mx}+Bn {e}^{nx}\\ Again differentiatingw.r.t.x,weget\\ \frac{{d}^{2}y}{d{x}^{2}}=Am\frac{d}{dx}{e}^{mx}+Bn \frac{d}{dx}{e}^{nx}\\ =A{m}^{2}{e}^{mx}+B{n}^{2}{e}^{nx}\\ L.H.S.=\frac{{d}^{2}y}{d{x}^{2}}-\left(m+n\right)\frac{dy}{dx}+mny\\ =\left(A{m}^{2}{e}^{mx}+B{n}^{2}{e}^{nx}\right)-\left(m+n\right)\left(Am{e}^{mx}+Bn {e}^{nx}\right)+mn\left(A{e}^{mx}+B {e}^{nx}\right)\\ =A{m}^{2}{e}^{mx}+B{n}^{2}{e}^{nx}-A{m}^{2}{e}^{mx}-Bmn {e}^{nx}-Amn{e}^{mx}-B{n}^{2} {e}^{nx}\\ +Amn{e}^{mx}+Bmn {e}^{nx}\\ =0=R.H.S.\end{array}$

Q.14

$\text{If}\phantom{\rule{0ex}{0ex}}\mathrm{y}=500{\mathrm{e}}^{7\mathrm{x}}+600\phantom{\rule{0ex}{0ex}}{\mathrm{e}}^{-7\mathrm{x}},\mathrm{show}\phantom{\rule{0ex}{0ex}}\mathrm{that}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{2}}=49\mathrm{y}$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{that}\\ \mathrm{y}=500{\mathrm{e}}^{7\mathrm{x}}+600\mathrm{ }{\mathrm{e}}^{–7\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{500}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{7\mathrm{x}}\mathrm{}+\mathrm{600}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{ }{\mathrm{e}}^{–7\mathrm{x}}\\ =\mathrm{500}×7{\mathrm{e}}^{7\mathrm{x}}+\mathrm{600}×\left(-7\right)\mathrm{ }{\mathrm{e}}^{–7\mathrm{x}}\\ =7\left(500{\mathrm{e}}^{7\mathrm{x}}-600 {\mathrm{e}}^{–7\mathrm{x}}\right)\\ \mathrm{Again}\mathrm{ }\mathrm{differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=7\left(\mathrm{500}×7{\mathrm{e}}^{7\mathrm{x}}-\mathrm{600}×\left(-7\right)\mathrm{ }{\mathrm{e}}^{–7\mathrm{x}}\right)\\ ⇒\mathrm{ }\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=49\left(500{\mathrm{e}}^{7\mathrm{x}}+600 {\mathrm{e}}^{–7\mathrm{x}}\right)\\ ⇒\mathrm{ }\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=49\mathrm{y}\\ \mathrm{Hence}\mathrm{proved}\mathrm{.}\end{array}$

Q.15

$\text{If}\phantom{\rule{0ex}{0ex}}{\mathrm{e}}^{\mathrm{y}}\left(\mathrm{x}+1\right)=1,\mathrm{show}\phantom{\rule{0ex}{0ex}}\mathrm{that}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{2}}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{\mathbf{2}}$

Ans

$\begin{array}{l}Given\text{that}\\ \text{}\text{}{e}^{y}\left(x+1\right)=1\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{e}^{y}=\frac{1}{x+1}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=\mathrm{log}\left(\frac{1}{x+1}\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=-\mathrm{log}\left(x+1\right)\\ Differentiating\text{w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{dy}{dx}=\frac{d}{dx}\left(-\mathrm{log}\left(x+1\right)\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{d}{dx}\mathrm{log}\left(x+1\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{1}{x+1}\frac{d}{dx}\left(x+1\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{1}{x+1}×\left(1+0\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{1}{x+1}\\ Again\text{\hspace{0.17em}}differentiating\text{w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}\left(-\frac{1}{x+1}\right)\\ ⇒\text{\hspace{0.17em}}\frac{{d}^{2}y}{d{x}^{2}}=-\frac{d}{dx}{\left(x+1\right)}^{-1}\\ ⇒\text{\hspace{0.17em}}\frac{{d}^{2}y}{d{x}^{2}}=-\left(-1\right){\left(x+1\right)}^{-2}\frac{d}{dx}\left(x+1\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{{\left(x+1\right)}^{2}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(-\frac{1}{x+1}\right)}^{2}\\ ⇒\text{\hspace{0.17em}}\frac{{d}^{2}y}{d{x}^{2}}={\left(\frac{dy}{dx}\right)}^{2}\\ Hence\text{proved}\text{.}\end{array}$

Q.16

$\text{If}\phantom{\rule{0ex}{0ex}}\mathrm{y}={\left(\mathrm{ta}{\mathrm{n}}^{-1}\mathrm{x}\right)}^{2},\phantom{\rule{0ex}{0ex}}\mathrm{show}\phantom{\rule{0ex}{0ex}}\mathrm{that}\phantom{\rule{0ex}{0ex}}\left({\mathrm{x}}^{2}+1\right){\mathrm{y}}_{2}+2\mathrm{x}\left({\mathrm{x}}^{2}+1\right){\mathrm{y}}_{1}=\mathbf{2}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{is}\\ \mathrm{ }\mathrm{y}={\left({\mathrm{tan}}^{-1}\mathrm{x}\right)}^{2}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\left({\mathrm{tan}}^{-1}\mathrm{x}\right)}^{2}\\ =2{\mathrm{tan}}^{-1}\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}\mathrm{x}\\ \mathrm{ }{\mathrm{y}}_{1}=2{\mathrm{tan}}^{-1}\mathrm{x}\left(\frac{1}{1+{\mathrm{x}}^{2}}\right)\\ \left(1+{\mathrm{x}}^{2}\right){\mathrm{y}}_{1}=2{\mathrm{tan}}^{-1}\mathrm{x}\\ \mathrm{Differentiating}\mathrm{ }\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left\{\left(1+{\mathrm{x}}^{2}\right){\mathrm{y}}_{1}\right\}=2\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}\mathrm{x}\\ \left(1+{\mathrm{x}}^{2}\right)\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}_{1}+{\mathrm{y}}_{1}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+{\mathrm{x}}^{2}\right)=2×\frac{1}{1+{\mathrm{x}}^{2}}\\ \mathrm{ }\left(1+{\mathrm{x}}^{2}\right){\mathrm{y}}_{2}+{\mathrm{y}}_{1}\left(0+2\mathrm{x}\right)=2×\frac{1}{1+{\mathrm{x}}^{2}}\\ \mathrm{ }\left(1+{\mathrm{x}}^{2}\right){\mathrm{y}}_{2}+2{\mathrm{xy}}_{1}=\frac{2}{1+{\mathrm{x}}^{2}}\\ {\left(1+{\mathrm{x}}^{2}\right)}^{2}{\mathrm{y}}_{2}+2\mathrm{x}\left(1+{\mathrm{x}}^{2}\right){\mathrm{y}}_{1}=2\\ ⇒{\left({\mathrm{x}}^{2}+1\right)}^{2}{\mathrm{y}}_{2}+2\mathrm{x}\left({\mathrm{x}}^{2}+1\right){\mathrm{y}}_{1}=2\\ \mathrm{Hence}\mathrm{proved}\mathrm{.}\end{array}$

Q.17

$\begin{array}{l}\mathrm{VerifyRollesTheoremforthe}function\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\\ \mathrm{f}\left(\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}=\phantom{\rule{0ex}{0ex}}{\mathrm{x}}^{2}+2\mathrm{x}-8,\phantom{\rule{0ex}{0ex}}\mathrm{x}\phantom{\rule{0ex}{0ex}}\in \phantom{\rule{0ex}{0ex}}\left[-4,2\right].\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function},\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{2}}+2\mathrm{x}-8,\mathrm{is}\mathrm{a}\mathrm{polynomial}.\mathrm{So},\mathrm{it}\\ \mathrm{is}\mathrm{continuous}\mathrm{in}\left[-4,\mathrm{ }2\right]\mathrm{and}\mathrm{is}\mathrm{differentiable}\mathrm{in}\left(-4,2\right).\\ \mathrm{f}\left(-4\right)={\left(-4\right)}^{\mathrm{2}}+\mathrm{2}\left(-4\right)-\mathrm{8}\\ \mathrm{ }=\mathrm{16}-8-8=0\\ \mathrm{f}\left(2\right)={\left(2\right)}^{\mathrm{2}}+\mathrm{2}\left(2\right)-\mathrm{8}\\ \mathrm{ }=\mathrm{4}-4-8=0\\ \therefore \mathrm{f}\left(-4\right)=\mathrm{f}\left(2\right)=0\\ ⇒\mathrm{The}\mathrm{value}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)\mathrm{at}-4\mathrm{and}2\mathrm{coincide}\mathrm{.}\\ \mathrm{Rolle}‘\mathrm{s}\mathrm{Theorem}\mathrm{states}\mathrm{that}\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{c}\in \mathrm{\left(}-4, 2\right)\mathrm{such}\\ \mathrm{that}\mathrm{ }\mathrm{f}‘\left(\mathrm{c}\right)=0\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{2}}+2\mathrm{x}-\mathrm{8}\\ \mathrm{Differentiating}\mathrm{f}\left(\mathrm{x}\right)\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{2}}+2\mathrm{x}-\mathrm{8}\right)\\ \mathrm{ }=2\mathrm{x}+2\\ \mathrm{f}‘\left(\mathrm{c}\right)=2\mathrm{c}+2\\ \because \mathrm{f}‘\left(\mathrm{c}\right)=0⇒2\mathrm{c}+2=0\\ ⇒\mathrm{c}=-1\in \mathrm{\left(}-4, 2\right)\\ \mathrm{Hence},\mathrm{Rolle}‘\mathrm{s}\mathrm{Theorem}\mathrm{is}\mathrm{verified}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{.}\end{array}$

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### 2. Why are the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 more important than others?

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