NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives (Ex 13.2) Exercise 13.2

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Limits And Derivatives Chapter of Class 11 is part of the calculus in Mathematics. Calculus is introduced to students in Class 11 through this chapter. The limits and derivatives chaptermight seem like a short chapter, to students if they look at the number of exercises in the chapter which is 2 exercises excluding the miscellaneous exercises. But this chapter is a really important one because it introduces new types of questions that students have not studied before. It is important for students to understand the importance of learning this chapter as they will have to study calculus in Class 12 as well. Hence students must have a basic understanding of this chapter by practising the Class 11 Maths Chapter 13 Exercise 13.2. The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 could be used for reference for this exercise.

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Students in Class 11 are introduced to many new topics, like Sets, three-dimensional geometry, complex numbers, limits and derivatives. It is important for students to study these topics thoroughly as it will help them solve questions later. Solving NCERT exercises of these chapters is very important for students to learn from the basics, this will help them solve advanced-level questions later. Hence, students must practise Chapter Limits And Derivatives thoroughly to gain knowledge. They can take help from NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 if they have any doubts; it is available on Extramarks’ website and app.NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives (Ex 13.2) Exercise 13.2[Include Download PDF Button]

Practising NCERT questions is important for anyone who is preparing for board exams or entrance exams. Practising NCERT questions will assist students in strengthening their understanding of a topic as well as laying the groundwork for concepts.The practise problems given in NCERT examples and NCERT exercises can help students. Students must remember that practising can help them eliminate their mistakes and can help them gain in-depth knowledge about the topic.

The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 can help students practise the NCERT exercises and understand the questions, which will help them answer questions from other books and test papers. The students should use the Extramarks study material for studying and the NCERT Solutions provided on their website and mobile application. The NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.2 are available to students via the Extramarks application and website, along with other study materials and resources.

The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 are for students who find solving questions difficult. The solutions provided by Extramarks will help them understand the solution to each question and find the correct answer. The correct answers to the questions in Class 11 Maths Exercise 13.2 can be found in the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2. This will help students in every way to get better at studying Mathematics and scoring good marks.

Access NCERT Solutions for Class-11 Maths Chapter- 13- Limits and Derivatives

Students can access the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 from the Extramarks’ website and app. The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 are available to students at any time while they are studying. Students must realise the value of practising Mathematics. It is possible that students’ responses contain careless errors. Even when students follow the proper technique for solving the questions, silly misunderstandings might lead to incorrect answers. Students can avoid making these errors if they practise enough. They must recognise that making mistakes is a typical part of learning, but in order to improve, they must acknowledge and fix them. Students can get assistance from the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2.

Practice is necessary to avoid errors since it enables students to recognise their faults and be attentive to them during the exam. The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 contains the proper solutions. Students should keep in mind that the purpose of the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 is to allow students to apply what they have learned in Class and during their study of these topics.

NCERT Solutions for Class 11 Maths Chapters

On Extramarks’ website and app, students can find NCERT solutions for Class 11 for every chapter. Students should keep in mind that working through NCERT books can help them perform better on entrance exams like JEE, CUET, and others as well as in board exams. NCERT textbooks can help students improve their fundamental understanding of the chapter as well as understand how the questions should be answered. Students can find NCERT solutions for different subjects and Classes like NCERT Solutions Class 10, NCERT Solutions Class 9 and NCERT Solutions Class 8.

To have a thorough understanding of the chapter, the students must look at all the example questions and exercise questions provided in the NCERT textbooks. Whether it is an entrance exam or an end-of-term exam, this is a crucial component of exam preparation. If students have trouble understanding the approaches and solutions that NCERT has given them through the example questions. Before beginning to answer the problems in chapter Limits and Derivatives, students must have a conceptual understanding of the idea. The questions cannot be answered by students using their existing knowledge of the fundamentals. To be able to handle complex questions for the entrance exam, they need to have an in-depth understanding of the chapter.

NCERT Solution Class 11 Maths of Chapter 13 All Exercises

Students can benefit from reading the NCERT textbooks. NCERT textbooks are particularly important to students who are getting ready for the CBSE board exams. For students preparing for any exam, including entrance examinations and board exams, especially the CBSE board exam, teachers advice practising NCERT questions. Additionally, some students might want to practise for tests by answering more questions. Extramarks offers course-specific study materials as well as entrance exam preparation for tests like JEE, NEET, CUET, etc. The SAT, PSAT, and TOEFL are just a few of the international examinations that Extramarks offers study materials for. Extramarks can provide assistance to students getting ready for any exam. The resources on Extramarks’ website and application must be utilised by students.

For other classes, Extramarks has also prepared various study materials. NCERT Solutions Class 7, NCERT Solutions Class 6, NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2 and NCERT Solutions Class 1 are also available to students through Extramarks’ website and app, both of which are available in Hindi and English, in addition to the NCERT Solutions Class 11 and NCERT Solutions Class 12.

All Topics and Important Formulae Under NCERT Class 11 Maths for Exercise 13.2

The Chapter Limits And Derivatives of NCERT Class 12 teach students some very important topics. Some of these include limits, left-hand limit and right-hand limit, the existence of limit, some properties of limits, some standard limits, derivatives, fundamental derivative rule of functions and some standard derivatives. These topics could be difficult for students to study and understand properly. Students find them very challenging, as it is a new topic in the subject. Students might need time to understand the fundamentals of this topic. Students must focus on studying all these topics from Chapter Limits And Derivatives.

The topics that Class 11 Maths Exercise 13.2 Solutions are based on are derivatives, the fundamental derivative rule of functions, and some standard derivatives. These topics can seem difficult to students, but with practise, theycan perform well in any test of this chapter. The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 can help students figure out how to understand these topics. The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 will help students because it provides explanations for the answers it provides to them.

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.2

The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 offer accurate solutions along with explanations to help students grasp the ideas presented in Exercise 13.2. Every NCERT exercise, including Exercise 13.2, which has solutions like the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2, should be remembered by students. The material covered in Exercise 13.2 is crucial for students to comprehend since it will enable them to decide how various questions should be answered. They can use this to their advantage when solving issues where the question only provides limited details. Knowing the process in different situations could benefit students. Students must therefore learn and retain this information to perform excellently in exams. They should get assistance from the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2. Reading through the Class 11 Maths Exercise 13.2 Solutions can clear up any questions students have about Exercise 13.2. Exercise 13.2’s concepts are essential for students’ board and entrance exams. Therefore, it is important for students to understand how different questions should be solved, as this can help with their academic performance.Students must use any help they can get from Extramarks if need be. They can refer to the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.2 for help with their examination.

Q.1 Find the derivative of x² – 2 at x = 10.

Ans.

$\begin{array}{l} We are given: \\ f (x) = x^{2} - 2 \\ Since, f ‘ (10) = \lim_{h \to 0} \frac{f (10 + h) - f (10)}{h} \\ ∴ f ‘ (10) = \lim_{h \to 0} \frac{{{(10 + h)}^{2} - 2} - {{(10)}^{2} - 2}}{h} \\ = \lim_{h \to 0} \frac{(100 + 20 h + h^{2} - 2) - (100 - 2)}{h} \\ = \lim_{h \to 0} \frac{98 + 20 h + h^{2} - 98}{h} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \lim_{h \to 0} \frac{h (20 + h)}{h} \\ f ‘ (10) = 20 \\ Thus, {the derivative of the function x}^{2} - 2 at x = 10 is 20. \end{array}$

Q.2 Find the derivative of 99x at x = l00.

Ans.

$\begin{array}{l} We are given: f (x) = 99x \\ Since, f ‘ (100) = \lim_{h \to 0} \frac{f (100 + h) - f (100)}{h} \\ ∴ f ‘ (100) = \lim_{h \to 0} \frac{99 (100 + h) - 99 (100)}{h} \\ = \lim_{h \to 0} \frac{9900 + 99 h - 9900}{h} \\ = \lim_{h \to 0} \frac{99 h}{h} \\ f ‘ (100) = 99 \\ Thus, the derivative of the function 99x at x= 100 is 99. \end{array}$

Q.3 Find the derivative of x at x = 1.

Ans.

$\begin{array}{l} We are given: \\ f (x) = x \\ Since, f ‘ (1) = \underset{}{\lim_{h \to 0}} \frac{f (1 + h) - f (1)}{h} \\ ∴ f ‘ (1) = \lim_{h \to 0} \frac{(1 + h) - (1)}{h} \\ = \lim_{h \to 0} \frac{1 + h - 1}{h} \\ = \lim_{h \to 0} \frac{h}{h} \\ f ‘ (1) = 1 \\ Thus, the derivative of the function x at x = 1 is 1. \end{array}$

Q.4 Find the derivative of the following functions from first principle.

$\begin{array}{l} (i) x^{3} - 27 (ii) (x - 1) (x - 2) \\ (iii) \frac{1}{x^{2}} (iv) \frac{x + 1}{x - 1} \end{array}$

Ans.

$\begin{array}{l} Since, \\ f ‘ (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ (i) f (x) = x^{3} - 27 \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{{{(x + h)}^{3} - 27} - (x^{3} - 27)}{h} \\ = \lim_{h \to 0} \frac{x^{3} + 3 x^{2} h + 3 {xh}^{2} + h^{3} - 27 - x^{3} + 27}{h} \\ = \lim_{h \to 0} \frac{3 x^{2} h + 3 {xh}^{2} + h^{3}}{h} \\ = \lim_{h \to 0} \frac{(3 x^{2} + 3 xh + h^{2}) h}{h} \\ = 3 x^{2} + 3 x (0) + {(0)}^{2} \\ f ‘ (x) = 3 x^{2} \end{array}$

$\begin{array}{l} (ii) f (x) = (x - 1) (x - 2) \\ = x^{2} - 3 x + 2 \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{{(x + h)}^{2} - 3 (x + h) + 2 - (x^{2} - 3 x + 2)}{h} \\ = \lim_{h \to 0} \frac{x^{2} + 2 xh + h^{2} - 3 x - 3 h + 2 - x^{2} + 3 x - 2}{h} \\ = \lim_{h \to 0} \frac{2 xh + h^{2} - 3 h}{h} \\ = \lim_{h \to 0} \frac{(2 x + h - 3) h}{h} \\ f ‘ (x) = 2 x - 3 \\ (iii) f (x) = \frac{1}{x^{2}} \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{\frac{1}{{(x + h)}^{2}} - \frac{1}{x^{2}}}{h} \\ = \lim_{h \to 0} \frac{x^{2} - {(x + h)}^{2}}{h {(x + h)}^{2} x^{2}} \\ = \lim_{h \to 0} \frac{x^{2} - x^{2} - 2 xh - h^{2}}{h {(x + h)}^{2} x^{2}} \\ = \lim_{h \to 0} \frac{- 2 xh - h^{2}}{h {(x + h)}^{2} x^{2}} \\ = \lim_{h \to 0} \frac{- h (2 x + h)}{h {(x + h)}^{2} x^{2}} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{- (2 x + 0)}{{(x + 0)}^{2} x^{2}} \\ = - \frac{2 x}{x^{4}} \\ f ‘ (x) = - \frac{2}{x^{3}} \\ (iv) f (x) = \frac{x + 1}{x - 1} \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{\frac{x + h + 1}{(x + h - 1)} - \frac{x + 1}{(x - 1)}}{h} \\ = \lim_{h \to 0} \frac{(x + h + 1) (x - 1) - (x + 1) (x + h - 1)}{h (x + h - 1) (x - 1)} \\ = \lim_{h \to 0} \frac{(x^{2} + xh + x - x - h - 1) - (x^{2} + xh - x + x + h - 1)}{h (x + h - 1) (x - 1)} \\ = \lim_{h \to 0} \frac{x^{2} + xh + x - x - h - 1 - x^{2} - xh + x - x - h + 1}{h (x + h - 1) (x - 1)} \\ = \lim_{h \to 0} \frac{- 2 h}{h (x + h - 1) (x - 1)} \\ = \frac{- 2}{(x + 0 - 1) (x - 1)} \\ f ‘ (x) = \frac{- 2}{{(x - 1)}^{2}} \end{array}$

Q.5

For the function f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . .. + \frac{x^{2}}{2} + x + 1 Prove that f ‘ (1) = 100 f ‘ (0) .

Ans.

$\begin{array}{l} We have, \\ f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . .. + \frac{x^{2}}{2} + x + 1 \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} (\frac{x^{100}}{100} + \frac{x^{99}}{99} + . .. + \frac{x^{2}}{2} + x + 1) \\ = \frac{d}{dx} \frac{x^{100}}{100} + \frac{d}{dx} \frac{x^{99}}{99} + . .. + \frac{d}{dx} \frac{x^{2}}{2} + \frac{d}{dx} x + \frac{d}{dx} 1 \\ = \frac{100 x^{99}}{100} + \frac{99 x^{98}}{99} + . .. + \frac{2 x^{1}}{2} + 1 + 0 [\frac{d}{dx} x^{n} = {nx}^{n - 1}] \\ f ‘ (x) = x^{99} + x^{98} + . .. + x + 1 \\ Substituting x = 1, we get \\ L . H . S . : f ‘ (1) = {(1)}^{99} + {(1)}^{98} + . .. + (1) + 1 \\ = 100 \\ Substituting x = 0, we get \\ f ‘ (0) = {(0)}^{99} + {(0)}^{98} + . .. + (0) + 1 \\ = 1 \\ R . H . S . : \\ 100 f ‘ (0) = 100 (1) \\ = 100 \end{array}$

$\begin{array}{l} Thus, f ‘ (1) = 100 f ‘ (0) . \\ Therefore, it is proved. \end{array}$

Q.6 Find the derivative of xⁿ + axⁿ⁻¹ + a²xⁿ⁻² + . . .+ aⁿ⁻¹x + aⁿ for some fixed real number a.

Ans.

$\begin{array}{l} We have, \\ f (x) = x^{n} + {ax}^{n}^{- 1} + a^{2} {x^{n}}^{- 2} + . . . + {a^{n}}^{- 1} x + a^{n} \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} (x^{n}) + \frac{d}{dx} ({ax}^{n}^{- 1}) + \frac{d}{dx} (a^{2} {x^{n}}^{- 2}) + . . . + \frac{d}{dx} ({a^{n}}^{- 1} x) \\ + \frac{d}{dx} a^{n} \\ = {nx}^{n - 1} + a (n - 1) x^{n - 2} + a^{2} (n - 2) x^{n - 3} + . .. + a^{n - 1} + 0 \\ = {nx}^{n - 1} + a (n - 1) x^{n - 2} + a^{2} (n - 2) x^{n - 3} + . .. + a^{n - 1} \end{array}$

Q.7

Find lim_{x \to 0} f (x), where f (x) = {\begin{matrix} \frac{x}{| x |}, x \neq 0 \\ 0, x = 0 \end{matrix}

Ans.

$\begin{array}{l} The given function is, \\ f (x) = {\begin{cases} \frac{x}{| x |}, x \neq 0 \\ 0, x = 0 \end{cases} \\ At x = 0, \\ L . H . L . = \lim_{x \to 0^{-}} f (x) \\ = \lim_{x \to 0^{-}} (\frac{x}{| x |}) \\ = \lim_{x \to 0} (\frac{x}{- x}) [when x < 0, | x | = - x] \\ = - 1 \\ R . H . L . = \lim_{x \to 0^{+}} f (x) \\ = \lim_{x \to 0^{+}} (\frac{x}{| x |}) \\ = \lim_{x \to 0} (\frac{x}{x}) [when x < 0, | x | = x] \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = 1 \end{array}$

$\begin{array}{l} Since, \lim_{x \to 0^{-}} f (x) \neq \lim_{x \to 0^{+}} f (x) \\ So, \lim_{x \to 0} f (x) does not exist. \end{array}$

Q.8

$\begin{array}{l} For some constants a and b, find the derivative of \\ (i) (x - a) (x - b) (ii) {({ax}^{2} + b)}^{2} (iii) \frac{x - a}{x - b} \end{array}$

Ans.

$\begin{array}{l} (i) We have, \\ f (x) = (x - a) (x - b) \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} {(x - a) (x - b)} \\ = (x - a) \frac{d}{dx} (x - b) + (x - b) \frac{d}{dx} (x - a) \end{array}$

$\begin{array}{l} [By Leibnitz rule] \\ = (x - a) (\frac{d}{dx} x - \frac{d}{dx} b) + (x - b) (\frac{d}{dx} x - \frac{d}{dx} a) \\ = (x - a) (1 - 0) + (x - b) (1 - 0) \\ = x - a + x - b \\ f ‘ (x) = 2 x - a - b \\ (ii) We have, \\ f (x) = {({ax}^{2} + b)}^{2} \\ = ({ax}^{2} + b) ({ax}^{2} + b) \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} {({ax}^{2} + b) ({ax}^{2} + b)} \\ [[By Leibnitz rule]] \\ f ‘ (x) = {\frac{d}{dx} ({ax}^{2} + b)} ({ax}^{2} + b) + ({ax}^{2} + b) {\frac{d}{dx} ({ax}^{2} + b)} \\ = (a \frac{d}{dx} x^{2} + \frac{d}{dx} b) ({ax}^{2} + b) + ({ax}^{2} + b) (a \frac{d}{dx} x^{2} + \frac{d}{dx} b) \\ = (2 ax + 0) ({ax}^{2} + b) + ({ax}^{2} + b) (2 ax + 0) \\ = 2 ax ({ax}^{2} + b) + 2 ax ({ax}^{2} + b) \\ = 4 ax ({ax}^{2} + b) \\ (iii) We have, \\ f (x) = \frac{x - a}{x - b} \\ Differentiating w.r.t. x, we get \end{array}$ $\begin{array}{l} f’ (x) = \frac{d}{dx} (\frac{x - a}{x - b}) \end{array}$

$\begin{array}{l} = \frac{(x - b) \frac{d}{dx} (x - a) - (x - a) \frac{d}{dx} (x - b)}{{(x - b)}^{2}} [By quotient rule] \\ = \frac{(x - b) (1 - 0) - (x - a) (1 - 0)}{{(x - b)}^{2}} \\ = \frac{x - b - x + a}{{(x - b)}^{2}} \\ f’ (x) = \frac{a - b}{{(x - b)}^{2}} \end{array}$

Q.9

Find the derivative of \frac{x^{n} - a^{n}}{x - a} for some constant a.

Ans.

$\begin{array}{l} We have, \\ f (x) = \frac{x^{n} - a^{n}}{x - a} \\ Differentiating w.r.t.x, we get \\ (x) = \frac{d}{dx} \{\frac{x^{n} - a^{n}}{x - a}\} \\ = \frac{(x - a) \frac{d}{dx} (x^{n} - a^{n}) - (x^{n} - a^{n}) \frac{d}{dx} (x - a)}{{(x - a)}^{2}} \\ [By Quotient Rule] \\ = \frac{(x - a) ({nx}^{n - 1} - 0) - (x^{n} - a^{n}) (1 - 0)}{{(x - a)}^{2}} \\ = \frac{{nx}^{n} - an x^{n - 1} - x^{n} + a^{n}}{{(x - a)}^{2}} \end{array}$

Q.10

$\begin{array}{l} Find the derivative of \\ (i) 2 x - \frac{3}{4} (ii) (5 x^{3} + 3 x - 1) (x - 1) \\ (iii) x^{- 3} (5 + 3 x) (iv) x^{5} (3 - 6 x^{- 9}) \\ (v) x^{- 4} (3 - 4 x^{- 5}) (vi) \frac{2}{x + 1} - \frac{x^{2}}{3 x - 1} \end{array}$

Ans.

$\begin{array}{l} (i) We have, f (x) = 2 x - \frac{3}{4} \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \frac{d}{dx} \{2 x - \frac{3}{4}\} \\ = \frac{d}{dx} (2 x) - \frac{d}{dx} (\frac{3}{4}) \\ = 2-0 \\ = 2 \\ (ii) We have, \\ f (x) = (5 x^{3} + 3 x -1) (x -1) \end{array}$

$\begin{array}{l} Differentiating w.r.t. x, we get \\ f’ (x) = \frac{d}{dx} \{({5x}^{3} +3x-1) (x-1)\} \\ = \{\frac{d}{dx} ({5x}^{3} +3x-1)\} (x-1) + ({5x}^{3} +3x-1) \{\frac{d}{dx} (x-1)\} \\ [By Leibnitz’s product rule] \\ = (5 \frac{d}{dx} x^{3} +3 \frac{d}{dx} x- \frac{d}{dx} 1) (x-1) + ({5x}^{3} +3x-1) (\frac{d}{dx} x- \frac{d}{dx} 1) \\ = ({15x}^{2} +3-0) (x-1) + ({5x}^{3} +3x-1) (1-0) \\ {= 15x}^{3} {+3x-15x}^{2} {-3+5x}^{3} +3x-1 \\ {= 20x}^{3} {-15x}^{2} +6x-4 \\ (iii) We have, \\ f (x) {= x}^{-3} (5+3x) \\ Differentiating w.r.t. x, we get \\ f’ (x) = \frac{d}{dx} \{x^{-3} (5+3x)\} \\ = \{\frac{d}{dx} x^{-3}\} (5+3x) {+x}^{-3} \{\frac{d}{dx} (5+3x)\} \\ [By Leibnitz’s product rule] \\ {= -3x}^{-4} (5+3x) {+x}^{-3} (0+3) \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = - 15 x^{- 4} - 9 x^{- 3} + 3 x^{- 3} \\ = - 15 x^{- 4} - 6 x^{- 3} \\ = - \frac{3}{x^{4}} (5 + 2 x) \\ (iv) We have, \\ f (x) = x^{5} (3 - 6 x^{- 9}) \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} \{x^{5} (3 - 6 x^{- 9})\} \\ = \{\frac{d}{dx} x^{5}\} (3 - 6 x^{- 9}) + x^{5} \{\frac{d}{dx} (3 - 6 x^{- 9})\} \\ [By Leibnitz’s product rule] \\ = 5 x^{4} (3 - 6 x^{- 9}) + x^{5} (0 + 54 x^{- 10}) \\ = 15 x^{4} - 30 x^{- 5} + 54 x^{- 5} \\ = 15 x^{4} + 24 x^{- 5} \\ = 15 x^{4} + \frac{24}{x^{5}} \\ (v) We have, \\ f (x) = x^{- 4} (3 - 4 x^{- 5}) \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} \{x^{- 4} (3 - 4 x^{- 5})\} \\ = \{\frac{d}{dx} x^{- 4}\} (3 - 4 x^{- 5}) + x^{- 4} \{\frac{d}{dx} (3 - 4 x^{- 5})\} \end{array}$

$\begin{array}{l} [By Leibnitz’s product rule] \\ = - 4 x^{- 5} (3 - 4 x^{- 5}) + x^{- 4} (0 + 20 x^{- 6}) \\ = - 12 x^{- 5} + 16 x^{- 10} + 20 x^{- 10} \\ = - 12 x^{- 5} + 36 x^{- 10} \\ = - \frac{12}{x^{5}} + \frac{36}{x^{10}} \\ (vi) We have, \\ f (x) = \frac{2}{x + 1} - \frac{x^{2}}{3 x - 1} \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} \{\frac{2}{x + 1} - \frac{x^{2}}{3 x - 1}\} \\ = \frac{\{(x + 1) \frac{d}{dx} (2) - 2 \frac{d}{dx} (x + 1)\}}{{(x + 1)}^{2}} - \frac{\{(3 x - 1) \frac{d}{dx} x^{2} - x^{2} \frac{d}{dx} (3 x - 1)\}}{{(3 x - 1)}^{2}} \\ [By Quotient Rule] \\ = \frac{\{(x + 1) (0) - 2 (1 + 0)\}}{{(x + 1)}^{2}} - \frac{\{(3 x - 1) (2 x) - x^{2} (3 - 0)\}}{{(3 x - 1)}^{2}} \\ = \frac{- 2}{{(x + 1)}^{2}} - \frac{6 x^{2} - 2 x - 3 x^{2}}{{(3 x - 1)}^{2}} \end{array}$

$= – \frac{2}{{(x+1)}^{2}} – \frac{x (3x-2)}{{(3x-1)}^{2}}$

Q.11 Find the derivative of cos x from first principle.

Ans.

$\begin{array}{l} Since, f (x) = \cos x \\ ∴ f ‘ (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = \lim_{h \to 0} \frac{\cos (x + h) - \cos (x)}{h} \\ = \lim_{h \to 0} \frac{- 2 \sin (\frac{x + h - x}{2}) \sin (\frac{x + h + x}{2})}{h} \\ = - \lim_{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \times \lim_{h \to 0} \sin (\frac{2 x + h}{2}) \\ = - 1 \times \sin (\frac{2 x + 0}{2}) \\ f ‘ (x) = - sinx \end{array}$

Q.12 Find the derivative of the following functions:

(i) sin x cos x
(ii) sec x
(iii) 5sec x + 4cos x
(iv) cosec x
(v) 3cot x + 5cosec x
(vi) 5sin x − 6cos x + 7
(vii) 2 tan x − 7sec x

Ans.

$\begin{array}{l} (i) We have, \\ f (x) = sinxcosx \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} {sinxcosx} \\ = (\frac{d}{dx} \sin x) \cos x + \sin x (\frac{d}{dx} \cos x) \\ [By Leibnitz’s product rule] \\ = \cos x \cos x + \sin x (- \sin x) \\ = \cos^{2} x - \sin^{2} x = \cos 2 x \\ (ii) We have, \\ f (x) = secx \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} (\sec x) \\ = \sec x \tan x \\ (iii) We have, \\ f (x) = 5 \sec x + 4 \cos x \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} (5 \sec x + 4 \cos x) \\ = 5 \frac{d}{dx} \sec x + 4 \frac{d}{dx} \cos x \\ = 5 (\sec x \tan x) + 4 (- \sin x) \\ = 5 \sec x \tan x - 4 \sin x \end{array}$

$\begin{array}{l} (iv) We have, \\ f (x) = \cos e c x \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} (cosec x) \\ = - cosec x \cot x \\ (v) We have, \\ f (x) = 3 \cot x + 5 cosec x \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} (3 \cot x + 5 cosec x) \\ = 3 \frac{d}{dx} \cot x + 5 \frac{d}{dx} cosec x \\ = 3 (- {cosec}^{2} x) - 5 cosec x \cot x \\ = - 3 {cosec}^{2} x - 5 cosec x \cot x \\ (vi) We have, \\ f (x) = 5 \sin x - 6 \cos x + 7 \\ Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} (5 \sin x - 6 \cos x + 7) \\ = 5 \frac{d}{dx} \sin x - 6 \frac{d}{dx} \cos x + \frac{d}{dx} 7 \\ = 5 \cos x + 6 \sin x + 0 \\ = 5 \cos x + 6 \sin x \\ (vii) We have, \\ f (x) = 2 \tan x - 7 \sec x \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} Differentiating w.r.t. x, we get \\ f ‘ (x) = \frac{d}{dx} (2 \tan x - 7 \sec x) \\ = 2 \frac{d}{dx} \tan x - 7 \frac{d}{dx} \sec x \\ = 2 \sec^{2} x - 7 secx \tan x \end{array}$

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NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives (Ex 13.2) Exercise 13.2

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