Last updated at March 9, 2017 by Teachoo

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Ex 13.2, 9 Find the derivative of (i) 2š„ ā 3ļ·®4ļ·Æ Let f(x) = 2š„ ā 3ļ·®4ļ·Æ fā(x) = š(2š„ ā 3ļ·®4ļ·Æ )ļ·®šš„ļ·Æ = 2 ā 0 = 2 ā“ fā(x) = 2 Ex 13.2, 9 Find the derivative of (ii) (5x3 + 3x ā 1) (x ā 1) Let f(x) = (5x3 + 3x ā 1) (x- 1) Let u = 5x3 + 3x ā 1 & v = x ā 1 ā“ f(x) = uv So, fā(x) = (uv)ā fā(x) = uāv + vāu Finding uā & vā u = 5x3 + 3x ā 1 uā = 5.3x3ā1 + 3.1.x1ā1 ā 0 = 15x2 + 3x0 = 15x2 + 3 v = x ā 1 vā = 1 .x1ā1 ā 0 = 1. x0 = 1 fā(x) = uāv + vāu = (15x2 + 3) (x ā 1) + (1) (5x3 + 3x ā 1) = 15x2 (x ā 1) + 3 (x ā 1) + 5x3 + 3x ā 1 = 15x3 ā 15x2 + 3x ā 3 + 5x3 + 3x ā 1 = 15x3 + 5x3 ā 15x2 + 3x + 3x ā 3 ā 1 = 20x3 ā 15x2 + 6x ā 4 Hence fā(x) = 20x3 ā 15x2 + 6x ā 4 Ex 13.2, 9 Find the derivative of (iii) xā3 (5 + 3x) Let f(x) = x ā 3 (5 + 3x) Let u = x ā 3 & v = 5 + 3x ā“ f(x) = uv So, fā(x) = (uv)ā fā(x) = uāv + vāu Finding uā & vā u = x ā 3 uā = ā 3x ā 3 ā 1 = ā 3x ā 4 v = 5 + 3x vā = 0 + 3 = 3 Now, fā(x) = (uv)ā = uāv + vāu = ā 3x ā 4 (5 + 3x) + 3 (x ā 3) = ā 15x ā 4 ā 9x ā 4+ 1 + 3x ā 3 = ā 15x ā 4 ā 6x ā 3 = ā15ļ·® š„ļ·®4ļ·Æļ·Æ ā 6ļ·® š„ļ·®3ļ·Æļ·Æ = ā3ļ·® š„ļ·®4ļ·Æļ·Æ (5 + 2x) Hence fā(x) = āšļ·® šļ·®šļ·Æļ·Æ (5 + 2x) Ex 13.2,9 Find the derivative of (iv) x5 (3 ā 6xā9 ). Let f (x) = x5 (3 ā 6xā9 ) Let u = x5 & v = 3 ā 6x ā 9 So, f(x) = uv ā“ fā (x) = (uv)ā fā (x) = uāv + vā u Finding uā & vā u = x5 uā = 5x5 ā 1 = 5x4 v = 3 ā 6xā9 vā = 0 ā 6( ā 9)x ā 9 ā 1 vā = 54x ā10 Now, fā(x) = (uv)ā = uāv + vā u = 5x4 (3 ā 6x ā 9) + 54x ā 10) (x5) = 15x4 ā 30x ā 9 + 4 + 54 ā 10 + 5 = 15x4 ā 30x ā5 + 54x ā 5 = 15x4 + 24x ā 5 = 15x4 + 24ļ·® š„ļ·®5ļ·Æļ·Æ Hence fā(x) = 15x4 + ššļ·® šļ·®šļ·Æļ·Æ Ex 13.2, 9 Find the derivative of (v) f (x) = xā4 (3 ā 4xā5) Let f(x) = x ā 4 (3 ā 4x ā 5) Let u = x ā 4 & v = 3 ā 4x ā 5 ā“ f(x) = uv So, fā(x) = (uv)ā fā(x) = uāv + vāu Finding uā & vā u = x ā 4 uā = ā 4x ā 4 ā 1 = ā 4x ā 5 v = 3 ā 4x ā 5 vā = 0 ā 4 ( ā 5) x ā 5 ā 1 = 20x ā 6 Now, fā(x) = (uv)ā = uāv + vāu = (ā 4xā5 ) (3 ā 4xā5) + (20xā6) (xā4) = ā 12xā5 + 16x ā5 ā 5 + 20 ā6 ā 4 = ā 12x ā5 + 16x ā10 + 20x ā 10 = ā 12x ā5 + 36x ā 10 = ā12ļ·® š„ļ·®5ļ·Æļ·Æ + 36ļ·® š„ļ·®10ļ·Æļ·Æ Hence fā(x) = āššļ·® šļ·®šļ·Æļ·Æ + ššļ·® šļ·®ššļ·Æļ·Æ Ex 13.2, 9 Find the derivative of (vi) f(x) = 2ļ·®x + 1ļ·Æ ā x2ļ·®3x ā 1ļ·Æ Let f (x) = 2ļ·®x + 1ļ·Æ ā x2ļ·®3x ā 1ļ·Æ Let f1 (x) = 2ļ·®x + 1ļ·Æ & f2 (x) = x2ļ·®3x ā 1ļ·Æ ā“ f(x) = f1(x) ā f2 (x) So, fā(x) = (f1(x) ā f2(x))ā fā(x) = fā1(x) ā fā2(x) Finding f1ā(x) f1 (x) = 2ļ·®š„ + 1ļ·Æ Let u = 2 & v = x + 1 ā“ f1(x) = š¢ļ·®š£ļ·Æ Now, f1ā(x) = š¢ļ·®š£ļ·Æļ·Æļ·®ā²ļ·Æ f1ā(x) = š¢ļ·®ā²ļ·Æ š£ ā š£ļ·®ā²ļ·Æ š¢ļ·® š£ļ·®2ļ·Æļ·Æ u = 2 uā = 0 v = x + 1 vā = 1 + 0 = 1 fā1(x) = š¢ļ·®š£ļ·Æļ·Æļ·®ā²ļ·Æ = š¢ļ·®ā²ļ·Æ š£ ā š£ļ·®ā²ļ·Æ š¢ļ·® š£ļ·®2ļ·Æļ·Æ = 0 š„ + 1ļ·Æ ā1 (2)ļ·® š„ + 1ļ·Æ2ļ·Æ = ā2ļ·® (š„ + 1)ļ·®2ļ·Æļ·Æ Hence, f1ā (x) = ā2ļ·® š„ + 1ļ·Æ2ļ·Æ Finding f2ā(x) f2 (x) = š„2ļ·®3š„ ā 1ļ·Æ Let u = x2 & v = 3x ā 1 ā“ f2 (x) = š¢ļ·®š£ļ·Æ Now, f2ā(x) = š¢ļ·®š£ļ·Æļ·Æļ·®ā²ļ·Æ f2ā(x) = š¢ļ·®ā²ļ·Æ š£ ā š£ļ·®ā²ļ·Æ š¢ļ·® š£ļ·®2ļ·Æļ·Æ Finding uā & vā u = x2 uā = 2x2 ā 1 = 2x & v = 3x ā 1 vā = 3(1) ā 0 = 3 fā2(x) = š¢ļ·®š£ļ·Æļ·Æļ·®ā²ļ·Æ = š¢ļ·®ā²ļ·Æ š£ ā š£ļ·®ā²ļ·Æ š¢ļ·® š£ļ·®2ļ·Æļ·Æ = 2š„ 3š„ ā 1ļ·Æ ā 3 (š„2)ļ·® 3š„ ā 1ļ·Æ2ļ·Æ = 6š„2 ā 2š„ ā 3š„2ļ·® (3š„ ā 1)ļ·®2ļ·Æļ·Æ = 3š„2 ā 2š„ ļ·® (3š„ ā 1)ļ·®2ļ·Æļ·Æ = š„(3š„ ā 2)ļ·® (3š„ ā 1)ļ·®2ļ·Æļ·Æ Hence fā2(x) = š„ (3š„ ā 2)ļ·® 3š„ + 1ļ·Æ2ļ·Æ Now fā (x) = f1ā(x) ā f2ā (x) = āšļ·® š + šļ·Æšļ·Æ ā š(šš ā š)ļ·®(šš ā š)ļ·Æ

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Ex 13.2, 9 Important You are here

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Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.