Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 7

Advertisement

Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 8

Advertisement

Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 9

  1. Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
  2. Serial order wise

Transcript

Ex 13.2, 4 Find the derivative of the following functions from first principle. (iii) 1/๐‘ฅ^2 Let f (x) = 1/๐‘ฅ^2 We need to find derivative of f(x) i.e. fโ€™ (x) We know that fโ€™(x) = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0) ๐‘“โกใ€–(๐‘ฅ + โ„Ž) โˆ’ ๐‘“(๐‘ฅ)ใ€—/โ„Ž Here, f (x) = 1/๐‘ฅ^2 So, f (x + h) = 1/ใ€–(๐‘ฅ + โ„Ž)ใ€—^2 Putting values fโ€™(x) = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–(1/ใ€–(๐‘ฅ + โ„Ž)ใ€—^2 โˆ’ 1/๐‘ฅ^2 )/โ„Žใ€— = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–(ใ€–ใ€–๐‘ฅ ใ€—^2 โˆ’ (๐‘ฅ + โ„Ž)ใ€—^2/(ใ€–(๐‘ฅ + โ„Ž)ใ€—^2 ๐‘ฅ^2 ))/โ„Žใ€— = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–( ๐‘ฅ2 โˆ’ (๐‘ฅ + โ„Ž)2)/(โ„Ž๐‘ฅ2 (๐‘ฅ + โ„Ž)2)ใ€— = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–((๐‘ฅ โˆ’ ( ๐‘ฅ + โ„Ž )) (๐‘ฅ + (๐‘ฅ + โ„Ž)))/(โ„Ž๐‘ฅ2 (๐‘ฅ + โ„Ž)2)ใ€— = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–( (๐‘ฅ โˆ’๐‘ฅ โˆ’ โ„Ž) (๐‘ฅ + ๐‘ฅ + โ„Ž))/(โ„Ž.๐‘ฅ2 (๐‘ฅ + โ„Ž)2)ใ€— (As a2 โ€“ b2 = (a โ€“ b) (a + b)) = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–((โˆ’โ„Ž)(2๐‘ฅ + โ„Ž))/(โ„Ž๐‘ฅ2 (๐‘ฅ + โ„Ž)2)ใ€— = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–((โˆ’1) 2๐‘ฅ + โ„Ž)/(๐‘ฅ2 (๐‘ฅ + โ„Ž)2)ใ€— Putting h = 0 = ((โˆ’1) 2๐‘ฅ + 0)/(๐‘ฅ2 (๐‘ฅ + 0)2) = ((โˆ’1) 2๐‘ฅ)/(๐‘ฅ2 (๐‘ฅ)2) = (โˆ’2๐‘ฅ)/๐‘ฅ4 = (โˆ’2)/๐‘ฅ^3 Thus, fโ€™(x) = (โˆ’๐Ÿ)/๐’™^๐Ÿ‘

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.