Derivative of cot-1 x (cot inverse x) - Teachoo  [with Video]

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Derivative of cot-1 x (cot inverse x) - Part 2

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  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Concept wise

Transcript

Derivative of ใ€–๐’„๐’๐’•ใ€—^(โˆ’๐Ÿ) ๐’™ ๐‘“ (๐‘ฅ)=ใ€–๐‘๐‘œ๐‘กใ€—^(โˆ’1) ๐‘ฅ Let ๐’š= ใ€–๐’„๐’๐’•ใ€—^(โˆ’๐Ÿ) ๐’™ cotโกใ€–๐‘ฆ=๐‘ฅใ€— ๐’™=๐œ๐จ๐ญโกใ€–๐’š ใ€— Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ = (๐‘‘ (cotโก๐‘ฆ ))/๐‘‘๐‘ฅ 1 = (๐‘‘ (cotโก๐‘ฆ ))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ 1 = (๐‘‘ (cotโก๐‘ฆ ))/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ 1 = โˆ’๐œ๐จใ€–๐ฌ๐ž๐œใ€—^๐Ÿ ๐’š . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ 1 = โˆ’(๐Ÿ +๐’„๐’๐’•๐Ÿ๐’š) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’1)/(1 + ใ€–๐œ๐จ๐ญใ€—^๐Ÿโก๐’š ) Putting ๐‘๐‘œ๐‘กโก๐‘ฆ = ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’1)/(๐’™^๐Ÿ + ๐Ÿ) Hence (๐’…(ใ€–๐œ๐จ๐ญใ€—^(โˆ’๐Ÿ)โกใ€–๐’™)ใ€—)/๐’…๐’™ = (โˆ’๐Ÿ)/(๐’™^๐Ÿ + ๐Ÿ) (๐ด๐‘  ใ€– ๐‘๐‘œ๐‘ ๐‘’๐‘ใ€—^2โกใ€–๐‘ฆ= ใ€–1+ใ€—โกใ€–๐‘๐‘œ๐‘กใ€—^2โก๐‘ฆ ใ€—) As ๐‘ฆ = ใ€–๐‘๐‘œ๐‘กใ€—^(โˆ’1) ๐‘ฅ So, ๐’„๐’๐’•โก๐’š = ๐’™

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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.