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

Transcript

Find the derivative of f given by f (x) = cotโ€“1 ๐‘ฅ assuming it exists. Let y = cotโ€“1 ๐‘ฅ co๐‘กโก๐‘ฆ= ๐‘ฅ ๐‘ฅ =co๐‘กโก๐‘ฆ Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ (๐‘‘(๐‘ฅ))/๐‘‘๐‘ฅ = (๐‘‘ (co๐‘กโก๐‘ฆ ))/๐‘‘ 1 = (๐‘‘ (co๐‘กโก๐‘ฆ ))/๐‘‘๐‘ฅ We need ๐‘‘๐‘ฆ in denominator, so multiplying & Dividing by ๐‘‘๐‘ฆ. 1 = (๐‘‘ (co๐‘กโก๐‘ฆ ))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ 1 = โˆ’cosec^2โก๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (โˆ’cosec^2โก๐‘ฆ ) = 1 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’1)/(ใ€–๐’„๐’๐’”๐’†๐’„ใ€—^๐Ÿ ๐’š) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’1)/(๐Ÿ + ใ€–๐œ๐จ๐ญใ€—^๐Ÿโก๐’š) (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€—) Putting ๐‘๐‘œ๐‘กโก๐‘ฆ = ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’1)/(๐’™^๐Ÿ + 1) As ๐‘ฆ = cot^(โˆ’1) ๐‘ฅ So, ๐‘๐‘œ๐‘กโก๐‘ฆ = ๐‘ฅ Hence, (๐’…(ใ€–๐œ๐จ๐’•ใ€—^(โˆ’๐Ÿ)โกใ€–๐’™)ใ€—)/๐’…๐’™ = (โˆ’๐Ÿ)/(๐’™^๐Ÿ + ๐Ÿ)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.