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Transcript

Derivative of ใ€–๐’”๐’†๐’„ใ€—^(โˆ’๐Ÿ) ๐’™ ๐‘“ (๐‘ฅ)=ใ€–๐‘ ๐‘’๐‘ใ€—^(โˆ’1) ๐‘ฅ Let ๐’š= ใ€–๐’”๐’†๐’„ใ€—^(โˆ’๐Ÿ) ๐’™ secโกใ€–๐‘ฆ=๐‘ฅใ€— ๐’™=๐ฌ๐ž๐œโกใ€–๐’š ใ€— Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ = (๐‘‘ (secโก๐‘ฆ ))/๐‘‘๐‘ฅ 1 = (๐‘‘ (secโก๐‘ฆ ))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ 1 = (๐‘‘ (secโก๐‘ฆ ))/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ 1 = ๐’•๐’‚๐’โก๐’š .๐’”๐’†๐’„โก๐’š. ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/(๐’•๐’‚๐’โก๐’š .ใ€– secใ€—โก๐‘ฆ ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/((โˆš(ใ€–๐ฌ๐ž๐œใ€—^๐Ÿโก๐’š โˆ’ ๐Ÿ)) .ใ€– secใ€—โก๐‘ฆ ) Putting value of ๐‘ ๐‘’๐‘โก๐‘ฆ = ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/((โˆš(๐‘ฅ^2 โˆ’ 1 ) ) . ๐‘ฅ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/(๐‘ฅ โˆš(๐‘ฅ^2 โˆ’ 1 ) ) Hence ๐’…(ใ€–๐’”๐’†๐’„ใ€—^(โ€“๐Ÿ) ๐’™)/๐’…๐’™ = ๐Ÿ/(๐’™ โˆš(๐’™^๐Ÿ โˆ’ ๐Ÿ ) ) As tan2 ฮธ = sec2 ฮธ โ€“ 1, tan ฮธ = โˆš("sec2 ฮธ โ€“ 1" ) As ๐‘ฆ = ใ€–๐‘ ๐‘’๐‘ใ€—^(โˆ’1) ๐‘ฅ So, ๐’”๐’†๐’„โก๐’š = ๐’™

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

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo