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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.3, 15 Find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ in, y = secโ€“1 (1/( 2๐‘ฅ2โˆ’1 )), 0 < x < 1/โˆš2 y = secโ€“1 (1/( 2๐‘ฅ^2 โˆ’ 1 )) ๐’”๐’†๐’„โก๐’š = 1/(2๐‘ฅ^2 โˆ’ 1) ๐Ÿ/๐œ๐จ๐ฌโก๐’š = 1/(2๐‘ฅ^2 โˆ’ 1) cosโก๐‘ฆ = 2๐‘ฅ2โˆ’1 y = cos โ€“1 (2๐‘ฅ2โˆ’1) Putting ๐‘ฅ = cosโกฮธ ๐‘ฆ = cos โ€“1 (2๐‘๐‘œ๐‘ 2๐œƒโˆ’1) ๐‘ฆ = cos โ€“1 (cosโก2 ๐œƒ) ๐‘ฆ = 2๐œƒ Putting value of ฮธ = cosโˆ’1 x ๐‘ฆ = 2 ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ . (๐‘‘(๐‘ฆ))/๐‘‘๐‘ฅ = (๐‘‘ (2ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2 (๐‘‘ (ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2 . ((โˆ’1)/โˆš(1 โˆ’ ๐‘ฅ^2 )) ๐’…๐’š/๐’…๐’™ = (โˆ’๐Ÿ)/โˆš(๐Ÿ โˆ’ ๐’™^๐Ÿ ) (๐‘๐‘œ๐‘ โก2๐œƒ " = 2 " ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐œƒโˆ’1) ((ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ")โ€˜ = " (โˆ’1)/โˆš(1 โˆ’ ๐‘ฅ^2 ))

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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.