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

Transcript

Ex 5.5, 2 Differentiate the functions in, โˆš(((๐‘ฅ โˆ’ 1)(๐‘ฅ โˆ’ 2))/((๐‘ฅ โˆ’ 3)(๐‘ฅ โˆ’ 4)(๐‘ฅ โˆ’ 5))) Let ๐‘ฆ=โˆš(((๐‘ฅ โˆ’ 1)(๐‘ฅ โˆ’ 2))/((๐‘ฅ โˆ’ 3)(๐‘ฅ โˆ’ 4)(๐‘ฅ โˆ’ 5))) ๐‘ฆ= (((๐‘ฅ โˆ’ 1)(๐‘ฅ โˆ’ 2))/((๐‘ฅ โˆ’ 3)(๐‘ฅ โˆ’ 4)(๐‘ฅ โˆ’ 5)))^(1/2) Taking log both sides logโก๐‘ฆ = log (((๐‘ฅ โˆ’ 1)(๐‘ฅ โˆ’ 2))/((๐‘ฅ โˆ’ 3)(๐‘ฅ โˆ’ 4)(๐‘ฅ โˆ’ 5)))^(1/2) logโก๐‘ฆ = 1/2 log (((๐‘ฅ โˆ’ 1)(๐‘ฅ โˆ’ 2))/((๐‘ฅ โˆ’ 3)(๐‘ฅ โˆ’ 4)(๐‘ฅ โˆ’ 5))) (As ๐‘™๐‘œ๐‘”โก(๐‘Ž^๐‘) = ๐‘ ๐‘™๐‘œ๐‘”โก๐‘Ž) logโก๐‘ฆ = 1/2 [logโกใ€–((๐‘ฅโˆ’1)(๐‘ฅโˆ’2))โˆ’logโก((๐‘ฅโˆ’3)(๐‘ฅโˆ’4)(๐‘ฅโˆ’5)) ใ€— ] logโก๐‘ฆ = 1/2 . [("log " (๐‘ฅ+1)" + log " (๐‘ฅโˆ’2))" โˆ’ " (logโก(๐‘ฅโˆ’3)+logโกใ€–(๐‘ฅโˆ’4)+logโก(๐‘ฅโˆ’5) ใ€— )] logโก๐‘ฆ = 1/2 . ["log " (๐‘ฅ+1)" + log " (๐‘ฅ+2)" โˆ’ " logโก(๐‘ฅโˆ’3)โˆ’logโกใ€–(๐‘ฅโˆ’4)โˆ’logโก(๐‘ฅโˆ’5) ใ€— ] Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘(logโก๐‘ฆ)/๐‘‘๐‘ฅ = (๐‘‘ (1/2 " ." logโกใ€–(๐‘ฅ+1)" +" logโกใ€– (๐‘ฅ+2)" โˆ’ " logโก(๐‘ฅโˆ’3)โˆ’logโกใ€–(๐‘ฅโˆ’4)โˆ’logโก(๐‘ฅโˆ’5) ใ€— ใ€— ใ€— ))/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ฆ)/๐‘‘๐‘ฅ (๐‘‘๐‘ฆ/๐‘‘๐‘ฆ) = 1/2 (๐‘‘(ใ€–log ใ€—โกใ€–(๐‘ฅ + 1)" +" logโกใ€–(๐‘ฅ + 2)" โˆ’ " logโก(๐‘ฅ โˆ’ 3) โˆ’ logโกใ€–(๐‘ฅ โˆ’ 4) โˆ’ logโก(๐‘ฅ โˆ’ 5) ใ€— ใ€— ใ€— )/๐‘‘๐‘ฅ) 1/๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/2 (1/(๐‘ฅ + 1)+1/(๐‘ฅ + 2)โˆ’1/(๐‘ฅ โˆ’ 3)โˆ’1/(๐‘ฅ โˆ’ 4)โˆ’1/(๐‘ฅ โˆ’ 5)) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/2 ๐‘ฆ(1/(๐‘ฅ โˆ’ 1)+1/(๐‘ฅ โˆ’ 2)โˆ’1/(๐‘ฅ โˆ’ 3)โˆ’1/(๐‘ฅ โˆ’ 4)โˆ’1/(๐‘ฅ โˆ’ 5)) ๐’…๐’š/๐’…๐’™ = ๐Ÿ/๐Ÿ โˆš(((๐’™ โˆ’ ๐Ÿ)(๐’™ โˆ’ ๐Ÿ))/((๐’™ โˆ’ ๐Ÿ‘)(๐’™ โˆ’ ๐Ÿ’)(๐’™ โˆ’ ๐Ÿ“))) (๐Ÿ/(๐’™ โˆ’ ๐Ÿ)+๐Ÿ/(๐’™ โˆ’ ๐Ÿ)โˆ’๐Ÿ/(๐’™ โˆ’ ๐Ÿ‘)โˆ’๐Ÿ/(๐’™ โˆ’ ๐Ÿ’)โˆ’๐Ÿ/(๐’™ โˆ’ 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.