Ex 5.5, 7 - Differentiate the function (log x)^x + x^log x

Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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

Transcript

Ex 5.5, 7 Differentiate the functions in, ใ€–(logโกใ€–๐‘ฅ)ใ€—ใ€—^๐‘ฅ + ๐‘ฅ^logโก๐‘ฅ Let ๐‘ฆ = ใ€–(logโกใ€–๐‘ฅ)ใ€—ใ€—^๐‘ฅ+ ๐‘ฅ^logโก๐‘ฅ Let ๐‘ข = ใ€–(logโกใ€–๐‘ฅ)ใ€—ใ€—^๐‘ฅ , ๐‘ฃ = ๐‘ฅ^logโก๐‘ฅ ๐‘ฆ = ๐‘ข+๐‘ฃ Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘ (๐‘ข + ๐‘ฃ))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ข/๐‘‘๐‘ฅ + ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ Calculating ๐’…๐’–/๐’…๐’™ ๐‘ข = ใ€–(logโกใ€–๐‘ฅ)ใ€—ใ€—^๐‘ฅ Taking log both sides logโก๐‘ข = log ใ€–(logโกใ€–๐‘ฅ)ใ€—ใ€—^๐‘ฅ logโก๐‘ข = ๐‘ฅ . log (logโกใ€–๐‘ฅ)ใ€— Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘(logโก๐‘ข )/๐‘‘๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (๐‘‘(๐‘ฅ . log (logโกใ€–๐‘ฅ)ใ€— ) )/๐‘‘๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" ๐‘‘(๐‘ฅ)/๐‘‘๐‘ฅ . log (log" " ๐‘ฅ) + ๐‘‘(log (log" " ๐‘ฅ))/๐‘‘๐‘ฅ ร— ๐‘ฅ Using product Rule As (๐‘ข๐‘ฃ)โ€™ = ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" 1 . log (log" " ๐‘ฅ) + (1/(log" " ๐‘ฅ) .๐‘‘(log" " ๐‘ฅ)/๐‘‘๐‘ฅ) ร— ๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (log" " ๐‘ฅ) + (1/(log" " ๐‘ฅ) . 1/๐‘ฅ) ร— ๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (log" " ๐‘ฅ) + 1/logโก๐‘ฅ ร— ๐‘ฅ/๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (log" " ๐‘ฅ) + 1/logโก๐‘ฅ ๐‘‘๐‘ข/๐‘‘๐‘ฅ " =" ๐‘ข (log (log" " ๐‘ฅ)" + " 1/logโก๐‘ฅ ) ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (logโก๐‘ฅ )^๐‘ฅ (log (log" " ๐‘ฅ)" + " 1/logโก๐‘ฅ ) ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (logโก๐‘ฅ )^๐‘ฅ ((logโกใ€–๐‘ฅ . ใ€–log ใ€—โก(logโก๐‘ฅ ) +ใ€— 1)/logโก๐‘ฅ ) ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (logโก๐‘ฅ )^๐‘ฅ/logโก๐‘ฅ (logโกใ€–๐‘ฅ . ใ€–log ใ€—โก(logโก๐‘ฅ )+ใ€— 1) ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (logโก๐‘ฅ )^(๐‘ฅ โˆ’1) (logโกใ€–๐‘ฅ . ใ€–log ใ€—โก(logโก๐‘ฅ )+ใ€— 1) Calculating ๐’…๐’—/๐’…๐’™ ๐‘ฃ = ๐‘ฅ^logโก๐‘ฅ Taking log both sides . logโก๐‘ฃ=logโกใ€– (๐‘ฅ^logโก๐‘ฅ )ใ€— logโก๐‘ฃ = log ๐‘ฅ . logโก๐‘ฅ logโก๐‘ฃ = (logโก๐‘ฅ )^2 (As ๐‘™๐‘œ๐‘”โก(๐‘Ž^๐‘ )=๐‘ . ๐‘™๐‘œ๐‘”โก๐‘Ž) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘(logโก๐‘ฃ )/๐‘‘๐‘ฅ = (๐‘‘(logโก๐‘ฅ )^2)/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ฃ )/๐‘‘๐‘ฅ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฃ = (๐‘‘(logโก๐‘ฅ )^2)/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ฃ )/๐‘‘๐‘ฃ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = (๐‘‘(logโก๐‘ฅ )^2)/๐‘‘๐‘ฅ 1/๐‘ฃ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = (๐‘‘(logโก๐‘ฅ )^2)/๐‘‘๐‘ฅ 1/๐‘ฃ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = 2 logโก๐‘ฅ . ๐‘‘(logโก๐‘ฅ )/๐‘‘๐‘ฅ 1/๐‘ฃ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = 2 logโก๐‘ฅ . 1/๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘ฃ ((2 logโก๐‘ฅ)/๐‘ฅ) ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘ฅ^logโก๐‘ฅ ((2 logโก๐‘ฅ)/๐‘ฅ) ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘ฅ^logโก๐‘ฅ /๐‘ฅ (2 logโก๐‘ฅ ) ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘ฅ^(logโก๐‘ฅ โˆ’ 1) . 2 logโก๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ใ€–2๐‘ฅใ€—^(logโก๐‘ฅ โˆ’ 1) . logโก๐‘ฅ Hence ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ข/๐‘‘๐‘ฅ + ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ Putting value of ๐‘‘๐‘ข/๐‘‘๐‘ฅ & ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ ๐’…๐’š/๐’…๐’™ = (๐ฅ๐จ๐ โก๐’™ )^(๐’™ โˆ’๐Ÿ) (๐Ÿ+๐ฅ๐จ๐ โก๐’™.๐ฅ๐จ๐ โก(๐ฅ๐จ๐ โก๐’™ ) ) + ใ€–๐Ÿ๐’™ใ€—^(๐’๐’๐’ˆ ๐’™ โˆ’๐Ÿ). ๐’๐’๐’ˆโก๐’™

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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.