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Ex 5.5, 6 - Differentiate (x + 1/x)^x + x^(1 + 1/x) - Teachoo

Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 7
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 8

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Transcript

Ex 5.5, 6 Differentiate the functions in, (๐‘ฅ+1/๐‘ฅ)^๐‘ฅ+ ๐‘ฅ^((1 + 1/๐‘ฅ) ) Let ๐‘ฆ= (๐‘ฅ+1/๐‘ฅ)^๐‘ฅ+ ๐‘ฅ^((1 + 1/๐‘ฅ) ) Let ๐‘ข = (๐‘ฅ+1/๐‘ฅ)^๐‘ฅ , ๐‘ฃ = ๐‘ฅ^((1 + 1/๐‘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 = (๐‘ฅ+1/๐‘ฅ)^๐‘ฅ Taking log both sides logโก๐‘ข = log (๐‘ฅ+1/๐‘ฅ)^๐‘ฅ logโก๐‘ข = ๐‘ฅ log (๐‘ฅ+1/๐‘ฅ) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘(logโก๐‘ข )/๐‘‘๐‘ฅ = (๐‘‘ (๐‘ฅ log" " (๐‘ฅ + 1/๐‘ฅ)))/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ข )/๐‘‘๐‘ฅ (๐‘‘๐‘ข/๐‘‘๐‘ข) = (๐‘‘ (๐‘ฅ log" " (๐‘ฅ + 1/๐‘ฅ)))/๐‘‘๐‘ฅ (As ๐‘™๐‘œ๐‘”โก(๐‘Ž^๐‘ )=๐‘ . ๐‘™๐‘œ๐‘”โก๐‘Ž) ๐‘‘(logโก๐‘ข )/๐‘‘๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" = " (๐‘‘ (๐‘ฅ log" " (๐‘ฅ + 1/๐‘ฅ)))/๐‘‘๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" = " (๐‘‘ (๐‘ฅ log" " (๐‘ฅ + 1/๐‘ฅ)))/๐‘‘๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" = " ๐‘‘(๐‘ฅ)/๐‘‘๐‘ฅ . log (๐‘ฅ + 1/๐‘ฅ) + ๐‘‘(log" " (๐‘ฅ + 1/๐‘ฅ))/๐‘‘๐‘ฅ . ๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" 1. log (๐‘ฅ + 1/๐‘ฅ) + ((1/(๐‘ฅ + 1/๐‘ฅ)).๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ + 1/๐‘ฅ)) . ๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (๐‘ฅ + 1/๐‘ฅ) + (1/(๐‘ฅ + 1/๐‘ฅ) . (๐‘‘(๐‘ฅ)/๐‘‘๐‘ฅ+(๐‘‘ (1/๐‘ฅ))/๐‘‘๐‘ฅ)) . ๐‘ฅ Using product rule in ๐‘ฅ ๐‘™๐‘œ๐‘”" " (๐‘ฅ + 1/๐‘ฅ) As (uv)โ€™ = uโ€™ v + vโ€™ u 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (๐‘ฅ+1/๐‘ฅ) + (1/(๐‘ฅ + 1/๐‘ฅ) . (1+(โˆ’1)/๐‘ฅ^2 " " )) . ๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (๐‘ฅ+1/๐‘ฅ) + (๐‘ฅ/(๐‘ฅ^2 + 1) . (1โˆ’1/๐‘ฅ^2 " " )) . ๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (๐‘ฅ+1/๐‘ฅ) + (๐‘ฅ/(๐‘ฅ^2 + 1) ((๐‘ฅ^2 โˆ’ 1)/๐‘ฅ^2 ).๐‘ฅ) 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (๐‘ฅ+1/๐‘ฅ) + (๐‘ฅ/(๐‘ฅ^2 + 1) ((๐‘ฅ^2 โˆ’ 1)/๐‘ฅ^2 ).๐‘ฅ) 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (๐‘ฅ+1/๐‘ฅ) + (๐‘ฅ^2/๐‘ฅ^2 ((๐‘ฅ^2 โˆ’ 1)/(๐‘ฅ^2 + 1))) 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (๐‘ฅ+1/๐‘ฅ) + ((๐‘ฅ^2 โˆ’ 1)/(๐‘ฅ^2+ 1)) ๐‘‘๐‘ข/๐‘‘๐‘ฅ "= " ๐‘ข (ใ€–log ใ€—โก(๐‘ฅ+1/๐‘ฅ)+((๐‘ฅ^2 โˆ’ 1)/(๐‘ฅ^2+ 1))) ๐‘‘๐‘ข/๐‘‘๐‘ฅ "=" (๐‘ฅ+1/๐‘ฅ)^๐‘ฅ (ใ€–log ใ€—โก(๐‘ฅ+1/๐‘ฅ)+((๐‘ฅ^2 โˆ’ 1)/(๐‘ฅ^2+ 1))) ๐’…๐’–/๐’…๐’™ "=" (๐’™+๐Ÿ/๐’™)^๐’™ ((๐’™^๐Ÿ โˆ’ ๐Ÿ)/(๐’™^๐Ÿ+ ๐Ÿ)โกใ€–+ ใ€–๐’๐’๐’ˆ ใ€—โก(๐’™+๐Ÿ/๐’™) ใ€— ) Calculating ๐’…๐’—/๐’…๐’™ ๐‘ฃ = ๐‘ฅ^(1 + 1/๐‘ฅ)" " Taking log both sides log ๐‘ฃ = log ๐‘ฅ^(1 + 1/๐‘ฅ)" " log ๐‘ฃ = (1 + 1/๐‘ฅ)log ๐‘ฅ^" " Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘(logโก๐‘ฃ )/๐‘‘๐‘ฅ = (๐‘‘ ((1 + 1/๐‘ฅ)" . " log ๐‘ฅ))/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ฃ )/๐‘‘๐‘ฅ (๐‘‘๐‘ฃ/๐‘‘๐‘ฃ) = (๐‘‘ ((1 + 1/๐‘ฅ)" . " log ๐‘ฅ))/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ฃ )/๐‘‘๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = (๐‘‘ ((1 + 1/๐‘ฅ)" . " log ๐‘ฅ))/๐‘‘๐‘ฅ 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = (๐‘‘ ((1 + 1/๐‘ฅ)" . " log ๐‘ฅ))/๐‘‘๐‘ฅ Using product rule in (๐‘ฅ+ 1/๐‘ฅ)" . " ๐‘™๐‘œ๐‘” ๐‘ฅ 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = ๐‘‘(1 + 1/๐‘ฅ)/๐‘‘๐‘ฅ . logโก๐‘ฅ + ๐‘‘(logโก๐‘ฅ )/๐‘‘๐‘ฅ . (1 + 1/๐‘ฅ) 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = (๐‘‘(1)/๐‘‘๐‘ฅ+๐‘‘(1/๐‘ฅ)/๐‘‘๐‘ฅ) . logโก๐‘ฅ + 1/๐‘ฅ (1 + 1/๐‘ฅ) 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = (0+((โˆ’1)/๐‘ฅ^2 )) . logโก๐‘ฅ + 1/๐‘ฅ (1 + 1/๐‘ฅ) 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = (โˆ’1)/๐‘ฅ^2 . logโก๐‘ฅ + 1/๐‘ฅ (1 + 1/๐‘ฅ) 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = (โˆ’logโก๐‘ฅ)/๐‘ฅ^2 + 1/๐‘ฅ + 1/๐‘ฅ^2 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = (โˆ’logโก๐‘ฅ)/๐‘ฅ^2 + 1/๐‘ฅ + 1/๐‘ฅ^2 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = ((โˆ’logโก๐‘ฅ + ๐‘ฅ + 1)/๐‘ฅ^2 ) ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘ฃ ((โˆ’logโก๐‘ฅ + ๐‘ฅ + 1)/๐‘ฅ^2 ) ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘ฅ^((1 + 1/๐‘ฅ) ) ((๐‘ฅ + 1 โˆ’ logโก๐‘ฅ )/๐‘ฅ^2 ) Now ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘‘๐‘ข/๐‘‘๐‘ฅ + ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ Putting values of ๐‘‘๐‘ข/๐‘‘๐‘ฅ & ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ ๐’…๐’š/๐’…๐’™ = (๐’™+๐Ÿ/๐’™)^๐’™ ((๐’™^๐Ÿ โˆ’ ๐Ÿ)/(๐’™^๐Ÿ+ ๐Ÿ)+๐ฅ๐จ๐ โก(๐’™+ ๐Ÿ/๐’™) ) + ๐’™^((๐Ÿ + ๐Ÿ/๐’™) ) ((๐’™ + ๐Ÿ โˆ’ ๐’๐’๐’ˆโก๐’™ )/๐’™^๐Ÿ )

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.