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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 11 Differentiate w.r.t. x the function, ๐‘ฅ^(๐‘ฅ^2โˆ’ 3)+(๐‘ฅโˆ’3)๐‘ฅ^2, for ๐‘ฅ > 3 Let ๐‘ฆ=๐‘ฅ^(๐‘ฅ^2โˆ’ 3)+(๐‘ฅโˆ’3)^(๐‘ฅ^2 ) And let ๐‘ข=๐‘ฅ^(๐‘ฅ^2โˆ’ 3) , ๐‘ฃ =(๐‘ฅโˆ’3)^(๐‘ฅ^2 ) Now, ๐’š = ๐’–+๐’— 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^(๐‘ฅ^2โˆ’ 3) Taking log on both sides log ๐‘ข=logโกใ€–๐‘ฅ^(๐‘ฅ^2โˆ’ 3) ใ€— log ๐‘ข=ใ€–(๐‘ฅใ€—^2โˆ’ 3). logโก๐‘ฅ Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘(logโก๐‘ข )/๐‘‘๐‘ฅ = ๐‘‘(ใ€–(๐‘ฅใ€—^2โˆ’ 3) logโก๐‘ฅ )/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ข )/๐‘‘๐‘ฅ . ๐‘‘๐‘ข/๐‘‘๐‘ข = ๐‘‘(ใ€–(๐‘ฅใ€—^2 โˆ’ 3) logโก๐‘ฅ )/๐‘‘๐‘ฅ " " ๐‘‘(logโก๐‘ข )/๐‘‘๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = ๐‘‘(ใ€–(๐‘ฅใ€—^2โˆ’ 3) logโก๐‘ฅ )/๐‘‘๐‘ฅ " " 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = ๐‘‘(ใ€–(๐‘ฅใ€—^2โˆ’ 3) logโก๐‘ฅ )/๐‘‘๐‘ฅ 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (๐‘‘ใ€–(๐‘ฅใ€—^2โˆ’ 3) )/๐‘‘๐‘ฅ . ใ€– logใ€—โก๐‘ฅ + ๐‘‘(logโก๐‘ฅ )/๐‘‘๐‘ฅ . ใ€–(๐‘ฅใ€—^2โˆ’ 3) 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (2๐‘ฅ โˆ’0) ใ€– logใ€—โก๐‘ฅ + 1/๐‘ฅ ร— ใ€–(๐‘ฅใ€—^2โˆ’ 3) 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = 2๐‘ฅ . logโก๐‘ฅ + (๐‘ฅ^2โˆ’ 3)/๐‘ฅ ๐‘‘๐‘ข/๐‘‘๐‘ฅ = u (2๐‘ฅ "." logโก๐‘ฅ "+ " (๐‘ฅ^2โˆ’ 3)/๐‘ฅ) ๐’…๐’–/๐’…๐’™ = ๐’™^(๐’™^๐Ÿโˆ’ ๐Ÿ‘) (๐Ÿ๐’™ "." ๐’๐’๐’ˆโก๐’™ "+ " (๐’™^๐Ÿโˆ’ ๐Ÿ‘)/๐’™) Calculating ๐’…๐’—/๐’…๐’™ ๐‘ฃ = (๐‘ฅโˆ’3)๐‘ฅ^2 Taking log on both sides log ๐‘ฃ=logโกใ€–(๐‘ฅโˆ’3)^(๐‘ฅ^2 ) ใ€— log ๐‘ฃ=ใ€–๐‘ฅ^2 . logใ€—โกใ€– (๐‘ฅโˆ’3)ใ€— Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘(logโก๐‘ฃ )/๐‘‘๐‘ฅ = (๐‘‘(ใ€–๐‘ฅ^2. logใ€—โกใ€– (๐‘ฅ โˆ’ 3)ใ€— ) )/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ฃ )/๐‘‘๐‘ฅ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฃ = (๐‘‘(ใ€–๐‘ฅ^2. logใ€—โกใ€– (๐‘ฅโˆ’3)ใ€— ) )/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ฃ )/๐‘‘๐‘ฃ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = (๐‘‘(ใ€–๐‘ฅ^2. logใ€—โกใ€– (๐‘ฅโˆ’3)ใ€— ) )/๐‘‘๐‘ฅ 1/๐‘ฃ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = (๐‘‘(ใ€–๐‘ฅ^2. logใ€—โกใ€– (๐‘ฅโˆ’3)ใ€— ) )/๐‘‘๐‘ฅ 1/๐‘ฃ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘‘(๐‘ฅ^2 )/๐‘‘๐‘ฅ . log (๐‘ฅโˆ’3) + ๐‘‘(log" " (๐‘ฅ โˆ’ 3))/๐‘‘๐‘ฅ . ๐‘ฅ^2 1/๐‘ฃ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = 2๐‘ฅ . log (๐‘ฅโˆ’3) + 1/((๐‘ฅ โˆ’ 3) ). (๐‘‘(๐‘ฅ โˆ’ 3)" " )/๐‘‘๐‘ฅ . ๐‘ฅ^2 1/๐‘ฃ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = 2๐‘ฅ . log (๐‘ฅโˆ’3) + 1/((๐‘ฅ โˆ’ 3) ) . ๐‘ฅ^2 1/๐‘ฃ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = 2๐‘ฅ. log (๐‘ฅโˆ’3) + ๐‘ฅ^2/(๐‘ฅ โˆ’3) ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘ฃ (2๐‘ฅ". " log" " (๐‘ฅโˆ’3)" + " ๐‘ฅ^2/(๐‘ฅ โˆ’3)) ๐’…๐’—/๐’…๐’™ = (๐’™โˆ’๐Ÿ‘)๐’™^๐Ÿ (๐Ÿ๐’™". " ๐ฅ๐จ๐ " " (๐’™โˆ’๐Ÿ‘)" + " ๐’™^๐Ÿ/(๐’™ โˆ’๐Ÿ‘)) Now, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ข/๐‘‘๐‘ฅ + ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐’™^(๐’™^๐Ÿโˆ’ ๐Ÿ‘) ((๐’™^๐Ÿโˆ’ ๐Ÿ‘)/๐’™+๐Ÿ๐’™ ๐ฅ๐จ๐ โก๐’™ ) + (๐’™โˆ’๐Ÿ‘)๐’™^๐Ÿ (๐’™^๐Ÿ/(๐’™ โˆ’๐Ÿ‘)+๐Ÿ๐’™ .๐ฅ๐จ๐ โก(๐’™ โˆ’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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.