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Misc  11 - Differentiate the function - x^(x^2-3) + (x-3)^(x^2)

Misc  11 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc  11 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Misc  11 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Misc  11 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Misc  11 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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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 > 3Let 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 " " (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‘‘(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) Using Product rule As (๐‘ข๐‘ฃ)โ€™ = ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข 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 (As logโก(๐‘Ž^๐‘) = ๐‘ logโก๐‘Ž) 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)Using product rule (๐‘ข๐‘ฃ)โ€™ = ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.