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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Concept wise

Transcript

Ex 5.5, 12 Find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ of the functions in, ๐‘ฅ^๐‘ฆ + ๐‘ฆ^๐‘ฅ = 1 ๐‘ฅ^๐‘ฆ + ๐‘ฆ^๐‘ฅ = 1 Let ๐‘ข = ๐‘ฅ^๐‘ฆ , ๐‘ฃ = ๐‘ฆ^๐‘ฅ Hence, ๐‘ข+๐‘ฃ=1 Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. (๐‘‘(๐‘ฃโกใ€–+ ๐‘ขใ€—))/๐‘‘๐‘ฅ = ๐‘‘(1)/๐‘‘๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + ๐‘‘๐‘ข/๐‘‘๐‘ฅ = 0 (Derivative of constant is 0) Calculating ๐’…๐’—/๐’…๐’™ ๐‘ฃ=๐‘ฅ^๐‘ฆ Taking log both sides logโก๐‘ฃ=logโกใ€– (๐‘ฅ^๐‘ฆ)" " ใ€— logโก๐‘ฃ=ใ€–๐‘ฆ. logใ€—โก๐‘ฅ" " Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ (๐‘‘(logโก๐‘ฃ))/๐‘‘๐‘ฅ = (๐‘‘(๐‘ฆ . logโก๐‘ฅ))/๐‘‘๐‘ฅ (๐‘‘(logโก๐‘ฃ))/๐‘‘๐‘ฅ (๐‘‘๐‘ฃ/๐‘‘๐‘ฃ) = ๐‘‘(๐‘ฆ logโก๐‘ฅ )/๐‘‘๐‘ฅ (๐‘‘(logโก๐‘ฃ))/๐‘‘๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = ๐‘‘(ใ€–๐‘ฆ logใ€—โก๐‘ฅ )/๐‘‘๐‘ฅ (๐ด๐‘  logโกใ€–(๐‘Ž^๐‘)ใ€—=๐‘ logโก๐‘Ž) 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = ( ๐‘‘(ใ€–๐‘ฆ logใ€—โก๐‘ฅ ))/๐‘‘๐‘ฅ 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = ( ๐‘‘(๐‘ฆ))/๐‘‘๐‘ฅ . logโก๐‘ฅ + (๐‘‘ (logโก๐‘ฅ))/๐‘‘๐‘ฅ . ๐‘ฆ 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = ( ๐‘‘๐‘ฆ)/๐‘‘๐‘ฅ . logโก๐‘ฅ + 1/๐‘ฅ . ๐‘ฆ 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = ( ๐‘‘๐‘ฆ)/๐‘‘๐‘ฅ logโก๐‘ฅ + ๐‘ฆ/๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = v (log ๐‘ฅ+๐‘ฆ/๐‘ฅ) Putting value of ๐‘ฃ = ๐‘ฅ^๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘ฅ^๐‘ฆ (logโกใ€–๐‘ฅ+ ๐‘ฆ/๐‘ฅใ€— ) ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘ฅ^๐‘ฆ.log ๐‘ฅ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ + ๐‘ฅ^๐‘ฆ. ๐‘ฆ/๐‘ฅ Using product Rule As (๐‘ข๐‘ฃ)โ€™ = ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข Calculating ๐’…๐’–/๐’…๐’™ ๐‘ข = ๐‘ฆ^๐‘ฅ Taking log both sides logโก๐‘ข=logโกใ€– (๐‘ฆ^๐‘ฅ)" " ใ€— logโก๐‘ข=ใ€–๐‘ฅ . logใ€—โก๐‘ฆ" " Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ (๐‘‘(logโก๐‘ข))/๐‘‘๐‘ฅ = (๐‘‘(๐‘ฅ . logโก๐‘ฆ))/๐‘‘๐‘ฅ (๐‘‘(logโก๐‘ข))/๐‘‘๐‘ฅ (๐‘‘๐‘ข/๐‘‘๐‘ข) = ๐‘‘(๐‘ฅ.logโก๐‘ฆ )/๐‘‘๐‘ฅ (๐ด๐‘  logโกใ€–(๐‘Ž^๐‘)ใ€—=๐‘ logโก๐‘Ž ) (๐‘‘(logโก๐‘ข))/๐‘‘๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ) = (๐‘‘ (๐‘ฅ . logโก๐‘ฆ))/๐‘‘๐‘ฅ 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (๐‘‘ (๐‘ฅ . logโก๐‘ฆ ))/๐‘‘๐‘ฅ 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (๐‘‘ (๐‘ฅ))/๐‘‘๐‘ฅ . logโก๐‘ฆ + (๐‘‘(logโก๐‘ฆ))/๐‘‘๐‘ฅ . ๐‘ฅ 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = 1 . logโก๐‘ฆ + ๐‘ฅ. ๐‘‘(logโก๐‘ฆ )/๐‘‘๐‘ฅ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = logโก๐‘ฆ + ๐‘ฅ. ๐‘‘(logโก๐‘ฆ )/๐‘‘๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = logโก๐‘ฆ + ๐‘ฅ. 1/๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Using product Rule As (๐‘ข๐‘ฃ)โ€™ = ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข ๐‘‘๐‘ข/๐‘‘๐‘ฅ = ๐‘ข(logโก๐‘ฆ " + " ๐‘ฅ/๐‘ฆ " . " (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ) ๐‘‘๐‘ข/๐‘‘๐‘ฅ " = " ๐‘ฆ^๐‘ฅ (logโก๐‘ฆ " + " ๐‘ฅ/๐‘ฆ " . " (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ) ๐‘‘๐‘ข/๐‘‘๐‘ฅ = ๐‘ฆ^๐‘ฅ logโก๐‘ฆ+๐‘ฆ^(๐‘ฅ โˆ’1). ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Now, ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + ๐‘‘๐‘ข/๐‘‘๐‘ฅ = 0 Putting value of ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ & ๐‘‘๐‘ข/๐‘‘๐‘ฅ (๐‘ฅ^๐‘ฆ logโกใ€–๐‘ฅ ใ€— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ +๐‘ฅ^๐‘ฆ ๐‘ฆ/๐‘ฅ) + (๐‘ฆ^๐‘ฅ logโกใ€–๐‘ฆ+๐‘ฆ^(๐‘ฅ โˆ’1) ใ€—.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = 0 ๐‘ฅ^๐‘ฆ logโกใ€–๐‘ฅ ใ€— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ +๐‘ฅ^(๐‘ฆโˆ’1) ๐‘ฆ + ๐‘ฆ^๐‘ฅ logโกใ€–๐‘ฆ+๐‘ฆ^(๐‘ฅ โˆ’1) ใ€—.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 ๐‘ฅ^๐‘ฆ logโกใ€–๐‘ฅ ใ€— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+๐‘ฆ^(๐‘ฅ โˆ’1).๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=โˆ’(๐‘ฅ^(๐‘ฆโˆ’1) ๐‘ฆ+๐‘ฆ^๐‘ฅ logโก๐‘ฆ ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ(๐‘ฅ^๐‘ฆ logโกใ€–๐‘ฅ ใ€—+๐‘ฆ^(๐‘ฅ โˆ’1).๐‘ฅ)=โˆ’(๐‘ฅ^(๐‘ฆโˆ’1) ๐‘ฆ+๐‘ฆ^๐‘ฅ logโก๐‘ฆ ) ๐’…๐’š/๐’…๐’™ = (โˆ’(๐’™^(๐’šโˆ’๐Ÿ) ๐’š + ๐’š^๐’™ ๐’๐’๐’ˆโก๐’š ))/((๐’™^๐’š ๐’๐’๐’ˆโกใ€–๐’™ ใ€—+ ๐’š^(๐’™ โˆ’๐Ÿ).๐’™))

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Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.