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

Transcript

Example 33 Find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ , if ๐‘ฆ^๐‘ฅ+๐‘ฅ^๐‘ฆ+๐‘ฅ^๐‘ฅ=๐‘Ž^๐‘. Let u = ๐‘ฆ๐‘ฅ, v = ๐‘ฅ๐‘ฆ & w = ๐‘ฅ^๐‘ฅ Now, ๐‘ข + ๐‘ฃ + ๐‘ค = ๐‘Ž^๐‘ Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ (๐‘‘ (๐‘ข + ๐‘ฃ + ๐‘ค))/๐‘‘๐‘ฅ = (๐‘‘(๐‘Ž^๐‘))/๐‘‘๐‘ฅ (๐‘‘(๐‘ข))/๐‘‘๐‘ฅ + (๐‘‘(๐‘ฃ))/๐‘‘๐‘ฅ + (๐‘‘(๐‘ค))/๐‘‘๐‘ฅ = 0 We will calculate derivative of u, v & w separately . (As ๐‘Ž^๐‘ is constant) โ€ฆ(1) Finding Derivative of 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 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (๐‘‘ (๐‘ฅ . logโก๐‘ฆ ))/๐‘‘๐‘ฅ (๐ด๐‘  logโกใ€–(๐‘Ž^๐‘)ใ€—=๐‘ logโก๐‘Ž) By product Rule (uv)โ€™ = uโ€™v + vโ€™u 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 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 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‘ฅ . 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 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = logโก๐‘ฆ + ๐‘ฅ/๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ 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) Finding derivative of v v = xy 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‘ฅ) = ๐‘‘(ใ€–๐‘ฆ logใ€—โก๐‘ฅ )/๐‘‘๐‘ฅ 1/๐‘ฃ (๐‘‘๐‘ฃ/๐‘‘๐‘ฅ) = ( ๐‘‘(ใ€–๐‘ฆ logใ€—โก๐‘ฅ ))/๐‘‘๐‘ฅ โ€ฆ(1) By product Rule (uv)โ€™ = uโ€™v + vโ€™u 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 + (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‘‘๐‘ฆ)/๐‘‘๐‘ฅ ๐‘ฅ+๐‘ฆ/๐‘ฅ) Putting values of ๐‘ฃ = ๐‘ฅ^๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = ๐‘ฅ^๐‘ฆ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ logโกใ€–๐‘ฅ+ ๐‘ฆ/๐‘ฅใ€— ) โ€ฆ(3) Calculating derivative of ๐’˜ ๐‘ค = ๐‘ฅ^๐‘ฅ Taking log both sides logโก๐‘ค=logโกใ€– (๐‘ฅ^๐‘ฅ)" " ใ€— logโก๐‘ค=ใ€–๐‘ฅ. logใ€—โก๐‘ฅ" " logโก๐‘ค=ใ€–๐‘ฅ. logใ€—โก๐‘ฅ" " Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ (๐‘‘(logโก๐‘ค))/๐‘‘๐‘ฅ = (๐‘‘(๐‘ฅ . logโก๐‘ฅ))/๐‘‘๐‘ฅ (๐‘‘(logโก๐‘ค))/๐‘‘๐‘ฅ (๐‘‘๐‘ค/๐‘‘๐‘ค) = ๐‘‘(๐‘ฅ logโก๐‘ฅ )/๐‘‘๐‘ฅ (๐‘‘(logโก๐‘ค))/๐‘‘๐‘ค . ๐‘‘๐‘ค/๐‘‘๐‘ฅ = ๐‘‘(๐‘ฅ logโก๐‘ฅ )/๐‘‘๐‘ฅ (๐ด๐‘  logโกใ€–(๐‘Ž^๐‘)ใ€—=๐‘ logโก๐‘Ž) 1/๐‘ค . ๐‘‘๐‘ค/๐‘‘๐‘ฅ = ๐‘‘(๐‘ฅ logโก๐‘ฅ )/๐‘‘๐‘ฅ logโก๐‘ค=ใ€–๐‘ฅ. logใ€—โก๐‘ฅ" " Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ (๐‘‘(logโก๐‘ค))/๐‘‘๐‘ฅ = (๐‘‘(๐‘ฅ . logโก๐‘ฅ))/๐‘‘๐‘ฅ (๐‘‘(logโก๐‘ค))/๐‘‘๐‘ฅ (๐‘‘๐‘ค/๐‘‘๐‘ค) = ๐‘‘(๐‘ฅ logโก๐‘ฅ )/๐‘‘๐‘ฅ (๐‘‘(logโก๐‘ค))/๐‘‘๐‘ค . ๐‘‘๐‘ค/๐‘‘๐‘ฅ = ๐‘‘(๐‘ฅ logโก๐‘ฅ )/๐‘‘๐‘ฅ 1/๐‘ค . ๐‘‘๐‘ค/๐‘‘๐‘ฅ = ๐‘‘(๐‘ฅ logโก๐‘ฅ )/๐‘‘๐‘ฅ By product Rule (uv)โ€™ = uโ€™v + vโ€™u 1/๐‘ค (๐‘‘๐‘ค/๐‘‘๐‘ฅ) = ( ๐‘‘(๐‘ฅ))/๐‘‘๐‘ฅ . logโก๐‘ฅ + (๐‘‘ (logโก๐‘ฅ))/๐‘‘๐‘ฅ . ๐‘ฅ 1/๐‘ค (๐‘‘๐‘ค/๐‘‘๐‘ฅ) = 1 . logโก๐‘ฅ + 1/๐‘ฅ . ๐‘ฅ 1/๐‘ค (๐‘‘๐‘ค/๐‘‘๐‘ฅ) = (logโกใ€–๐‘ฅ+1ใ€—) ๐‘‘๐‘ค/๐‘‘๐‘ฅ = ๐‘ค(logโกใ€–๐‘ฅ+1ใ€—) ๐‘‘๐‘ค/๐‘‘๐‘ฅ = ๐‘ฅ^๐‘ฅ (logโกใ€–๐‘ฅ+1ใ€— ) From (1) ๐‘‘๐‘ข/๐‘‘๐‘ฅ + ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + ๐‘‘๐‘ค/๐‘‘๐‘ฅ = 0 Putting values from (2), (3) & (4) (๐‘ฆ^๐‘ฅ logโกใ€–๐‘ฆ+๐‘ฆ^(๐‘ฅโˆ’1). ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ใ€— ) + (๐‘ฅ^๐‘ฆ logโกใ€–๐‘ฅ.๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+๐‘ฅ^๐‘ฆ.๐‘ฆ/๐‘ฅ ใ€— ) + (๐‘ฅ^๐‘ฅ (logโกใ€–๐‘ฅ+1ใ€—))=0 โ€ฆ(4) (๐‘ฆ^๐‘ฅ logโกใ€–๐‘ฆ+๐‘ฅ^๐‘ฆ. ๐‘ฆ/๐‘ฅ+๐‘ฅ^๐‘ฅ (logโกใ€–๐‘ฅ+1ใ€—)ใ€— ) + (๐‘ฆ^(๐‘ฅโˆ’1) .โกใ€–๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+๐‘ฅ^๐‘ฆ logโกใ€–๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅใ€— ใ€— ) = 0 (๐‘ฆ^(๐‘ฅโˆ’1) .โกใ€–๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+๐‘ฅ^๐‘ฆ logโกใ€–๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅใ€— ใ€— ) = โˆ’ (๐‘ฆ^๐‘ฅ logโกใ€–๐‘ฆ+๐‘ฅ^๐‘ฆ. ๐‘ฆ/๐‘ฅ+๐‘ฅ^๐‘ฅ (logโกใ€–๐‘ฅ+1ใ€—)ใ€— ) (๐‘ฆ^(๐‘ฅโˆ’1) .โกใ€–๐‘ฅ +๐‘ฅ^๐‘ฆ logโกใ€–๐‘ฅ ใ€— ใ€— ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’ (๐‘ฆ^๐‘ฅ logโกใ€–๐‘ฆ+๐‘ฅ^๐‘ฆ. ๐‘ฆ/๐‘ฅ+๐‘ฅ^๐‘ฅ (logโกใ€–๐‘ฅ+1ใ€—)ใ€— ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = "โˆ’" (๐‘ฆ^๐‘ฅ ๐‘™๐‘œ๐‘”โกใ€–๐‘ฆ + ๐‘ฅ^๐‘ฆ. ๐‘ฆ/๐‘ฅ + ๐‘ฅ^๐‘ฅ (1 + ๐‘™๐‘œ๐‘”โก๐‘ฅ)ใ€— )/((ใ€–๐‘ฅ๐‘ฆใ€—^(๐‘ฅโˆ’1) +โกใ€–๐‘ฅ^๐‘ฆ ๐‘™๐‘œ๐‘”โกใ€–๐‘ฅ ใ€— ใ€—)) ๐’…๐’š/๐’…๐’™ = "โˆ’" (๐’š^๐’™ ๐’๐’๐’ˆโกใ€–๐’š + ๐’™^(๐’š โˆ’ ๐Ÿ) ๐’š + ๐’™^๐’™ (๐Ÿ + ๐’๐’๐’ˆโก๐’™)ใ€— )/((ใ€–๐’™๐’šใ€—^(๐’™โˆ’๐Ÿ) +โกใ€–๐’™^๐’š ๐’๐’๐’ˆโกใ€–๐’™ ใ€— ใ€—))

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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.