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


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Example 30 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’– + ๐’— + ๐’˜ = ๐’‚^๐’ƒ 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 . 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 . 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’–/๐’…๐’™ = ๐’š^๐’™ (๐’๐’๐’ˆโก๐’š "+ " ๐’™/๐’š " " ๐’…๐’š/๐’…๐’™) 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 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’—/๐’…๐’™ = ๐’™^๐’š (๐’…๐’š/๐’…๐’™ ๐’๐’๐’ˆโกใ€–๐’™+ ๐’š/๐’™ใ€— ) 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" " 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’˜/๐’…๐’™ = ๐’™^๐’™ (๐’๐’๐’ˆโกใ€–๐’™+๐Ÿใ€— ) 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(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 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.