Ex 5.5, 12 - Find dy/dx, xy + yx = 1 - Class 12 CBSE NCERT

Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 4 Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 5 Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 6 Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 7

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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 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 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.