




Last updated at Sept. 17, 2019 by Teachoo
Transcript
Ex 5.5, 12 Find 𝑑𝑦𝑑𝑥 of the functions in, 𝑥𝑦 + 𝑦𝑥 = 1 𝑥𝑦 + 𝑦𝑥 = 1 Let 𝑢 = 𝑥𝑦 , 𝑣 = 𝑦𝑥 Hence, 𝑢+𝑣=1 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑( 𝑣+ 𝑢)𝑑𝑥 = 𝑑 1𝑑𝑥 𝑑(𝑣)𝑑𝑥 + 𝑑𝑢𝑑𝑥 = 0 Calculating 𝒅𝒗𝒅𝒙 𝑣= 𝑥𝑦 Taking log both sides log𝑣= log ( 𝑥𝑦) log𝑣= 𝑦. log𝑥 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑( log𝑣)𝑑𝑥 = 𝑑(𝑦 . log𝑥)𝑑𝑥 𝑑( log𝑣)𝑑𝑥 𝑑𝑣𝑑𝑣 = 𝑑 𝑦 log𝑥𝑑𝑥 𝑑( log𝑣)𝑑𝑣 𝑑𝑣𝑑𝑥 = 𝑑 𝑦 log𝑥𝑑𝑥 1𝑣 𝑑𝑣𝑑𝑥 = 𝑑 𝑦 log𝑥𝑑𝑥 1𝑣 𝑑𝑣𝑑𝑥 = 𝑑 𝑦𝑑𝑥 . log𝑥 + 𝑑 ( log𝑥)𝑑𝑥 . 𝑦 1𝑣 𝑑𝑣𝑑𝑥 = 𝑑𝑦𝑑𝑥 . log𝑥 + 1𝑥 . 𝑦 1𝑣 𝑑𝑣𝑑𝑥 = 𝑑𝑦𝑑𝑥 log𝑥 + 𝑦𝑥 𝑑𝑣𝑑𝑥 = v log 𝑥+ 𝑦𝑥 Putting value of 𝑣 = 𝑥𝑦 𝑑𝑣𝑑𝑥 = 𝑥𝑦 log𝑥+ 𝑦𝑥 𝑑𝑣𝑑𝑥 = 𝑥𝑦.log 𝑥 . 𝑑𝑦𝑑𝑥 + 𝑥𝑦. 𝑦𝑥 Calculating 𝒅𝒖𝒅𝒙 𝑢 = 𝑦𝑥 Taking log both sides log𝑢= log ( 𝑦𝑥) log𝑢= 𝑥 . log𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑( log𝑢)𝑑𝑥 = 𝑑(𝑥 . log𝑦)𝑑𝑥 𝑑( log𝑢)𝑑𝑥 𝑑𝑢𝑑𝑢 = 𝑑 𝑥. log𝑦𝑑𝑥 𝑑( log𝑢)𝑑𝑢 𝑑𝑢𝑑𝑥 = 𝑑 (𝑥 . log𝑦)𝑑𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = 𝑑 𝑥 . log𝑦𝑑𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = 𝑑 𝑥𝑑𝑥 . log𝑦 + 𝑑( log𝑦)𝑑𝑥 . 𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = 1 . log𝑦 + 𝑥. 𝑑 log𝑦𝑑𝑥 . 𝑑𝑦𝑑𝑦 1𝑢 . 𝑑𝑢𝑑𝑥 = log𝑦 + 𝑥. 𝑑 log𝑦𝑑𝑦 . 𝑑𝑦𝑑𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = log𝑦 + 𝑥. 1𝑦 . 𝑑𝑦𝑑𝑥 𝑑𝑢𝑑𝑥 = 𝑢 log𝑦 + 𝑥𝑦 . 𝑑𝑦 𝑑𝑥 𝑑𝑢𝑑𝑥 = 𝑦𝑥 log𝑦 + 𝑥𝑦 . 𝑑𝑦 𝑑𝑥 𝑑𝑢𝑑𝑥 = 𝑦𝑥 log𝑦+ 𝑦𝑥 −1. 𝑥 𝑑𝑦𝑑𝑥 Now, 𝑑𝑢𝑑𝑥 + 𝑑𝑣𝑑𝑥 = 0 Putting value of 𝑑𝑣𝑑𝑥 & 𝑑𝑢𝑑𝑥 𝑥𝑦 log𝑥 𝑑𝑦𝑑𝑥 + 𝑥𝑦 𝑦𝑥 + 𝑦𝑥 log𝑦+ 𝑦𝑥 −1 𝑑𝑦𝑑𝑥 = 0 𝑥𝑦 log𝑥 𝑑𝑦𝑑𝑥 + 𝑦𝑥 −1 𝑑𝑦𝑑𝑥 + 𝑥𝑦 . 𝑦𝑥 + 𝑦𝑥 log 𝑦 = 0 𝑑𝑦𝑑𝑥 𝑥𝑦 log𝑥 + 𝑦𝑥 −1 = − 𝑥𝑦 −1×𝑦+ 𝑦𝑥 log 𝑦 𝒅𝒚𝒅𝒙 = − 𝒚 𝒙𝒚 −𝟏 + 𝒚𝒙 𝐥𝐨𝐠 𝒚 𝒙𝒚 𝒍𝒐𝒈𝒙 + 𝒚𝒙 −𝟏
Ex 5.5
Ex 5.5, 2
Ex 5.5, 3 Important
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Ex 5.5,6 Important
Ex 5.5, 7 Important
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Ex 5.5, 9 Important
Ex 5.5, 10 Important
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Ex 5.5, 12 You are here
Ex 5.5, 13
Ex 5.5, 14 Important
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Ex 5.5, 16 Important
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Ex 5.5, 18
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