# Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.