# Ex 5.5, 14 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.5, 14 Find 𝑑𝑦𝑑𝑥 of the functions in, ( cos𝑥 )𝑦 = ( cos𝑦 )𝑥 Given ( cos𝑥)𝑦 = ( cos𝑦)𝑥 Taking log both sides log ( cos𝑥)𝑦 = log ( cos𝑦)𝑥 𝑦 . log ( cos𝑥)=𝑥. log( cos𝑦) Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑(𝑦 . log ( cos𝑥))𝑑𝑥 = 𝑑 𝑥. log( cos𝑦)𝑑𝑥 Finding 𝒅(𝒚 . 𝒍𝒐𝒈 ( 𝒄𝒐𝒔𝒙))𝒅𝒙 𝑑(𝑦 . 𝑙𝑜𝑔 ( 𝑐𝑜𝑠𝑥))𝑑𝑥 = 𝑑(𝑦)𝑑𝑥 . log co𝑠𝑥 + 𝑑(𝑙𝑜𝑔 ( 𝑐𝑜𝑠𝑥))𝑑𝑥 . 𝑦 = 𝑑𝑦𝑑𝑥 . log co𝑠𝑥 + 1 𝑐𝑜𝑠𝑥 . 𝑑 𝑐𝑜𝑠𝑥𝑑𝑥 . 𝑦 = 𝑑𝑦𝑑𝑥 . log co𝑠𝑥 + 1 𝑐𝑜𝑠𝑥 . − sin𝑥 . 𝑦 = 𝑑𝑦𝑑𝑥 . log co𝑠𝑥 + − sin𝑥 𝑐𝑜𝑠𝑥 . 𝑦 = 𝑑𝑦𝑑𝑥 . log co𝑠𝑥− tan𝑥. 𝑦 Finding 𝒅 𝒙. 𝒍𝒐𝒈( 𝒄𝒐𝒔𝒚)𝒅𝒙 𝑑 𝑥. 𝑙𝑜𝑔( 𝑐𝑜𝑠𝑦)𝑑𝑥 = 𝑑(𝑥)𝑑𝑥 . log co𝑠𝑦 + 𝑑(𝑙𝑜𝑔 ( 𝑐𝑜𝑠𝑦))𝑑𝑥 . 𝑥 = log co𝑠𝑦 + 1 𝑐𝑜𝑠𝑦 . 𝑑 𝑐𝑜𝑠𝑦𝑑𝑥 . 𝑥 = log co𝑠𝑦 + 1 𝑐𝑜𝑠𝑦 . 𝑑 𝑐𝑜𝑠𝑦𝑑𝑥 . 𝑑𝑦𝑑𝑦 . 𝑥 = log co𝑠𝑦 + 1 𝑐𝑜𝑠𝑦 . 𝑑 𝑐𝑜𝑠𝑦𝑑𝑦 . 𝑑𝑦𝑑𝑥 . 𝑥 = log co𝑠𝑦 + 1 𝑐𝑜𝑠𝑦 . (− sin𝑦) . 𝑑𝑦𝑑𝑥 . 𝑥 = log co𝑠𝑦 + − tan𝑦 . 𝑥 . 𝑑𝑦𝑑𝑥 Now , 𝑑(𝑦 . log ( cos𝑥))𝑑𝑥 = 𝑑 𝑥. log( cos𝑦)𝑑𝑥 𝑑𝑦𝑑𝑥 log co𝑠𝑥− tan𝑥. 𝑦 = log co𝑠𝑦 − tan𝑦 . 𝑥 . 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 log co𝑠𝑥−𝑦 . tan𝑥 = log co𝑠𝑦 − 𝑥 . tan𝑦 . 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 log co𝑠𝑥+𝑥 tan 𝑑𝑦𝑑𝑥 = log co𝑠𝑦 + 𝑦 tan𝑥 𝑑𝑦𝑑𝑥 log co𝑠𝑥+𝑥 tan 𝑦 = log co𝑠𝑦 + 𝑦 tan𝑥 𝒅𝒚𝒅𝒙 = 𝐥𝐨𝐠 𝒄𝒐𝒔𝒚 + 𝒚 𝒕𝒂𝒏𝒙𝐥𝐨𝐠 𝒄𝒐𝒔𝒙 + 𝒙 𝐭𝐚𝐧 𝒚

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Davneet Singh

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.