Ex 5.5, 14 - Find dy/dx of (cos x)y = (cos y)x - Logarithmic Differentiation - Type 1

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Ex 5.5, 14 Find 𝑑𝑦﷮𝑑𝑥﷯ of the functions in, ( cos﷮𝑥 ﷯)﷮𝑦﷯ = ( cos﷮𝑦 ﷯)﷮𝑥﷯ Given ( cos﷮𝑥﷯)﷮𝑦﷯ = ( cos﷮𝑦﷯)﷮𝑥﷯ Taking log both sides log ( cos﷮𝑥﷯)﷮𝑦﷯ = log ( cos﷮𝑦﷯)﷮𝑥﷯ 𝑦 . log ( cos﷮𝑥﷯)=𝑥. log﷮( cos﷮𝑦﷯)﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑(𝑦 . log ( cos﷮𝑥﷯))﷮𝑑𝑥﷯ = 𝑑 𝑥. log﷮( cos﷮𝑦﷯)﷯﷯﷮𝑑𝑥﷯ Finding 𝒅(𝒚 . 𝒍𝒐𝒈 ( 𝒄𝒐𝒔﷮𝒙﷯))﷮𝒅𝒙﷯ 𝑑(𝑦 . 𝑙𝑜𝑔 ( 𝑐𝑜𝑠﷮𝑥﷯))﷮𝑑𝑥﷯ = 𝑑(𝑦)﷮𝑑𝑥﷯ . log co𝑠﷮𝑥﷯ + 𝑑(𝑙𝑜𝑔 ( 𝑐𝑜𝑠﷮𝑥﷯))﷮𝑑𝑥﷯ . 𝑦 = 𝑑𝑦﷮𝑑𝑥﷯ . log co𝑠﷮𝑥﷯ + 1﷮ 𝑐𝑜𝑠﷮𝑥﷯﷯ . 𝑑 𝑐𝑜𝑠﷮𝑥﷯﷯﷮𝑑𝑥﷯ . 𝑦 = 𝑑𝑦﷮𝑑𝑥﷯ . log co𝑠﷮𝑥﷯ + 1﷮ 𝑐𝑜𝑠﷮𝑥﷯﷯ . − sin﷮𝑥﷯﷯ . 𝑦 = 𝑑𝑦﷮𝑑𝑥﷯ . log co𝑠﷮𝑥﷯ + − sin﷮𝑥﷯﷯﷮ 𝑐𝑜𝑠﷮𝑥﷯﷯ . 𝑦 = 𝑑𝑦﷮𝑑𝑥﷯ . log co𝑠﷮𝑥﷯− tan﷮𝑥﷯. 𝑦 Finding 𝒅 𝒙. 𝒍𝒐𝒈﷮( 𝒄𝒐𝒔﷮𝒚﷯)﷯﷯﷮𝒅𝒙﷯ 𝑑 𝑥. 𝑙𝑜𝑔﷮( 𝑐𝑜𝑠﷮𝑦﷯)﷯﷯﷮𝑑𝑥﷯ = 𝑑(𝑥)﷮𝑑𝑥﷯ . log co𝑠﷮𝑦﷯ + 𝑑(𝑙𝑜𝑔 ( 𝑐𝑜𝑠﷮𝑦﷯))﷮𝑑𝑥﷯ . 𝑥 = log co𝑠﷮𝑦﷯ + 1﷮ 𝑐𝑜𝑠﷮𝑦﷯﷯ . 𝑑 𝑐𝑜𝑠﷮𝑦﷯﷯﷮𝑑𝑥﷯ . 𝑥 = log co𝑠﷮𝑦﷯ + 1﷮ 𝑐𝑜𝑠﷮𝑦﷯﷯ . 𝑑 𝑐𝑜𝑠﷮𝑦﷯﷯﷮𝑑𝑥﷯ . 𝑑𝑦﷮𝑑𝑦﷯ . 𝑥 = log co𝑠﷮𝑦﷯ + 1﷮ 𝑐𝑜𝑠﷮𝑦﷯﷯ . 𝑑 𝑐𝑜𝑠﷮𝑦﷯﷯﷮𝑑𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ . 𝑥 = log co𝑠﷮𝑦﷯ + 1﷮ 𝑐𝑜𝑠﷮𝑦﷯﷯ . (− sin﷮𝑦﷯) . 𝑑𝑦﷮𝑑𝑥﷯ . 𝑥 = log co𝑠﷮𝑦﷯ + − tan﷮𝑦﷯ . 𝑥 . 𝑑𝑦﷮𝑑𝑥﷯ Now , 𝑑(𝑦 . log ( cos﷮𝑥﷯))﷮𝑑𝑥﷯ = 𝑑 𝑥. log﷮( cos﷮𝑦﷯)﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ log co𝑠﷮𝑥﷯− tan﷮𝑥﷯. 𝑦 = log co𝑠﷮𝑦﷯ − tan﷮𝑦﷯ . 𝑥 . 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ log co𝑠﷮𝑥﷯−𝑦 . tan﷮𝑥﷯ = log co𝑠﷮𝑦﷯ − 𝑥 . tan﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ log co𝑠﷮𝑥﷯+𝑥 tan 𝑑𝑦﷮𝑑𝑥﷯ = log co𝑠﷮𝑦﷯ + 𝑦 tan﷮𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ log co𝑠﷮𝑥﷯+𝑥 tan 𝑦﷯ = log co𝑠﷮𝑦﷯ + 𝑦 tan﷮𝑥﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝐥𝐨𝐠 𝒄𝒐𝒔﷮𝒚﷯ + 𝒚 𝒕𝒂𝒏﷮𝒙﷯﷮𝐥𝐨𝐠 𝒄𝒐𝒔﷮𝒙﷯ + 𝒙 𝐭𝐚𝐧 𝒚﷯

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