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Example 33 - Find dy/dx, if yx + xy + xx = ab - Class 12 - Logarithmic Differentiation - Type 2

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Example 33 Find 𝑑𝑦﷮𝑑𝑥﷯ , if 𝑦𝑥 + 𝑥𝑦 + 𝑥﷮𝑥﷯=𝑎𝑏 . Let u = 𝑦𝑥, v = 𝑥𝑦 & w = 𝑥﷮𝑥﷯ Now, 𝑢 + 𝑣 + 𝑤 = 𝑎﷮𝑏﷯ Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑 (𝑢 + 𝑣 + 𝑤)﷮𝑑𝑥﷯ = 𝑑( 𝑎﷮𝑏﷯)﷮𝑑𝑥﷯ 𝑑(𝑢)﷮𝑑𝑥﷯ + 𝑑(𝑣)﷮𝑑𝑥﷯ + 𝑑(𝑤)﷮𝑑𝑥﷯ = 0 We will calculate derivative of u, v & w separately . Finding Derivative of 𝒖 . 𝑢 = 𝑦﷮𝑥﷯ Taking log both sides log﷮𝑢﷯= log﷮ ( 𝑦﷮𝑥﷯) ﷯ log﷮𝑢﷯= 𝑥 . log﷮𝑦 ﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑( log﷮𝑢﷯)﷮𝑑𝑥﷯ = 𝑑(𝑥 . log﷮𝑦﷯)﷮𝑑𝑥﷯ 𝑑( log﷮𝑢﷯)﷮𝑑𝑥﷯ 𝑑𝑢﷮𝑑𝑢﷯﷯ = 𝑑 𝑥. log﷮𝑦﷯﷯﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 𝑥 . log﷮𝑦﷯﷯﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 𝑥﷯﷮𝑑𝑥﷯ . log﷮𝑦﷯ + 𝑑( log﷮𝑦﷯)﷮𝑑﷯ . 𝑥 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 𝑥﷯﷮𝑑𝑥﷯ . log﷮𝑦﷯ + 𝑑( log﷮𝑦﷯)﷮𝑑﷯ . 𝑥 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 1 . log⁡𝑦 + 𝑥. 𝑑 log﷮𝑦﷯﷯﷮𝑑𝑥﷯ . 𝑑𝑦﷮𝑑𝑦﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = log⁡𝑦 + 𝑥. 𝑑 log﷮𝑦﷯﷯﷮𝑑𝑥﷯ . 𝑑𝑦﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = log⁡𝑦 + 𝑥. 1﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = log⁡𝑦 + 𝑥﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑢﷮𝑑𝑥﷯ = 𝑢 log⁡𝑦 + 𝑥﷮𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ 𝑑𝑢﷮𝑑𝑥﷯ = 𝑦﷮𝑥﷯ log⁡𝑦 + 𝑥﷮𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ Finding derivative of v v = xy Taking log both sides log﷮𝑣﷯= log﷮ ( 𝑥﷮𝑦﷯) ﷯ log﷮𝑣﷯= 𝑦. log﷮𝑥 ﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑( log﷮𝑣﷯)﷮𝑑𝑥﷯ = 𝑑(𝑦 . log﷮𝑥﷯)﷮𝑑𝑥﷯ 𝑑( log﷮𝑣﷯)﷮𝑑𝑥﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑 𝑦 log﷮𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑 𝑦 log﷮𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑 𝑦﷯﷮𝑑𝑥﷯ . log⁡𝑥 + 𝑑 ( log﷮𝑥﷯)﷮𝑑𝑥﷯ . 𝑦 1﷮𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑 𝑦﷯﷮𝑑𝑥﷯ . log⁡𝑥 + 𝑑 ( log﷮𝑥﷯)﷮𝑑𝑥﷯ . 𝑦 1﷮𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑𝑦﷮𝑑𝑥﷯ . log⁡𝑥 + 1﷮𝑥﷯ . 𝑦 1﷮𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑𝑦﷮𝑑𝑥﷯ log⁡𝑥 + 𝑦﷮𝑥﷯ 𝑑𝑣﷮𝑑𝑥﷯ = v log 𝑑𝑦﷮𝑑𝑥﷯ 𝑥+ 𝑦﷮𝑥﷯﷯ Putting values of 𝑣 = 𝑥﷮𝑦﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥﷮𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯ log﷮𝑥+ 𝑦﷮𝑥﷯﷯﷯ Calculating derivative of 𝒘 𝑤 = 𝑥﷮𝑥﷯ Taking log both sides log﷮𝑤﷯= log﷮ ( 𝑥﷮𝑥﷯) ﷯ log﷮𝑤﷯= 𝑥. log﷮𝑥 ﷯ log﷮𝑤﷯= 𝑥. log﷮𝑥 ﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑( log﷮𝑤﷯)﷮𝑑𝑥﷯ = 𝑑(𝑥 . log﷮𝑥﷯)﷮𝑑𝑥﷯ 𝑑( log﷮𝑤﷯)﷮𝑑𝑥﷯ 𝑑𝑤﷮𝑑𝑤﷯﷯ = 𝑑 𝑥 log﷮𝑥﷯﷯﷮𝑑𝑥﷯ 𝑑( log﷮𝑤﷯)﷮𝑑𝑤﷯ . 𝑑𝑤﷮𝑑𝑥﷯ = 𝑑 𝑥 log﷮𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑤﷯ . 𝑑𝑤﷮𝑑𝑥﷯ = 𝑑 𝑥 log﷮𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑤﷯ 𝑑𝑤﷮𝑑𝑥﷯﷯ = 𝑑 𝑥﷯﷮𝑑𝑥﷯ . log⁡𝑥 + 𝑑 ( log﷮𝑥﷯)﷮𝑑𝑥﷯ . 𝑥 1﷮𝑤﷯ 𝑑𝑤﷮𝑑𝑥﷯﷯ = 1 . log⁡𝑥 + 1﷮𝑥﷯ . 𝑥 1﷮𝑤﷯ 𝑑𝑤﷮𝑑𝑥﷯﷯ = ( log﷮𝑥+1﷯) 𝑑𝑤﷮𝑑𝑥﷯ = 𝑤( log﷮𝑥+1﷯) 𝑑𝑤﷮𝑑𝑥﷯ = 𝑥﷮𝑥﷯ log﷮𝑥+1﷯﷯ From (1) 𝑑𝑢﷮𝑑𝑥﷯ + 𝑑𝑣﷮𝑑𝑥﷯ + 𝑑𝑤﷮𝑑𝑥﷯ = 0 Putting values from (2), (3) & (4) 𝑦﷮𝑥﷯ log﷮𝑦+ 𝑦﷮𝑥−1﷯. 𝑥 𝑑𝑦﷮𝑑𝑥﷯ ﷯﷯ + 𝑥﷮𝑦﷯ log﷮𝑥. 𝑑𝑦﷮𝑑𝑥﷯+ 𝑥﷮𝑦﷯. 𝑦﷮𝑥﷯ ﷯﷯ + 𝑥﷮𝑥﷯( log﷮𝑥+1﷯)﷯=0 𝑦﷮𝑥﷯ log﷮𝑦+ 𝑥﷮𝑦﷯. 𝑦﷮𝑥﷯+ 𝑥﷮𝑥﷯ ( log﷮𝑥+1﷯)﷯﷯ + 𝑦﷮𝑥−1﷯ .﷮𝑥 𝑑𝑦﷮𝑑𝑥﷯+ 𝑥﷮𝑦﷯ log﷮𝑥 𝑑𝑦﷮𝑑𝑥﷯﷯﷯﷯ = 0 𝑦﷮𝑥−1﷯ .﷮𝑥 𝑑𝑦﷮𝑑𝑥﷯+ 𝑥﷮𝑦﷯ log﷮𝑦 𝑑𝑦﷮𝑑𝑥﷯﷯﷯﷯ = − 𝑦﷮𝑥﷯ log﷮𝑦+ 𝑥﷮𝑦﷯. 𝑦﷮𝑥﷯+ 𝑥﷮𝑥﷯ ( log﷮𝑥+1﷯)﷯﷯ 𝑦﷮𝑥−1﷯ .﷮𝑥 + 𝑥﷮𝑦﷯ log﷮𝑥 ﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − 𝑦﷮𝑥﷯ log﷮𝑦+ 𝑥﷮𝑦﷯. 𝑦﷮𝑥﷯+ 𝑥﷮𝑥﷯ ( log﷮𝑥+1﷯)﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = − 𝒚﷮𝒙﷯ 𝒍𝒐𝒈﷮𝒚 + 𝒙﷮𝒚﷯. 𝒚﷮𝒙﷯ + 𝒙﷮𝒙﷯ (𝟏 + 𝒍𝒐𝒈﷮𝒙﷯)﷯﷯﷮( 𝒙𝒚﷮𝒙−𝟏﷯ +﷮ 𝒙﷮𝒚﷯ 𝒍𝒐𝒈﷮𝒙 ﷯﷯)﷯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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