Ex 5.5, 18 - Chapter 5 Class 12 Continuity and Differentiability

Transcript

Ex 5.5, 18 If 𝑢 , 𝑣 and 𝑤 are functions of 𝑥, then show that 𝑑﷮𝑑𝑥﷯ (𝑢 . 𝑣 . 𝑤 ) = 𝑑𝑢﷮𝑑𝑥﷯ 𝑣. 𝑤+𝑢 . 𝑑𝑣﷮𝑑𝑥﷯ . 𝑤+𝑢 . 𝑣 𝑑𝑤﷮𝑑𝑥﷯ in two ways − first by repeated application of product rule, second by logarithmic differentiation. By product Rule Let 𝑦=𝑢𝑣𝑤 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦 ﷮𝑑𝑥﷯ = 𝑑 𝑢 𝑣 𝑤﷯﷮𝑑𝑥﷯ 𝑑𝑦 ﷮𝑑𝑥﷯ = 𝑑 𝑢𝑣﷯ 𝑤﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑢𝑣﷯ ﷮𝑑𝑥﷯ . 𝑤 + 𝑑 𝑤﷯ ﷮𝑑𝑥﷯ . 𝑢𝑣﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑢﷯ ﷮𝑑𝑥﷯ . 𝑣+ 𝑑 𝑣﷯ ﷮𝑑𝑥﷯ 𝑢﷯𝑤 + 𝑑 𝑤﷯﷮𝑑𝑥﷯ . 𝑢𝑣 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑢﷮𝑑𝑥﷯ . 𝑣.𝑤+ 𝑑𝑣﷮𝑑𝑥﷯ . 𝑢.𝑤 + 𝑑𝑤﷮𝑑𝑥﷯ . 𝑢𝑣 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑢﷮𝑑𝑥﷯ . 𝑣.𝑤+𝑢 𝑑𝑣﷮𝑑𝑥﷯ .𝑤+𝑢.𝑣. 𝑑𝑤﷮𝑑𝑥﷯ Hence , 𝒅(𝒖 . 𝒗 . 𝒘)﷮𝒅𝒙﷯ = 𝒅𝒖﷮𝒅𝒙﷯ . 𝒗.𝒘+𝒖 𝒅𝒗﷮𝒅𝒙﷯ .𝒘+𝒖.𝒗. 𝒅𝒘﷮𝒅𝒙﷯ Using logarithmic differentiation. Let 𝑦=𝑢𝑣𝑤 Taking log both sides log 𝑦 = log 𝑢𝑣𝑤﷯ log 𝑦=log 𝑢+ log ﷮𝑣﷯+ log﷮𝑤﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑 log﷮𝑦﷯﷯﷮𝑑𝑥﷯ = 𝑑 log 𝑢 + log ﷮𝑣﷯ + log﷮𝑤﷯﷯﷮𝑑𝑥﷯ 𝑑 log﷮𝑦﷯﷯﷮𝑑𝑥﷯ . 𝑑𝑦﷮𝑑𝑦﷯ = 𝑑 log 𝑢﷯﷮𝑑𝑥﷯ + 𝑑 log ﷮𝑣﷯﷯﷮𝑑𝑥﷯ + 𝑑 log﷮𝑤﷯﷯﷮𝑑𝑥﷯ 𝑑 log﷮𝑦﷯﷯﷮𝑑𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 log 𝑢﷯﷮𝑑𝑥﷯ + 𝑑 log ﷮𝑣﷯﷯﷮𝑑𝑥﷯ + 𝑑 log﷮𝑤﷯﷯﷮𝑑𝑥﷯ 1﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮𝑢﷯ . 𝑑 𝑢﷯﷮𝑑𝑥﷯ + 1﷮𝑣﷯. 𝑑 𝑣﷯﷮𝑑𝑥﷯ + 1﷮𝑤﷯ . 𝑑 𝑤﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑦 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ + 1﷮𝑣﷯. 𝑑𝑣﷮𝑑𝑥﷯ + 1﷮𝑤﷯ . 𝑑𝑤﷮𝑑𝑥﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑢𝑣𝑤 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ + 1﷮𝑣﷯. 𝑑𝑣﷮𝑑𝑥﷯ + 1﷮𝑤﷯ . 𝑑𝑤﷮𝑑𝑥﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑢𝑣𝑤 . 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ + 𝑢𝑣𝑤 . 1﷮𝑣﷯. 𝑑𝑣﷮𝑑𝑥﷯ + 𝑢𝑣𝑤 . 1﷮𝑤﷯. 𝑑𝑤﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑣𝑤 . 𝑑𝑢﷮𝑑𝑥﷯ + 𝑢𝑤 . 𝑑𝑣﷮𝑑𝑥﷯ + 𝑢𝑣 . 𝑑𝑤﷮𝑑𝑥﷯ 𝒅(𝒖 . 𝒗 . 𝒘)﷮𝒅𝒙﷯ = 𝒅𝒖﷮𝒅𝒙﷯ . 𝒗𝒘+𝒖 . 𝒅𝒗﷮𝒅𝒙﷯ .𝒘+𝒖.𝒗. 𝒅𝒘﷮𝒅𝒙﷯