Ex 5.5, 18 - Show that d/dx (u.v.w)=du/dx v.w + u dv/dx w

Ex 5.5, 18 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 18 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.5, 18 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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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 Using product Rule in (𝑢𝑣). 𝑤 (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 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 Again Using product Rule in 𝑢𝑣 (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 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 (As log (ab) = log a + log b) 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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.