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

  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.5, 18 If ๐‘ข , ๐‘ฃ and ๐‘ค are functions of ๐‘ฅ, then show that ๐‘‘/๐‘‘๐‘ฅ (๐‘ข . ๐‘ฃ . ๐‘ค ) = ๐‘‘๐‘ข/๐‘‘๐‘ฅ ๐‘ฃ. ๐‘ค+๐‘ข . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ . ๐‘ค+๐‘ข . ๐‘ฃ ๐‘‘๐‘ค/๐‘‘๐‘ฅ in two ways โˆ’ first by repeated application of product rule, second by logarithmic differentiation. By product Rule Let ๐‘ฆ=๐‘ข๐‘ฃ๐‘ค Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = ๐‘‘(๐‘ข ๐‘ฃ ๐‘ค)/๐‘‘๐‘ฅ (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = ๐‘‘((๐‘ข๐‘ฃ) ๐‘ค)/๐‘‘๐‘ฅ Using product Rule in (๐‘ข๐‘ฃ). ๐‘ค (๐‘ข๐‘ฃ)โ€™ = ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘‘(๐‘ข๐‘ฃ)" " /๐‘‘๐‘ฅ . ๐‘ค + ๐‘‘(๐‘ค)" " /๐‘‘๐‘ฅ . (๐‘ข๐‘ฃ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= (๐‘‘(๐‘ข)" " /๐‘‘๐‘ฅ " . " ๐‘ฃ+ ๐‘‘(๐‘ฃ)" " /๐‘‘๐‘ฅ ๐‘ข)๐‘ค + ๐‘‘(๐‘ค)/๐‘‘๐‘ฅ . ๐‘ข๐‘ฃ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ข/๐‘‘๐‘ฅ . ๐‘ฃ.๐‘ค+ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ . ๐‘ข.๐‘ค + ๐‘‘๐‘ค/๐‘‘๐‘ฅ . ๐‘ข๐‘ฃ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ข/๐‘‘๐‘ฅ . ๐‘ฃ.๐‘ค+๐‘ข ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ .๐‘ค+๐‘ข.๐‘ฃ. ๐‘‘๐‘ค/๐‘‘๐‘ฅ Hence , (๐’…"(" ๐’– . ๐’—" . " ๐’˜")" )/๐’…๐’™ = ๐’…๐’–/๐’…๐’™ . ๐’—.๐’˜+๐’– ๐’…๐’—/๐’…๐’™ .๐’˜+๐’–.๐’—. ๐’…๐’˜/๐’…๐’™ Again Using product Rule in ๐‘ข๐‘ฃ (๐‘ข๐‘ฃ)โ€™ = ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข Using logarithmic differentiation. Let ๐‘ฆ=๐‘ข๐‘ฃ๐‘ค Taking log both sides log ๐‘ฆ = log (๐‘ข๐‘ฃ๐‘ค) log ๐‘ฆ=log ๐‘ข+ใ€–log ใ€—โก๐‘ฃ+logโก๐‘ค Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘(logโก๐‘ฆ )/๐‘‘๐‘ฅ = ๐‘‘(log ๐‘ข + ใ€–log ใ€—โก๐‘ฃ + logโก๐‘ค )/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ฆ )/๐‘‘๐‘ฅ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ = ๐‘‘(log ๐‘ข)/๐‘‘๐‘ฅ + ๐‘‘(ใ€–log ใ€—โก๐‘ฃ )/๐‘‘๐‘ฅ + ๐‘‘(logโก๐‘ค )/๐‘‘๐‘ฅ (As log (ab) = log a + log b) ๐‘‘(logโก๐‘ฆ )/๐‘‘๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(log ๐‘ข)/๐‘‘๐‘ฅ + ๐‘‘(ใ€–log ใ€—โก๐‘ฃ )/๐‘‘๐‘ฅ + ๐‘‘(logโก๐‘ค )/๐‘‘๐‘ฅ 1/๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/๐‘ข . ๐‘‘(๐‘ข)/๐‘‘๐‘ฅ + 1/๐‘ฃ. ๐‘‘(๐‘ฃ)/๐‘‘๐‘ฅ + 1/๐‘ค . ๐‘‘(๐‘ค)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฆ (1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ + 1/๐‘ฃ. ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + 1/๐‘ค . ๐‘‘๐‘ค/๐‘‘๐‘ฅ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ข๐‘ฃ๐‘ค (1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ + 1/๐‘ฃ. ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + 1/๐‘ค . ๐‘‘๐‘ค/๐‘‘๐‘ฅ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ข๐‘ฃ๐‘ค . 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ + ๐‘ข๐‘ฃ๐‘ค . 1/๐‘ฃ. ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + ๐‘ข๐‘ฃ๐‘ค . 1/๐‘ค. ๐‘‘๐‘ค/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฃ๐‘ค . ๐‘‘๐‘ข/๐‘‘๐‘ฅ + ๐‘ข๐‘ค . ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + ๐‘ข๐‘ฃ . ๐‘‘๐‘ค/๐‘‘๐‘ฅ (๐’…"(" ๐’– . ๐’—" . " ๐’˜")" )/๐’…๐’™ = ๐’…๐’–/๐’…๐’™ . ๐’—๐’˜+๐’– . ๐’…๐’—/๐’…๐’™ .๐’˜+๐’–.๐’—. ๐’…๐’˜/๐’…๐’™

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.