Ex 5.7, 5 - Find second order derivatives of x3 logx - Ex 5.7

Ex 5.7, 5 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.7, 5 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Transcript

Ex 5.7, 5 Find the second order derivatives of the function ๐‘ฅ^3 logโก๐‘ฅ Let y = ๐‘ฅ^3 logโก๐‘ฅ Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘(๐‘ฅ^3 " " logโก๐‘ฅ))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(๐‘ฅ^3 )/๐‘‘๐‘ฅ .logโก๐‘ฅ + (๐‘‘(logโก๐‘ฅ))/๐‘‘๐‘ฅ .๐‘ฅ^3 using product rule in ๐‘ฅ^3 ๐‘™๐‘œ๐‘”โก๐‘ฅ . As (๐‘ข๐‘ฃ)โ€™= ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข where u = x3 & v = log x ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 3๐‘ฅ2 . logโก๐‘ฅ + 1/๐‘ฅ . ๐‘ฅ^3 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 3๐‘ฅ2 . logโก๐‘ฅ + ๐‘ฅ^2 Again Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = (๐‘‘ (3๐‘ฅ2 . logโก๐‘ฅ "+ " ๐‘ฅ^2))/๐‘‘๐‘ฅ (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = (๐‘‘ (3๐‘ฅ2 . logโก๐‘ฅ))/๐‘‘๐‘ฅ + (๐‘‘ (๐‘ฅ^2))/๐‘‘๐‘ฅ (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = (3 ๐‘‘(๐‘ฅ2 . logโก๐‘ฅ))/๐‘‘๐‘ฅ + 2 ๐‘ฅ using product rule in ๐‘ฅ^2 ๐‘™๐‘œ๐‘”โก๐‘ฅ . As (๐‘ข๐‘ฃ)โ€™= ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = 3 . ((๐‘‘(๐‘ฅ)^2)/๐‘‘๐‘ฅ .logโกใ€–๐‘ฅ+(๐‘‘(logโกใ€–๐‘ฅ)ใ€—)/๐‘‘๐‘ฅใ€—. ๐‘ฅ^2 ) + 2 ๐‘ฅ = 3(2๐‘ฅ . logโกใ€–๐‘ฅ+ 1/๐‘ฅใ€—. ๐‘ฅ^2 ) + 2 ๐‘ฅ = 3 (2๐‘ฅ . logโก๐‘ฅ+ ๐‘ฅ) + 2 ๐‘ฅ = 6 ๐‘ฅ logโก๐‘ฅ + 3๐‘ฅ + 2๐‘ฅ = 6 ๐‘ฅ log โก๐‘ฅ+5๐‘ฅ = x (6 logโกใ€–๐‘ฅ+5ใ€— ) = ๐‘ฅ (5+6 logโก๐‘ฅ ) Hence , (๐’…^๐Ÿ ๐’š)/(๐’…๐’™^๐Ÿ ) = ๐’™ (๐Ÿ“+๐Ÿ” ๐’๐’๐’ˆโก๐’™ )

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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.