Ex 5.7, 5 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 5.7, 5 Find the second order derivatives of the function ๐ฅ^3 logโก๐ฅ Let y = ๐ฅ^3 logโก๐ฅ Differentiating ๐ค.๐.๐ก.๐ฅ . ๐๐ฆ/๐๐ฅ = (๐(๐ฅ^3 " " logโก๐ฅ))/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐(๐ฅ^3 )/๐๐ฅ .logโก๐ฅ + (๐(logโก๐ฅ))/๐๐ฅ .๐ฅ^3 using product rule in ๐ฅ^3 ๐๐๐โก๐ฅ . As (๐ข๐ฃ)โ= ๐ขโ๐ฃ + ๐ฃโ๐ข where u = x3 & v = log x ๐๐ฆ/๐๐ฅ = 3๐ฅ2 . logโก๐ฅ + 1/๐ฅ . ๐ฅ^3 ๐๐ฆ/๐๐ฅ = 3๐ฅ2 . logโก๐ฅ + ๐ฅ^2 Again Differentiating ๐ค.๐.๐ก.๐ฅ ๐/๐๐ฅ (๐๐ฆ/๐๐ฅ) = (๐ (3๐ฅ2 . logโก๐ฅ "+ " ๐ฅ^2))/๐๐ฅ (๐^2 ๐ฆ)/(๐๐ฅ^2 ) = (๐ (3๐ฅ2 . logโก๐ฅ))/๐๐ฅ + (๐ (๐ฅ^2))/๐๐ฅ (๐^2 ๐ฆ)/(๐๐ฅ^2 ) = (3 ๐(๐ฅ2 . logโก๐ฅ))/๐๐ฅ + 2 ๐ฅ using product rule in ๐ฅ^2 ๐๐๐โก๐ฅ . As (๐ข๐ฃ)โ= ๐ขโ๐ฃ + ๐ฃโ๐ข (๐^2 ๐ฆ)/(๐๐ฅ^2 ) = 3 . ((๐(๐ฅ)^2)/๐๐ฅ .logโกใ๐ฅ+(๐(logโกใ๐ฅ)ใ)/๐๐ฅใ. ๐ฅ^2 ) + 2 ๐ฅ = 3(2๐ฅ . logโกใ๐ฅ+ 1/๐ฅใ. ๐ฅ^2 ) + 2 ๐ฅ = 3 (2๐ฅ . logโก๐ฅ+ ๐ฅ) + 2 ๐ฅ = 6 ๐ฅ logโก๐ฅ + 3๐ฅ + 2๐ฅ = 6 ๐ฅ log โก๐ฅ+5๐ฅ = x (6 logโกใ๐ฅ+5ใ ) = ๐ฅ (5+6 logโก๐ฅ ) Hence , (๐ ^๐ ๐)/(๐ ๐^๐ ) = ๐ (๐+๐ ๐๐๐โก๐ )
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo