# Ex 5.7, 5 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by Teachoo

Last updated at March 11, 2021 by Teachoo

Transcript

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

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.