Question 16 (OR 2 nd question)

If y = logโก (1 + 2t + t 4 ), x = tan -1 โกt, find d 2 y/dx 2

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Question 16 (OR 2nd question) If y = logโกใ€–(1+2๐‘ก^2+๐‘ก^4)ใ€—, x = tan^(โˆ’1)โก๐‘ก, find (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) Finding ๐’…๐’š/๐’…๐’• y = logโกใ€–(1+2๐‘ก^2+๐‘ก^4)ใ€— ๐‘‘๐‘ฆ/๐‘‘๐‘ก = 1/((1 + 2๐‘ก^2 + ๐‘ก^4 ) ) ร— ๐‘‘(1 + 2๐‘ก^2 + ๐‘ก^4 )/๐‘‘๐‘ก ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ((4๐‘ก^3 + 4๐‘ก))/((1 + 2๐‘ก^2 + ๐‘ก^4 ) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ก = (4๐‘ก(๐‘ก^2 + 1))/(((๐‘ก^2 )^2 + 2๐‘ก^2 + 1^2 ) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ก = (4๐‘ก(๐‘ก^2 + 1))/(๐‘ก^2 + 1)^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ก = 4๐‘ก/((๐‘ก^2 + 1) ) Finding ๐’…๐’™/๐’…๐’• x = tan^(โˆ’1)โก๐‘ก ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 1/(1 + ๐‘ก^2 ) Now, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘๐‘ฆ/๐‘‘๐‘ก)/(๐‘‘๐‘ฅ/๐‘‘๐‘ก) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (4๐‘ก/((๐‘ก^2 + 1) ))/(1/(1 + ๐‘ก^2 )) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 4๐‘ก/((๐‘ก^2 + 1) )ร—((1 + ๐‘ก^2))/1 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 4t Now, (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = (๐‘‘(4๐‘ก))/๐‘‘๐‘ฅ = (๐‘‘(4๐‘ก))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ก/๐‘‘๐‘ก = (๐‘‘(4๐‘ก))/๐‘‘๐‘ก ร— ๐‘‘๐‘ก/๐‘‘๐‘ฅ = 4 ร— ๐‘‘๐‘ก/๐‘‘๐‘ฅ = 4 ร— 1/(๐‘‘๐‘ฅ/๐‘‘๐‘ก) = 4 (1 + t2) As ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 1/(1 + ๐‘ก^2 ) ๐‘‘๐‘ฅ/๐‘‘๐‘ก = (1 + t2)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.