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Question 16 (OR 2 nd question)

If y = log⁡ (1 + 2t + t 4 ), x = tan -1 ⁡t, find d 2 y/dx 2

If y = log⁡ (1 + 2t^2 + t^4), x = tan^-1 ⁡t, find y'' - Teachoo

Question 16 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 16 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards - Part 3

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Transcript

Question 16 (OR 2nd question) If y = log⁡〖(1+2𝑡^2+𝑡^4)〗, x = tan^(−1)⁡𝑡, find (𝑑^2 𝑦)/(𝑑𝑥^2 ) Finding 𝒅𝒚/𝒅𝒕 y = log⁡〖(1+2𝑡^2+𝑡^4)〗 𝑑𝑦/𝑑𝑡 = 1/((1 + 2𝑡^2 + 𝑡^4 ) ) × 𝑑(1 + 2𝑡^2 + 𝑡^4 )/𝑑𝑡 𝑑𝑦/𝑑𝑡 = ((4𝑡^3 + 4𝑡))/((1 + 2𝑡^2 + 𝑡^4 ) ) 𝑑𝑦/𝑑𝑡 = (4𝑡(𝑡^2 + 1))/(((𝑡^2 )^2 + 2𝑡^2 + 1^2 ) ) 𝑑𝑦/𝑑𝑡 = (4𝑡(𝑡^2 + 1))/(𝑡^2 + 1)^2 𝑑𝑦/𝑑𝑡 = 4𝑡/((𝑡^2 + 1) ) Finding 𝒅𝒙/𝒅𝒕 x = tan^(−1)⁡𝑡 𝑑𝑥/𝑑𝑡 = 1/(1 + 𝑡^2 ) Now, 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) 𝑑𝑦/𝑑𝑥 = (4𝑡/((𝑡^2 + 1) ))/(1/(1 + 𝑡^2 )) 𝑑𝑦/𝑑𝑥 = 4𝑡/((𝑡^2 + 1) )×((1 + 𝑡^2))/1 𝑑𝑦/𝑑𝑥 = 4t Now, (𝑑^2 𝑦)/(𝑑𝑥^2 ) = (𝑑(4𝑡))/𝑑𝑥 = (𝑑(4𝑡))/𝑑𝑥 × 𝑑𝑡/𝑑𝑡 = (𝑑(4𝑡))/𝑑𝑡 × 𝑑𝑡/𝑑𝑥 = 4 × 𝑑𝑡/𝑑𝑥 = 4 × 1/(𝑑𝑥/𝑑𝑡) = 4 (1 + t2) As 𝑑𝑥/𝑑𝑡 = 1/(1 + 𝑡^2 ) 𝑑𝑡/𝑑𝑥 = (1 + t2)

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Davneet Singh

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.