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Transcript

Example 42 For a positive constant a find 𝑑𝑦/𝑑𝑥 , where 𝑦 = 𝑎^(𝑡+1/𝑡) , and 𝑥 =(𝑡+1/𝑡)^2 Here 𝒅𝒚/𝒅𝒙 = (𝒅𝒚/𝒅𝒕)/(𝒅𝒙/𝒅𝒕) Calculating 𝒅𝒚/𝒅𝒕 𝑦=𝑎^(𝑡 + 1/𝑡) Differentiating 𝑤.𝑟.𝑡. t 𝒅𝒚/𝒅𝒕 = 𝒅(𝒂^((𝒕 + 𝟏/𝒕) ) )/𝒅𝒕 𝑑𝑦/𝑑𝑡 = 𝑎^((𝑡 + 1/𝑡) ) .log⁡𝑎.𝑑(𝑡 + 1/𝑡)/𝑑𝑡 𝑑𝑦/𝑑𝑡 = 𝑎^((𝑡 + 1/𝑡) ) .log⁡𝑎.(1+(−1) 𝑡^(−2) ) 𝒅𝒚/𝒅𝒕 = 𝒂^((𝒕 + 𝟏/𝒕) ) .𝒍𝒐𝒈⁡𝒂.(𝟏−𝟏/𝒕^𝟐 ) "As " 𝑑(𝑎^𝑥 )/𝑑𝑥 " = " 𝑎^𝑥.𝑙𝑜𝑔⁡𝑎 Calculating 𝒅𝒙/𝒅𝒕 𝑥=(𝑡+1/𝑡)^𝑎 Differentiating 𝑤.𝑟.𝑡. t 𝑑𝑥/𝑑𝑡 = 𝑑((𝑡 + 1/𝑡)^(𝑎 ) )/𝑑𝑡 𝑑𝑥/𝑑𝑡 = a (𝑡+1/𝑡)^(𝑎 −1 ) . 𝑑(𝑡 + 1/𝑡)/𝑑𝑡 𝑑𝑥/𝑑𝑡 = a (𝑡+1/𝑡)^(𝑎 −1 ) . (𝑑(𝑡)/𝑑𝑡 + 𝑑(1/𝑡)/𝑑𝑡) 𝑑𝑥/𝑑𝑡 = a (𝑡+1/𝑡)^(𝑎 −1 ) . (1+ 𝑑(𝑡^(−1) )/𝑑𝑡) 𝑑𝑥/𝑑𝑡 = a 𝑝^(𝑎 −1 ) . 𝑑(𝑝)/𝑑𝑡 𝑑𝑥/𝑑𝑡 = a (𝑡+1/𝑡)^(𝑎 −1 ) . (1+(−1) 〖 𝑡〗^(−2) ) 𝑑𝑥/𝑑𝑡 = a (𝑡+1/𝑡)^(𝑎 −1 ) . (1− 1/𝑡^2 ) Calculating 𝒅𝒚/𝒅𝒙 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) 𝑑𝑦/𝑑𝑥 = (𝑎^(𝑡 + 1/𝑡) . log⁡〖𝑎 〗 × (1 − 1/𝑡^2 ))/(𝑎(𝑡 + 1/𝑡)^(𝑎 − 1) (1 − 1/𝑡^2 ).) 𝒅𝒚/𝒅𝒙 = (𝒂^(𝒕 + 𝟏/𝒕) . 𝒍𝒐𝒈⁡〖𝒂 〗)/(𝒂(𝒕 + 𝟏/𝒕)^(𝒂 − 𝟏) )

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.