Example 47 - Chapter 5 Class 12 Continuity and Differentiability

Transcript

Example 47 For a positive constant a find 𝑑𝑦/𝑑𝑥 , where 𝑦 = 𝑎^(𝑡+1/𝑡) , and 𝑥 =(𝑡+1/𝑡)^2 We need to find 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑑𝑦/𝑑𝑥 × 𝑑𝑡/𝑑𝑡 𝑑𝑦/𝑑𝑥 = 𝑑𝑦/𝑑𝑡 × 𝑑𝑡/𝑑𝑥 𝒅𝒚/𝒅𝒙 = (𝒅𝒚/𝒅𝒕)/(𝒅𝒙/𝒅𝒕) Calculating 𝒅𝒚/𝒅𝒕 𝑦=𝑎^(𝑡+1/𝑡) Let (𝑡+1/𝑡) = 𝑝 𝑦=𝑎^𝑝 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦/𝑑𝑡 = 𝑑(𝑎^𝑝 )/𝑑𝑡 𝑑𝑦/𝑑𝑡 = 𝑑(𝑎^𝑝 )/𝑑𝑡 . 𝑑𝑝/𝑑𝑝 𝑑𝑦/𝑑𝑡 = 𝑑(𝑎^𝑝 )/𝑑𝑝 . 𝑑𝑝/𝑑𝑡 𝑑𝑦/𝑑𝑡 = 𝑎^𝑝 .log⁡𝑎. 𝑑(𝑝)/𝑑𝑡 "As " 𝑑(𝑎^𝑥 )/𝑑𝑥 " = " 𝑎^𝑥.𝑙𝑜𝑔⁡𝑎 Calculated in eg31 Putting values of p = 𝑡+1/𝑡 𝑑𝑦/𝑑𝑡 = 𝑎^((𝑡 + 1/𝑡) ) .log⁡𝑎. 𝑑(𝑡 + 1/𝑡)/𝑑𝑡 𝑑𝑦/𝑑𝑡 = 𝑎^((𝑡 + 1/𝑡) ) .log⁡𝑎. (𝑑(𝑡)/𝑑𝑡 + 𝑑(1/𝑡)/𝑑𝑡) 𝑑𝑦/𝑑𝑡 = 𝑎^((𝑡 + 1/𝑡) ) .log⁡𝑎. (1+ 𝑑(𝑡^(−1) )/𝑑𝑡) 𝑑𝑦/𝑑𝑡 = 𝑎^((𝑡 + 1/𝑡) ) .log⁡𝑎.(1+(−1) 𝑡^(−1−1) ) 𝑑𝑦/𝑑𝑡 = 𝑎^((𝑡 + 1/𝑡) ) .log⁡𝑎.(1+(−1) 𝑡^(−2) ) 𝒅𝒚/𝒅𝒕 = 𝒂^((𝒕 + 𝟏/𝒕) ) .𝒍𝒐𝒈⁡𝒂.(𝟏−𝟏/𝒕^𝟐 ) Calculating 𝒅𝒙/𝒅𝒕 𝑥=(𝑡+1/𝑡)^𝑎 Let (𝑡+1/𝑡)=𝑝 𝑥=𝑝^𝑎 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑥/𝑑𝑡 = 𝑑(𝑝^𝑎 )/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 𝑑(𝑝^𝑎 )/𝑑𝑡 . 𝑑𝑝/𝑑𝑝 𝑑𝑥/𝑑𝑡 = 𝑑(𝑝^𝑎 )/𝑑𝑝 . 𝑑𝑝/𝑑𝑡 𝑑𝑥/𝑑𝑡 = a 𝑝^(𝑎 −1 ) . 𝑑(𝑝)/𝑑𝑡 "As " 𝑑(𝑥^𝑛 )/𝑑𝑥 " = " 〖𝑛𝑥〗^(𝑛 − 1) 𝑑(𝑝^𝑎 )/𝑑𝑝 = 〖𝑎𝑝〗^(𝑎 − 1) Putting value of p = 𝑡+1/𝑡 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 (𝑡+1/𝑡)^(𝑎 −1 ) . (1+(−1) 〖 𝑡〗^(−2) ) 𝑑𝑥/𝑑𝑡 = a (𝑡+1/𝑡)^(𝑎 −1 ) . (1− 1/𝑡^2 ) Now, 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) )