Example 47 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Transcript

Example 47 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) )