Slide42.JPG

Slide43.JPG
Slide44.JPG Slide45.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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) )

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.