Check sibling questions

Example 47 - Find dy/dx, where y = at + 1/t, x = (t + 1/t)2

Example 47 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Example 47 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Example 47 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Example 47 For a positive constant a find 𝑑𝑦/𝑑π‘₯ , where 𝑦 = π‘Ž^(𝑑+1/𝑑) , and π‘₯ =(𝑑+1/𝑑)^2 Here π’…π’š/𝒅𝒙 = (π’…π’š/𝒅𝒕)/(𝒅𝒙/𝒅𝒕) Calculating π’…π’š/𝒅𝒕 𝑦=π‘Ž^(𝑑 + 1/𝑑) Differentiating 𝑀.π‘Ÿ.𝑑. t π’…π’š/𝒅𝒕 = 𝒅(𝒂^((𝒕 + 𝟏/𝒕) ) )/𝒅𝒕 𝑑𝑦/𝑑𝑑 = π‘Ž^((𝑑 + 1/𝑑) ) .logβ‘π‘Ž.𝑑(𝑑 + 1/𝑑)/𝑑𝑑 𝑑𝑦/𝑑𝑑 = π‘Ž^((𝑑 + 1/𝑑) ) .logβ‘π‘Ž.(1+(βˆ’1) 𝑑^(βˆ’2) ) π’…π’š/𝒅𝒕 = 𝒂^((𝒕 + 𝟏/𝒕) ) .π’π’π’ˆβ‘π’‚.(πŸβˆ’πŸ/𝒕^𝟐 ) "As " 𝑑(π‘Ž^π‘₯ )/𝑑π‘₯ " = " π‘Ž^π‘₯.π‘™π‘œπ‘”β‘π‘Ž Calculating 𝒅𝒙/𝒅𝒕 π‘₯=(𝑑+1/𝑑)^π‘Ž Differentiating 𝑀.π‘Ÿ.𝑑. t 𝑑π‘₯/𝑑𝑑 = 𝑑((𝑑 + 1/𝑑)^(π‘Ž ) )/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = a (𝑑+1/𝑑)^(π‘Ž βˆ’1 ) . 𝑑(𝑑 + 1/𝑑)/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = a (𝑑+1/𝑑)^(π‘Ž βˆ’1 ) . (𝑑(𝑑)/𝑑𝑑 + 𝑑(1/𝑑)/𝑑𝑑) 𝑑π‘₯/𝑑𝑑 = a (𝑑+1/𝑑)^(π‘Ž βˆ’1 ) . (1+ 𝑑(𝑑^(βˆ’1) )/𝑑𝑑) 𝑑π‘₯/𝑑𝑑 = a 𝑝^(π‘Ž βˆ’1 ) . 𝑑(𝑝)/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = a (𝑑+1/𝑑)^(π‘Ž βˆ’1 ) . (1+(βˆ’1) γ€– 𝑑〗^(βˆ’2) ) 𝑑π‘₯/𝑑𝑑 = a (𝑑+1/𝑑)^(π‘Ž βˆ’1 ) . (1βˆ’ 1/𝑑^2 ) Calculating π’…π’š/𝒅𝒙 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) 𝑑𝑦/𝑑π‘₯ = (π‘Ž^(𝑑 + 1/𝑑) . logβ‘γ€–π‘Ž γ€— Γ— (1 βˆ’ 1/𝑑^2 ))/(π‘Ž(𝑑 + 1/𝑑)^(π‘Ž βˆ’ 1) (1 βˆ’ 1/𝑑^2 ).) π’…π’š/𝒅𝒙 = (𝒂^(𝒕 + 𝟏/𝒕) . π’π’π’ˆβ‘γ€–π’‚ γ€—)/(𝒂(𝒕 + 𝟏/𝒕)^(𝒂 βˆ’ 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.