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


Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.