Ex 5.6, 11 - Show that dy/dx = - y/x, if x = root a sin-1 t

Ex 5.6, 11 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.6, 11 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.6, 11 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.6, 11 - Chapter 5 Class 12 Continuity and Differentiability - Part 5

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Ex 5.6, 11 If π‘₯=√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) ) , y=√(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) ), show that 𝑑𝑦/𝑑π‘₯ = βˆ’ 𝑦/π‘₯Here 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Calculating π’…π’š/𝒅𝒕 𝑦 = √(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) ) 𝑑𝑦/𝑑𝑑 " "= 𝑑(√(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) ))/𝑑𝑑 𝑑𝑦/𝑑𝑑 " "= 1/(2√(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) )) . 𝑑(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) )/𝑑𝑑 𝑑𝑦/𝑑𝑑 " "= 1/(2√(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) )) . π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑). logβ‘π‘Ž Γ— 𝑑(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑)/𝑑𝑑 𝑑𝑦/𝑑𝑑 " "= 1/(2√(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) )) . π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑). logβ‘π‘Ž Γ— (βˆ’1)/√(1 βˆ’ 𝑑^2 ) (𝐴𝑠" " 𝑑(π‘Ž^π‘₯ )/π‘‘πœƒ=π‘Ž^π‘₯.logβ‘π‘Ž ) 𝑑𝑦/𝑑𝑑 " "= (βˆ’ π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) ". " logβ‘π‘Ž" " )/(2√(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) ) . √(1 βˆ’ 𝑑^2 )) Calculating 𝒅𝒙/𝒅𝒕 π‘₯=√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) ) 𝑑π‘₯/𝑑𝑑 = 𝑑(√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) ))/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = 1/(2√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) )) Γ— 𝑑(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) )/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = 1/(2√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) )) Γ— π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑). logβ‘π‘Ž . 𝑑(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑)/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = 1/(2√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) )) Γ— π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑). logβ‘π‘Ž . 1/√(1 βˆ’ 𝑑^2 ) 𝑑π‘₯/𝑑𝑑 = " " π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑)/(2√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) )) Γ— logβ‘π‘Ž/√(1 βˆ’ 𝑑^2 ) 𝑑π‘₯/𝑑𝑑 = " " (√(𝒂^(γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) 𝒕) ) . π’π’π’ˆβ‘π’‚)/(𝟐√(𝟏 βˆ’ 𝒕^𝟐 )) Finding π’…π’š/𝒅𝒙 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) 𝑑𝑦/𝑑π‘₯ = ((βˆ’ √(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) ) . logβ‘π‘Ž" " )/(2√(1 βˆ’ 𝑑^2 )))/((√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) ) . logβ‘π‘Ž)/(2√(1 βˆ’ 𝑑^2 ))) 𝑑𝑦/𝑑π‘₯ = (βˆ’ √(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) ) . βˆ’ logβ‘π‘Ž" " )/(2√(1 βˆ’ 𝑑^2 )) Γ— (2√(1 βˆ’ 𝑑^2 ))/(√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) ) . logβ‘π‘Ž ) 𝑑𝑦/𝑑π‘₯ = (βˆ’βˆš(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) )⁑)/√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) ) Γ— (logβ‘π‘Ž Γ— 2√(1 βˆ’ 𝑑^2 ))/(logβ‘π‘Ž Γ— 2√(1 βˆ’ 𝑑^2 )) 𝑑𝑦/𝑑π‘₯ = (βˆ’βˆš(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) )⁑)/√(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) ) π’…π’š/𝒅𝒙 = (βˆ’π’šβ‘)/𝒙 Hence proved. 𝐴𝑠 √(π‘Ž^(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) 𝑑) )=𝑦 √(π‘Ž^(〖𝑠𝑖𝑛〗^(βˆ’1) 𝑑) )=π‘₯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.