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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

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


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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 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.