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

Transcript

Ex 5.5, 8 Differentiate the functions in, γ€–(sin⁑π‘₯)γ€—^π‘₯+ sin^(βˆ’1) √π‘₯ Let 𝑦=(sin⁑π‘₯ )^π‘₯ + sin^(βˆ’1)⁑√π‘₯ Let 𝑒 = (sin⁑π‘₯ )^π‘₯ & 𝑣 = sin^(βˆ’1)⁑√π‘₯ 𝑦 = 𝑒 + 𝑣 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = (𝑑 (𝑒 + 𝑣))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Calculating 𝒅𝒖/𝒅𝒙 𝑒 =(sin⁑π‘₯ )^π‘₯ Taking log both sides log⁑𝑒=log⁑〖(sin⁑π‘₯ )^π‘₯ γ€— log⁑𝑒=π‘₯ .log⁑sin⁑〖π‘₯ γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. (𝑑(log⁑𝑒))/𝑑π‘₯ = 𝑑(π‘₯.γ€– log〗⁑sin⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑π‘₯ . 𝑑𝑒/𝑑𝑒 = 𝑑(π‘₯ .γ€– log〗⁑sin⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(π‘₯ .γ€– log〗⁑sin⁑π‘₯ )/𝑑π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 . π‘™π‘œπ‘”β‘π‘Ž) 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(π‘₯ .γ€– log〗⁑sin⁑π‘₯ )/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑π‘₯/𝑑π‘₯ . log⁑sin⁑π‘₯ + 𝑑(log⁑sin⁑π‘₯ )/𝑑π‘₯ Γ— π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 1 . log⁑sin⁑π‘₯ + (1/sin⁑π‘₯ . 𝑑(sin⁑π‘₯ )/𝑑π‘₯) π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = log⁑sin⁑π‘₯ + (1/sin⁑π‘₯ . cos⁑π‘₯ )π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = log⁑sin⁑π‘₯ + π‘₯ (cos⁑π‘₯/sin⁑π‘₯ ) 𝑑𝑒/𝑑π‘₯ = u(log⁑sin⁑〖π‘₯+π‘₯ cot⁑π‘₯ γ€— ) 𝑑𝑒/𝑑π‘₯ = (sin π‘₯)^π‘₯ (log⁑sin⁑〖π‘₯+π‘₯ cot⁑π‘₯ γ€— ) Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 Calculating 𝒅𝒗/𝒅𝒙 𝑣=sin^(βˆ’1)⁑√π‘₯ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑣/𝑑π‘₯ = 𝑑(sin^(βˆ’1)⁑√π‘₯ )/𝑑π‘₯ 𝑑𝑣/𝑑π‘₯ = 1/√(1 βˆ’(√π‘₯)^2 ) . (π‘‘βˆšπ‘₯)/𝑑π‘₯ 𝑑𝑣/𝑑π‘₯ = 1/√(1 βˆ’ π‘₯) . 1/(2√π‘₯) 𝑑𝑣/𝑑π‘₯ = 1/(2√π‘₯ √(1 βˆ’ π‘₯)) 𝑑𝑣/𝑑π‘₯ = 1/(2√(π‘₯(1 βˆ’π‘₯) ) ) 𝑑𝑣/𝑑π‘₯ = 1/(2√(π‘₯ βˆ’π‘₯^2 ) ) Now , 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Putting value of 𝑑𝑒/𝑑π‘₯ & 𝑑𝑣/𝑑π‘₯ π’…π’š/𝒅𝒙 = (𝐬𝐒𝐧⁑𝒙 )^𝒙 (𝒙 πœπ¨π­β‘π’™+π₯𝐨𝐠⁑𝐬𝐒𝐧⁑𝒙 ) + 𝟏/(𝟐 √(𝒙 βˆ’ 𝒙^𝟐 ) )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.