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Misc 13 Find 𝑑𝑦/𝑑π‘₯ , if 𝑦=〖𝑠𝑖𝑛〗^(βˆ’πŸ) π‘₯+〖𝑠𝑖𝑛〗^(βˆ’1) √(1βˆ’π‘₯2), – 1 ≀ π‘₯ ≀ 1 𝑦=〖𝑠𝑖𝑛〗^(βˆ’πŸ) π‘₯+〖𝑠𝑖𝑛〗^(βˆ’1) √(1βˆ’π‘₯^2 ) , – 1 ≀ π‘₯ ≀ 1 Putting 𝒙 = π’”π’Šπ’β‘πœ½ 𝑦=〖𝑠𝑖𝑛〗^(βˆ’πŸ) (sinβ‘πœƒ)+〖𝑠𝑖𝑛〗^(βˆ’1) √(1βˆ’sin^2 πœƒ ) 𝑦=𝜽+〖𝑠𝑖𝑛〗^(βˆ’1) √(γ€–πœπ¨π¬γ€—^𝟐 πœƒ ) 𝑦=πœƒ+〖𝑠𝑖𝑛〗^(βˆ’1) (cos πœƒ) 𝑦=πœƒ+〖𝑠𝑖𝑛〗^(βˆ’1) (sin⁑(𝝅/𝟐 βˆ’πœ½) ) 𝑦=πœƒ+ (πœ‹/2 βˆ’πœƒ) 𝑦=πœƒβˆ’πœƒ + πœ‹/2 π’š= 𝝅/𝟐 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = 𝑑(πœ‹/2)/𝑑π‘₯ π’…π’š/𝒅𝒙 = 0

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.