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Misc 16 If cos⁑𝑦=π‘₯ cos⁑(π‘Ž + 𝑦), with cosβ‘π‘Ž β‰  Β± 1, prove that 𝑑𝑦/𝑑π‘₯ = (γ€–π‘π‘œπ‘ γ€—^2 (π‘Ž + 𝑦))/sinβ‘γ€–π‘Ž γ€— Given cos⁑𝑦 = π‘₯ cos⁑(π‘Ž + 𝑦) cos⁑𝑦/(cos⁑(π‘Ž + 𝑦)) = π‘₯ 𝒙 = π’„π’π’”β‘π’š/(𝒄𝒐𝒔⁑(𝒂 + π’š)) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(π‘₯)/𝑑π‘₯ = 𝑑/𝑑π‘₯ (cos⁑𝑦/cos⁑(π‘Ž + 𝑦) ) 1 = 𝑑/𝑑π‘₯ (cos⁑𝑦/cos⁑(π‘Ž + 𝑦) ) . 𝑑𝑦/𝑑𝑦 1 = 𝑑/𝑑𝑦 (cos⁑𝑦/cos⁑(π‘Ž + 𝑦) ) . 𝑑𝑦/𝑑π‘₯ 1 = ((𝑑(cos⁑𝑦 )/𝑑𝑦 . γ€– cos〗⁑(π‘Ž + 𝑦) βˆ’ 𝑑(γ€– cos〗⁑(π‘Ž + 𝑦) )/𝑑𝑦 . cos⁑𝑦)/(cos⁑(π‘Ž + 𝑦) )^2 ) . 𝑑𝑦/𝑑π‘₯ 1 = ((βˆ’sin⁑𝑦 . γ€– cos〗⁑(π‘Ž + 𝑦) βˆ’(γ€–βˆ’sin 〗⁑(π‘Ž + 𝑦) ) 𝑑(π‘Ž + 𝑦)/𝑑𝑦 . cos⁑𝑦)/γ€–cos^2 〗⁑(π‘Ž + 𝑦) ) . 𝑑𝑦/𝑑π‘₯ Using quotient rule As (𝑒/𝑣)β€² = (𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 where u = cos y & v = cos (π‘Ž + y) 1 = ((βˆ’sin⁑𝑦 . γ€– cos〗⁑(π‘Ž + 𝑦) + γ€–sin 〗⁑(π‘Ž + 𝑦) (0 + 1) . cos⁑𝑦)/γ€–cos^2 〗⁑(π‘Ž + 𝑦) ) . 𝑑𝑦/𝑑π‘₯ 1 = ((sin⁑(π‘Ž + 𝑦) . γ€– cos〗⁑𝑦 βˆ’ γ€–cos 〗⁑(π‘Ž + 𝑦) . sin⁑𝑦)/γ€–cos^2 〗⁑(π‘Ž + 𝑦) ) . 𝑑𝑦/𝑑π‘₯ 1 = π’”π’Šπ’β‘((𝒂 + π’š) βˆ’ π’š)/γ€–cos^2 〗⁑(π‘Ž + 𝑦) . 𝑑𝑦/𝑑π‘₯ 1 = sin⁑(π‘Ž)/γ€–cos^2 〗⁑(π‘Ž + 𝑦) . 𝑑𝑦/𝑑π‘₯ γ€–cos^2 〗⁑(π‘Ž + 𝑦)/sin⁑(π‘Ž) = 𝑑𝑦/𝑑π‘₯ π’…π’š/𝒅𝒙 = 〖〖𝒄𝒐𝒔〗^𝟐 〗⁑(𝒂 + π’š)/π’”π’Šπ’β‘(𝒂) We know that π’”π’Šπ’β‘(𝒙 βˆ’π’š)=𝑠𝑖𝑛⁑π‘₯ γ€– π‘π‘œπ‘ γ€—β‘π‘¦βˆ’γ€–π‘π‘œπ‘  〗⁑π‘₯ . 𝑠𝑖𝑛⁑𝑦

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo