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

Transcript

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‘₯ π‘₯ = cos⁑𝑦/(cos⁑(π‘Ž + 𝑦)) 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) 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 = sin⁑((π‘Ž + 𝑦) βˆ’ 𝑦)/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 𝑠𝑖𝑛⁑(π‘₯ βˆ’π‘¦)=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 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.