Slide31.JPG

Slide32.JPG
Slide33.JPG

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


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‘₯ 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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.