Misc 16 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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Misc 16 If cos⁡𝑦=𝑥 cos⁡(𝑎 + 𝑦), with cos⁡𝑎 ≠ ± 1, prove that 𝑑𝑦/𝑑𝑥 = (〖𝑐𝑜𝑠〗^2 (𝑎 + 𝑦))/sin⁡〖𝑎 〗 Given cos⁡𝑦 = 𝑥 cos⁡(𝑎 + 𝑦) cos⁡𝑦/(cos⁡(𝑎 + 𝑦)) = 𝑥 𝒙 = 𝒄𝒐𝒔⁡𝒚/(𝒄𝒐𝒔⁡(𝒂 + 𝒚)) Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑(𝑥)/𝑑𝑥 = 𝑑/𝑑𝑥 (cos⁡𝑦/cos⁡(𝑎 + 𝑦) ) 1 = 𝑑/𝑑𝑥 (cos⁡𝑦/cos⁡(𝑎 + 𝑦) ) . 𝑑𝑦/𝑑𝑦 1 = 𝑑/𝑑𝑦 (cos⁡𝑦/cos⁡(𝑎 + 𝑦) ) . 𝑑𝑦/𝑑𝑥 1 = ((𝑑(cos⁡𝑦 )/𝑑𝑦 . 〖 cos〗⁡(𝑎 + 𝑦) − 𝑑(〖 cos〗⁡(𝑎 + 𝑦) )/𝑑𝑦 . cos⁡𝑦)/(cos⁡(𝑎 + 𝑦) )^2 ) . 𝑑𝑦/𝑑𝑥 1 = ((−sin⁡𝑦 . 〖 cos〗⁡(𝑎 + 𝑦) −(〖−sin 〗⁡(𝑎 + 𝑦) ) 𝑑(𝑎 + 𝑦)/𝑑𝑦 . cos⁡𝑦)/〖cos^2 〗⁡(𝑎 + 𝑦) ) . 𝑑𝑦/𝑑𝑥 Using quotient rule As (𝑢/𝑣)′ = (𝑢^′ 𝑣 − 𝑣^′ 𝑢)/𝑣^2 where u = cos y & v = cos (𝑎 + y) 1 = ((−sin⁡𝑦 . 〖 cos〗⁡(𝑎 + 𝑦) + 〖sin 〗⁡(𝑎 + 𝑦) (0 + 1) . cos⁡𝑦)/〖cos^2 〗⁡(𝑎 + 𝑦) ) . 𝑑𝑦/𝑑𝑥 1 = ((sin⁡(𝑎 + 𝑦) . 〖 cos〗⁡𝑦 − 〖cos 〗⁡(𝑎 + 𝑦) . sin⁡𝑦)/〖cos^2 〗⁡(𝑎 + 𝑦) ) . 𝑑𝑦/𝑑𝑥 1 = 𝒔𝒊𝒏⁡((𝒂 + 𝒚) − 𝒚)/〖cos^2 〗⁡(𝑎 + 𝑦) . 𝑑𝑦/𝑑𝑥 1 = sin⁡(𝑎)/〖cos^2 〗⁡(𝑎 + 𝑦) . 𝑑𝑦/𝑑𝑥 〖cos^2 〗⁡(𝑎 + 𝑦)/sin⁡(𝑎) = 𝑑𝑦/𝑑𝑥 𝒅𝒚/𝒅𝒙 = 〖〖𝒄𝒐𝒔〗^𝟐 〗⁡(𝒂 + 𝒚)/𝒔𝒊𝒏⁡(𝒂) We know that 𝒔𝒊𝒏⁡(𝒙 −𝒚)=𝑠𝑖𝑛⁡𝑥 〖 𝑐𝑜𝑠〗⁡𝑦−〖𝑐𝑜𝑠 〗⁡𝑥 . 𝑠𝑖𝑛⁡𝑦