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


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Misc 9 Differentiate w.r.t. x the function, (sin⁡𝑥−cos⁡𝑥 )^((sin⁡〖𝑥−cos⁡〖𝑥)〗 〗 ), 𝜋/4 <𝑥< 3𝜋/4 Let y = (sin⁡𝑥−cos⁡𝑥 )^((sin⁡〖𝑥−cos⁡〖𝑥)〗 〗 ) Taking log on both sides log⁡𝑦 = log (sin⁡𝑥−cos⁡𝑥 )^((sin⁡〖𝑥−cos⁡〖𝑥)〗 〗 ) log⁡𝑦 = (sin⁡𝑥−cos⁡𝑥 ). 〖 log〗⁡〖 (sin⁡𝑥−cos⁡𝑥 )〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑(log⁡𝑦 )/𝑑𝑥 = 𝑑((sin⁡𝑥 − cos⁡𝑥 ). 〖 log〗⁡(sin⁡𝑥 − cos⁡𝑥 ) )/𝑑𝑥 𝑑(log⁡𝑦 )/𝑑𝑥 (𝑑𝑦/𝑑𝑦) = 𝑑((sin⁡𝑥−cos⁡𝑥 ). 〖 log〗⁡〖 (sin⁡𝑥−cos⁡𝑥 )〗 )/𝑑𝑥 𝑑(log⁡𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑𝑥) = 𝑑((sin⁡𝑥−cos⁡𝑥 ). 〖 log〗⁡〖 (sin⁡𝑥−cos⁡𝑥 )〗 )/𝑑𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 𝑑((sin⁡𝑥−cos⁡𝑥 ). 〖 log〗⁡〖 (sin⁡𝑥−cos⁡𝑥 )〗 " " )/𝑑𝑥 " " 1/𝑦. 𝑑𝑦/𝑑𝑥 = 𝑑(sin⁡𝑥 − cos⁡𝑥 )/𝑑𝑥 . 〖 log 〗⁡(sin⁡𝑥−cos⁡𝑥 ) + 𝑑(〖 log〗⁡〖 (sin⁡𝑥 − cos⁡𝑥 )〗 )/𝑑𝑥 .(sin⁡𝑥−cos⁡𝑥 ) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = (cos⁡𝑥−(−sin⁡𝑥 )). log⁡〖 (sin⁡𝑥−〖 cos〗⁡𝑥 )〗 + 1/((sin⁡𝑥 − cos⁡𝑥 ) ) . 𝑑(sin⁡𝑥 − cos⁡𝑥 )/𝑑𝑥 . (sin⁡𝑥−〖 cos〗⁡𝑥 ) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = (cos⁡𝑥+sin⁡𝑥 ) . log⁡〖 (sin⁡𝑥−cos⁡𝑥 )〗 + 1/((sin⁡𝑥−cos⁡𝑥 ) ) . (cos⁡𝑥−(−sin⁡𝑥 )) . (sin⁡𝑥−cos⁡𝑥 ) Using product rule (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 where u = sin x − cos x & v = log (sin x − cos x) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = (cos⁡𝑥+sin⁡𝑥 ) . log⁡〖 (sin⁡𝑥−cos⁡𝑥 )〗 + 1/((sin⁡𝑥−cos⁡𝑥 ) ) . (cos⁡𝑥+sin⁡𝑥 ) . (sin⁡𝑥−cos⁡𝑥 ) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = (cos⁡𝑥+sin⁡𝑥 ) . log⁡〖 (sin⁡𝑥−cos⁡𝑥 )〗 + (cos⁡𝑥+sin⁡𝑥 ) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = (cos⁡𝑥+sin⁡𝑥 ) . (log⁡〖 (sin⁡𝑥−cos⁡𝑥 )+1〗 ) 𝑑𝑦/𝑑𝑥 = 𝑦(cos⁡𝑥+sin⁡𝑥 ) . (log⁡〖 (sin⁡𝑥−cos⁡𝑥 )+1〗 ) 𝒅𝒚/𝒅𝒙 = (𝒔𝒊𝒏⁡𝒙−𝒄𝒐𝒔⁡𝒙 )^((𝒔𝒊𝒏⁡〖𝒙−𝒄𝒐𝒔⁡〖𝒙)〗 〗 ) (𝒄𝒐𝒔⁡𝒙+𝒔𝒊𝒏⁡𝒙 )(𝒍𝒐𝒈⁡〖 (𝒔𝒊𝒏⁡𝒙−𝒄𝒐𝒔⁡𝒙 )+𝟏〗 )

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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.