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


Transcript

Example 30 Find 𝑑𝑦/𝑑𝑥 , if 𝑦^𝑥+𝑥^𝑦+𝑥^𝑥=𝑎^𝑏. Let u = 𝑦𝑥, v = 𝑥𝑦 & w = 𝑥^𝑥 Now, 𝒖 + 𝒗 + 𝒘 = 𝒂^𝒃 Differentiating 𝑤.𝑟.𝑡.𝑥 (𝑑 (𝑢 + 𝑣 + 𝑤))/𝑑𝑥 = (𝑑(𝑎^𝑏))/𝑑𝑥 (𝑑(𝑢))/𝑑𝑥 + (𝑑(𝑣))/𝑑𝑥 + (𝑑(𝑤))/𝑑𝑥 = 0 We will calculate derivative of u, v & w separately . Finding Derivative of 𝒖 . 𝑢 = 𝑦^𝑥 Taking log both sides log⁡𝑢=log⁡〖 (𝑦^𝑥)" " 〗 log⁡𝑢=〖𝑥 . log〗⁡𝑦" " Differentiating both sides 𝑤.𝑟.𝑡.𝑥 (𝑑(log⁡𝑢))/𝑑𝑥 = (𝑑(𝑥 . log⁡𝑦))/𝑑𝑥 (𝑑(log⁡𝑢))/𝑑𝑥 (𝑑𝑢/𝑑𝑢) = 𝑑(𝑥.log⁡𝑦 )/𝑑𝑥 1/𝑢 . 𝑑𝑢/𝑑𝑥 = (𝑑 (𝑥 . log⁡𝑦 ))/𝑑𝑥 (𝐴𝑠 log⁡〖(𝑎^𝑏)〗=𝑏 log⁡𝑎) By product Rule (uv)’ = u’v + v’u 1/𝑢 . 𝑑𝑢/𝑑𝑥 = 𝑑𝑥/𝑑𝑥 . log⁡𝑦 + (𝑑(log⁡𝑦))/𝑑𝑥 . 𝑥 1/𝑢 . 𝑑𝑢/𝑑𝑥 = 1 . log⁡𝑦 + 𝑥. 𝑑(log⁡𝑦 )/𝑑𝑥 . 𝑑𝑦/𝑑𝑦 1/𝑢 . 𝑑𝑢/𝑑𝑥 = log⁡𝑦 + 𝑥. 𝑑(log⁡𝑦 )/𝑑𝑥 . 𝑑𝑦/𝑑𝑥 1/𝑢 . 𝑑𝑢/𝑑𝑥 = log⁡𝑦 + 𝑥. 1/𝑦 . 𝑑𝑦/𝑑𝑥 1/𝑢 . 𝑑𝑢/𝑑𝑥 = log⁡𝑦 + 𝑥/𝑦 . 𝑑𝑦/𝑑𝑥 𝑑𝑢/𝑑𝑥 = 𝑢 (log⁡𝑦 "+ " 𝑥/𝑦 " " 𝑑𝑦/𝑑𝑥) 𝒅𝒖/𝒅𝒙 = 𝒚^𝒙 (𝒍𝒐𝒈⁡𝒚 "+ " 𝒙/𝒚 " " 𝒅𝒚/𝒅𝒙) Finding derivative of v v = xy Taking log both sides log⁡𝑣=log⁡〖 (𝑥^𝑦)" " 〗 log⁡𝑣=〖𝑦. log〗⁡𝑥" " Differentiating both sides 𝑤.𝑟.𝑡.𝑥 (𝑑(log⁡𝑣))/𝑑𝑥 = (𝑑(𝑦 . log⁡𝑥))/𝑑𝑥 (𝑑(log⁡𝑣))/𝑑𝑥 (𝑑𝑣/𝑑𝑥) = 𝑑(〖𝑦 log〗⁡𝑥 )/𝑑𝑥 1/𝑣 (𝑑𝑣/𝑑𝑥) = ( 𝑑(〖𝑦 log〗⁡𝑥 ))/𝑑𝑥 By product Rule (uv)’ = u’v + v’u 1/𝑣 (𝑑𝑣/𝑑𝑥) = ( 𝑑(𝑦))/𝑑𝑥 . log⁡𝑥 + (𝑑 (log⁡𝑥))/𝑑𝑥 . 𝑦 1/𝑣 (𝑑𝑣/𝑑𝑥) = ( 𝑑(𝑦))/𝑑𝑥 . log⁡𝑥 + (𝑑 (log⁡𝑥))/𝑑𝑥 . 𝑦 1/𝑣 (𝑑𝑣/𝑑𝑥) = ( 𝑑𝑦)/𝑑𝑥 . log⁡𝑥 + 1/𝑥 . 𝑦 1/𝑣 (𝑑𝑣/𝑑𝑥) = ( 𝑑𝑦)/𝑑𝑥 log⁡𝑥 + 𝑦/𝑥 𝑑𝑣/𝑑𝑥 = v (log ( 𝑑𝑦)/𝑑𝑥 𝑥+𝑦/𝑥) Putting values of 𝑣 = 𝑥^𝑦 𝒅𝒗/𝒅𝒙 = 𝒙^𝒚 (𝒅𝒚/𝒅𝒙 𝒍𝒐𝒈⁡〖𝒙+ 𝒚/𝒙〗 ) Calculating derivative of 𝒘 𝑤 = 𝑥^𝑥 Taking log both sides log⁡𝑤=log⁡〖 (𝑥^𝑥)" " 〗 log⁡𝑤=〖𝑥. log〗⁡𝑥" " Differentiating both sides 𝑤.𝑟.𝑡.𝑥 (𝑑(log⁡𝑤))/𝑑𝑥 = (𝑑(𝑥 . log⁡𝑥))/𝑑𝑥 (𝑑(log⁡𝑤))/𝑑𝑥 (𝑑𝑤/𝑑𝑤) = 𝑑(𝑥 log⁡𝑥 )/𝑑𝑥 (𝑑(log⁡𝑤))/𝑑𝑤 . 𝑑𝑤/𝑑𝑥 = 𝑑(𝑥 log⁡𝑥 )/𝑑𝑥 (𝐴𝑠 log⁡〖(𝑎^𝑏)〗=𝑏 log⁡𝑎) 1/𝑤 . 𝑑𝑤/𝑑𝑥 = 𝑑(𝑥 log⁡𝑥 )/𝑑𝑥 log⁡𝑤=〖𝑥. log〗⁡𝑥" " Differentiating both sides 𝑤.𝑟.𝑡.𝑥 (𝑑(log⁡𝑤))/𝑑𝑥 = (𝑑(𝑥 . log⁡𝑥))/𝑑𝑥 (𝑑(log⁡𝑤))/𝑑𝑥 (𝑑𝑤/𝑑𝑤) = 𝑑(𝑥 log⁡𝑥 )/𝑑𝑥 (𝑑(log⁡𝑤))/𝑑𝑤 . 𝑑𝑤/𝑑𝑥 = 𝑑(𝑥 log⁡𝑥 )/𝑑𝑥 1/𝑤 . 𝑑𝑤/𝑑𝑥 = 𝑑(𝑥 log⁡𝑥 )/𝑑𝑥 By product Rule (uv)’ = u’v + v’u 1/𝑤 (𝑑𝑤/𝑑𝑥) = ( 𝑑(𝑥))/𝑑𝑥 . log⁡𝑥 + (𝑑 (log⁡𝑥))/𝑑𝑥 . 𝑥 1/𝑤 (𝑑𝑤/𝑑𝑥) = 1 . log⁡𝑥 + 1/𝑥 . 𝑥 1/𝑤 (𝑑𝑤/𝑑𝑥) = (log⁡〖𝑥+1〗) 𝑑𝑤/𝑑𝑥 = 𝑤(log⁡〖𝑥+1〗) 𝒅𝒘/𝒅𝒙 = 𝒙^𝒙 (𝒍𝒐𝒈⁡〖𝒙+𝟏〗 ) From (1) 𝑑𝑢/𝑑𝑥 + 𝑑𝑣/𝑑𝑥 + 𝑑𝑤/𝑑𝑥 = 0 Putting values from (2), (3) & (4) (𝑦^𝑥 log⁡〖𝑦+𝑦^(𝑥−1). 𝑥 𝑑𝑦/𝑑𝑥 〗 ) + (𝑥^𝑦 log⁡〖𝑥.𝑑𝑦/𝑑𝑥+𝑥^𝑦.𝑦/𝑥 〗 ) + (𝑥^𝑥 (log⁡〖𝑥+1〗))=0(𝑦^𝑥 log⁡〖𝑦+𝑥^𝑦. 𝑦/𝑥+𝑥^𝑥 (log⁡〖𝑥+1〗)〗 ) + (𝑦^(𝑥−1) .⁡〖𝑥 𝑑𝑦/𝑑𝑥+𝑥^𝑦 log⁡〖𝑥 𝑑𝑦/𝑑𝑥〗 〗 ) = 0 (𝑦^(𝑥−1) .⁡〖𝑥 𝑑𝑦/𝑑𝑥+𝑥^𝑦 log⁡〖𝑦 𝑑𝑦/𝑑𝑥〗 〗 ) = − (𝑦^𝑥 log⁡〖𝑦+𝑥^𝑦. 𝑦/𝑥+𝑥^𝑥 (log⁡〖𝑥+1〗)〗 ) (𝑦^(𝑥−1) .⁡〖𝑥 +𝑥^𝑦 log⁡〖𝑥 〗 〗 ) 𝑑𝑦/𝑑𝑥 = − (𝑦^𝑥 log⁡〖𝑦+𝑥^𝑦. 𝑦/𝑥+𝑥^𝑥 (log⁡〖𝑥+1〗)〗 ) 𝑑𝑦/𝑑𝑥 = "−" (𝑦^𝑥 𝑙𝑜𝑔⁡〖𝑦 + 𝑥^𝑦. 𝑦/𝑥 + 𝑥^𝑥 (1 + 𝑙𝑜𝑔⁡𝑥)〗 )/((〖𝑥𝑦〗^(𝑥−1) +⁡〖𝑥^𝑦 𝑙𝑜𝑔⁡〖𝑥 〗 〗)) 𝒅𝒚/𝒅𝒙 = "−" (𝒚^𝒙 𝒍𝒐𝒈⁡〖𝒚 + 𝒙^(𝒚 − 𝟏) 𝒚 + 𝒙^𝒙 (𝟏 + 𝒍𝒐𝒈⁡𝒙)〗 )/((〖𝒙𝒚〗^(𝒙−𝟏) +⁡〖𝒙^𝒚 𝒍𝒐𝒈⁡〖𝒙 〗 〗))

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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.