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Ex 5.5, 12 - Find dy/dx, xy + yx = 1 - Class 12 CBSE NCERT

Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 7

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Transcript

Ex 5.5, 12 Find 𝑑𝑦/𝑑π‘₯ of the functions in, π‘₯^𝑦 + 𝑦^π‘₯ = 1 π‘₯^𝑦 + 𝑦^π‘₯ = 1 Let 𝑒 = π‘₯^𝑦 , 𝑣 = 𝑦^π‘₯ Hence, 𝑒+𝑣=1 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. (𝑑(𝑣⁑〖+ 𝑒〗))/𝑑π‘₯ = 𝑑(1)/𝑑π‘₯ 𝑑𝑣/𝑑π‘₯ + 𝑑𝑒/𝑑π‘₯ = 0 (Derivative of constant is 0) Calculating 𝒅𝒗/𝒅𝒙 𝑣=π‘₯^𝑦 Taking log both sides log⁑𝑣=log⁑〖 (π‘₯^𝑦)" " γ€— log⁑𝑣=〖𝑦. log〗⁑π‘₯" " Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑(log⁑𝑣))/𝑑π‘₯ = (𝑑(𝑦 . log⁑π‘₯))/𝑑π‘₯ (𝑑(log⁑𝑣))/𝑑π‘₯ (𝑑𝑣/𝑑𝑣) = 𝑑(𝑦 log⁑π‘₯ )/𝑑π‘₯ (𝑑(log⁑𝑣))/𝑑𝑣 (𝑑𝑣/𝑑π‘₯) = 𝑑(〖𝑦 log〗⁑π‘₯ )/𝑑π‘₯ 1/𝑣 (𝑑𝑣/𝑑π‘₯) = ( 𝑑(〖𝑦 log〗⁑π‘₯ ))/𝑑π‘₯ 1/𝑣 (𝑑𝑣/𝑑π‘₯) = ( 𝑑(𝑦))/𝑑π‘₯ . log⁑π‘₯ + (𝑑 (log⁑π‘₯))/𝑑π‘₯ . 𝑦 1/𝑣 (𝑑𝑣/𝑑π‘₯) = ( 𝑑𝑦)/𝑑π‘₯ . log⁑π‘₯ + 1/π‘₯ . 𝑦 1/𝑣 (𝑑𝑣/𝑑π‘₯) = ( 𝑑𝑦)/𝑑π‘₯ log⁑π‘₯ + 𝑦/π‘₯ 𝑑𝑣/𝑑π‘₯ = v (log π‘₯+𝑦/π‘₯) Putting value of 𝑣 = π‘₯^𝑦 𝑑𝑣/𝑑π‘₯ = π‘₯^𝑦 (log⁑〖π‘₯+ 𝑦/π‘₯γ€— ) 𝑑𝑣/𝑑π‘₯ = π‘₯^𝑦.log π‘₯ . 𝑑𝑦/𝑑π‘₯ + π‘₯^𝑦. 𝑦/π‘₯ Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 Calculating 𝒅𝒖/𝒅𝒙 𝑒 = 𝑦^π‘₯ Taking log both sides log⁑𝑒=log⁑〖 (𝑦^π‘₯)" " γ€— log⁑𝑒=γ€–π‘₯ . log〗⁑𝑦" " Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑(log⁑𝑒))/𝑑π‘₯ = (𝑑(π‘₯ . log⁑𝑦))/𝑑π‘₯ (𝑑(log⁑𝑒))/𝑑π‘₯ (𝑑𝑒/𝑑𝑒) = 𝑑(π‘₯.log⁑𝑦 )/𝑑π‘₯ (𝐴𝑠 log⁑〖(π‘Ž^𝑏)γ€—=𝑏 logβ‘π‘Ž ) (𝑑(log⁑𝑒))/𝑑𝑒 (𝑑𝑒/𝑑π‘₯) = (𝑑 (π‘₯ . log⁑𝑦))/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = (𝑑 (π‘₯ . log⁑𝑦 ))/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = (𝑑 (π‘₯))/𝑑π‘₯ . log⁑𝑦 + (𝑑(log⁑𝑦))/𝑑π‘₯ . π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 1 . log⁑𝑦 + π‘₯. 𝑑(log⁑𝑦 )/𝑑π‘₯ . 𝑑𝑦/𝑑𝑦 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = log⁑𝑦 + π‘₯. 𝑑(log⁑𝑦 )/𝑑𝑦 . 𝑑𝑦/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = log⁑𝑦 + π‘₯. 1/𝑦 . 𝑑𝑦/𝑑π‘₯ Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 𝑑𝑒/𝑑π‘₯ = 𝑒(log⁑𝑦 " + " π‘₯/𝑦 " . " (𝑑𝑦 )/𝑑π‘₯) 𝑑𝑒/𝑑π‘₯ " = " 𝑦^π‘₯ (log⁑𝑦 " + " π‘₯/𝑦 " . " (𝑑𝑦 )/𝑑π‘₯) 𝑑𝑒/𝑑π‘₯ = 𝑦^π‘₯ log⁑𝑦+𝑦^(π‘₯ βˆ’1). π‘₯ 𝑑𝑦/𝑑π‘₯ Now, 𝑑𝑣/𝑑π‘₯ + 𝑑𝑒/𝑑π‘₯ = 0 Putting value of 𝑑𝑣/𝑑π‘₯ & 𝑑𝑒/𝑑π‘₯ (π‘₯^𝑦 log⁑〖π‘₯ γ€— 𝑑𝑦/𝑑π‘₯ +π‘₯^𝑦 𝑦/π‘₯) + (𝑦^π‘₯ log⁑〖𝑦+𝑦^(π‘₯ βˆ’1) γ€—.π‘₯ 𝑑𝑦/𝑑π‘₯) = 0 π‘₯^𝑦 log⁑〖π‘₯ γ€— 𝑑𝑦/𝑑π‘₯ +π‘₯^(π‘¦βˆ’1) 𝑦 + 𝑦^π‘₯ log⁑〖𝑦+𝑦^(π‘₯ βˆ’1) γ€—.π‘₯ 𝑑𝑦/𝑑π‘₯ = 0 π‘₯^𝑦 log⁑〖π‘₯ γ€— 𝑑𝑦/𝑑π‘₯+𝑦^(π‘₯ βˆ’1).π‘₯ 𝑑𝑦/𝑑π‘₯=βˆ’(π‘₯^(π‘¦βˆ’1) 𝑦+𝑦^π‘₯ log⁑𝑦 ) 𝑑𝑦/𝑑π‘₯(π‘₯^𝑦 log⁑〖π‘₯ γ€—+𝑦^(π‘₯ βˆ’1).π‘₯)=βˆ’(π‘₯^(π‘¦βˆ’1) 𝑦+𝑦^π‘₯ log⁑𝑦 ) π’…π’š/𝒅𝒙 = (βˆ’(𝒙^(π’šβˆ’πŸ) π’š + π’š^𝒙 π’π’π’ˆβ‘π’š ))/((𝒙^π’š π’π’π’ˆβ‘γ€–π’™ γ€—+ π’š^(𝒙 βˆ’πŸ).𝒙))

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.