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Misc 14 If 𝑥 √(1+𝑦)+𝑦 √(1+𝑥) = 0 , for –1 < 𝑥 < 1, prove that 𝑑𝑦/𝑑𝑥 = (−1)/(1 + 𝑥)2 𝑥 √(1+𝑦)+𝑦 √(1+𝑥) = 0 𝑥 √(1+𝑦) = – 𝑦 √(1+𝑥) Squaring both sides (𝑥√(1+𝑦) )^2 = (−𝑦 √(1+𝑥))^2 𝑥^2 (√(1+𝑦 ) )^2 = (−𝑦)^2 (√(1+𝑥))^2 𝑥^2 (1+𝑦) = 𝑦^2 (1+𝑥) 𝑥^2+𝑥^2 𝑦 = 𝑦^2 + 𝑦^2 𝑥 𝑥^2 − 𝑦^2 = xy2 − x2y (𝒙 −𝒚) (𝑥+𝑦)=𝑥𝑦 (𝑦 −𝑥) −(𝒚 −𝒙) (𝑥+𝑦)=𝑥𝑦 (𝑦 −𝑥) −(𝑥+𝑦) = 𝑥𝑦 −𝑥 −𝑦 = 𝑥𝑦 −𝑥 = 𝑥𝑦+𝑦 −𝑥 = (𝑥+1) 𝑦 𝒚 = (−𝒙)/(𝒙 + 𝟏) Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦/𝑑𝑥 = 𝑑/𝑑𝑥 ((−𝑥)/(𝑥 + 1)) Using quotient rule As (𝑢/𝑣)′ = (𝑢^′ 𝑣 − 𝑣^′ 𝑢)/𝑣^2 where u = −x & v = x + 1 𝑑𝑦/𝑑𝑥 = (𝑑(−𝑥)/𝑑𝑥 (𝑥 + 1) − 𝑑(𝑥 + 1)/𝑑𝑥. (−𝑥))/(𝑥 + 1)^2 𝑑𝑦/𝑑𝑥 = (−1 (𝑥 + 1) + (1 + 0) 𝑥)/(𝑥 + 1)^2 𝑑𝑦/𝑑𝑥 = (−𝑥 − 1 + 𝑥)/(𝑥 + 1)^2 𝒅𝒚/𝒅𝒙 = (−𝟏)/(𝒙 + 𝟏)^𝟐

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.