Slide22.JPG

Slide23.JPG
Slide24.JPG

Go Ad-free

Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

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 and Computer Science at Teachoo