Check sibling questions

Misc 14 - if x = root 1+y + y root 1+x = 0, prove dy/dx - Miscellaneou

Misc  14 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc  14 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


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