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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

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Misc 14 If π‘₯ √(1+𝑦)+𝑦 √(1+π‘₯) = 0 , for –1 < π‘₯ < 1, prove that 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/(1 + π‘₯)2 π‘₯ √(1+𝑦)+𝑦 √(1+π‘₯) = 0 π‘₯ √(1+𝑦) = – 𝑦 √(1+π‘₯) Squaring both sides (π‘₯√(1+𝑦) )^2 = (βˆ’π‘¦ √(1+π‘₯))^2 π‘₯^2 (√(1+𝑦 ) )^2 = (βˆ’π‘¦)^2 (√(1+π‘₯))^2 π‘₯^2 (1+𝑦) = 𝑦^2 (1+π‘₯) π‘₯^2+π‘₯^2 𝑦 = 𝑦^2 + 𝑦^2 π‘₯ π‘₯^2 βˆ’ 𝑦^2 = xy2 βˆ’ x2y (𝒙 βˆ’π’š) (π‘₯+𝑦)=π‘₯𝑦 (𝑦 βˆ’π‘₯) βˆ’(π’š βˆ’π’™) (π‘₯+𝑦)=π‘₯𝑦 (𝑦 βˆ’π‘₯) βˆ’(π‘₯+𝑦) = π‘₯𝑦 βˆ’π‘₯ βˆ’π‘¦ = π‘₯𝑦 βˆ’π‘₯ = π‘₯𝑦+𝑦 βˆ’π‘₯ = (π‘₯+1) 𝑦 π’š = (βˆ’π’™)/(𝒙 + 𝟏) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = 𝑑/𝑑π‘₯ ((βˆ’π‘₯)/(π‘₯ + 1)) Using quotient rule As (𝑒/𝑣)β€² = (𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 where u = βˆ’x & v = x + 1 𝑑𝑦/𝑑π‘₯ = (𝑑(βˆ’π‘₯)/𝑑π‘₯ (π‘₯ + 1) βˆ’ 𝑑(π‘₯ + 1)/𝑑π‘₯. (βˆ’π‘₯))/(π‘₯ + 1)^2 𝑑𝑦/𝑑π‘₯ = (βˆ’1 (π‘₯ + 1) + (1 + 0) π‘₯)/(π‘₯ + 1)^2 𝑑𝑦/𝑑π‘₯ = (βˆ’π‘₯ βˆ’ 1 + π‘₯)/(π‘₯ + 1)^2 π’…π’š/𝒅𝒙 = (βˆ’πŸ)/(𝒙 + 𝟏)^𝟐

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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.