Ex 5.3, 6 - Find dy/dx in x3 + x2y + xy2 + y3 = 81 - CBSE

Ex 5.3, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.3, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Transcript

Ex 5.3, 6 Find 𝑑𝑦/𝑑𝑥 in, 𝑥3 + 𝑥2𝑦 + 𝑥𝑦2 + 𝑦3 = 81 𝑥3 + 𝑥2𝑦 + 𝑥𝑦2 + 𝑦3 = 81 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 . 𝑑(𝑥3 + 𝑥2𝑦 + 𝑥𝑦2 + 𝑦3)/𝑑𝑥 = (𝑑 (81))/𝑑𝑥 𝑑(𝑥3)/𝑑𝑥 + 𝑑(𝑥2𝑦)/𝑑𝑥 + (𝑑(𝑥𝑦2))/𝑑𝑥 + 𝑑(𝑦3)/𝑑𝑥 =0 3𝑥^(3−1) + (𝑑(𝑥2𝑦))/𝑑𝑥 + (𝑑(𝑥𝑦2))/𝑑𝑥 + 𝑑(𝑦3)/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 = 0 3𝑥^2 + (𝑑(𝑥2𝑦))/𝑑𝑥 + (𝑑(𝑥𝑦2))/𝑑𝑥 + 𝑑(𝑦3)/𝑑𝑥 × 𝑑𝑦/𝑑𝑥 = 0 3𝑥^2+ (𝑑(𝑥2𝑦))/𝑑𝑥 + (𝑑(𝑥𝑦2))/𝑑𝑥 +3𝑦^(3−1) . 𝑑𝑦/𝑑𝑥 = 0 3𝑥2 + (𝑑(𝑥2𝑦))/𝑑𝑥 + (𝑑(𝑥𝑦2))/𝑑𝑥 +3𝑦^2 𝑑𝑦/𝑑𝑥 = 0 Using product rule in 𝑥2𝑦 & 𝑥𝑦2 (uv)’ = u’v + v’u 3𝑥2 + ((𝑑(𝑥2))/𝑑𝑥.𝑦+𝑥2 .(𝑑(𝑦))/𝑑𝑥)+((𝑑(𝑥))/𝑑𝑥.𝑦2+ .(𝑑(𝑦2))/𝑑𝑥 𝑥 )+ 3y2 𝑑𝑦/( 𝑑𝑥) = 0 3𝑥2 + (2𝑥.𝑦+𝑥2 (𝑑(𝑦))/𝑑𝑥) + (1.𝑦2+𝑥 .(𝑑(𝑦2))/𝑑𝑥 )+ 3y2 𝑑𝑦/( 𝑑𝑥) = 0 3𝑥2 + (2𝑥.𝑦+𝑥2 (𝑑(𝑦))/𝑑𝑥) + (𝑦2+𝑥 .(𝑑(𝑦2))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦)+ 3y2 𝑑𝑦/( 𝑑𝑥) = 0 3𝑥2 + 2𝑥𝑦+𝑥2 𝑑𝑦/𝑑𝑥+𝑦2+𝑥.(𝑑(𝑦2))/𝑑𝑦 ×𝑑𝑦/𝑑𝑥+3𝑦2 𝑑𝑦/( 𝑑𝑥) = 0 3𝑥2 + 2𝑥𝑦+𝑥2 𝑑𝑦/𝑑𝑥+𝑦2 + 𝑥. 2𝑦^(2−1) (𝑑𝑦/𝑑𝑥)+3𝑦2 𝑑𝑦/𝑑𝑥=0 3𝑥2 + 2𝑥𝑦+𝑦2 + x2 𝑑𝑦/𝑑𝑥+𝑥.2𝑦 𝑑𝑦/𝑑𝑥+3𝑦2 𝑑𝑦/𝑑𝑥=0 "(3" 𝑥"2 "+" " 2𝑥𝑦+𝑦2")" + 𝑑𝑦/𝑑𝑥 (𝑥2+2𝑥𝑦+3𝑦2)=0 𝑑𝑦/𝑑𝑥 (𝑥2+2𝑥𝑦+3𝑦2)=− "(3" 𝑥"2 "+" " 2𝑥𝑦+𝑦2")" 𝒅𝒚/𝒅𝒙= (− "(" 𝟑𝒙"2 " +" " 𝟐𝒙𝒚 + 𝒚𝟐")" )/((𝒙𝟐 + 𝟐𝒙𝒚 + 𝟑𝒚𝟐) )

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.