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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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Ex 5.3, 6 Find 𝑑𝑦/𝑑π‘₯ in, π‘₯3 + π‘₯2𝑦 + π‘₯𝑦2 + 𝑦3 = 81 π‘₯3 + π‘₯2𝑦 + π‘₯𝑦2 + 𝑦3 = 81 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ . 𝑑(π‘₯3 + π‘₯2𝑦 + π‘₯𝑦2 + 𝑦3)/𝑑π‘₯ = (𝑑 (81))/𝑑π‘₯ 𝑑(π‘₯3)/𝑑π‘₯ + 𝑑(π‘₯2𝑦)/𝑑π‘₯ + (𝑑(π‘₯𝑦2))/𝑑π‘₯ + 𝑑(𝑦3)/𝑑π‘₯ =0 3π‘₯^(3βˆ’1) + (𝑑(π‘₯2𝑦))/𝑑π‘₯ + (𝑑(π‘₯𝑦2))/𝑑π‘₯ + 𝑑(𝑦3)/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 = 0 3π‘₯^2 + (𝑑(π‘₯2𝑦))/𝑑π‘₯ + (𝑑(π‘₯𝑦2))/𝑑π‘₯ + 𝑑(𝑦3)/𝑑π‘₯ Γ— 𝑑𝑦/𝑑π‘₯ = 0 3π‘₯^2+ (𝑑(π‘₯2𝑦))/𝑑π‘₯ + (𝑑(π‘₯𝑦2))/𝑑π‘₯ +3𝑦^(3βˆ’1) . 𝑑𝑦/𝑑π‘₯ = 0 3π‘₯2 + (𝑑(π‘₯2𝑦))/𝑑π‘₯ + (𝑑(π‘₯𝑦2))/𝑑π‘₯ +3𝑦^2 𝑑𝑦/𝑑π‘₯ = 0 Using product rule in π‘₯2𝑦 & π‘₯𝑦2 (uv)’ = u’v + v’u 3π‘₯2 + ((𝑑(π‘₯2))/𝑑π‘₯.𝑦+π‘₯2 .(𝑑(𝑦))/𝑑π‘₯)+((𝑑(π‘₯))/𝑑π‘₯.𝑦2+ .(𝑑(𝑦2))/𝑑π‘₯ π‘₯ )+ 3y2 𝑑𝑦/( 𝑑π‘₯) = 0 3π‘₯2 + (2π‘₯.𝑦+π‘₯2 (𝑑(𝑦))/𝑑π‘₯) + (1.𝑦2+π‘₯ .(𝑑(𝑦2))/𝑑π‘₯ )+ 3y2 𝑑𝑦/( 𝑑π‘₯) = 0 3π‘₯2 + (2π‘₯.𝑦+π‘₯2 (𝑑(𝑦))/𝑑π‘₯) + (𝑦2+π‘₯ .(𝑑(𝑦2))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦)+ 3y2 𝑑𝑦/( 𝑑π‘₯) = 0 3π‘₯2 + 2π‘₯𝑦+π‘₯2 𝑑𝑦/𝑑π‘₯+𝑦2+π‘₯.(𝑑(𝑦2))/𝑑𝑦 ×𝑑𝑦/𝑑π‘₯+3𝑦2 𝑑𝑦/( 𝑑π‘₯) = 0 3π‘₯2 + 2π‘₯𝑦+π‘₯2 𝑑𝑦/𝑑π‘₯+𝑦2 + π‘₯. 2𝑦^(2βˆ’1) (𝑑𝑦/𝑑π‘₯)+3𝑦2 𝑑𝑦/𝑑π‘₯=0 3π‘₯2 + 2π‘₯𝑦+𝑦2 + x2 𝑑𝑦/𝑑π‘₯+π‘₯.2𝑦 𝑑𝑦/𝑑π‘₯+3𝑦2 𝑑𝑦/𝑑π‘₯=0 "(3" π‘₯"2 "+" " 2π‘₯𝑦+𝑦2")" + 𝑑𝑦/𝑑π‘₯ (π‘₯2+2π‘₯𝑦+3𝑦2)=0 𝑑𝑦/𝑑π‘₯ (π‘₯2+2π‘₯𝑦+3𝑦2)=βˆ’ "(3" π‘₯"2 "+" " 2π‘₯𝑦+𝑦2")" π’…π’š/𝒅𝒙= (βˆ’ "(" πŸ‘π’™"2 " +" " πŸπ’™π’š + π’šπŸ")" )/((π’™πŸ + πŸπ’™π’š + πŸ‘π’šπŸ) )

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