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

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

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

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Ex 5.3, 6 Find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ in, ๐‘ฅ3 + ๐‘ฅ2๐‘ฆ + ๐‘ฅ๐‘ฆ2 + ๐‘ฆ3 = 81 ๐‘ฅ3 + ๐‘ฅ2๐‘ฆ + ๐‘ฅ๐‘ฆ2 + ๐‘ฆ3 = 81 Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ . ๐‘‘(๐‘ฅ3 + ๐‘ฅ2๐‘ฆ + ๐‘ฅ๐‘ฆ2 + ๐‘ฆ3)/๐‘‘๐‘ฅ = (๐‘‘ (81))/๐‘‘๐‘ฅ ๐‘‘(๐‘ฅ3)/๐‘‘๐‘ฅ + ๐‘‘(๐‘ฅ2๐‘ฆ)/๐‘‘๐‘ฅ + (๐‘‘(๐‘ฅ๐‘ฆ2))/๐‘‘๐‘ฅ + ๐‘‘(๐‘ฆ3)/๐‘‘๐‘ฅ =0 3๐‘ฅ^(3โˆ’1) + (๐‘‘(๐‘ฅ2๐‘ฆ))/๐‘‘๐‘ฅ + (๐‘‘(๐‘ฅ๐‘ฆ2))/๐‘‘๐‘ฅ + ๐‘‘(๐‘ฆ3)/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ = 0 3๐‘ฅ^2 + (๐‘‘(๐‘ฅ2๐‘ฆ))/๐‘‘๐‘ฅ + (๐‘‘(๐‘ฅ๐‘ฆ2))/๐‘‘๐‘ฅ + ๐‘‘(๐‘ฆ3)/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 3๐‘ฅ^2+ (๐‘‘(๐‘ฅ2๐‘ฆ))/๐‘‘๐‘ฅ + (๐‘‘(๐‘ฅ๐‘ฆ2))/๐‘‘๐‘ฅ +3๐‘ฆ^(3โˆ’1) . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 3๐‘ฅ2 + (๐‘‘(๐‘ฅ2๐‘ฆ))/๐‘‘๐‘ฅ + (๐‘‘(๐‘ฅ๐‘ฆ2))/๐‘‘๐‘ฅ +3๐‘ฆ^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 Using product rule in ๐‘ฅ2๐‘ฆ & ๐‘ฅ๐‘ฆ2 (uv)โ€™ = uโ€™v + vโ€™u 3๐‘ฅ2 + ((๐‘‘(๐‘ฅ2))/๐‘‘๐‘ฅ.๐‘ฆ+๐‘ฅ2 .(๐‘‘(๐‘ฆ))/๐‘‘๐‘ฅ)+((๐‘‘(๐‘ฅ))/๐‘‘๐‘ฅ.๐‘ฆ2+ .(๐‘‘(๐‘ฆ2))/๐‘‘๐‘ฅ ๐‘ฅ )+ 3y2 ๐‘‘๐‘ฆ/( ๐‘‘๐‘ฅ) = 0 3๐‘ฅ2 + (2๐‘ฅ.๐‘ฆ+๐‘ฅ2 (๐‘‘(๐‘ฆ))/๐‘‘๐‘ฅ) + (1.๐‘ฆ2+๐‘ฅ .(๐‘‘(๐‘ฆ2))/๐‘‘๐‘ฅ )+ 3y2 ๐‘‘๐‘ฆ/( ๐‘‘๐‘ฅ) = 0 3๐‘ฅ2 + (2๐‘ฅ.๐‘ฆ+๐‘ฅ2 (๐‘‘(๐‘ฆ))/๐‘‘๐‘ฅ) + (๐‘ฆ2+๐‘ฅ .(๐‘‘(๐‘ฆ2))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ)+ 3y2 ๐‘‘๐‘ฆ/( ๐‘‘๐‘ฅ) = 0 3๐‘ฅ2 + 2๐‘ฅ๐‘ฆ+๐‘ฅ2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+๐‘ฆ2+๐‘ฅ.(๐‘‘(๐‘ฆ2))/๐‘‘๐‘ฆ ร—๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+3๐‘ฆ2 ๐‘‘๐‘ฆ/( ๐‘‘๐‘ฅ) = 0 3๐‘ฅ2 + 2๐‘ฅ๐‘ฆ+๐‘ฅ2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+๐‘ฆ2 + ๐‘ฅ. 2๐‘ฆ^(2โˆ’1) (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)+3๐‘ฆ2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=0 3๐‘ฅ2 + 2๐‘ฅ๐‘ฆ+๐‘ฆ2 + x2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+๐‘ฅ.2๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+3๐‘ฆ2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=0 "(3" ๐‘ฅ"2 "+" " 2๐‘ฅ๐‘ฆ+๐‘ฆ2")" + ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (๐‘ฅ2+2๐‘ฅ๐‘ฆ+3๐‘ฆ2)=0 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (๐‘ฅ2+2๐‘ฅ๐‘ฆ+3๐‘ฆ2)=โˆ’ "(3" ๐‘ฅ"2 "+" " 2๐‘ฅ๐‘ฆ+๐‘ฆ2")" ๐’…๐’š/๐’…๐’™= (โˆ’ "(" ๐Ÿ‘๐’™"2 " +" " ๐Ÿ๐’™๐’š + ๐’š๐Ÿ")" )/((๐’™๐Ÿ + ๐Ÿ๐’™๐’š + ๐Ÿ‘๐’š๐Ÿ) )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.