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

Transcript

Ex 5.3, 4 Find 𝑑𝑦/𝑑π‘₯ in, π‘₯𝑦 + 𝑦2 = tan⁑π‘₯ + 𝑦 π‘₯𝑦 + 𝑦2 = tan⁑π‘₯ + 𝑦 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑(π‘₯𝑦 + 𝑦2)/𝑑π‘₯ = (𝑑 (tan⁑π‘₯ + 𝑦" " ))/𝑑π‘₯ 𝑑(π‘₯𝑦)/𝑑π‘₯ +𝑑(𝑦2)/𝑑π‘₯= (𝑑 (tan⁑π‘₯))/𝑑π‘₯ + (𝑑 (𝑦))/𝑑π‘₯ Using product rule in (π‘₯𝑦)^β€²=π‘₯^β€² 𝑦+𝑦^β€² π‘₯ ((𝑑(π‘₯))/𝑑π‘₯. 𝑦+ (𝑑(𝑦))/𝑑π‘₯.π‘₯) + 𝑑(𝑦2)/𝑑π‘₯= (𝑑 (tan⁑π‘₯))/𝑑π‘₯ + 𝑑𝑦/𝑑π‘₯ 1.𝑦 + π‘₯.𝑑𝑦/𝑑π‘₯ + (𝑑(𝑦2) )/𝑑π‘₯ ×𝑑𝑦/𝑑𝑦= sec2 π‘₯ + 𝑑𝑦/𝑑π‘₯ 𝑦 + π‘₯.𝑑𝑦/𝑑π‘₯ + (𝑑(𝑦2) )/𝑑𝑦 ×𝑑𝑦/𝑑π‘₯ = sec2 π‘₯ + 𝑑𝑦/𝑑π‘₯ 𝑦 + π‘₯.𝑑𝑦/𝑑π‘₯ + 2y .𝑑𝑦/𝑑π‘₯ =sec2 π‘₯ + 𝑑𝑦/𝑑π‘₯ π‘₯.𝑑𝑦/𝑑π‘₯ + 2y .𝑑𝑦/𝑑π‘₯ " "βˆ’ 𝑑𝑦/𝑑π‘₯ =sec2 π‘₯βˆ’π‘¦ 𝑑𝑦/𝑑π‘₯ (π‘₯+2π‘¦βˆ’1)=sec2 π‘₯βˆ’π‘¦ π’…π’š/𝒅𝒙 = (π’”π’†π’„πŸ 𝒙 βˆ’ π’š" " )/(𝒙 + πŸπ’š βˆ’ 𝟏)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.