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Ex 5.3, 4 - Find dy/dx in, xy + y2 = tan x + y - Chapter 5

Ex 5.3, 4 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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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 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.