Ex 5.3, 3 - Find dy/dx in, ax+by2 = cos y - Chapter 5 NCERT

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

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Ex 5.3, 3 Find 𝑑𝑦/𝑑π‘₯ in, π‘Žπ‘₯ + 𝑏𝑦2 = cos⁑𝑦 π‘Žπ‘₯ + 𝑏𝑦2 = cos⁑𝑦 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑(π‘Žπ‘₯ + 𝑏𝑦2)/𝑑π‘₯ = (𝑑 (cπ‘œπ‘ β‘π‘¦ ))/𝑑π‘₯ 𝑑(π‘Žπ‘₯)/𝑑π‘₯ +𝑑(𝑏𝑦2)/𝑑π‘₯= (𝑑 γ€–(cos〗⁑𝑦))/𝑑π‘₯ a 𝑑π‘₯/𝑑π‘₯ +𝑏 𝑑(𝑦2)/𝑑π‘₯= (𝑑 )/𝑑π‘₯ cos y a + b (𝑑 (𝑦2))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦= (𝑑 (cos⁑𝑦 ))/𝑑π‘₯ ×𝑑𝑦/𝑑𝑦 a + b (𝑑 (𝑦2))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯= (𝑑 (cos⁑𝑦 ))/𝑑𝑦 ×𝑑𝑦/𝑑π‘₯ π‘Ž + 𝑏 .2𝑦× 𝑑𝑦/𝑑π‘₯= βˆ’sin⁑〖𝑦 γ€— 𝑑𝑦/𝑑π‘₯ π‘Ž + 2𝑏𝑦 . 𝑑𝑦/𝑑π‘₯= βˆ’sin⁑〖𝑦 γ€— 𝑑𝑦/𝑑π‘₯ 2𝑏𝑦 . 𝑑𝑦/𝑑π‘₯ + sin⁑〖𝑦 γ€— 𝑑𝑦/𝑑π‘₯ = 0 βˆ’ π‘Ž 𝑑𝑦/𝑑π‘₯ (2𝑏𝑦 + sin⁑〖𝑦 γ€—) = βˆ’π‘Ž π’…π’š/𝒅𝒙 = (βˆ’π’‚)/(πŸπ’ƒπ’š " + " π’”π’Šπ’β‘γ€–π’š γ€— )

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo