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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

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.