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

Transcript

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 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.