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Ex 5.3, 1 Find 𝑑𝑦/𝑑π‘₯ in, 2π‘₯ + 3𝑦 = sin⁑π‘₯ Given 2π‘₯+3𝑦 = sin⁑π‘₯ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑 (2π‘₯ + 3𝑦))/(𝑑π‘₯ ) = (𝑑(sin⁑〖π‘₯)γ€—)/(𝑑π‘₯ ) (𝑑(2π‘₯))/(𝑑π‘₯ ) +(𝑑(3𝑦))/(𝑑π‘₯ )=(𝑑(sin⁑〖π‘₯)γ€—)/(𝑑π‘₯ ) 2 (𝑑(π‘₯))/(𝑑π‘₯ ) +"3" (𝑑(𝑦))/(𝑑π‘₯ )=(𝑑(sin⁑〖π‘₯)γ€—)/(𝑑π‘₯ ) 2 + 3 𝑑𝑦/(𝑑π‘₯ ) = 𝒄𝒐𝒔⁑𝒙 3 𝑑𝑦/(𝑑π‘₯ ) = cos⁑π‘₯ βˆ’ 2 π’…π’š/(𝒅𝒙 ) = 𝟏/πŸ‘ (𝒄𝒐𝒔⁑𝒙 βˆ’ 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.