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

Last updated at April 16, 2024 by

Ex 5.3, 2 - Find dy/dx in, 2x + 3y = sin y - Class 12 NCERT

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

Transcript

Ex 5.3, 2 Find 𝑑𝑦/𝑑π‘₯ in , 2π‘₯ + 3𝑦 = sin⁑𝑦 . 2π‘₯ + 3𝑦 = sin⁑𝑦 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑(2π‘₯ + 3𝑦)/𝑑π‘₯ = (𝑑 (sin⁑𝑦 ))/𝑑π‘₯ 𝑑(2π‘₯)/𝑑π‘₯ +𝑑(3𝑦)/𝑑π‘₯= (𝑑 (sin⁑𝑦 ))/𝑑π‘₯ 2 𝑑π‘₯/𝑑π‘₯ +3𝑑(𝑦)/𝑑π‘₯= (𝑑 (sin⁑𝑦 ))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ 2 + 3 𝑑𝑦/𝑑π‘₯ = cos⁑𝑦 "Γ— " 𝑑𝑦/𝑑π‘₯ cos⁑𝑦 "Γ— " 𝑑𝑦/𝑑π‘₯ – 3 𝑑𝑦/𝑑π‘₯ = 2 (Derivative of 𝑠𝑖𝑛⁑π‘₯ is π‘π‘œπ‘ β‘π‘₯) 𝑑𝑦/𝑑π‘₯ (cos y – 3) = 2 π’…π’š/𝒅𝒙 = 𝟐/((π’„π’π’”β‘γ€–π’šβˆ’πŸ‘)γ€— )

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