Ex 5.3, 8 - Find dy/dx in, sin2 x + cos2 y = 1 - Class 12

Ex 5.3, 8 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.3, 8 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Ex 5.3, 8 Find 𝑑𝑦/𝑑π‘₯ in, sin2 π‘₯ + cos2 𝑦 = 1 sin2 π‘₯ + cos2 𝑦 = 1 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ . (𝑑 (sin2 π‘₯ + cos2 𝑦))/𝑑π‘₯ = (𝑑 (1))/𝑑π‘₯ (𝑑 (sin2 π‘₯))/𝑑π‘₯ + (𝑑 (cos⁑2 𝑦))/𝑑π‘₯ = 0 Calculating Derivative of sin2 π‘₯ & cos^2 𝑦 sepretaly Finding Derivative of π’”π’Šπ’πŸ 𝒙 (𝑑 (sin2 π‘₯))/𝑑π‘₯ =2γ€– 𝑠𝑖𝑛〗^(2βˆ’1) π‘₯ . (𝑑(sin^2⁑π‘₯))/𝑑π‘₯ =2 sin⁑π‘₯ . (𝑑(sin⁑π‘₯))/𝑑π‘₯ (Derivative of constant is 0) =2 sin⁑〖π‘₯ γ€– cos〗⁑π‘₯ γ€— Finding Derivative of 〖𝒄𝒐𝒔〗^𝟐 π’š (𝑑 (cos2 𝑦))/𝑑π‘₯ =2γ€–cos⁑𝑦〗^(2βˆ’1) ". " 𝑑/𝑑π‘₯ " "(cos⁑𝑦) =2 cos⁑𝑦 . (βˆ’sin⁑𝑦) . (𝑑(𝑦))/𝑑π‘₯ =βˆ’ 2 cos⁑𝑦 sin⁑𝑦 . 𝑑𝑦/𝑑π‘₯ Now, (𝑑 (sin2 π‘₯))/𝑑π‘₯+ (𝑑 (cos2 𝑦))/𝑑π‘₯ = 0 2 sin⁑π‘₯ .cos⁑π‘₯ + (βˆ’ 2 cos⁑𝑦 sin⁑𝑦 ". " 𝑑𝑦/𝑑π‘₯)= 0 2 sin⁑π‘₯ .cos⁑π‘₯ βˆ’ 2 sin⁑𝑦⁑〖 .γ€— cos⁑𝑦 . 𝑑𝑦/𝑑π‘₯ = 0 βˆ’ 2 sin⁑𝑦⁑〖 .γ€— cos⁑𝑦 . 𝑑𝑦/𝑑π‘₯ = βˆ’ 2 sin⁑π‘₯ cos⁑π‘₯ βˆ’ sin⁑2𝑦⁑〖 .γ€— 𝑑𝑦/𝑑π‘₯ = βˆ’ sin⁑2π‘₯ 𝑑𝑦/𝑑π‘₯ = γ€–βˆ’ sin〗⁑2π‘₯/(βˆ’sin⁑2𝑦 ) π’…π’š/𝒅𝒙 = π’”π’Šπ’β‘πŸπ’™/π’”π’Šπ’β‘πŸπ’š (2 sin x cos x = sin 2x)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.