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

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

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

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

Transcript

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
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.