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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Transcript

Ex 5.3, 8 Find 𝑑𝑦/𝑑𝑥 in, sin2 𝑥 + cos2 𝑦 = 1 sin2 𝑥 + cos2 𝑦 = 1 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 . (𝑑 (sin2 𝑥 + cos2 𝑦))/𝑑𝑥 = (𝑑 (1))/𝑑𝑥 (𝑑 (sin2 𝑥))/𝑑𝑥 + (𝑑 (cos⁡2 𝑦))/𝑑𝑥 = 0 Calculating Derivative of sin2 𝑥 & cos^2 𝑦 sepretaly Finding Derivative of 𝒔𝒊𝒏𝟐 𝒙 (𝑑 (sin2 𝑥))/𝑑𝑥 =2〖 𝑠𝑖𝑛〗^(2−1) 𝑥 . (𝑑(sin^2⁡𝑥))/𝑑𝑥 =2 sin⁡𝑥 . (𝑑(sin⁡𝑥))/𝑑𝑥 (Derivative of constant is 0) =2 sin⁡〖𝑥 〖 cos〗⁡𝑥 〗 Finding Derivative of 〖𝒄𝒐𝒔〗^𝟐 𝒚 (𝑑 (cos2 𝑦))/𝑑𝑥 =2〖cos⁡𝑦〗^(2−1) ". " 𝑑/𝑑𝑥 " "(cos⁡𝑦) =2 cos⁡𝑦 . (−sin⁡𝑦) . (𝑑(𝑦))/𝑑𝑥 =− 2 cos⁡𝑦 sin⁡𝑦 . 𝑑𝑦/𝑑𝑥 Now, (𝑑 (sin2 𝑥))/𝑑𝑥+ (𝑑 (cos2 𝑦))/𝑑𝑥 = 0 2 sin⁡𝑥 .cos⁡𝑥 + (− 2 cos⁡𝑦 sin⁡𝑦 ". " 𝑑𝑦/𝑑𝑥)= 0 2 sin⁡𝑥 .cos⁡𝑥 − 2 sin⁡𝑦⁡〖 .〗 cos⁡𝑦 . 𝑑𝑦/𝑑𝑥 = 0 − 2 sin⁡𝑦⁡〖 .〗 cos⁡𝑦 . 𝑑𝑦/𝑑𝑥 = − 2 sin⁡𝑥 cos⁡𝑥 − sin⁡2𝑦⁡〖 .〗 𝑑𝑦/𝑑𝑥 = − sin⁡2𝑥 𝑑𝑦/𝑑𝑥 = 〖− sin〗⁡2𝑥/(−sin⁡2𝑦 ) 𝒅𝒚/𝒅𝒙 = 𝒔𝒊𝒏⁡𝟐𝒙/𝒔𝒊𝒏⁡𝟐𝒚 (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.