Ex 5.3, 7 - Find dy/dx in sin2 y + cos xy = pi - Class 12 - Finding derivative of Implicit functions

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Ex 5.3, 7 Find 𝑑𝑦﷮𝑑𝑥﷯ in, sin2 𝑦 +cos⁡ 𝑥𝑦 =𝜋 sin2 𝑦 +cos⁡ 𝑥𝑦 =𝜋 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 . 𝑑 sin2 𝑦 + cos⁡ 𝑥𝑦﷯﷮𝑑𝑥﷯ = 𝑑 𝜋﷯﷮𝑑𝑥﷯ 𝑑 (sin2 𝑦)﷮𝑑𝑥﷯ + 𝑑 ( cos﷮ 𝑥﷯𝑦)﷮𝑑𝑥﷯= 0 Calculating Derivative of sin2 𝑦 & cos (x𝑦) separately Calculating Derivative of 𝒔𝒊𝒏𝟐 𝒚 𝑑 (sin2 𝑦)﷮𝑑𝑥﷯= 𝑑 (sin2 𝑦)﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑦﷯ = 𝑑 (sin2(𝑦))﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑥﷯ =2 𝑠𝑖𝑛﷮2−1﷯ 𝑥. 𝑑( sin﷮𝑦)﷯﷮𝑑𝑦﷯ × 𝑑𝑦﷮𝑑𝑥﷯ = 2 sin﷮𝑦 . cos﷮𝑦 × ﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ Calculating Derivative of 𝒄𝒐𝒔 (𝒙𝒚) 𝑑( cos﷮(𝑥𝑦))﷯﷮𝑑𝑥﷯= − sin⁡(𝑥𝑦) × 𝑑﷮𝑑𝑥﷯(𝑥𝑦) = − sin﷮ 𝑥𝑦﷯﷯× 𝑑 𝑥﷯﷮𝑑𝑥﷯.𝑦+ 𝑑 𝑦﷯﷮𝑑𝑥﷯.𝑥﷯ = − sin﷮ 𝑥𝑦﷯﷯ . 1.𝑦+𝑥 𝑑𝑦﷮𝑑𝑥﷯﷯ = − sin﷮ 𝑥𝑦﷯﷯ 𝑦+𝑥 𝑑𝑦﷮𝑑𝑥﷯﷯ = − sin﷮ 𝑥𝑦﷯﷯ . 𝑦− sin﷮ 𝑥𝑦﷯ . 𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = −𝑦 sin﷮ 𝑥𝑦﷯﷯ − sin﷮ 𝑥𝑦﷯ . 𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ Now, 𝑑 (𝑠𝑖𝑛2𝑦)﷮𝑑𝑥﷯+ 𝑑 ( cos﷮(𝑥𝑦)) ﷯﷮𝑑𝑥﷯ = 0 Putting values 2 sin﷮𝑦﷯. cos﷮𝑦 .﷯ 𝑑𝑦﷮𝑑𝑥﷯ + − 𝑦 sin﷮ 𝑥𝑦﷯﷯−𝑥 sin﷮ 𝑥𝑦﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 0 2 sin﷮𝑦﷯. cos﷮𝑦 .﷯ 𝑑𝑦﷮𝑑𝑥﷯ − 𝑦 sin⁡(𝑥𝑦) − 𝑥 sin⁡(𝑥𝑦) 𝑑𝑦﷮𝑑𝑥﷯ = 0 2 sin﷮𝑦﷯ cos﷮𝑦 ﷯ 𝑑𝑦﷮𝑑𝑥﷯ −𝑥 sin⁡(𝑥𝑦) 𝑑𝑦﷮𝑑𝑥﷯ = 𝑦 sin⁡(𝑥𝑦) 𝑑𝑦﷮𝑑𝑥﷯ (2 sin⁡𝑦 cos⁡𝑦 − 𝑥 sin⁡(𝑥𝑦) = 𝑦 sin⁡(𝑥𝑦) 𝑑𝑦﷮𝑑𝑥﷯ = 𝑦 sin﷮(𝑥𝑦)﷯﷮2 sin 𝑦﷮ cos﷮𝑦﷯ − 𝑥 sin﷮𝑥𝑦﷯ ﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒚 𝒔𝒊𝒏﷮(𝒙𝒚)﷯﷮ 𝒔𝒊𝒏 𝟐𝒚﷮− 𝒙 𝒔𝒊𝒏﷮𝒙𝒚﷯ ﷯﷯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.