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

Transcript

Ex 5.3, 7 Find 𝑑𝑦/𝑑π‘₯ in, sin2 𝑦 +cos⁑ π‘₯𝑦 =πœ‹ sin2 𝑦 +cos⁑ π‘₯𝑦 =πœ‹ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ . (𝑑 (sin2 𝑦 + cos⁑ π‘₯𝑦))/𝑑π‘₯ = (𝑑 (πœ‹))/𝑑π‘₯ (𝑑 (sin2 𝑦))/𝑑π‘₯ + (𝑑 (cos⁑〖 π‘₯γ€— 𝑦))/𝑑π‘₯= 0 Calculating Derivative of sin2 𝑦 & cos (x𝑦) separately (Derivative of constant is zero) Calculating Derivative of π’”π’Šπ’πŸ π’š (𝑑 (sin2 𝑦))/𝑑π‘₯= (𝑑 (sin2 𝑦))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 =(𝑑 (sin2(𝑦)))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ =2 𝑠𝑖𝑛 𝑦 Γ— (𝑑(sin⁑〖𝑦)γ€—)/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ = 2 sin⁑〖𝑦 cos⁑〖𝑦 Γ— γ€— γ€— 𝑑𝑦/𝑑π‘₯ Calculating Derivative of 𝒄𝒐𝒔 (π’™π’š) (𝑑(cos⁑〖(π‘₯𝑦))γ€—)/𝑑π‘₯= βˆ’ sin⁑(π‘₯𝑦) Γ— 𝑑/𝑑π‘₯(π‘₯𝑦) = βˆ’sin⁑(π‘₯𝑦)Γ—(𝑑(π‘₯)/𝑑π‘₯.𝑦+𝑑(𝑦)/𝑑π‘₯.π‘₯) = βˆ’sin⁑(π‘₯𝑦) . (1.𝑦+π‘₯ 𝑑𝑦/𝑑π‘₯) = βˆ’sin⁑(π‘₯𝑦) (𝑦+π‘₯ 𝑑𝑦/𝑑π‘₯) = βˆ’sin⁑(π‘₯𝑦) . π‘¦βˆ’sin⁑〖(π‘₯𝑦) . π‘₯γ€— 𝑑𝑦/𝑑π‘₯ = βˆ’π‘¦ sin⁑〖 (π‘₯𝑦)γ€— βˆ’ sin⁑〖(π‘₯𝑦) . π‘₯γ€— 𝑑𝑦/𝑑π‘₯ Now, (𝑑 (𝑠𝑖𝑛2𝑦))/𝑑π‘₯+(𝑑 (cos⁑〖(π‘₯𝑦)) γ€—)/𝑑π‘₯ = 0 Putting values 2 sin⁑𝑦.cos⁑〖𝑦 .γ€— 𝑑𝑦/𝑑π‘₯ + (βˆ’ 𝑦 sin⁑(π‘₯𝑦)βˆ’π‘₯ sin⁑(π‘₯𝑦) 𝑑𝑦/𝑑π‘₯) = 0 2 sin⁑𝑦.cos⁑〖𝑦 .γ€— 𝑑𝑦/𝑑π‘₯ βˆ’ 𝑦 sin⁑(π‘₯𝑦) βˆ’ π‘₯ sin⁑(π‘₯𝑦) 𝑑𝑦/𝑑π‘₯ = 0 2 sin⁑𝑦 cos⁑〖𝑦 γ€— 𝑑𝑦/𝑑π‘₯ βˆ’π‘₯ sin⁑(π‘₯𝑦) 𝑑𝑦/𝑑π‘₯ = 𝑦 sin⁑(π‘₯𝑦) 𝑑𝑦/𝑑π‘₯ (2 sin⁑𝑦 cos⁑𝑦 βˆ’ π‘₯ sin⁑(π‘₯𝑦) = 𝑦 sin⁑(π‘₯𝑦) 𝑑𝑦/𝑑π‘₯ = (𝑦 sin⁑〖(π‘₯𝑦)γ€—)/(2 γ€–sin 𝑦〗⁑〖cos⁑𝑦 βˆ’ π‘₯ sin⁑π‘₯𝑦 γ€— ) π’…π’š/𝒅𝒙 = (π’š π’”π’Šπ’β‘γ€–(π’™π’š)γ€—)/γ€–π’”π’Šπ’ πŸπ’šγ€—β‘γ€–βˆ’ 𝒙 π’”π’Šπ’β‘π’™π’š γ€— (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 9 years. He provides courses for Maths and Science at Teachoo.