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Ex 5.3, 7 - Find dy/dx in sin2 y + cos xy = pi - Class 12

Ex 5.3, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.3, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.3, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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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 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€—

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.