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

Transcript

Misc 12 Find 𝑑𝑦/𝑑π‘₯, if 𝑦=12 (1 –cos⁑𝑑 ), π‘₯=10 (𝑑 –sin⁑𝑑 ),βˆ’πœ‹/2 " "<π‘₯< πœ‹/2 Here, 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Calculating π’…π’š/𝒅𝒕 𝑦=12 (1 –cos⁑𝑑 ) 𝑦=12 βˆ’12 cos⁑𝑑 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑𝑑 = 𝑑(12 βˆ’ 12 cos⁑𝑑 )/𝑑𝑑 𝑑𝑦/𝑑𝑑 = 𝑑(12)/𝑑𝑑 βˆ’ 12 𝑑(cos⁑𝑑 )/𝑑𝑑 𝑑𝑦/𝑑𝑑 = 0 βˆ’ 12 (βˆ’sin⁑𝑑 ) π’…π’š/𝒅𝒕 = 𝟏𝟐 𝐬𝐒𝐧⁑𝒕 Calculating 𝒅𝒙/𝒅𝒕 π‘₯=10 (𝑑 –sin⁑𝑑 ) π‘₯=10𝑑 βˆ’10 sin⁑𝑑 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑π‘₯/𝑑𝑑 = 𝑑(10 βˆ’ 10 sin⁑𝑑 )/𝑑𝑑 𝑑/𝑑𝑑 (10t βˆ’ 10 sint) 𝑑π‘₯/𝑑𝑑 = 𝑑(10 𝑑)/𝑑𝑑 βˆ’ 𝑑(10 sin⁑𝑑 )/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = 10βˆ’10 cos⁑𝑑 𝒅𝒙/𝒅𝒕 = 𝟏𝟎(πŸβˆ’π’„π’π’”β‘π’• ) Therefore 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) 𝑑𝑦/𝑑π‘₯ = (12 sin⁑𝑑)/10(1 βˆ’γ€– cos〗⁑𝑑 ) 𝑑𝑦/𝑑π‘₯ = (6 π’”π’Šπ’β‘π’•)/(5 (𝟏 βˆ’γ€– 𝒄𝒐𝒔〗⁑𝒕 ) ) 𝑑𝑦/𝑑π‘₯ = (6 . 𝟐 〖𝐬𝐒𝐧 〗⁑〖𝒕/πŸγ€— 𝒄𝒐𝒔⁑〖 𝒕/πŸγ€—)/(5 (𝟐 γ€–π’”π’Šπ’γ€—^πŸβ‘γ€–π’•/πŸγ€— ) ) 𝑑𝑦/𝑑π‘₯ = (6 cos⁑〖 𝑑/2γ€—)/(5 γ€–sin 〗⁑〖𝑑/2γ€— ) π’…π’š/𝒅𝒙 = πŸ”/πŸ“ 𝒄𝒐𝒕 𝒕/𝟐 𝑑𝑦/𝑑π‘₯ = 6(2 sin⁑〖𝑑/2γ€— βˆ’ cos⁑〖 𝑑/2γ€— )/(5 (1 βˆ’(1 βˆ’ 2 sin^2⁑〖𝑑/2γ€— )) ) We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by πœƒ/2 sin ΞΈ = 2 π’”π’Šπ’β‘γ€–πœ½/πŸγ€— π’„π’π’”β‘γ€–πœ½/πŸγ€— and cos 2ΞΈ = 1 – 2sin2 ΞΈ Replacing ΞΈ by πœƒ/2 cos ΞΈ = 1 – 2sin2 πœƒ/2 1 – cos ΞΈ = 2sin2 𝜽/𝟐

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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.