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Ex 5.7, 12 - If y = cos^-1 x, Find d^2y/dx^2 in terms of y alone

Ex 5.7, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.7, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.7, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Transcript

Ex 5.7, 12 If y= γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ , Find 𝑑2𝑦/𝑑π‘₯2 in terms of 𝑦 alone.Let y = γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/√(1 βˆ’ π‘₯^2 ) Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = 𝑑/𝑑π‘₯ ((βˆ’1)/√(1 βˆ’ π‘₯^2 )) ("As" 𝑑(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯)/𝑑π‘₯=(βˆ’1)/√(1 βˆ’ π‘₯^2 )) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑑/𝑑π‘₯ ((βˆ’1)/√(1 βˆ’ π‘₯^2 )) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’π‘‘/𝑑π‘₯ (1βˆ’π‘₯^2 )^((βˆ’1)/2) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’((βˆ’1)/2) (1βˆ’π‘₯^2 )^((βˆ’1)/2 βˆ’ 1)Γ— (1βˆ’π‘₯^2 )^β€² (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 1/2 (1βˆ’π‘₯^2 )^((βˆ’3)/2) Γ— (βˆ’2π‘₯) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’π‘₯(1βˆ’π‘₯^2 )^((βˆ’3)/2) (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = (βˆ’ 𝒙)/( (𝟏 βˆ’ 𝒙^𝟐 )^(πŸ‘/𝟐 ) ) But we need to calculate (𝑑^2 𝑦)/(𝑑π‘₯^2 ) in terms of y . y = γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ cos⁑𝑦 = π‘₯ π‘₯ = cos⁑𝑦 Putting π‘₯ = cos⁑𝑦 in equation (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’ π‘₯)/( (1 βˆ’ π‘₯^2 )^(3/2 ) ) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’cos⁑𝑦 " " )/( (1 βˆ’ γ€–(cos⁑𝑦)γ€—^2 )^(3/2 ) ) = (βˆ’ cos⁑𝑦 " " )/( γ€–(sin2⁑〖𝑦) γ€—γ€—^(3/2 ) ) (As 1βˆ’cos⁑〖2 γ€— πœƒ = sin⁑2 πœƒ) = (βˆ’ cos⁑𝑦 " " )/( γ€–(sin⁑〖𝑦) γ€—γ€—^(2 Γ— 3/2 ) ) = (βˆ’ π‘π‘œπ‘ β‘π‘¦)/( γ€–(𝑠𝑖𝑛⁑〖𝑦) γ€—γ€—^3 ) = (βˆ’ π‘π‘œπ‘ β‘π‘¦)/sin⁑𝑦 Γ—1/( γ€–(𝑠𝑖𝑛⁑𝑦)γ€—^2 ) = βˆ’πœπ¨π­β‘π’š 𝒄𝒐𝒔𝒆𝒄^𝟐 π’š

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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.