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Differentiation of cos inverse x (cos^-1 x) - Teachoo [with Video]

Derivative of cos-1 x (Cos inverse x) - Part 2

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Derivative of 〖𝒄𝒐𝒔〗^(βˆ’πŸ) 𝒙Derivative of 〖𝒄𝒐𝒔〗^(βˆ’πŸ) 𝒙 𝑓 (π‘₯)=γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ Let π’š= 〖𝒄𝒐𝒔〗^(βˆ’πŸ) 𝒙 cos⁑〖𝑦=π‘₯γ€— 𝒙=πœπ¨π¬β‘γ€–π’š γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑π‘₯/𝑑π‘₯ = (𝑑 (cos⁑𝑦 ))/𝑑π‘₯ 1 = (𝑑 (cos⁑𝑦 ))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 1 = (𝑑 (cos⁑𝑦 ))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ 1 = (βˆ’sin⁑𝑦) 𝑑𝑦/𝑑π‘₯ (βˆ’1)/sin⁑𝑦 =𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/π’”π’Šπ’β‘π’š 𝑑𝑦/𝑑π‘₯= (βˆ’1)/√(𝟏 βˆ’ 〖𝒄𝒐𝒔〗^𝟐 π’š) Putting π‘π‘œπ‘ β‘γ€–π‘¦=π‘₯γ€— 𝑑𝑦/𝑑π‘₯= (βˆ’1)/√(1 βˆ’ 𝒙^𝟐 ) Hence, (𝒅(〖𝒄𝒐𝒔〗^(βˆ’πŸ) 𝒙" " ))/𝒅𝒙 = (βˆ’πŸ)/√(𝟏 βˆ’ 𝒙^𝟐 ) "We know that" 〖𝑠𝑖𝑛〗^2 πœƒ+γ€–π‘π‘œπ‘ γ€—^2 πœƒ=1 〖𝑠𝑖𝑛〗^2 πœƒ=1βˆ’γ€–π‘π‘œπ‘ γ€—^2 πœƒ π’”π’Šπ’β‘πœ½=√(πŸβˆ’γ€–π’„π’π’”γ€—^𝟐 𝜽) " " As 𝑦 = γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ So, π’„π’π’”β‘π’š = 𝒙

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.