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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

  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Concept wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo