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

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

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

Transcript

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

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
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.