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What is the Differentiation of sin inverse x (sin^-1 x)? - Teachoo

Example 26 - Chapter 5 Class 12 Continuity and Differentiability - Part 2


Example 26 Find the derivative of f given by 𝑓 (π‘₯)=〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯ assuming it exists. 𝑓 (π‘₯)=〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯ Let π’š= γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) 𝒙 sin⁑〖𝑦=π‘₯γ€— 𝒙=π¬π’π§β‘γ€–π’š γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑π‘₯/𝑑π‘₯ = (𝑑 (sin⁑𝑦 ))/𝑑π‘₯ 1 = (𝑑 (sin⁑𝑦 ))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 1 = (𝑑 (sin⁑𝑦 ))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ 1 = cos⁑𝑦 𝑑𝑦/𝑑π‘₯ 1/cos⁑𝑦 =𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/π’„π’π’”β‘π’š 𝑑𝑦/𝑑π‘₯= 1/√(𝟏 βˆ’ γ€–π’”π’Šπ’γ€—^𝟐 π’š) Putting 𝑠𝑖𝑛⁑〖𝑦=π‘₯γ€— 𝑑𝑦/𝑑π‘₯= 1/√(1 βˆ’ 𝒙^𝟐 ) Hence, (𝒅(γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) 𝒙" " ))/𝒅𝒙 = 𝟏/√(𝟏 βˆ’ 𝒙^𝟐 ) "We know that" 〖𝑠𝑖𝑛〗^2 πœƒ+γ€–π‘π‘œπ‘ γ€—^2 πœƒ=1 γ€–π‘π‘œπ‘ γ€—^2 πœƒ=1βˆ’γ€–π‘ π‘–π‘›γ€—^2 πœƒ π’„π’π’”β‘πœ½=√(πŸβˆ’γ€–π’”π’Šπ’γ€—^𝟐 𝜽) " " As 𝑦 = 〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯ So, π’”π’Šπ’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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.