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

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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.