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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Example 24 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.