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The derivative of sinβˆ’1 (2x √(1 - x 2 )) w.r.t sinβˆ’1 x, 1/√2Β  < π‘₯ < 1, is:

(a) 2Β  Β  Β  Β  Β  Β  Β  Β  Β (b) Ο€/2 βˆ’ 2
(c) Ο€/2Β  Β  Β  Β  Β  Β  Β  (d) βˆ’2



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Question 24 The derivative of sinβˆ’1 ("2π‘₯" √("1 βˆ’ " π‘₯^2 )) w.r.t sinβˆ’1 x, 1/√2 < π‘₯ < 1, is: (a) 2 (b) πœ‹/2 βˆ’ 2 (c) πœ‹/2 (d) βˆ’2 Let y = sinβˆ’1 x sin y = x x = sin y We need to find the derivative of sinβˆ’1 ("2π‘₯" √("1 βˆ’ " π‘₯^2 )) w.r.t sinβˆ’1 x i.e. sinβˆ’1 ("2π‘₯" √("1 βˆ’ " π‘₯^2 )) w.r.t y i.e. (𝒅(〖𝐬𝐒𝐧〗^(βˆ’πŸ)⁑〖(πŸπ’™βˆš(𝟏 βˆ’ 𝒙^𝟐 ))) γ€—)/π’…π’š Now, (𝑑(sin^(βˆ’1)⁑〖(2π‘₯√(1 βˆ’ π‘₯^2 ))) γ€—)/𝑑𝑦 Putting x = sin y = (𝑑(sin^(βˆ’1)⁑〖(2 sin⁑𝑦 √(1 βˆ’ 〖𝑠𝑖𝑛〗^2 𝑦))) γ€—)/𝑑𝑦 = (𝑑(sin^(βˆ’1)⁑〖(2 sin⁑𝑦 √(cos^2⁑𝑦 )))γ€—)/𝑑𝑦 = (𝑑(sin^(βˆ’1)⁑〖(2 sin⁑𝑦 cos⁑𝑦 ))γ€—)/𝑑𝑦 = (𝑑(sin^(βˆ’1)⁑〖(sin⁑2𝑦 ))γ€—)/𝑑𝑦 = 𝑑(2𝑦)/𝑑𝑦 = 2 So, the correct answer is (A)

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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.