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Class 12
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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 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.