The derivative of sin−1 (2x √(1 - x 2 )) w.r.t sin−1 x, 1/√2 < π₯ < 1, is:
(a) 2 (b) π/2 − 2
(c) π/2 (d) −2
![Slide62.JPG](https://d1avenlh0i1xmr.cloudfront.net/4f2f4278-1792-4c9f-bb03-2346fdac76f8/slide62.jpg)
![Slide63.JPG](https://d1avenlh0i1xmr.cloudfront.net/e7009e77-866f-49f8-a41f-b400d5ebaa01/slide63.jpg)
CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
Last updated at April 16, 2024 by Teachoo
Question 24 The derivative of sinβ1 ("2π₯" β("1 β " π₯^2 )) w.r.t sinβ1 x, 1/β2 < π₯ < 1, is: (a) 2 (b) π/2 β 2 (c) π/2 (d) β2 Let y = sinβ1 x sin y = x x = sin y We need to find the derivative of sinβ1 ("2π₯" β("1 β " π₯^2 )) w.r.t sinβ1 x i.e. sinβ1 ("2π₯" β("1 β " π₯^2 )) w.r.t y i.e. (π (γπ¬π’π§γ^(βπ)β‘γ(ππβ(π β π^π ))) γ)/π π Now, (π(sin^(β1)β‘γ(2π₯β(1 β π₯^2 ))) γ)/ππ¦ Putting x = sin y = (π(sin^(β1)β‘γ(2 sinβ‘π¦ β(1 β γπ ππγ^2 π¦))) γ)/ππ¦ = (π(sin^(β1)β‘γ(2 sinβ‘π¦ β(cos^2β‘π¦ )))γ)/ππ¦ = (π(sin^(β1)β‘γ(2 sinβ‘π¦ cosβ‘π¦ ))γ)/ππ¦ = (π(sin^(β1)β‘γ(sinβ‘2π¦ ))γ)/ππ¦ = π(2π¦)/ππ¦ = 2 So, the correct answer is (A)