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Question
Find `bb(dy/dx)` in the following:
y = `sin^(-1)(2xsqrt(1-x^2)), -1/sqrt2 < x < 1/sqrt2`
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Solution
y = `sin^-1 (2x sqrt(1 - x^2))`
Let, x = sin θ
⇒ θ = sin−1 x
∴ y = `sin^-1 (2 sin theta sqrt(1 - sin^2 theta))`
= sin−1 (2 sin θ × cos θ)
= sin−1 (sin 2 θ) ....(โต sin 2 θ = 2 sin × cos θ)
= 2 θ
= 2 sin−1 x
On differentiating with respect to x,
`dy/dx = 2 d/dx sin^-1 x`
`dy/dx = 2 xx 1/(sqrt(1 - x^2))`
`dy/dx= 2/sqrt(1 - x^2)`
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