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Now given that,
sin−1x+sin−1y=π2
∴sin−1x=π2−sin−1y
∴sin−1x=cos−1y
∴y=cossin−1x
∴dydx=−sinsin−1x.ddx(sin−1x)
∴dydx=−x1√1−x2
∴dydx=−x√1−x2.
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