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B
cosx2y−1
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C
sinx1−2y
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D
sinx2y−1
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Solution
The correct option is Bcosx2y−1 Given that, y=√sinx+ySquaringbothsides,weget⇒y2=sinx+yDifferentiatingw.r.t.xbothsides,weget⇒2ydydx=cosx+dydx⇒(2y−1)dydx=cosx⇒dydx=cosx(2y−1)