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$\sqrt {\large\frac{y}{x}}+\sqrt {\large\frac{x}{y}}$$=2=>\large\frac{dy}{dx}=$
\[(a)\;\frac{x^2+y^2}{x+y}\quad (b)\;\frac{x^2-y^2}{x+y} \quad (c)\;1 \quad (d)\;2 \]
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