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Home  >>  EAMCET  >>  Mathematics
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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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1 Answer

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$(c)\;1$
answered Nov 7, 2013 by pady_1
 
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