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If $x^y=y^x$. then $x(x-y \log x) \large\frac{dy}{dx}$ is equal to :

\[\begin {array} {1 1} (a)\;y(y- x \log y) & \quad (b)\;y(y+ x \log y)\\ (c)\;x(x+y \log x) & \quad (d)\;x(y-x \log y) \end {array}\]

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$(a)\;y(y- x \log y) $
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