If u(x,y) = y log x + x log y,then {u_x}{u_y} | vedclass

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(C) Given $u(x,y) = y \log x + x \log y$.First,find the partial derivatives:${u_x} = \frac{\partial u}{\partial x} = \frac{y}{x} + \log y$${u_y} = \fr

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(C) Given $u(x,y) = y \log x + x \log y$.First,find the partial derivatives:${u_x} = \frac{\partial u}{\partial x} = \frac{y}{x} + \log y$${u_y} = \frac{\partial u}{\partial y} = \log x + \frac{x}{y}$Now,substitute these into the expression ${u_x}{u_y} - {u_x} \log x - {u_y} \log y + \log x \log y$:$= (\frac{y}{x} + \log y)(\log x + \frac{x}{y}) - (\frac{y}{x} + \log y)\log x - (\log x + \frac{x}{y})\log y + \log x \log y$Expand the terms:$= (\frac{y}{x} \log x + 1 + \log y \log x + \frac{x}{y} \log y) - (\frac{y}{x} \log x + \log y \log x) - (\log x \log y + \frac{x}{y} \log y) + \log x \log y$Cancel the terms:$= \frac{y}{x} \log x + 1 + \log y \log x + \frac{x}{y} \log y - \frac{y}{x} \log x - \log y \log x - \log x \log y - \frac{x}{y} \log y + \log x \log y$$= 1$.

If $u(x,y) = y \log x + x \log y$,then ${u_x}{u_y} - {u_x} \log x - {u_y} \log y + \log x \log y = $

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