If u(x, y)=y log x+x log y,then u_x u_y-u_x l | vedclass

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

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(C) Given,$u(x, y)=y \log x+x \log y$. First,we find the partial derivatives $u_x$ and $u_y$: $u_x = \frac{\partial}{\partial x}(y \log x + x \log y) = \frac{y}{x} + \log y$. $u_y = \frac{\partial}{\partial y}(y \log x + x \log y) = \log x + \frac{x}{y}$. Now,consider the expression $E = u_x u_y - u_x \log x - u_y \log y + \log x \log y$. We can factor this expression as: $E = u_x(u_y - \log x) - \log y(u_y - \log x)$. $E = (u_x - \log y)(u_y - \log x)$. Substitute the values of $u_x$ and $u_y$: $u_x - \log y = (\frac{y}{x} + \log y) - \log y = \frac{y}{x}$. $u_y - \log x = (\log x + \frac{x}{y}) - \log x = \frac{x}{y}$. Therefore,$E = (\frac{y}{x}) 	imes (\frac{x}{y}) = 1$.

If $u(x, y)=y \log x+x \log y$,then $u_x u_y-u_x \log x-u_y \log y+\log x \log y$ is equal to :

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