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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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The correct answer is:
1
We have,
$u(x, y)=y \log x+x \log y$
On differentiating partially w.r.t. $x$ and $y$ respectively,
$u_x=\frac{y}{x}+\log y, u_y=\log x+\frac{x}{y}$
Now, $u_x u_y-u_x \log x-u_y \log y+\log x \log y$
$=\left(\frac{y}{x}+\log y\right)\left(\log x+\frac{x}{y}\right)-\left(\frac{y}{x}+\log y\right) \log x$ $-\left(\log x+\frac{x}{y}\right) \log y+\log x \log y$
$=1+\frac{y}{x} \log x+\frac{x}{y} \log y+\log x \log y$
$-\frac{y}{x} \log x-\log x \log y-\log x \log y$
$-\frac{x}{y} \log y+\log x \log y$
$=1$
$u(x, y)=y \log x+x \log y$
On differentiating partially w.r.t. $x$ and $y$ respectively,
$u_x=\frac{y}{x}+\log y, u_y=\log x+\frac{x}{y}$
Now, $u_x u_y-u_x \log x-u_y \log y+\log x \log y$
$=\left(\frac{y}{x}+\log y\right)\left(\log x+\frac{x}{y}\right)-\left(\frac{y}{x}+\log y\right) \log x$ $-\left(\log x+\frac{x}{y}\right) \log y+\log x \log y$
$=1+\frac{y}{x} \log x+\frac{x}{y} \log y+\log x \log y$
$-\frac{y}{x} \log x-\log x \log y-\log x \log y$
$-\frac{x}{y} \log y+\log x \log y$
$=1$
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