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

Question

Dr. Abubacker · Accepted Answer

(c)
On differentiating the given equation partially w.r.t. $x$ and $y$ respectively
$u_{x} = \frac{y}{x} + \log{y,\ \ \ \ u_{y} = \log{x + \frac{x}{y}}}$
Now, $u_{x}u_{y} - u_{x}\log{x - u_{y}\log{y + \log{x\log y\ }}}$
$= \left( \frac{y}{x} + \log y \right)\left( \log{x + \frac{x}{y}} \right) - \left( \frac{y}{x} + \log{y\log{x\ }} \right)$
$- \left( \log{x + \frac{x}{y}} \right)\log{y + \log{x\log{y = 1}}}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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