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If \(u = log\left( {\frac{{{x^2} + {y^2}}}{{x + y}}} \right)\), what is the value of \(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}}?\)

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Correct Answer - Option 2 : 1

Concept:

A function f(x, y) is said to be homogeneous of degree n in x and y, it can be written in the form

f(λx, λy) = λn f(x, y)

Euler’s theorem:

If f(x, y) is a homogeneous function of degree n in x and y and has continuous first and second-order partial derivatives, then

\(x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} = nf\)

\({x^2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} + 2xy\frac{{{\partial ^2}f}}{{\partial x\partial y}} + {y^2}\frac{{\partial f}}{{\partial y}} = n\left( {n - 1} \right)f\)

If z is homogeneous function of x & y of degree n and z = f(u), then

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = n\frac{{f\left( u \right)}}{{f'\left( u \right)}}\)

Calculation:

Given, \(u = log\left( {\frac{{{x^2} + {y^2}}}{{x + y}}} \right)\)

\(z = \frac{{{x^2} + {y^2}}}{{x + y}}\)

z is a homogenous function of x & y with a degree 1.

Now, z = eu

Thus, by Euler’s theorem:

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = \frac{{{e^u}}}{{{e^u}}}\)

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 1\)

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