If z = frac{(x^4 + y^4)^{1/3}}{(x^3 + y^3)^{1 | vedclass

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(D) The given function is $z = \frac{(x^4 + y^4)^{1/3}}{(x^3 + y^3)^{1/4}}$.This is a homogeneous function of $x$ and $y$.Let $f(x, y) = (x^4 + y^4)^{

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$\frac{7}{12}z$

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(D) The given function is $z = \frac{(x^4 + y^4)^{1/3}}{(x^3 + y^3)^{1/4}}$.This is a homogeneous function of $x$ and $y$.Let $f(x, y) = (x^4 + y^4)^{1/3}$ and $g(x, y) = (x^3 + y^3)^{1/4}$.The degree of $f(x, y)$ is $n_1 = \frac{4}{3}$ and the degree of $g(x, y)$ is $n_2 = \frac{3}{4}$.Since $z = \frac{f}{g}$,the degree of $z$ is $n = n_1 - n_2 = \frac{4}{3} - \frac{3}{4} = \frac{16 - 9}{12} = \frac{7}{12}$.By Euler's Theorem on homogeneous functions,if $z$ is a homogeneous function of degree $n$,then $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = nz$.Substituting $n = \frac{7}{12}$,we get $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \frac{7}{12}z$.

If $z = \frac{(x^4 + y^4)^{1/3}}{(x^3 + y^3)^{1/4}}$,then $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = $

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