If u = sin^{-1} sqrt{frac{x^2 + y^2}{x + y}}, | vedclass

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(C) Given $u = \sin^{-1} \sqrt{\frac{x^2 + y^2}{x + y}}$.Let $f(u) = \sin u = \sqrt{\frac{x^2 + y^2}{x + y}}$.Let $z = f(u) = \sqrt{\frac{x^2 + y^2}{x

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$\frac{1}{2} 	an u$

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(C) Given $u = \sin^{-1} \sqrt{\frac{x^2 + y^2}{x + y}}$.Let $f(u) = \sin u = \sqrt{\frac{x^2 + y^2}{x + y}}$.Let $z = f(u) = \sqrt{\frac{x^2 + y^2}{x + y}}$.We check the degree of homogeneity of $z$. Replacing $x$ with $tx$ and $y$ with $ty$:$z(tx, ty) = \sqrt{\frac{(tx)^2 + (ty)^2}{tx + ty}} = \sqrt{\frac{t^2(x^2 + y^2)}{t(x + y)}} = t^{1/2} \sqrt{\frac{x^2 + y^2}{x + y}} = t^{1/2} z$.Thus,$z$ is a homogeneous function of degree $n = 1/2$.By Euler's theorem for homogeneous functions,$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = n z$.Since $z = \sin u$,we have $x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = \frac{1}{2} \sin u$.$x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = \frac{1}{2} \sin u$.Dividing by $\cos u$,we get $x u_x + y u_y = \frac{1}{2} \frac{\sin u}{\cos u} = \frac{1}{2} 	an u$.

If $u = \sin^{-1} \sqrt{\frac{x^2 + y^2}{x + y}}$,then $x u_x + y u_y$ is

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