If z = sin^{-1}left( frac{x+y}{sqrt{x} + sqrt | vedclass

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(B) Let $u = \sin z = \frac{x+y}{\sqrt{x} + \sqrt{y}}$.We check if $u$ is a homogeneous function of $x$ and $y$.$u(tx, ty) = \frac{tx+ty}{\sqrt{tx} +

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

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(B) Let $u = \sin z = \frac{x+y}{\sqrt{x} + \sqrt{y}}$.We check if $u$ is a homogeneous function of $x$ and $y$.$u(tx, ty) = \frac{tx+ty}{\sqrt{tx} + \sqrt{ty}} = \frac{t(x+y)}{\sqrt{t}(\sqrt{x} + \sqrt{y})} = t^{1/2} u(x, y)$.Thus,$u$ is a homogeneous function of degree $n = 1/2$.By Euler's Theorem on homogeneous functions,$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = n u$.Substituting $u = \sin z$ and $n = 1/2$,we get:$x\frac{\partial}{\partial x}(\sin z) + y\frac{\partial}{\partial y}(\sin z) = \frac{1}{2} \sin z$.Applying the chain rule,$x(\cos z \frac{\partial z}{\partial x}) + y(\cos z \frac{\partial z}{\partial y}) = \frac{1}{2} \sin z$.Dividing both sides by $\cos z$,we obtain:$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \frac{1}{2} \frac{\sin z}{\cos z} = \frac{1}{2} 	an z$.

If $z = \sin^{-1}\left( \frac{x+y}{\sqrt{x} + \sqrt{y}} \right)$,then $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y}$ is equal to

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