If u=sin ^{-1}left(frac{x^4+y^4}{x+y}right),t | vedclass

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(D) Given $u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}\right)$.Let $v=\sin u=\frac{x^4+y^4}{x+y}$.Here,$v$ is a homogeneous function of $x$ and $y$ with deg

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$3 	an u$

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(D) Given $u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}ight)$.Let $v=\sin u=\frac{x^4+y^4}{x+y}$.Here,$v$ is a homogeneous function of $x$ and $y$ with degree $n = 4 - 1 = 3$.According to Euler's theorem for homogeneous functions,$x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = n v$.Substituting $v = \sin u$ and $n = 3$,we get $x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = 3 \sin u$.Applying the chain rule,$x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = 3 \sin u$.Dividing both sides by $\cos u$,we get $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 \frac{\sin u}{\cos u}$.Therefore,$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 	an u$.

If $u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}\right)$,then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to

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