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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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The correct answer is:
$3 \tan u$
$u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}\right)$
Let $v=\sin u=\frac{x^4+y^4}{x+y}$, here degree is homogeneous, so $n=4-1=3$ By Euler's theorem, we have to prove that,
$x \frac{\partial v}{\partial x}+y \frac{\partial v}{\partial y}=3 v$
$x \frac{\partial}{\partial x}(\sin u)+y \frac{\partial v}{\partial y}(\sin u)=3 \sin u$
$x \cos u \cdot \frac{\partial u}{\partial x}+y \cos u \frac{\partial u}{\partial y}=3 \sin u$
$x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 \frac{\sin u}{\cos u}$
$x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 \tan u$
Let $v=\sin u=\frac{x^4+y^4}{x+y}$, here degree is homogeneous, so $n=4-1=3$ By Euler's theorem, we have to prove that,
$x \frac{\partial v}{\partial x}+y \frac{\partial v}{\partial y}=3 v$
$x \frac{\partial}{\partial x}(\sin u)+y \frac{\partial v}{\partial y}(\sin u)=3 \sin u$
$x \cos u \cdot \frac{\partial u}{\partial x}+y \cos u \frac{\partial u}{\partial y}=3 \sin u$
$x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 \frac{\sin u}{\cos u}$
$x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 \tan u$
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