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If $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$ then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ equal to :
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Verified Answer
The correct answer is:
$\tan u$
$u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$
Here $u$ is not a homogeneous function. But $f(x, y)=\sin u=\frac{x^2+y^2}{x+y}$ is a homogeneous function of degree one. Here by Euler's theorem
$x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=f$
$\Rightarrow \quad x \frac{\partial}{\partial x}(\sin u)+y \frac{\partial}{\partial y}(\sin u)=\sin u$
$\Rightarrow \quad x \cos u \frac{\partial u}{\partial x}+y \cos u \frac{\partial u}{\partial y}=\sin u$
$\Rightarrow \quad x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=\tan u$
Here $u$ is not a homogeneous function. But $f(x, y)=\sin u=\frac{x^2+y^2}{x+y}$ is a homogeneous function of degree one. Here by Euler's theorem
$x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=f$
$\Rightarrow \quad x \frac{\partial}{\partial x}(\sin u)+y \frac{\partial}{\partial y}(\sin u)=\sin u$
$\Rightarrow \quad x \cos u \frac{\partial u}{\partial x}+y \cos u \frac{\partial u}{\partial y}=\sin u$
$\Rightarrow \quad x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=\tan u$
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