If u=sin ^{-1}left(frac{x^2+y^2}{x+y}right) t | vedclass

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(B) Given $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$.This implies $\sin u = \frac{x^2+y^2}{x+y}$.Let $f(x, y) = \sin u = \frac{x^2+y^2}{x+y}$.Here

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$	an u$

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(B) Given $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}ight)$.This implies $\sin u = \frac{x^2+y^2}{x+y}$.Let $f(x, y) = \sin u = \frac{x^2+y^2}{x+y}$.Here, $f(x, y)$ is a homogeneous function of degree $n=1$ because $f(tx, ty) = \frac{(tx)^2+(ty)^2}{tx+ty} = t \cdot \frac{x^2+y^2}{x+y} = t^1 f(x, y)$.According to Euler's theorem for homogeneous functions, $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f$.Substituting $f = \sin u$ and $n = 1$, we get $x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = 1 \cdot \sin u$.Applying the chain rule, $x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = \sin u$.Dividing both sides by $\cos u$, we get $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{\sin u}{\cos u} = 	an u$.

If $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$ then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to:

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