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

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(A) Given $u = \sin^{-1}\left(\frac{x}{y}ight) + 	an^{-1}\left(\frac{y}{x}ight)$.First,we find the partial derivative with respect to $x$:$\frac{

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(A) Given $u = \sin^{-1}\left(\frac{x}{y}ight) + 	an^{-1}\left(\frac{y}{x}ight)$.First,we find the partial derivative with respect to $x$:$\frac{\partial u}{\partial x} = \frac{1}{\sqrt{1 - (x/y)^2}} \cdot \frac{1}{y} + \frac{1}{1 + (y/x)^2} \cdot \left(-\frac{y}{x^2}ight)$$= \frac{1}{\sqrt{(y^2 - x^2)/y^2}} \cdot \frac{1}{y} - \frac{x^2}{x^2 + y^2} \cdot \frac{y}{x^2}$$= \frac{y}{\sqrt{y^2 - x^2}} \cdot \frac{1}{y} - \frac{y}{x^2 + y^2} = \frac{1}{\sqrt{y^2 - x^2}} - \frac{y}{x^2 + y^2}$Multiplying by $x$,we get:$x \frac{\partial u}{\partial x} = \frac{x}{\sqrt{y^2 - x^2}} - \frac{xy}{x^2 + y^2} \quad \dots(i)$Next,we find the partial derivative with respect to $y$:$\frac{\partial u}{\partial y} = \frac{1}{\sqrt{1 - (x/y)^2}} \cdot \left(-\frac{x}{y^2}ight) + \frac{1}{1 + (y/x)^2} \cdot \left(\frac{1}{x}ight)$$= \frac{y}{\sqrt{y^2 - x^2}} \cdot \left(-\frac{x}{y^2}ight) + \frac{x^2}{x^2 + y^2} \cdot \frac{1}{x}$$= -\frac{x}{y \sqrt{y^2 - x^2}} + \frac{x}{x^2 + y^2}$Multiplying by $y$,we get:$y \frac{\partial u}{\partial y} = -\frac{x}{\sqrt{y^2 - x^2}} + \frac{xy}{x^2 + y^2} \quad \dots(ii)$Adding equations $(i)$ and $(ii)$:$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \left(\frac{x}{\sqrt{y^2 - x^2}} - \frac{xy}{x^2 + y^2}ight) + \left(-\frac{x}{\sqrt{y^2 - x^2}} + \frac{xy}{x^2 + y^2}ight) = 0$.

If $u=\sin ^{-1}\left(\frac{x}{y}\right)+\tan ^{-1}\left(\frac{y}{x}\right)$,then the value of $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is

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