If u = log_e(x^2 + y^2) + tan^{-1}left(frac{y | vedclass

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(A) Given $u = \log_e(x^2 + y^2) + 	an^{-1}\left(\frac{y}{x}ight)$.First,find the partial derivatives with respect to $x$:$\frac{\partial u}{\parti

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(A) Given $u = \log_e(x^2 + y^2) + 	an^{-1}\left(\frac{y}{x}ight)$.First,find the partial derivatives with respect to $x$:$\frac{\partial u}{\partial x} = \frac{2x}{x^2 + y^2} + \frac{1}{1 + (y/x)^2} \cdot \left(-\frac{y}{x^2}ight) = \frac{2x}{x^2 + y^2} - \frac{y}{x^2 + y^2} = \frac{2x - y}{x^2 + y^2}$.Now,find the second partial derivative with respect to $x$:$\frac{\partial^2 u}{\partial x^2} = \frac{(x^2 + y^2)(2) - (2x - y)(2x)}{(x^2 + y^2)^2} = \frac{2x^2 + 2y^2 - 4x^2 + 2xy}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2 + 2xy}{(x^2 + y^2)^2}$.Next,find the partial derivatives with respect to $y$:$\frac{\partial u}{\partial y} = \frac{2y}{x^2 + y^2} + \frac{1}{1 + (y/x)^2} \cdot \left(\frac{1}{x}ight) = \frac{2y}{x^2 + y^2} + \frac{x}{x^2 + y^2} = \frac{2y + x}{x^2 + y^2}$.Now,find the second partial derivative with respect to $y$:$\frac{\partial^2 u}{\partial y^2} = \frac{(x^2 + y^2)(2) - (2y + x)(2y)}{(x^2 + y^2)^2} = \frac{2x^2 + 2y^2 - 4y^2 - 2xy}{(x^2 + y^2)^2} = \frac{2x^2 - 2y^2 - 2xy}{(x^2 + y^2)^2}$.Adding the two second partial derivatives:$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2y^2 - 2x^2 + 2xy + 2x^2 - 2y^2 - 2xy}{(x^2 + y^2)^2} = \frac{0}{(x^2 + y^2)^2} = 0$.

If $u = \log_e(x^2 + y^2) + \tan^{-1}\left(\frac{y}{x}\right)$,then $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = $

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