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if tan^-1(y/x)= log( root of x^2 +y^2 , prove that dy/dx = x+y/x-y

\tan^{- 1} (\frac{y}{x}) = \log (\sqrt{x^{2} + y^{2}}) = \frac{1}{2} \log [x^{2} + y^{2}] Differentiating, we get \frac{1}{1 + {(\frac{y}{x})}^{2}} \times \frac{d}{dx} (\frac{y}{x}) = \frac{1}{2 [x^{2} + y^{2}]} \times \frac{d}{dx} [x^{2} + y^{2}] \frac{x^{2}}{x^{2} + y^{2}} \times \frac{x \frac{dy}{dx} - y}{x^{2}} = \frac{1}{2 [x^{2} + y^{2}]} \times [2 x + 2 y \frac{dy}{dx}] x \frac{dy}{dx} - y = x + y \frac{dy}{dx} x + y = (y - x) \frac{dy}{dx} \frac{dy}{dx} = \frac{x + y}{(y - x)}

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