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

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(B) Given $u = x^2 	an^{-1}(\frac{y}{x}) - y^2 	an^{-1}(\frac{x}{y})$.Using the identity $	an^{-1}(\frac{x}{y}) = \frac{\pi}{2} - 	an^{-1}(\frac{y

Vedclass · Accepted Answer

$\frac{x^2 - y^2}{x^2 + y^2}$

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(B) Given $u = x^2 	an^{-1}(\frac{y}{x}) - y^2 	an^{-1}(\frac{x}{y})$.Using the identity $	an^{-1}(\frac{x}{y}) = \frac{\pi}{2} - 	an^{-1}(\frac{y}{x})$ for $y > 0$,we have:$u = x^2 	an^{-1}(\frac{y}{x}) - y^2 (\frac{\pi}{2} - 	an^{-1}(\frac{y}{x}))$$u = (x^2 + y^2) 	an^{-1}(\frac{y}{x}) - \frac{\pi}{2} y^2$Now,differentiate $u$ partially with respect to $y$:$\frac{\partial u}{\partial y} = (x^2 + y^2) \cdot \frac{1}{1 + (y/x)^2} \cdot \frac{1}{x} + 2y 	an^{-1}(\frac{y}{x}) - \pi y$$\frac{\partial u}{\partial y} = (x^2 + y^2) \cdot \frac{x^2}{x^2 + y^2} \cdot \frac{1}{x} + 2y 	an^{-1}(\frac{y}{x}) - \pi y$$\frac{\partial u}{\partial y} = x + 2y 	an^{-1}(\frac{y}{x}) - \pi y$Next,differentiate $\frac{\partial u}{\partial y}$ partially with respect to $x$:$\frac{\partial^2 u}{\partial x \partial y} = 1 + 2y \cdot \frac{1}{1 + (y/x)^2} \cdot (-\frac{y}{x^2}) - 0$$\frac{\partial^2 u}{\partial x \partial y} = 1 + 2y \cdot \frac{x^2}{x^2 + y^2} \cdot (-\frac{y}{x^2})$$\frac{\partial^2 u}{\partial x \partial y} = 1 - \frac{2y^2}{x^2 + y^2} = \frac{x^2 + y^2 - 2y^2}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}$.

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

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