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

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(B) दिया गया है $u = x^2 	an^{-1}(\frac{y}{x}) - y^2 	an^{-1}(\frac{x}{y})$.सर्वसमिका $	an^{-1}(\frac{x}{y}) = \frac{\pi}{2} - 	an^{-1}(\frac{y}{x

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

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(B) दिया गया है $u = x^2 	an^{-1}(\frac{y}{x}) - y^2 	an^{-1}(\frac{x}{y})$.सर्वसमिका $	an^{-1}(\frac{x}{y}) = \frac{\pi}{2} - 	an^{-1}(\frac{y}{x})$ का उपयोग करने पर:$u = x^2 	an^{-1}(\frac{y}{x}) - y^2 (\frac{\pi}{2} - 	an^{-1}(\frac{y}{x}))$$u = (x^2 + y^2) 	an^{-1}(\frac{y}{x}) - \frac{\pi}{2} y^2$अब,$u$ का $y$ के सापेक्ष आंशिक अवकलन करने पर:$\frac{\partial u}{\partial y} = (x^2 + y^2) \cdot \frac{1}{1 + (y/x)^2} \cdot \frac{1}{x} + 2y 	an^{-1}(\frac{y}{x}) - \pi y$$\frac{\partial u}{\partial y} = x + 2y 	an^{-1}(\frac{y}{x}) - \pi y$अब,$\frac{\partial u}{\partial y}$ का $x$ के सापेक्ष आंशिक अवकलन करने पर:$\frac{\partial^2 u}{\partial x \partial y} = 1 + 2y \cdot \frac{1}{1 + (y/x)^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}{x^2 + y^2}$.

यदि $u = x^2 \tan^{-1}(\frac{y}{x}) - y^2 \tan^{-1}(\frac{x}{y})$ है,तो $\frac{\partial^2 u}{\partial x \partial y} = $

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