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

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Question

(B) આપેલ છે કે $u = x^2 	an^{-1}(\frac{y}{x}) - y^2 	an^{-1}(\frac{x}{y})$.નિત્યસમ $	an^{-1}(\frac{x}{y}) = \frac{\pi}{2} - 	an^{-1}(\frac{y}{x})$

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) આપેલ છે કે $u = x^2 	an^{-1}(\frac{y}{x}) - y^2 	an^{-1}(\frac{x}{y})$.નિત્યસમ $	an^{-1}(\frac{x}{y}) = \frac{\pi}{2} - 	an^{-1}(\frac{y}{x})$ નો ઉપયોગ કરતા:$u = x^2 	an^{-1}(\frac{y}{x}) - y^2 (\frac{\pi}{2} - 	an^{-1}(\frac{y}{x}))$$u = (x^2 + y^2) 	an^{-1}(\frac{y}{x}) - \frac{\pi}{2} y^2$હવે,$u$ નું $y$ ની સાપેક્ષમાં આંશિક વિકલન કરતા:$\frac{\partial u}{\partial y} = (x^2 + y^2) \cdot \frac{1}{1 + (y/x)^2} \cdot \frac{1}{x} + 2y 	an^{-1}(\frac{y}{x}) - \pi y$$\frac{\partial u}{\partial y} = x + 2y 	an^{-1}(\frac{y}{x}) - \pi y$હવે,$\frac{\partial u}{\partial y}$ નું $x$ ની સાપેક્ષમાં આંશિક વિકલન કરતા:$\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}$.

જો $u = x^2 \tan^{-1}(\frac{y}{x}) - y^2 \tan^{-1}(\frac{x}{y})$ હોય,તો $\frac{\partial^2 u}{\partial x \partial y} = $

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