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

Question

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

Accepted Answer

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

Answer

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

Answer

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

Answer

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

differentiation