STD 12 - 5. continuity and differentiation — JEE कक्षा 11 साइन्स — Question

$\therefore \frac{{\partial u}}{{\partial x}} = \frac{1}{{{x^2} + {y^2}}}.2x$

$\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{{({x^2} + {y^2}).2 - 2x.2x}}{{{{({x^2} + {y^2})}^2}}}$$ = \frac{{2({y^2} - {x^2})}}{{{{({x^2} + {y^2})}^2}}}$’

$\frac{{\partial u}}{{\partial y}} = \frac{1}{{{x^2} + {y^2}}}.2y$

$\therefore \frac{{{\partial ^2}u}}{{\partial {y^2}}} = \frac{{({x^2} + {y^2})\,.\,2 - 2y.2y}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2({x^2} - {y^2})}}{{{{({x^2} + {y^2})}^2}}}$

$\therefore $ $\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0$.