MCQ
If $u = {\log _e}({x^2} + {y^2}) + {\tan ^{ - 1}}\left( {{y \over x}} \right)$, then ${{{\partial ^2}u} \over {\partial {x^2}}} + {{{\partial ^2}u} \over {\partial {y^2}}} = $
- ✓$0$
- B$2u$
- C$1/u$
- D$u$
$\frac{{\partial u}}{{\partial x}} = \frac{{2x}}{{{x^2} + {y^2}}} + \frac{1}{{1 + \frac{{{y^2}}}{{{x^2}}}}}.\left( { - \frac{y}{{{x^2}}}} \right)$ $ = \frac{{2x - y}}{{{x^2} + {y^2}}}$
$\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{{({x^2} + {y^2}).2 - (2x - y)2x}}{{{{({x^2} + {y^2})}^2}}}$ $ = \frac{{2{y^2} - 2{x^2} + 2xy}}{{{{({x^2} + {y^2})}^2}}}$
$\frac{{\partial u}}{{\partial y}} = \frac{{2y}}{{{x^2} + {y^2}}} + \frac{1}{{1 + \frac{{{y^2}}}{{{x^2}}}}}.\frac{1}{x} = \frac{{2y + x}}{{{x^2} + {y^2}}}$
$\frac{{{\partial ^2}u}}{{\partial {y^2}}} = \frac{{({x^2} + {y^2}).2 - (2y + x)2y}}{{{{({x^2} + {y^2})}^2}}}$= $\frac{{2{x^2} - 2{y^2} - 2xy}}{{{{({x^2} + {y^2})}^2}}}$
$\therefore $ $\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0$.
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