If u = log ({x^2} + {y^2}), then frac{{partia | vedclass

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(B) Given $u = \log ({x^2} + {y^2})$.First,find the partial derivative with respect to $x$:$\frac{{\partial u}}{{\partial x}} = \frac{1}{{{x^2} + {y^2

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(B) Given $u = \log ({x^2} + {y^2})$.First,find the partial derivative with respect to $x$:$\frac{{\partial u}}{{\partial x}} = \frac{1}{{{x^2} + {y^2}}} \cdot 2x = \frac{{2x}}{{{x^2} + {y^2}}}$.Now,find the second partial derivative with respect to $x$ using the quotient rule:$\frac{{\partial ^2}u}{\partial {x^2}} = \frac{{({x^2} + {y^2}) \cdot 2 - 2x \cdot 2x}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2{x^2} + 2{y^2} - 4{x^2}}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2({y^2} - {x^2})}}{{{{({x^2} + {y^2})}^2}}}$.Similarly,find the partial derivative with respect to $y$:$\frac{{\partial u}}{{\partial y}} = \frac{1}{{{x^2} + {y^2}}} \cdot 2y = \frac{{2y}}{{{x^2} + {y^2}}}$.Now,find the second partial derivative with respect to $y$:$\frac{{\partial ^2}u}{\partial {y^2}} = \frac{{({x^2} + {y^2}) \cdot 2 - 2y \cdot 2y}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2{x^2} + 2{y^2} - 4{y^2}}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2({x^2} - {y^2})}}{{{{({x^2} + {y^2})}^2}}}$.Adding these two results:$\frac{{\partial ^2}u}{\partial {x^2}} + \frac{{\partial ^2}u}{\partial {y^2}} = \frac{{2({y^2} - {x^2})}}{{{{({x^2} + {y^2})}^2}}} + \frac{{2({x^2} - {y^2})}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2{y^2} - 2{x^2} + 2{x^2} - 2{y^2}}}{{{{({x^2} + {y^2})}^2}}} = 0$.

If $u = \log ({x^2} + {y^2}),$ then $\frac{{\partial ^2}u}{\partial {x^2}} + \frac{{\partial ^2}u}{\partial {y^2}} = $

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