数学题在线解答

问题:验证函数ln√(x^2+y^2) 对x和y的二级偏导数之和为0

解答

验证函数\ln \sqrt{x^2+y^2} 满足方程 \frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0

\begin{aligned}
&\frac{\partial^2 z}{\partial x^2} \\
=& \frac{\partial }{\partial x} (\frac{2x}{2(x^2+y^2) } )\\
=& \frac{1 \cdot (x^2+y^2) – x \cdot 2x }{(x^2+ y^2)^2}\\
=& -\frac{x^2 – y^2 }{x^2 + y^2}
\end{aligned}

同理

\frac{\partial^2 z}{\partial y^2} = \frac{ x^2 – y^2 }{x^2 + y^2}

故而

\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0

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