题号:705    题型:填空题    来源:1994 年全国硕士研究生入学统一考试数学一试题解
设 $u=\mathrm{e}^{-x} \sin \frac{x}{y}$, 则 $\frac{\partial^{2} u}{\partial x \partial y}$ 在点 $\left(2, \frac{1}{\pi}\right)$ 处的值为
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答案:
$\frac{\pi^{2}}{e^{2}}$

解析:

由于混合偏导数在连续条件下与求导次序无关, 为了简化运算, 所以本题可以先
求 $\frac{\partial u}{\partial y}$, 再求 $\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right)$.
$$
\begin{aligned}
&\frac{\partial u}{\partial y}=-\frac{x}{y^{2}} e^{-x} \cos \frac{x}{y} \\
&\left.\frac{\partial^{2} u}{\partial x \partial y}\right|_{\left(2, \frac{1}{\pi}\right)}=\left.\frac{\partial^{2} u}{\partial y \partial x}\right|_{\left(2, \frac{1}{\pi}\right)}=\left.\frac{\partial}{\partial x}\left(\left.\frac{\partial u}{\partial y}\right|_{y=\frac{1}{\pi}}\right)\right|_{x=2}=\left.\frac{\partial}{\partial x}\left(-\pi^{2} x e^{-x} \cos \pi x\right)\right|_{x=2}
\end{aligned}
$$
$=\left.\left(-\pi^{2} e^{-x}(1-x) \cos \pi x\right)\right|_{x=2}+0=\frac{\pi^{2}}{e^{2}}$

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