题号:1063    题型:填空题    来源:1998年全国硕士研究生招生考试试题
设 $z=\frac{1}{x} f(x y)+y \varphi(x+y), f, \varphi$ 具有二阶连续导数, 则 $\frac{\partial^{2} z}{\partial x \partial y}=$
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答案:
$y f^{\prime \prime}(x y)+\varphi^{\prime}(x+y)+y \varphi^{\prime \prime}(x+y)$

解析:

方法1:先求 $\frac{\partial z}{\partial x}$.
$$
\begin{aligned}
\frac{\partial z}{\partial x}=& \frac{\partial}{\partial x}\left[\frac{1}{x} f(x y)+y \varphi(x+y)\right]=-\frac{1}{x^{2}} f(x y)+\frac{y}{x} f^{\prime}(x y)+y \varphi^{\prime}(x+y) \\
\frac{\partial^{2} z}{\partial x \partial y} &=\frac{\partial}{\partial y}\left(-\frac{1}{x^{2}} f(x y)+\frac{y}{x} f^{\prime}(x y)+y \varphi^{\prime}(x+y)\right) \\
&=-\frac{1}{x^{2}} f^{\prime}(x y) x+\frac{1}{x} f^{\prime}(x y)+\frac{y}{x} f^{\prime \prime}(x y) x+\varphi^{\prime}(x+y)+y \varphi^{\prime \prime}(x+y) \\
&=-\frac{1}{x} f^{\prime}(x y)+\frac{1}{x} f^{\prime}(x y)+y f^{\prime \prime}(x y)+\varphi^{\prime}(x+y)+y \varphi^{\prime \prime}(x+y) \\
&=y f^{\prime \prime}(x y)+\varphi^{\prime}(x+y)+y \varphi^{\prime \prime}(x+y)
\end{aligned}
$$
方法2: 先求 $\frac{\partial z}{\partial y}$.
$$
\begin{aligned}
\frac{\partial z}{\partial y} &=\frac{\partial}{\partial y}\left[\frac{1}{x} f(x y)+y \varphi(x+y)\right]=\frac{1}{x} f^{\prime}(x y) x+\varphi(x+y)+y \varphi^{\prime}(x+y) \\
&=f^{\prime}(x y)+\varphi(x+y)+y \varphi^{\prime}(x+y) \\
\frac{\partial^{2} z}{\partial x \partial y} &=\frac{\partial^{2} z}{\partial y \partial x}=\frac{\partial}{\partial x}\left[f^{\prime}(x y)+\varphi(x+y)+y \varphi^{\prime}(x+y)\right] \\
&=y f^{\prime \prime}(x y)+\varphi^{\prime}(x+y)+y \varphi^{\prime \prime}(x+y)
\end{aligned}
$$

方法3:对两项分别采取不同的顺序更简单些:
$$
\begin{aligned}
\frac{\partial^{2} z}{\partial x \partial y} &=\frac{\partial}{\partial x}\left[\frac{\partial}{\partial y}\left(\frac{1}{x} f(x y)\right)\right]+\frac{\partial}{\partial y}\left[\frac{\partial}{\partial x}(y \varphi(x+y))\right] \\
&=\frac{\partial}{\partial x}\left[\frac{1}{x} f^{\prime}(x y) x\right]+\frac{\partial}{\partial y}\left[y \varphi^{\prime}(x+y)\right] \\
&=\frac{\partial}{\partial x}\left[f^{\prime}(x y)\right]+\frac{\partial}{\partial y}\left[y \varphi^{\prime}(x+y)\right] \\
&=y f^{\prime \prime}(x y)+\varphi^{\prime}(x+y)+y \varphi^{\prime \prime}(x+y)
\end{aligned}
$$

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