人邮高数 第6章 第6-3-24题

[AI解答]

[AI解答]
已知
$$
u = f(x, y, z),\quad y = \varphi(x, t),\quad t = \psi(x, z)
$$
且 $f,\varphi,\psi$ 均可微。要求 $\displaystyle{\frac{\partial u}{\partial x}}$。

首先，$u$ 最终依赖于 $x$ 和 $z$，因为 $y$ 和 $t$ 是中间变量。
由链式法则：

$$
\frac{\partial u}{\partial x}
= \frac{\partial f}{\partial x}
+ \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial x}
+ \frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial x}
$$

但注意 $z$ 与 $x$ 是独立的自变量吗？题目中 $u$ 最终是 $x$ 和 $z$ 的函数，因此 $\displaystyle{\frac{\partial z}{\partial x} = 0}$。于是上式简化为：

$$
\frac{\partial u}{\partial x}
= \frac{\partial f}{\partial x}
+ \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial x}
$$

现在求 $\displaystyle{\frac{\partial y}{\partial x}}$。由于 $y = \varphi(x, t)$，而 $t = \psi(x, z)$，所以 $y$ 对 $x$ 求偏导时，$t$ 也依赖于 $x$，但 $z$ 视为常数：

$$
\frac{\partial y}{\partial x}
= \frac{\partial \varphi}{\partial x}
+ \frac{\partial \varphi}{\partial t} \cdot \frac{\partial t}{\partial x}
$$

而 $\displaystyle{\frac{\partial t}{\partial x}}$ 由 $t = \psi(x, z)$ 得：

$$
\frac{\partial t}{\partial x} = \frac{\partial \psi}{\partial x}
$$

代入得：

$$
\frac{\partial y}{\partial x}
= \frac{\partial \varphi}{\partial x}
+ \frac{\partial \varphi}{\partial t} \cdot \frac{\partial \psi}{\partial x}
$$

最后代回 $\displaystyle{\frac{\partial u}{\partial x}}$ 的表达式：

$$
\boxed{\displaystyle{\frac{\partial u}{\partial x}
= \frac{\partial f}{\partial x}
+ \frac{\partial f}{\partial y}
\left( \frac{\partial \varphi}{\partial x}
+ \frac{\partial \varphi}{\partial t} \cdot \frac{\partial \psi}{\partial x} \right)}}
$$

难度：★★☆☆☆