人邮高数 第6章 第6-3-7题

[AI解答]

[AI解答]

设 $u = x + xy + xyz$，则 $w = f(u)$，其中 $f$ 为可微函数。

首先求 $\displaystyle \frac{\partial w}{\partial x}$：
由链式法则，
$$
\frac{\partial w}{\partial x} = f'(u) \cdot \frac{\partial u}{\partial x}.
$$
计算 $\displaystyle \frac{\partial u}{\partial x}$：
$$
u = x + xy + xyz,
$$
对 $x$ 求偏导时，$y$ 和 $z$ 视为常数：
$$
\frac{\partial u}{\partial x} = 1 + y + yz.
$$
因此，
$$
\frac{\partial w}{\partial x} = f'(u) \cdot (1 + y + yz).
$$

其次求 $\displaystyle \frac{\partial w}{\partial y}$：
$$
\frac{\partial w}{\partial y} = f'(u) \cdot \frac{\partial u}{\partial y}.
$$
计算 $\displaystyle \frac{\partial u}{\partial y}$：
$$
\frac{\partial u}{\partial y} = 0 + x + xz = x(1+z).
$$
因此，
$$
\frac{\partial w}{\partial y} = f'(u) \cdot x(1+z).
$$

最后求 $\displaystyle \frac{\partial w}{\partial z}$：
$$
\frac{\partial w}{\partial z} = f'(u) \cdot \frac{\partial u}{\partial z}.
$$
计算 $\displaystyle \frac{\partial u}{\partial z}$：
$$
\frac{\partial u}{\partial z} = 0 + 0 + xy = xy.
$$
因此，
$$
\frac{\partial w}{\partial z} = f'(u) \cdot xy.
$$

综上，结果为：
$$
\frac{\partial w}{\partial x} = f'(x+xy+xyz)(1+y+yz),
$$
$$
\frac{\partial w}{\partial y} = f'(x+xy+xyz) \cdot x(1+z),
$$
$$
\frac{\partial w}{\partial z} = f'(x+xy+xyz) \cdot xy.
$$

难度：★☆☆☆☆