人邮高数 第6章 第6-3-3题

[AI解答]

[AI解答]
已知函数
$$
z = e^{u} \sin v, \quad u = xy, \quad v = x + y.
$$
我们需要求偏导数 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$。根据多元复合函数求导法则（链式法则），有

$$
\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}.
$$

先计算各偏导数：
$$
\frac{\partial z}{\partial u} = e^{u} \sin v, \quad \frac{\partial z}{\partial v} = e^{u} \cos v,
$$
$$
\frac{\partial u}{\partial x} = y, \quad \frac{\partial v}{\partial x} = 1.
$$

代入得
$$
\frac{\partial z}{\partial x} = e^{u} \sin v \cdot y + e^{u} \cos v \cdot 1
= e^{u} \big( y \sin v + \cos v \big).
$$
将 $u = xy, v = x + y$ 代回：
$$
\frac{\partial z}{\partial x} = e^{xy} \big( y \sin(x+y) + \cos(x+y) \big).
$$

同理，对于 $y$ 的偏导：
$$
\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y},
$$
其中
$$
\frac{\partial u}{\partial y} = x, \quad \frac{\partial v}{\partial y} = 1,
$$
所以
$$
\frac{\partial z}{\partial y} = e^{u} \sin v \cdot x + e^{u} \cos v \cdot 1
= e^{xy} \big( x \sin(x+y) + \cos(x+y) \big).
$$

因此最终结果为：
$$
\boxed{\frac{\partial z}{\partial x} = e^{xy} \big( y \sin(x+y) + \cos(x+y) \big)},
$$
$$
\boxed{\frac{\partial z}{\partial y} = e^{xy} \big( x \sin(x+y) + \cos(x+y) \big)}.
$$

难度：★☆☆☆☆