人邮高数 第6章 第6-3-3题
📝 题目
3．设 $z=\mathrm{e}^{u} \sin v$ ，而 $u=x y, v=x+y$ ，求 $\displaystyle \frac{\partial z}{\partial x}$ 和 $\displaystyle \frac{\partial z}{\partial y}$ ．
💡 答案解析
[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：★☆☆☆☆