同济高数 第9章 第9-4-1题
📝 题目
1．设 $z=u^{2}+v^{2}$ ，而 $u=x+y, v=x-y$ ，求 $\displaystyle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$ ．
💡 答案解析
[AI解答]
[AI解答]
已知
$$
z = u^2 + v^2,\quad u = x + y,\quad v = x - y.
$$
先计算中间变量对自变量的偏导数：
$$
\frac{\partial u}{\partial x} = 1,\quad \frac{\partial u}{\partial y} = 1,\quad
\frac{\partial v}{\partial x} = 1,\quad \frac{\partial v}{\partial y} = -1.
$$
再计算 $z$ 对 $u, v$ 的偏导数：
$$
\frac{\partial z}{\partial u} = 2u,\quad \frac{\partial z}{\partial v} = 2v.
$$
由链式法则：
$$
\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}
= 2u \cdot 1 + 2v \cdot 1 = 2(u+v).
$$
代入 $u = x+y,\ v = x-y$，得
$$
\frac{\partial z}{\partial x} = 2\big[(x+y)+(x-y)\big] = 2(2x) = 4x.
$$
同理，
$$
\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}
= 2u \cdot 1 + 2v \cdot (-1) = 2(u - v).
$$
代入得
$$
\frac{\partial z}{\partial y} = 2\big[(x+y)-(x-y)\big] = 2(2y) = 4y.
$$
因此，
$$
\boxed{\frac{\partial z}{\partial x} = 4x,\quad \frac{\partial z}{\partial y} = 4y}.
$$
难度：★☆☆☆☆