📝 题目
1．设 $z=u^{2}+v^{2}$ ，而 $u=x+y, v=x-y$ ，求 $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$ ．
💡 答案与解析
[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：★☆☆☆☆