同济高数 第9章 第9-4-12题
📝 题目
12．设 $z=f\left(x^{2}+y^{2}\right)$ ，其中 $f$ 具有二阶导数，求 $\displaystyle \frac{\partial^{2} z}{\partial x^{2}}, \frac{\partial^{2} z}{\partial x \partial y}, \frac{\partial^{2} z}{\partial y^{2}}$ ．
💡 答案解析
[AI解答]
[AI解答]
已知 $ z = f(x^2 + y^2) $，其中 $ f $ 具有二阶导数。令
$$
u = x^2 + y^2,
$$
则
$$
z = f(u).
$$
---
第一步：求一阶偏导数
对 $ x $ 求偏导：
$$
\frac{\partial z}{\partial x} = f'(u) \cdot \frac{\partial u}{\partial x} = f'(u) \cdot 2x.
$$
对 $ y $ 求偏导：
$$
\frac{\partial z}{\partial y} = f'(u) \cdot \frac{\partial u}{\partial y} = f'(u) \cdot 2y.
$$
---
第二步：求二阶偏导数
1. 求 $\frac{\partial^2 z}{\partial x^2}$：
由 $\frac{\partial z}{\partial x} = 2x f'(u)$，再对 $x$ 求偏导：
$$
\frac{\partial^2 z}{\partial x^2} = 2 f'(u) + 2x \cdot f''(u) \cdot \frac{\partial u}{\partial x}.
$$
而 $\frac{\partial u}{\partial x} = 2x$，所以
$$
\frac{\partial^2 z}{\partial x^2} = 2 f'(u) + 2x \cdot f''(u) \cdot 2x = 2 f'(u) + 4x^2 f''(u).
$$
2. 求 $\frac{\partial^2 z}{\partial y^2}$：
同理，由 $\frac{\partial z}{\partial y} = 2y f'(u)$，再对 $y$ 求偏导：
$$
\frac{\partial^2 z}{\partial y^2} = 2 f'(u) + 2y \cdot f''(u) \cdot \frac{\partial u}{\partial y} = 2 f'(u) + 2y \cdot f''(u) \cdot 2y = 2 f'(u) + 4y^2 f''(u).
$$
3. 求 $\frac{\partial^2 z}{\partial x \partial y}$：
由 $\frac{\partial z}{\partial x} = 2x f'(u)$，再对 $y$ 求偏导：
$$
\frac{\partial^2 z}{\partial x \partial y} = 2x \cdot f''(u) \cdot \frac{\partial u}{\partial y} = 2x \cdot f''(u) \cdot 2y = 4xy f''(u).
$$
---
最终结果：
$$
\boxed{\frac{\partial^{2} z}{\partial x^{2}} = 2 f'(x^2+y^2) + 4x^2 f''(x^2+y^2)}
$$
$$
\boxed{\frac{\partial^{2} z}{\partial y^{2}} = 2 f'(x^2+y^2) + 4y^2 f''(x^2+y^2)}
$$
$$
\boxed{\frac{\partial^{2} z}{\partial x \partial y} = 4xy f''(x^2+y^2)}
$$