同济高数 第9章 第9-4-*13题

[AI解答]

[AI解答]

我们逐小题计算二阶偏导数，假设 $f$ 具有二阶连续偏导数，因此混合偏导可交换次序。

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（1）$z = f(xy, y)$

令 $u = xy$，$v = y$，则
$$
z = f(u, v)
$$

一阶偏导：
$$
\frac{\partial z}{\partial x} = f_u \cdot y + f_v \cdot 0 = y f_u
$$
$$
\frac{\partial z}{\partial y} = f_u \cdot x + f_v \cdot 1 = x f_u + f_v
$$

二阶偏导：

对 $x$ 再求导：
$$
\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}(y f_u) = y \left( f_{uu} \cdot y + f_{uv} \cdot 0 \right) = y^2 f_{uu}
$$

先对 $x$ 后对 $y$：
$$
\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}(y f_u) = f_u + y \left( f_{uu} \cdot x + f_{uv} \cdot 1 \right) = f_u + xy f_{uu} + y f_{uv}
$$

对 $y$ 再求导：
$$
\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}(x f_u + f_v)
= x \left( f_{uu} \cdot x + f_{uv} \cdot 1 \right) + \left( f_{vu} \cdot x + f_{vv} \cdot 1 \right)
$$
由连续性 $f_{uv}=f_{vu}$，得：
$$
= x^2 f_{uu} + x f_{uv} + x f_{uv} + f_{vv} = x^2 f_{uu} + 2x f_{uv} + f_{vv}
$$

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（2）$z = f\left(x, \frac{x}{y}\right)$

令 $u = x$，$v = \frac{x}{y}$，则
$$
z = f(u, v)
$$

一阶偏导：
$$
\frac{\partial z}{\partial x} = f_u \cdot 1 + f_v \cdot \frac{1}{y} = f_u + \frac{1}{y} f_v
$$
$$
\frac{\partial z}{\partial y} = f_u \cdot 0 + f_v \cdot \left(-\frac{x}{y^2}\right) = -\frac{x}{y^2} f_v
$$

二阶偏导：

对 $x$ 再求导：
$$
\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}\left( f_u + \frac{1}{y} f_v \right)
= \left( f_{uu} \cdot 1 + f_{uv} \cdot \frac{1}{y} \right) + \frac{1}{y} \left( f_{vu} \cdot 1 + f_{vv} \cdot \frac{1}{y} \right)
$$
$$
= f_{uu} + \frac{2}{y} f_{uv} + \frac{1}{y^2} f_{vv}
$$

先对 $x$ 后对 $y$：
$$
\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}\left( f_u + \frac{1}{y} f_v \right)
= \left( f_{uu} \cdot 0 + f_{uv} \cdot \left(-\frac{x}{y^2}\right) \right) + \left( -\frac{1}{y^2} f_v + \frac{1}{y} \left[ f_{vu} \cdot 0 + f_{vv} \cdot \left(-\frac{x}{y^2}\right) \right] \right)
$$
$$
= -\frac{x}{y^2} f_{uv} - \frac{1}{y^2} f_v - \frac{x}{y^3} f_{vv}
$$

对 $y$ 再求导：
$$
\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}\left( -\frac{x}{y^2} f_v \right)
= \frac{2x}{y^3} f_v - \frac{x}{y^2} \left( f_{vu} \cdot 0 + f_{vv} \cdot \left(-\frac{x}{y^2}\right) \right)
= \frac{2x}{y^3} f_v + \frac{x^2}{y^4} f_{vv}
$$

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（3）$z = f(x y^2, x^2 y)$

令 $u = x y^2$，$v = x^2 y$，则
$$
z = f(u, v)
$$

一阶偏导：
$$
\frac{\partial z}{\partial x} = f_u \cdot y^2 + f_v \cdot 2xy = y^2 f_u + 2xy f_v
$$
$$
\frac{\partial z}{\partial y} = f_u \cdot 2xy + f_v \cdot x^2 = 2xy f_u + x^2 f_v
$$

二阶偏导：

对 $x$ 再求导：
$$
\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}(y^2 f_u + 2xy f_v)
= y^2 ( f_{uu} \cdot y^2 + f_{uv} \cdot 2xy ) + 2y f_v + 2xy ( f_{vu} \cdot y^2 + f_{vv} \cdot 2xy )
$$
$$
= y^4 f_{uu} + 2x y^3 f_{uv} + 2y f_v + 2x y^3 f_{uv} + 4x^2 y^2 f_{vv}
$$
$$
= y^4 f_{uu} + 4x y^3 f_{uv} + 4x^2 y^2 f_{vv} + 2y f_v
$$

先对 $x$ 后对 $y$：
$$
\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}(y^2 f_u + 2xy f_v)
= 2y f_u + y^2 ( f_{uu} \cdot 2xy + f_{uv} \cdot x^2 ) + 2x f_v + 2xy ( f_{vu} \cdot 2xy + f_{vv} \cdot x^2 )
$$
$$
= 2y f_u + 2x y^3 f_{uu} + x^2 y^2 f_{uv} + 2x f_v + 4x^2 y^2 f_{uv} + 2x^3 y f_{vv}
$$
$$
= 2y f_u + 2x f_v + 2x y^3 f_{uu} + (x^2 y^2 + 4x^2 y^2) f_{uv} + 2x^3 y f_{vv}
$$
$$
= 2y f_u + 2x f_v + 2x y^3 f_{uu} + 5x^2 y^2 f_{uv} + 2x^3 y f_{vv}
$$

对 $y$ 再求导：
$$
\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}(2xy f_u + x^2 f_v)
= 2x f_u + 2xy ( f_{uu} \cdot 2xy + f_{uv} \cdot x^2 ) + x^2 ( f_{vu} \cdot 2xy + f_{vv} \cdot x^2 )
$$
$$
= 2x f_u + 4x^2 y^2 f_{uu} + 2x^3 y f_{uv} + 2x^3 y f_{uv} + x^4 f_{vv}
$$
$$
= 2x f_u + 4x^2 y^2 f_{uu} + 4x^3 y f_{uv} + x^4 f_{vv}
$$

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（4）$z = f(\sin x, \cos y, e^{x+y})$

令 $u = \sin x$，$v = \cos y$，$w = e^{x+y}$，则
$$
z = f(u, v, w)
$$

一阶偏导：
$$
\frac{\partial z}{\partial x} = f_u \cos x + f_w e^{x+y}
$$
$$
\frac{\partial z}{\partial y} = f_v (-\sin y) + f_w e^{