方企勤 第五章 多元函数微分学 第6题
📝 题目
例 6 设函数 $u = u\left( {x,y}\right)$ 由方程
$$ u = f\left( {x,y,z,t}\right) ,\;g\left( {y,z,t}\right) = 0,\;h\left( {z,t}\right) = 0 $$
定义,求 $\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}$ .
💡 答案解析
解 函数 $u = u\left( {x,y}\right)$ 可按如下步骤得出: 先由后两个方程把 $t$ , $z$ 解为 $y$ 的函数,再代入前式而得 $u = u\left( {x,y}\right)$ . 这样求导过程如下:
$$ \frac{\partial u}{\partial x} = \frac{\partial f}{\partial x},\;\frac{\partial u}{\partial y} = \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z}\frac{\mathrm{d}z}{\mathrm{\;d}y} + \frac{\partial f}{\partial t}\frac{\mathrm{d}t}{\mathrm{\;d}y}. \tag{3.5} $$
再由后两式求 $\frac{\mathrm{d}z}{\mathrm{\;d}y},\frac{\mathrm{d}t}{\mathrm{\;d}y}$ . 因
$$ \left\{ \begin{array}{l} \frac{\partial g}{\partial y} + \frac{\partial g}{\partial z}\frac{\mathrm{d}z}{\mathrm{\;d}y} + \frac{\partial g}{\partial t}\frac{\mathrm{d}t}{\mathrm{\;d}y} = 0, \\ \frac{\partial h}{\partial z}\frac{\mathrm{d}z}{\mathrm{\;d}y} + \frac{\partial h}{\partial t}\frac{\mathrm{d}t}{\mathrm{\;d}y} = 0, \end{array}\right. $$
解出
$$ \frac{\mathrm{d}z}{\mathrm{\;d}y} = - \frac{\frac{\partial g}{\partial y}\frac{\partial h}{\partial t}}{\frac{\partial \left( {g,h}\right) }{\partial \left( {z,t}\right) }},\;\frac{\mathrm{d}t}{\mathrm{\;d}y} = \frac{\frac{\partial g}{\partial y}\frac{\partial h}{\partial z}}{\frac{\partial \left( {g,h}\right) }{\partial \left( {z,t}\right) }}. $$
把结果代入 (3.5) 式, 即得
$$ \frac{\partial u}{\partial y} = \frac{\partial f}{\partial y} + \frac{-\frac{\partial f}{\partial z}\frac{\partial g}{\partial y}\frac{\partial h}{\partial t} + \frac{\partial f}{\partial t}\frac{\partial g}{\partial y}\frac{\partial h}{\partial z}}{\frac{\partial \left( {g,h}\right) }{\partial \left( {z,t}\right) }} = \frac{\frac{\partial \left( {f,g,h}\right) }{\partial \left( {y,z,t}\right) }}{\frac{\partial \left( {g,h}\right) }{\partial \left( {z,t}\right) }}. $$
评注 (1) 若函数 $u = f\left( {x,y,z,t}\right)$ 记作 $u = u\left( {x,y,z,t}\right)$ ,则 (3.5)式中 $\frac{\partial f}{\partial y}$ 相应地记为 $\frac{\partial u}{\partial y}$ . 这时等式两边都有 $\frac{\partial u}{\partial y}$ ,左边的 $\frac{\partial u}{\partial y}$ 表示把 $z,t$ 看成 $y$ 的函数时求出的偏导数,右边的 $\frac{\partial u}{\partial y}$ 表示把 $z,t$ 看成自变量时求出的偏导数. 当然我们不希望引起记号的混淆,故将函数与因变量采用不同记号来表示.
(2)求导前先要分析函数关系,明确谁是因变量,谁是中间变量, 谁是自变量. 这题也可看成由方程组
$$ \left\{ \begin{array}{l} f\left( {x,y,z,t}\right) - u = 0, \\ g\left( {y,z,t}\right) = 0, \\ h\left( {z,t}\right) = 0 \end{array}\right. $$
解出 $u,z,t$ 是 $x,y$ 的函数. 现在 $z,t$ 不是看成中间变量,而是看成因变量. 由隐函数求导得
$$ \left\{ \begin{array}{l} \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial y} + \frac{\partial f}{\partial t}\frac{\partial t}{\partial y} - 1 \cdot \frac{\partial u}{\partial y} = 0, \\ \frac{\partial g}{\partial y} + \frac{\partial g}{\partial z}\frac{\partial z}{\partial y} + \frac{\partial g}{\partial t} \cdot \frac{\partial t}{\partial y} + 0 \cdot \frac{\partial u}{\partial y} = 0, \\ \frac{\partial h}{\partial z}\frac{\partial z}{\partial y} + \frac{\partial h}{\partial t}\frac{\partial t}{\partial y} + 0 \cdot \frac{\partial u}{\partial y} = 0. \end{array}\right. $$
把 $\frac{\partial z}{\partial y},\frac{\partial t}{\partial y},\frac{\partial u}{\partial y}$ 当作未知数,即可求出 $\frac{\partial u}{\partial y}$ .