三、多元函数的微分法则
（一）全微分四则运算法则
设 \(u = u(x,y),v = v(x,y)\) 在 \((x,y)\) 可微，则
\[\mathrm {d} (u \pm v) = \mathrm {d} u \pm \mathrm {d} v; \quad \mathrm {d} (c u) = c \mathrm {d} u, c \text {为常数};\] \[\mathrm {d} (u v) = v \mathrm {d} u + u \mathrm {d} v; \quad \mathrm {d} \left(\frac {u}{v}\right) = \frac {1}{v ^ {2}} (v \mathrm {d} u - u \mathrm {d} v) \quad (v \neq 0).\]对 \(n(n > 2)\) 元函数有类似结论。
（二）多元复合函数的微分法则
1. 多元函数与一元函数的复合
【定理8.7】 设 \(x = x(t),y = y(t),z = z(t)\) 在 \(t\) 可导，\(u = f(x,y,z)\) 在对应点 \((x,y,z) = (x(t),y(t),z(t))\) 处可微，则复合函数 \(u = f(x(t),y(t),z(t))\) 在 \(t\) 可导，且
\[\frac {\mathrm {d} u}{\mathrm {d} t} = \frac {\partial f}{\partial x} \frac {\mathrm {d} x}{\mathrm {d} t} + \frac {\partial f}{\partial y} \frac {\mathrm {d} y}{\mathrm {d} t} + \frac {\partial f}{\partial z} \frac {\mathrm {d} z}{\mathrm {d} t}, \tag {8.2}\]这里也把 \(\frac{\mathrm{d}u}{\mathrm{d}t}\) 称为全导数。
2. 多元函数与多元函数的复合
【定理8.8】（链锁法则）设 \(u = \varphi (x,y),v = \psi (x,y)\) 在点 \((x,y)\) 处有对 \(x,y\) 的偏导数，\(z = f(u,v)\) 在对应点 \((u,v) = (\varphi (x,y),\psi (x,y))\) 处可微，则 \(z = f(\varphi (x,y),\psi (x,y))\) 在点 \((x,y)\) 处有对 \(x,y\) 的偏导数，且
\[\frac {\partial z}{\partial x} = \frac {\partial f}{\partial u} \frac {\partial u}{\partial x} + \frac {\partial f}{\partial v} \frac {\partial v}{\partial x}, \quad \frac {\partial z}{\partial y} = \frac {\partial f}{\partial u} \frac {\partial u}{\partial y} + \frac {\partial f}{\partial v} \frac {\partial v}{\partial y}. \tag {8.3}\]类似地，设 \(z = f(u,v,w),u = u(x,y),v = v(x,y),w = w(x,y)\)，则它们的复合函数 \(z = f(u(x,y),v(x,y),w(x,y))\) 在点 \((x,y)\) 的偏导数为
\[\frac {\partial z}{\partial x} = \frac {\partial f}{\partial u} \frac {\partial u}{\partial x} + \frac {\partial f}{\partial v} \frac {\partial v}{\partial x} + \frac {\partial f}{\partial w} \frac {\partial w}{\partial x}, \quad \frac {\partial z}{\partial y} = \frac {\partial f}{\partial u} \frac {\partial u}{\partial y} + \frac {\partial f}{\partial v} \frac {\partial v}{\partial y} + \frac {\partial f}{\partial w} \frac {\partial w}{\partial y}. \tag {8.4}\]设 \(z = f(u, v, w)\)，为了方便，我们常用 \(f_{1}'\) 表示 \(f(u, v, w)\) 对第一个变量 \(u\) 的偏导数，类似有 \(f_{2}' = \frac{\partial f}{\partial v}\)，\(f_{3}' = \frac{\partial f}{\partial w}\)，\(f_{11}' = \frac{\partial^{2} f}{\partial u^{2}}\)，\(f_{12}' = \frac{\partial^{2} f}{\partial u \partial v}\)，\(f_{23}' = \frac{\partial^{2} f}{\partial v \partial w}\) 等等。
利用这样的记号，上述复合函数求导公式(8.4)可写成
\[z _ {x} ^ {\prime} = f _ {1} ^ {\prime} u _ {x} ^ {\prime} + f _ {2} ^ {\prime} v _ {x} ^ {\prime} + f _ {3} ^ {\prime} w _ {x} ^ {\prime}, \quad z _ {y} ^ {\prime} = f _ {1} ^ {\prime} u _ {y} ^ {\prime} + f _ {2} ^ {\prime} v _ {y} ^ {\prime} + f _ {3} ^ {\prime} w _ {y} ^ {\prime}.\]一阶全微分形式不变性 设 \(z = f(u, v), u = \varphi(x, y), v = \psi(x, y)\) 都是可微函数，利用(8.3)可得函数 \(z = f(u, v) = f(\varphi(x, y), \psi(x, y))\) 的全微分
\[\mathrm {d} z = \frac {\partial z}{\partial x} \mathrm {d} x + \frac {\partial z}{\partial y} \mathrm {d} y = \frac {\partial f}{\partial u} \left(\frac {\partial u}{\partial x} \mathrm {d} x + \frac {\partial u}{\partial y} \mathrm {d} y\right) + \frac {\partial f}{\partial v} \left(\frac {\partial v}{\partial x} \mathrm {d} x + \frac {\partial v}{\partial y} \mathrm {d} y\right) = \frac {\partial f}{\partial u} \mathrm {d} u + \frac {\partial f}{\partial v} \mathrm {d} v.\]这表明当 \(f, u, v\) 都是可微函数时，尽管 \(u, v\) 不是自变量，但函数 \(z = f(u, v)\) 的一阶全微分也具有与 \(u, v\) 是自变量时的一阶全微分的相同形式。这种性质称为一阶全微分形式不变性。
在应用中，首先直接写出 \(\mathrm{dz} = \frac{\partial f}{\partial u}\mathrm{du} + \frac{\partial f}{\partial v}\mathrm{dv}\)，然后再计算出 \(\mathrm{du}\) 和 \(\mathrm{dv}\) 的表达式并代入前式，经过整理，在所得式中 \(\mathrm{dx}\) 的系数就是 \(\frac{\partial z}{\partial x}\)，\(\mathrm{dy}\) 的系数就是 \(\frac{\partial z}{\partial y}\)。这种做法不仅可同时求出多元复合函数的所有的一阶偏导数，而且不容易出错，应熟练掌握。
在求多元复合函数与隐函数的偏导数或全微分时，一阶全微分形式不变性与全微分的四则运算法则是必不可少的重要工具。
【例8.6】 设 \(z = f(u,v,x),u = \varphi (x,y),v = \psi (y)\) 都是可微函数，求复合函数 \(z = f(\varphi (x,y),\psi (y),x)\) 的偏导数 \(\frac{\partial z}{\partial x}\) 与 \(\frac{\partial z}{\partial y}\)。
【解】 由复合函数求导法可得
\[\begin{aligned} \frac {\partial z}{\partial x} &= f _ {1} ^ {\prime} \frac {\partial \varphi}{\partial x} + f _ {2} ^ {\prime} \frac {\partial \psi}{\partial x} + f _ {3} ^ {\prime} = f _ {1} ^ {\prime} \frac {\partial \varphi}{\partial x} + f _ {3} ^ {\prime}, \\ \frac {\partial z}{\partial y} &= f _ {1} ^ {\prime} \frac {\partial \varphi}{\partial y} + f _ {2} ^ {\prime} \psi^ {\prime} (y). \tag {*} \end{aligned}\]评注 在该题的情况下，记号 \(\frac{\partial f}{\partial x}\) 的含意是不清楚的。\(f(u, v, x)\) 作为 \(u, v, x\) 的三元函数求 \(\frac{\partial f(u, v, x)}{\partial x}\)，与 \(f(\varphi(x, y), \psi(y), x)\) 作为 \(x, y\) 的二元函数求 \(\frac{\partial f(\varphi(x, y), \psi(y), x)}{\partial x}\) 的含意是不同的。因此，这里应避免使用记号 \(\frac{\partial f}{\partial x}\)，若要使用它，则必须对其含意加以说明。
若 \(\frac{\partial f}{\partial x}\) 表示 \(f(u, v, x)\) 对 \(x\) 的偏导数，则该例中 \(z = f(\varphi(x, y), \psi(y), x)\) 的偏导数 \(\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}\) 也可表为 \(\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial \varphi}{\partial x} + \frac{\partial f}{\partial x}, \frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial \varphi}{\partial y} + \frac{\partial f}{\partial v} \psi'(y)\)。
应用复合函数求导法则应注意以下几点：
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关键问题是弄清复合函数的结构，分清中间变量与自变量。
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偏导数公式的构成。复合函数对指定自变量的偏导数
原则上函数有几个中间变量，偏导数公式中就有几项的和；函数有几重复合，偏导数公式每项中就有几个因子的乘积。一定要注意对中间变量求导不要漏项。有时在求得的复合函数的偏导数公式中所含的项数比中间变量个数少，那是因为某些中间变量与求偏导数的自变量无关，从而相应的导数为零。如【例8.6】中的(*)式。有时一个变量既是自变量又是中间变量，如【例8.6】的 \(x\)，这时中间变量 \(x\) 对自变量 \(x\) 的导数为1。
- 在复合函数求导公式中，函数对中间变量的偏导数仍然是中间变量的函数，如设 \(z = f(u, v)\)，\(u = \varphi(x, y)\)，\(v = \psi(x, y)\)，则 \(\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial \varphi}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial \psi}{\partial x}\)。
这里 \(\frac{\partial f}{\partial u}, \frac{\partial f}{\partial v}\) 均是 \(u, v\) 的函数，而 \(u = \varphi(x, y), v = \psi(x, y)\)，它们的复合仍是 \(x, y\) 的函数，求高阶偏导数时要特别注意这一点。求 \(\frac{\partial}{\partial x} \left( \frac{\partial f}{\partial u} \right)\) 与 \(\frac{\partial}{\partial x} (f)\) 时，\(\frac{\partial f}{\partial u}\) 与 \(f\) 的地位是相同的。理解这一点，对求复合函数的二阶偏导数至关重要。
（三）复合函数的二阶偏导数
【例8.7】 设 \(z = f(u,v),u = \varphi (x,y),v = \psi (x,y)\) 具有二阶连续偏导数，求复合函数 \(z = f[\varphi (x,y),\psi (x,y)]\) 的一阶与二阶偏导数。
【解】 已求得 \(\frac{\partial z}{\partial x} = f_1' \frac{\partial \varphi}{\partial x} + f_2' \frac{\partial \psi}{\partial x}, \quad \frac{\partial z}{\partial y} = f_1' \frac{\partial \varphi}{\partial y} + f_2' \frac{\partial \psi}{\partial y}\)。下面进一步求 \(\frac{\partial^2 z}{\partial x^2}\)。
第一步，先对 \(\frac{\partial z}{\partial x}\) 的表达式用求导的四则运算法则得
\[\frac {\partial^ {2} z}{\partial x ^ {2}} = \frac {\partial}{\partial x} \left(f _ {1} ^ {\prime}\right) \frac {\partial \varphi}{\partial x} + f _ {1} ^ {\prime} \frac {\partial^ {2} \varphi}{\partial x ^ {2}} + \frac {\partial}{\partial x} \left(f _ {2} ^ {\prime}\right) \frac {\partial \psi}{\partial x} + f _ {2} ^ {\prime} \frac {\partial^ {2} \psi}{\partial x ^ {2}}. \tag {*}\]第二步，再求 \(\frac{\partial}{\partial x} (f_1')\) 与 \(\frac{\partial}{\partial x} (f_2')\)。这里 \(f(u, v)\) 对中间变量 \(u, v\) 的导数 \(f_1' = \frac{\partial f}{\partial u}\)，\(f_2' = \frac{\partial f}{\partial v}\) 仍然是 \(u, v\) 的函数，而 \(u, v\) 还是 \(x, y\) 的函数，它们的复合仍是 \(x, y\) 的函数，因而还要用复合函数求导法求 \(\frac{\partial}{\partial x} (f_{1}^{\prime})\)，\(\frac{\partial}{\partial x} (f_{2}^{\prime})\)。即 \(\frac{\partial}{\partial x} (f_{1}^{\prime}) = \left(f_{11}^{\prime \prime}\frac{\partial\varphi}{\partial x} + f_{12}^{\prime \prime}\frac{\partial\psi}{\partial x}\right), \quad \frac{\partial}{\partial x} (f_{2}^{\prime}) = \left(f_{21}^{\prime \prime}\frac{\partial\varphi}{\partial x} + f_{22}^{\prime \prime}\frac{\partial\psi}{\partial x}\right)\)。
第三步，将它们代入 \((*)\) 式得
\[\frac {\partial^ {2} z}{\partial x ^ {2}} = f _ {1 1} ^ {\prime \prime} \left(\frac {\partial \varphi}{\partial x}\right) ^ {2} + 2 f _ {1 2} ^ {\prime \prime} \frac {\partial \varphi}{\partial x} \frac {\partial \psi}{\partial x} + f _ {2 2} ^ {\prime \prime} \left(\frac {\partial \psi}{\partial x}\right) ^ {2} + f _ {1} ^ {\prime} \frac {\partial^ {2} \varphi}{\partial x ^ {2}} + f _ {2} ^ {\prime} \frac {\partial^ {2} \psi}{\partial x ^ {2}}. \tag {**}\]用类似方法可求得 \(\frac{\partial^2z}{\partial y^2},\quad \frac{\partial^2z}{\partial x\partial y}\)。
评注
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因为 \(f(x, y)\) 有二阶连续偏导数，故 \(f_{12}^{\prime \prime} = f_{21}^{\prime \prime}\)，从而可在 \((**)\) 式中把 \(f_{12}^{\prime \prime}\) 与 \(f_{21}^{\prime \prime}\) 合并。
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在抽象的多元函数的二阶偏导数计算中，最容易出错的地方是：对一阶偏导数 \(\frac{\partial f}{\partial u} = f_u'(u, v)\) 再求偏导数这一步。出错的原因常是忽视为 \(f_u'(u, v)\) 与 \(f_v'(u, v)\) 仍然是与 \(z = f(u, v)\) 保持相同复合结构的复合函数，而容易被误解为仅仅是 \(u\) 或 \(v\) 的函数，从而导致如下漏项的错误
漏掉了对中间变量求偏导的项 \(\frac{\partial^2 f}{\partial u \partial v}\) 与 \(\frac{\partial^2 f}{\partial v \partial u}\)。
因此，求 \(\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial u}\right), \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial v}\right)\) 时必须再用复合函数求导法则，即
\[\frac {\partial}{\partial x} \left(\frac {\partial f}{\partial u}\right) = \frac {\partial}{\partial u} \left(\frac {\partial f}{\partial u}\right) \frac {\partial \varphi}{\partial x} + \frac {\partial}{\partial v} \left(\frac {\partial f}{\partial u}\right) \frac {\partial \psi}{\partial x} = \frac {\partial^ {2} f}{\partial u ^ {2}} \frac {\partial \varphi}{\partial x} + \frac {\partial^ {2} f}{\partial u \partial v} \frac {\partial \psi}{\partial x},\] \[\frac {\partial}{\partial x} \left(\frac {\partial f}{\partial v}\right) = \frac {\partial}{\partial u} \left(\frac {\partial f}{\partial v}\right) \frac {\partial \varphi}{\partial x} + \frac {\partial}{\ --- # 练习题 ### 例题1 设 $ z = f(u, v) $，其中 $ u = x^2 + y^2 $，$ v = xy $，且 $ f(u, v) $ 具有一阶连续偏导数。求复合函数 $ z = f(x^2 + y^2, xy) $ 的偏导数 $ \frac{\partial z}{\partial x} $ 和 $ \frac{\partial z}{\partial y} $。 **解答** 根据链锁法则（定理8.8）：\]\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} = f_1’ \cdot (2x) + f_2’ \cdot y,
\[\]\frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} = f_1’ \cdot (2y) + f_2’ \cdot x.
\[其中 $ f_1' = \frac{\partial f}{\partial u} $，$ f_2' = \frac{\partial f}{\partial v} $。 --- ### 例题2 设 $ u = f(x, y, z) $，其中 $ x = t $，$ y = t^2 $，$ z = t^3 $，且 $ f $ 在对应点可微。求全导数 $ \frac{\mathrm{d}u}{\mathrm{d}t} $。 **解答** 根据多元函数与一元函数复合的定理8.7：\]\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{\partial f}{\partial x} \cdot \frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial f}{\partial y} \cdot \frac{\mathrm{d}y}{\mathrm{d}t} + \frac{\partial f}{\partial z} \cdot \frac{\mathrm{d}z}{\mathrm{d}t} = f_x \cdot 1 + f_y \cdot 2t + f_z \cdot 3t^2.
\[其中 $ f_x, f_y, f_z $ 分别表示 $ f $ 对 $ x, y, z $ 的偏导数。 --- ### 例题3 设 $ z = f(u, v, w) $，其中 $ u = x + y $，$ v = x - y $，$ w = xy $，且 $ f $ 具有一阶连续偏导数。利用一阶全微分形式不变性，求 $ \mathrm{d}z $ 并导出 $ \frac{\partial z}{\partial x} $ 和 $ \frac{\partial z}{\partial y} $。 **解答** 由一阶全微分形式不变性：\]\mathrm{d}z = \frac{\partial f}{\partial u} \mathrm{d}u + \frac{\partial f}{\partial v} \mathrm{d}v + \frac{\partial f}{\partial w} \mathrm{d}w.
\[计算 $ \mathrm{d}u, \mathrm{d}v, \mathrm{d}w $：\]\mathrm{d}u = \mathrm{d}x + \mathrm{d}y, \quad \mathrm{d}v = \mathrm{d}x - \mathrm{d}y, \quad \mathrm{d}w = y\mathrm{d}x + x\mathrm{d}y.
\[代入得：\]\mathrm{d}z = f_1’ (\mathrm{d}x + \mathrm{d}y) + f_2’ (\mathrm{d}x - \mathrm{d}y) + f_3’ (y\mathrm{d}x + x\mathrm{d}y) = (f_1’ + f_2’ + y f_3’) \mathrm{d}x + (f_1’ - f_2’ + x f_3’) \mathrm{d}y.
\[因此，\]\frac{\partial z}{\partial x} = f_1’ + f_2’ + y f_3’, \quad \frac{\partial z}{\partial y} = f_1’ - f_2’ + x f_3’.
\[ --- ### 例题4 设 $ z = f(u, v) $，其中 $ u = x^2 - y^2 $，$ v = 2xy $，且 $ f $ 具有二阶连续偏导数。求 $ \frac{\partial^2 z}{\partial x^2} $。 **解答** 先求一阶偏导数：\]\frac{\partial z}{\partial x} = f_1’ \cdot (2x) + f_2’ \cdot (2y).
\[再求二阶偏导数：\]\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left( f_1’ \cdot 2x + f_2’ \cdot 2y \right).
\[应用乘积法则：\]\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}(f_1’) \cdot 2x + f_1’ \cdot 2 + \frac{\partial}{\partial x}(f_2’) \cdot 2y + f_2’ \cdot 0.
\[计算 $ \frac{\partial}{\partial x}(f_1') $ 和 $ \frac{\partial}{\partial x}(f_2') $：\]\frac{\partial}{\partial x}(f_1’) = f_{11}” \cdot (2x) + f_{12}” \cdot (2y), \quad \frac{\partial}{\partial x}(f_2’) = f_{21}” \cdot (2x) + f_{22}” \cdot (2y).
\[代入得：\]\frac{\partial^2 z}{\partial x^2} = \left( f_{11}” \cdot 2x + f_{12}” \cdot 2y \right) \cdot 2x + 2f_1’ + \left( f_{21}” \cdot 2x + f_{22}” \cdot 2y \right) \cdot 2y.
\[整理（注意 $ f_{12}'' = f_{21}'' $）：\]\frac{\partial^2 z}{\partial x^2} = 4x^2 f_{11}” + 8xy f_{12}” + 4y^2 f_{22}” + 2f_1’.
\[ --- ### 例题5 设 $ z = f(u, v, x) $，其中 $ u = \varphi(x, y) $，$ v = \psi(y) $，且所有函数可微。求 $ \frac{\partial z}{\partial x} $ 和 $ \frac{\partial z}{\partial y} $，并解释记号 $ \frac{\partial f}{\partial x} $ 的歧义。 **解答** 由复合函数求导法则：\]\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} + \frac{\partial f}{\partial x} = f_1’ \cdot \frac{\partial \varphi}{\partial x} + f_2’ \cdot 0 + f_3’ = f_1’ \cdot \frac{\partial \varphi}{\partial x} + f_3’,
\[\]\frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \cdot \frac{\mathrm{d}v}{\mathrm{d}y} + \frac{\partial f}{\partial x} \cdot 0 = f_1’ \cdot \frac{\partial \varphi}{\partial y} + f_2’ \cdot \psi’(y).
\[其中 $ f_3' = \frac{\partial f}{\partial x} $ 表示 $ f(u, v, x) $ 对第三个自变量 $ x $ 的偏导数。注意：若 $ \frac{\partial f}{\partial x} $ 表示复合函数对 $ x $ 的偏导数，则含义不同，容易混淆，因此需明确说明。 --- ### 例题6 设 $ w = f(x, y, z) $，其中 $ x = u + v $，$ y = u - v $，$ z = uv $，且 $ f $ 具有二阶连续偏导数。求 $ \frac{\partial^2 w}{\partial u^2} $。 **解答** 先求一阶偏导数：\]\frac{\partial w}{\partial u} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial u} + \frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial u} = f_x \cdot 1 + f_y \cdot 1 + f_z \cdot v.
\[再求二阶偏导数：\]\frac{\partial^2 w}{\partial u^2} = \frac{\partial}{\partial u}(f_x) + \frac{\partial}{\partial u}(f_y) + \frac{\partial}{\partial u}(f_z \cdot v).
\[计算各项：\]\frac{\partial}{\partial u}(f_x) = f_{xx} \cdot 1 + f_{xy} \cdot 1 + f_{xz} \cdot v, \quad \frac{\partial}{\partial u}(f_y) = f_{yx} \cdot 1 + f_{yy} \cdot 1 + f_{yz} \cdot v,
\[\]\frac{\partial}{\partial u}(f_z \cdot v) = \frac{\partial}{\partial u}(f_z) \cdot v + f_z \cdot 1 = (f_{zx} \cdot 1 + f_{zy} \cdot 1 + f_{zz} \cdot v) \cdot v + f_z.
\[代入并整理（利用 $ f_{xy} = f_{yx} $, $ f_{xz} = f_{zx} $, $ f_{yz} = f_{zy} $）：\]\frac{\partial^2 w}{\partial u^2} = f_{xx} + 2f_{xy} + f_{yy} + 2v(f_{xz} + f_{yz}) + v^2 f_{zz} + f_z.
\[\]