📝 题目
例 10 设 $f,g : {\mathbf{R}}^{m} \rightarrow {\mathbf{R}}^{n}$ 是可微函数. 试用复合函数求导法则来证明向量内积的求导公式:
$$ \mathrm{D}\left( {\mathbf{f}\left( \mathbf{x}\right) \cdot \mathbf{g}\left( \mathbf{x}\right) }\right) = {\mathbf{f}}^{\mathrm{T}}\left( \mathbf{x}\right) \mathrm{D}\mathbf{g}\left( \mathbf{x}\right) + {\mathbf{g}}^{\mathrm{T}}\left( \mathbf{x}\right) \mathrm{D}\mathbf{f}\left( \mathbf{x}\right) . $$
💡 答案与解析
证 令 $F : {\mathbf{R}}^{2n} \rightarrow \mathbf{R}$ ,
$$ F\left( \mathbf{u}\right) = F\left( {{u}_{1},\cdots ,{u}_{n},{u}_{n + 1},\cdots ,{u}_{2n}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{u}_{i} \cdot {u}_{n + 1}. $$
令
$$ G : {\mathbf{R}}^{m} \rightarrow {\mathbf{R}}^{2n} $$
$$ \mathbf{u} = \mathbf{G}\left( \mathbf{x}\right) = \left( {{f}_{1}\left( \mathbf{x}\right) ,\cdots ,{f}_{n}\left( \mathbf{x}\right) ,{g}_{1}\left( \mathbf{x}\right) ,\cdots ,{g}_{n}\left( \mathbf{x}\right) }\right) , $$
则
$$ \left( {F \circ G}\right) \left( x\right) = f\left( x\right) \cdot g\left( x\right) . $$
因 $F$ 的 ${2n}$ 个偏导数连续,所以 $F$ 可微. 又因 $\mathbf{G}$ 的每个分量可微,所以 $\mathbf{G}$ 也可微. 这样由复合函数求导法则,得
$$ \mathrm{D}\left( {\mathbf{f}\left( \mathbf{x}\right) \cdot \mathbf{g}\left( \mathbf{x}\right) }\right) = \mathrm{D}\left( {F \circ \mathbf{G}}\right) \left( \mathbf{x}\right) = \mathrm{D}F\left( \mathbf{u}\right) \mathrm{D}\mathbf{G}\left( \mathbf{x}\right) $$
$$ = \left( {{u}_{n + 1},\cdots ,{u}_{2n},{u}_{1},\cdots ,{u}_{n}}\right) \left( \begin{array}{l} \mathrm{D}\mathbf{f}\left( \mathbf{x}\right) \\ \mathrm{D}\mathbf{g}\left( \mathbf{x}\right) \end{array}\right) . $$
利用矩阵分块相乘得
$$ \mathrm{D}\left( {\mathbf{f}\left( \mathbf{x}\right) \cdot \mathbf{g}\left( \mathbf{x}\right) }\right) = \left( {{u}_{n + 1},\cdots ,{u}_{2n}}\right) \mathrm{D}\mathbf{f}\left( \mathbf{x}\right) + \left( {{u}_{1},\cdots ,{u}_{n}}\right) \mathrm{D}\mathbf{g}\left( \mathbf{x}\right) $$
$$ = {\mathbf{g}}^{\mathrm{T}}\left( \mathbf{x}\right) \mathrm{D}\mathbf{f}\left( \mathbf{x}\right) + {\mathbf{f}}^{\mathrm{T}}\left( \mathbf{x}\right) \mathrm{D}\mathbf{g}\left( \mathbf{x}\right) . $$
评注 对多元函数来说, 求导法则只需复合函数求导一条法则, 其余求导法则皆可由它推出.