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  • 多元链法则(3)

    设$E$是$\mathbf{R}^n$的子集,$F$是$\mathbf{R}^m$的子集,设$f:E\to F$是函数,$g:F\to \mathbf{R}^p(p\in\mathbf{N}^+)$是另一个函数.设$x_0$是$E$的内点,假设$f$在$x_0$处可微,且$f(x_0)$是$F$的内点,还假设$g$在$f(x_0)$处可微,那么$g\circ f:E\to\mathbf{R}^p$在$x_0$处可微,且
    \begin{equation}
    \label{eq:16.13.222}
    (g\circ f)'(x_0)=g'(f(x_0))f'(x_0)
    \end{equation}

    证明:设点$x_0$的坐标是$(a_1,\cdots,a_n)$.则
    \begin{equation}
    \label{eq:21.16.1}
    (g\circ f)'(x_0)=\begin{pmatrix}
    \frac{\partial g\circ f}{\partial a_1}(x_0)&\cdots&\frac{\partial g\circ f}{\partial a_n}(x_0)\\
    \end{pmatrix}
    \end{equation}
    根据多元链法则(2),可得
    \begin{equation}
    \label{eq:21.16.3}
    \frac{\partial g\circ f}{\partial a_i}(x_0)=g'(f(x_0))\frac{\partial f}{\partial a_i}(x_0)
    \end{equation}
    把\ref{eq:21.16.3}代入\ref{eq:21.16.1},可得
    \begin{equation}
    \label{eq:21.17.4}
    (g\circ f)'(x_0)=\begin{pmatrix}
    g'(f(x_0))\frac{\partial f}{\partial a_1}(x_{0})&\cdots&g'(f(x_0))\frac{\partial f}{\partial a_n}(x_0)
    \end{pmatrix}
    \end{equation}
    我们知道,
    \begin{equation}
    \label{eq:21.17.5}
    f'(x_0)=\begin{pmatrix}
    \frac{\partial f}{\partial a_1}(x_0)&\cdots&\frac{\partial f}{\partial a_n}(x_{0})
    \end{pmatrix}
    \end{equation}
    因此\ref{eq:16.13.222}和\ref{eq:21.17.4}是相等的(根据的是分块矩阵的乘法).

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  • 原文地址:https://www.cnblogs.com/yeluqing/p/3828237.html
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