합성함수의 자코비안
정리
두 함수 $f : \mathbb{R}^{n} \to \mathbb{R}^{m}$, $g : \mathbb{R}^{m} \to \mathbb{R}^{k}$가 주어졌다고 하자. $f$의 자코비안를 $J(f)$와 같이 표기하자. 그러면 다음이 성립한다.
$$ J(g \circ f) = J(g) J(f) $$
설명
자코비안은 가장 일반화된 도함수이므로, 위 정리는 연쇄법칙의 일반화이다.
증명
자코비안의 정의에 의해
$$ J(g \circ f) = \begin{bmatrix} \dfrac{\partial (g \circ f)_{1}}{\partial x_{1}} & \cdots & \dfrac{\partial (g \circ f)_{1}}{\partial x_{n}} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial (g \circ f)_{k}}{\partial x_{1}} & \cdots & \dfrac{\partial (g \circ f)_{k}}{\partial x_{n}} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial g_{1}}{\partial x_{1}} & \cdots & \dfrac{\partial g_{1}}{\partial x_{n}} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial g_{k}}{\partial x_{1}} & \cdots & \dfrac{\partial g_{k}}{\partial x_{n}} \end{bmatrix} $$
이때 $g_{i} = g_{i}(f_{1}(\mathbf{x}), \dots, f_{m}(\mathbf{x})))$이므로,
$$ \dfrac{\partial g_{i}}{\partial x_{j}} = \sum \limits_{\ell=1}^{m} \dfrac{\partial g_{i}}{\partial f_{\ell}} \dfrac{\partial f_{\ell}}{\partial x_{j}} $$
따라서
$$ \begin{align*} J(g \circ f) =&\ \begin{bmatrix} \sum \limits_{\ell=1}^{m} \dfrac{\partial g_{1}}{\partial f_{\ell}} \dfrac{\partial f_{\ell}}{\partial x_{1}} & \cdots & \sum \limits_{\ell=1}^{m} \dfrac{\partial g_{1}}{\partial f_{\ell}} \dfrac{\partial f_{\ell}}{\partial x_{n}} \\ \vdots & \ddots & \vdots \\ \sum \limits_{\ell=1}^{m} \dfrac{\partial g_{k}}{\partial f_{\ell}} \dfrac{\partial f_{\ell}}{\partial x_{1}} & \cdots & \sum \limits_{\ell=1}^{m} \dfrac{\partial g_{k}}{\partial f_{\ell}} \dfrac{\partial f_{\ell}}{\partial x_{m}} \end{bmatrix} \\ =&\ \begin{bmatrix} \dfrac{\partial g_{1}}{\partial f_{1}} & \cdots & \dfrac{\partial g_{1}}{\partial f_{m}} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial g_{k}}{\partial f_{1}}& \cdots & \dfrac{\partial g_{k}}{\partial f_{m}} \end{bmatrix} \begin{bmatrix} \dfrac{\partial f_{1}}{\partial x_{1}} & \cdots & \dfrac{\partial f_{1}}{\partial x_{n}} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial f_{m}}{\partial x_{1}} & \cdots & \dfrac{\partial f_{m}}{\partial x_{n}} \end{bmatrix} \\ =&\ J(g) J(f) \end{align*} $$
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