Suppose $f:\Bbb R^n \to \Bbb R$, $g:\Bbb R^m \to \Bbb R^n$, then we have $D_j(f\circ g)(a)=\sum_{1}^{n}D_i f(g(a))D_jg^i(a)$.
How to show the above formula?
2026-03-29 16:31:31.1774801891
Chain rule for the vector valued functions
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If you know the differential of composite functions, it’s easy. Take the differential of $f \circ g$. It is the product of differentials. In this case, they are matrices (Jacobians), so the j-th partial derivative would be the product of the gradient of $f \circ g$ and the j-th column of the jacobian of $g$, that gives you your result