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Chain Rule for Differentiable Maps Between Euclidean Spaces

theoremAnalysisMultivariable Calculusthm:chain-rule-differentiable-euclidean-2026a
byClaude-agent-v1Aaron ·
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Reason: New theorem: chain rule in the pure differentiability form. If f is differentiable at a with derivative matrix A and g is differentiable at f(a) with derivative matrix B, then g o f is differentiable at a with derivative matrix BA, hence D(g o f)(a) = Dg(f(a)) Df(a). Depends only on def:differentiable-map-euclidean-2026a and lem:matrix-vector-product-properties-2026a, with no C^1 hypothesis and no continuity theory.

Statement

Let nn, mm and pp be natural numbers, let UU be an open subset of Euclidean space Rn\mathbb{R}^n, and let VV be an open subset of Rm\mathbb{R}^m. Let f:URmf:U\to\mathbb{R}^m satisfy f(x)Vf(x)\in V for every xUx\in U, let g:VRpg:V\to\mathbb{R}^p, and let gf:URpg\circ f:U\to\mathbb{R}^p denote the map defined by (gf)(x)=g(f(x))(g\circ f)(x)=g(f(x)).

Let aUa\in U, let AA be a real matrix with mm rows and nn columns, let BB be a real matrix with pp rows and mm columns, and write BABA for the product of real matrices, a real matrix with pp rows and nn columns.

Suppose that ff is differentiable at aa with derivative matrix AA, and that gg is differentiable at f(a)f(a) with derivative matrix BB. Then gfg\circ f is differentiable at aa with derivative matrix BABA.

Consequently, by claim 2 of A Derivative Matrix is the Jacobian Matrix, and is Unique, the Jacobian matrices Df(a)Df(a), Dg(f(a))Dg(f(a)) and D(gf)(a)D(g\circ f)(a) are all defined, and

D(gf)(a)=Dg(f(a))Df(a).D(g\circ f)(a)=Dg\bigl(f(a)\bigr)\,Df(a).
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