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How can I prove the usual chain rule, but in the context of smooth manifolds? I mean: let $f$ and $g$ be two differentiable maps (from $M$ to $N$ and from $N$ to $P$, respectively) and where all are smooth manifolds. Also, let $p$ be a point of $M$. I want to show that the usual chain rule: $d(g \circ f)(p) = (dg)(f(p)) \circ (df)(p)$, where $d$ denotes the differential, still applies. Even if its really easy, or straight from the definitions. I'm still getting used to differential geometry notation and I have only problems and no solutions or examples to "practice the language". Thank you a lot for the patience.

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Since any point on a manifold has a neighborhood that is diffeomorphic to $\Bbb R^n$ and differentiation is only really interested in very local behaviour, the chain rule applies as it would in $\Bbb R^n$. –  Arthur Sep 25 '12 at 22:23
@DonAntonio It is a composition, just consider the case $M=N=P=\mathbb{R}^n$ with $n \ge 2$. –  Mercy Sep 25 '12 at 23:08
Yes, I realized that when I stopped misreading (first I thought those were real valued functions). Thanks –  DonAntonio Sep 25 '12 at 23:20

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