The derivative of a smooth map between smooth submanifolds of Euclidean spaces: Is it a best linear approximation? Let $M \subset \mathbb R^m$ and $N \subset \mathbb R^n$ be smooth submanifolds and $f : M \to N$ be a smooth map. We can define the (Euclidean) tangent space $\tilde T_pM$ at $p \in M$ as the set of all derivatives $u'(0)  \in \mathbb R^m$ of smooth curves $u : (-a,a)  \to  \mathbb R^M$ such that $p = u(0)$ and $u((-a,a)) \subset M$. It is a linear subspace of $\mathbb R^m$ having the same dimension as $M$. The affine subspace $p + \tilde T_pM$ of $\mathbb R^m$ is tangent to $M$ at $p$.
$f$ induces a linear map $df_p : \tilde T_pM \to \tilde T_{f(p)}N$ via $df_p(u'(0)) = (f \circ u)'(0)$.
Can we understand this map as the best linear approximation of $f$ around $p$ in the following sense:
Let $\pi : \mathbb R^m \to \tilde T_pM$ denote the orthogonal projection. Then

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*$\lim_{x \to p \text{ with } x \in M}\dfrac{\lVert f(x) - f(p) - df_p(\pi(x-p))\rVert}{\lVert x - p \rVert} = 0$


*No other linear map $\tilde T_pM \to \tilde T_{f(p)}N$ has this property.
Remark:
I approximate $f$ by
$$\tilde f : M \to \tilde T_{f(p)}N \subset \mathbb R^n, \tilde f(x) = f(p) + df_p(\pi(x-p)).$$
I called this the "best linear approximation of $f$ around $p$", but Ted Shifrin correctly states that it is the best affine approximation of $f$ around $p$.
The idea to introduce $\tilde f$ is this:

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*Project $M$ orthogonally to $p + \tilde T_pM$. This is described via $x \mapsto \omega(x)= p + \pi(x-p)$.


*Map $p + \tilde T_pM$ affinely to $f(p) + \tilde T_{f(p)}N$. This described via $p + v \mapsto f(p) + df_p(v)$.
The map $\tilde f$ does in general not take values in $N$, but since $N$ and $f(p) + \tilde T_{f(p)}N$ are subsets of $\mathbb R^n$, we can nevertheless consider the difference $f - \tilde f$ in $\mathbb R^n$.
Alternatively one could project $N$ orthogonally to $f(p) + \tilde T_{f(p)}N$ which is described via $y \mapsto \rho(y)= f(p) + \sigma(y -f(p))$, where $\sigma : \mathbb R^n \to \tilde T_{f(p)}N$ denotes the orthogonal projection. Then we can consider the map $\hat f = \rho \circ f: M \to  f(p) + \tilde T_{f(p)}N$ which is a good approximation of $f$ in $\mathbb R^n$ and compare it with $\tilde f$.
 A: Yes, it's correct, except for the technical point that the derivative map acts on (and maps to) the tangent space, not the affine tangent space. The tangent vector $u'(0)$ is based at the origin, not at $u(0)$, unless you are totally reinventing notation. But it's easy enough to work with the usual tangent space and just say that its affine translation is the best (affine) linear approximation at $p$.
You can prove the result by taking a local smooth extension of $f$ (in a neighborhood of $p$) to an open set in $\Bbb R^M$, as is quite standard. The main point then is that
$$x-p = \pi(x-p) + (x-p)^\perp,$$
where $u^\perp\in (T_pM)^\perp$, and $\|(x-p)^\perp\| = o(\|x-p\|)$. You can verify this by assuming $T_pM = \Bbb R^m\times\{0\}\subset\Bbb R^M$ [as I said in the comments, horrendous, horrendous choice of notation] and writing $M$ locally as the graph of a smooth function $\phi\colon U\to\Bbb R^{M-m}$, $0\in U\subset\Bbb R^m$ open, $p=0\in\Bbb R^M$; the claim then follows by considering the second-order Taylor polynomial of $\phi$ at $0$.
The uniqueness result follows from the fact that the derivative $df_p$ is equal to the restriction of $dF_p$ for any local smooth extension $F$ of $f$.
