Differential of a smooth extension of a map Let  $X\subset R^N$, $Y\subset R^M$ be manifolds in the sense of Guillemin-Pollack (every point in $x\in X$ has a neighborhood locally diffeomorphic to an open subset of $R^k$, and similarly for $Y$). A map $f: X\rightarrow Y$ is smooth if it extends to a smooth map $F: U\rightarrow R^M$ where $X\subset U$. ($F$ must agree with $f$ on $X$.) 
Probably this is a silly question, but is it true that $dF_x=df_x$? I don't see directly from definitions why this must hold. The definition of $df_x$ is as follows:
If $\phi: V\rightarrow X, x'\rightarrow x, \psi: W\rightarrow Y, y'\mapsto y$ are local parametrizations around $x$ and $y$ and if $h=\psi^{-1}\circ f\circ \phi$, then $df_x=d\psi_{y'}\circ dh_{x'}\circ d(\phi^{-1})_x$ 
 A: They are not exactly equal, as $dF_x$ is defined on $\mathbb{R}^N$ and $df_x$ is defined only on the tangent space of $X$ at the point $x$, which I believe Guillemin and Pollack denote by $T_x(X)$.
But if you restrict $dF_x$ to $T_x(X)$, then the maps are the same. To see this, notice that as $F$ is a local smooth extension of $f$ at the point $p$, it is defined on an open subset $W \subset \mathbb{R}^N$ containing $x$ such that $F$ agrees with $f$ along $W \cap M$. Note that $W \cap M$ is an open subset of $M$ (we are considering the induced topology on $M$), so by taking $V$ and $U$ sufficiently small, we may assume that $\phi(U) \subset W \cap M$ and that the composition $\psi^{-1} \circ F \circ \phi: U \to V$ is well-defined and is actually equal to $h = \psi^{-1} \circ f \circ \phi$ (I am using the same notation as you did in your question).
Now, if we want to apply $dF_x$ to a vector $v \in T_x(X)$, by yhe definition of the tangent space we have that $v = d\phi_{x'}(z)$ for some $z \in U$. Then
$$
dF_x(v) = dF_x(d\phi_{x'}(z)) = d(F \circ \phi)_{x'}(z),
$$
where I have used the usual Chain Rule for maps between open subsets of the Euclidean space above. But we have the following commutative diagram:
$$
\require{AMScd}
\begin{CD}
W \cap M @>F>> Y\\
@A \phi AA @A\psi AA\\
U @>h>> V
\end{CD}
$$
from which we can conclude that $F \circ \phi = \psi \circ h$. Taking differentials at the point $x'$, evaluating at $z$ and using the Chain Rule once again yields
$$
d(F \circ g)_{x'}(z) = d(\psi \circ h)_{x'}(z) = d\psi_{y'} \circ dh_{x'}(z).
$$
Now since $v=d\phi_{x'}(z)$, we have $v=(d\phi_{x'})^{-1}(v)$. Pluggin this in the equation above gives us
$$
dF_x(v) = d(F \circ g)_{x'}(z) = d\psi_{y'} \circ dh_{x'} \circ d\phi^{-1}_{x'}(v) = df_x(v),
$$
so that $dF_x(v)=df_x(v)$ in the sense of the defintion given.
In resume, the differentials $df_x$ and $dF_x$ are equal if we restrict $dF_x$ to $T_x(X)$. In Milnor's book Topology from the Differential Viewpoint, the differential of a smooth map $f$ between manifolds is actually defined as the differential of any smooth local extension $F$.
