Is the map assigning to $x$ its tangent space smooth? I'm having some difficulties understanding smooth maps between manifolds, and in particular I would like to know if the map $x \mapsto T_{x}M$ is a smooth map $M\to Gr(k,n)$? Here $x \in M$, where $M$ is a $k$-dimensional manifold in $\mathbb{R}^{n}$. 
From Lee's Introduction to Smooth Manifolds, we have the definition:

Let $M, N$ be smooth manifolds and let $F : M \to N$ be any map. $F$ is smooth if for every $p \in M$, there exist smooth chart $(U, \phi)$ containing $p$ and $(V,\psi)$ containing $F(p)$ s.t. $F(U) \subseteq V$ and the composite map $\psi \circ F \circ \phi^{-1}$ is smooth from $\phi(U)$ to $\psi(V)$. 

Is it possible to show that the map assigning $x$ to its tangent space is smooth directly from the definition?
 A: Sure, this is almost immediate from the definition of charts on the Grassmannian.  Let $U\subseteq\mathbb{R}^{n\times k}$ be the open set consisting of $n\times k$ matrices of rank $k$.  There is an obvious map $f:U\to Gr(k,n)$ taking a matrix to the span of its columns.  The derivative of any local parametrization of $M$ defines a smooth map to $U$ whose composition with $f$ is exactly the map $x\mapsto T_xM$.  So, it suffices to show that $f$ is smooth.
Now fix $i_1<i_2<\dots<i_k$ and let $V\subseteq U$ be the open subset consisting of matrices whose rows $i_1,\dots,i_k$ are linearly independent, and let $C\subseteq V$ be the set of matrices whose rows $i_1,\dots,i_k$ form the $k\times k$ identity matrix (which can be identified with $\mathbb{R}^{k(n-k)}$ since there are $n-k$ rows whose entries are unrestricted).  By definition of the smooth structure on $Gr(k,n)$, $f$ restricted to $C$ is smooth (it is the inverse of one of the coordinate charts on $Gr(k,n)$).  Now let $g:V\to C$ be the smooth map defined by $g(A)=AB^{-1}$ where $B$ is the $k\times k$ matrix formed by rows $i_1,\dots,i_k$ of $A$.  The columns of $g(A)$ have the same span as the columns of $A$ since we just multiplied on the right by an invertible matrix, so $f=f\circ g$ on $V$.  But $g$ maps to $C$ and $f$ is smooth on $C$, so we conclude that $f$ is smooth on $V$.  Since open subsets $V$ of this form cover $U$, this proves $f:U\to Gr(k,n)$ is smooth.
