Tangent vectors to coadjoint orbits Let $O_x:=\{Ad^*_g (x); g \in G\}$ be the orbit of $x \in \mathfrak{g}^*$ and $Ad$ the adjoint map.
Now take $\xi \in \mathfrak{g}$ then $g(t):=Ad^*_{e^{t \xi}}(x)$ defines a map $g: I \rightarrow O_x$ with $g(0)=x.$ Thus, $g'(0) = ad_{\xi}^*(x) \in T_xO_x.$
Now, I was wondering whether this construction is also surjective, so has every $\eta \in T_xO_x$ a representation of the form $ad_{\xi}^*(x)$?
 A: Yes. It's easy to see why this "should" be so: by definition, the map $G\to O_x$ is surjective, so it's not surprising that the map on tangent spaces is surjective.  Of course, one has to be careful about arguments like this.  For example, $f(x)=x^3$ is a bijection from $\mathbb{R}$ to itself, but is not surjective on the tangent space at 0 (it's not a diffeomorphism).  
The important extra thing to observe is that if $\mathrm{ad}^*_{\xi}(x)=0$, then the exponential $e^{t\xi}$ fixes $x$;  thus if we let $\mathfrak{g}_x=\{\xi \in  \mathfrak{g}\mid \mathrm{ad}^*_{\xi}(x)=0\}$, then this is a Lie subalgebra, and the stabilizer group $G_x=\{g\in G \mid \mathrm{Ad}^*_g x=x\}$ contains its exponential.  On the other hand, any vector tangent to $G_x$ obviously lies in $\mathfrak{g}_x$, so $\mathfrak{g}_x$ is the Lie algebra of $G_x$.  In particular, we have an injective map $\mathfrak{g}/\mathfrak{g}_x\hookrightarrow T_xO_x$, but $O_x\cong G/G_x$, so these spaces have the same dimension, and necessarily $\mathfrak{g}/\mathfrak{g}_x\cong T_xO_x$.
