I'm having trouble verifying that $dg_x : T_x(X) \to \mathbb{R}^l$ is surjective iff the $l$ functionals $d(g_1)_x, ... , d(g_l)_x$ are linearly independent on $T_x(X)$.
Suppose $\dim(X) = n$, then we have $T_x(X) \cong \mathbb{R}^n$. I know that each $d(g_i)_x \in (T_x(X))^*$, and that if the $l$ functionals $d(g_i)_x$ are linearly independent on $T_x(X)$, then they span a $l$-dimensional subspace of $(T_x(X))^* \cong T_x(X) \cong \mathbb{R}^n$. But I need to show that this $l$-dimensional subspace is $\mathbb{R}^l$ (and from that I'm assuming I can conclude that $dg_x$ is surjective?)
For the forward direction, I'm not sure where to start. For example supposing $$\sum_{i=o}^l c_i \cdot d(g_i)_x = 0$$
for $i \in [1, ... ,l]$ and scalars $c_i \in \mathbb{R}$ doesn't really lead anywhere (if you try to do a proof by contridction, all you end up contradicting in the end is that $dg_x$ is not injective)
I'm a bit stuck here, any help would be appreciated.
The image $V := \operatorname{im} dg_x$ is a linear subspace of $\mathbb{R}^l$. Now recall that we have a correspondence between linear subspaces of $\mathbb{R}^l$ and their annihilators. We have $V = \mathbb{R}^l \iff V^{\perp} = \{0\}$.
On the other hand, we have a correspondence between the linear relations that the $(dg_i)_x$ satisfy and the annihilator of $V$, namely
\begin{align} \sum_{i = 1}^l c_i\cdot (dg_i)_x = 0 &\iff (\forall v\in T_xX)\Biggl(\sum_{i = 1}^l c_i (dg_i)_x(v) = 0\Biggr)\\ &\iff (\forall w\in V)\Biggl(\sum_{i = 1}^l c_i w_i = 0\Biggr) \\ &\iff (c_1,\dotsc, c_l) \in V^{\perp}. \end{align}
Thus the $(dg_i)_x$ are linearly independent if and only if they satisfy only the trivial linear relation, if and only if $V^{\perp} = \{0\}$, if and only if $V = \mathbb{R}^l$.