On page 23 of Guillemin and Pollack they say the following.
Suppose $g_1,\dots,g_l$ are smooth, real-valued functions on a manifold $X$ of dimension $k\geq l$. Under what conditions is the set $Z$ of common zeroes a reasonable geometric object? WE can answers this question easily by considering the map $g=(g_1,\dots,g_l):X\to\mathbb{R}^l$.
Since each $g_i$ is a smooth map of $X$ into $\mathbb{R}$, its derivative at a point $x$ is a linear map $d(g_i)_x:T_x(X)\to\mathbb{R}$, i.e., a linear functional on $T_x(X)$. You can quickly verify that $dg_x:T_x(X)\to\mathbb{R}^l$ is surjective iff the $l$ functionals $d(g_1)_x,\dots,d(g_l)_x$ are linearly independent on $T_x(X)$.
I could prove that if $dg_x$ is surjective, then the functionals are linearly independent, but I can't prove the converse. How do you see the converse? If the functionals are linearly independent, none is identically zero, so each is surjective onto $\mathbb{R}$. If $(a_1,\dots,a_l)$ is an arbitrary point of $\mathbb{R}^l$, then there must exist $b_i\in T_xX$ such that $d(g_i)_x(b_i)=a_i$ for each $i$. But how can you get a common point which is sent to the desired $a_i$?
Note that $$ g = (g_1,\dots,g_l)\implies dg_x = (d(g_1)_x,\dots,d(g_l)_x). $$ Since $k\ge l$, $dg_x$ is surjective if and only if its rank is $l$, if and only if its row rank is $l$, if and only if its $l$ rows are linearly independent, if and only if the $d(g_i)_x$ are independent.