Munkres Analysis on Manifolds, Differentiation Question Munkres (pg. 45). Munkres states that if a function $g:\mathbb{R}^2 \to \mathbb{R}$ is differentiable then its matrix $Dg(0) = [a \text{ } b]$ and that $g'(0;u) = ah + bk$. The latter part makes sense since if a function is differentiable then its derivative is given by $Dg(a) \cdot u$, in this case $0$. However, how does he know that $Dg(0) = [a \text{ } b]$. I tried going back to the defintion of the derivative but we don't know anything about $g(h)$ or $g(0)$, so can someone provide an explanation of how $Dg(0)$ was found? 
Or is he saying that it must be a $1$ by $2$ matrix, where $a$ and $b$ don't have to be constants (i.e. can be dependent on $x$ and $y$)? But, that doesn't quite make sense he just says $a$ and $b$. Is that because $x = 0$ and $y=0$ so any terms dependent on $x$ or $y$ are removed?
 A: $ Dg(0)$ (by which Munkres means $Dg(0,0)$) is meant to denote the derivative of $g$ at the point $(0,0)$. 
It may be meant to denote a linear transformation, or perhaps the matrix of that transformation with respect to the standard basis. Either way, it's a constant. 
By analogy, think of the function
$$
f: \Bbb R \to \Bbb R : x \mapsto x^3.
$$
Then 
$Df(2)$ is (in a Calculus 1 class) the number $3 \cdot 2^2 = 12$, and in a more sophisticated view, it's the linear transformation from the reals to the reals given by "multiply by 12". And in a still more sophisticated case, it's a linear transformation from "the tangent space to $\Bbb R$ at $2$" to "the tangent space to $\Bbb R$ at $8$", again given by multiplication by $12$ (with respect to the natural bases on each). In all cases, however, it's a constant. 
A: If $f:\mathbf{R}^d \to \mathbf{R}$ is differentiable at a point $p,$ then $f(p + he_i)=f(p)+h f'(p)\cdot e_i+o(h).$ We know that $f'(p) \cdot e_i$ is the $i$th entry of the vector representation of the linear form $f'(p)$ and also $f'(p)\cdot e_i = \partial_i f.$ Thus, the result you were asking.
