I am asking for help interpreting an example in a textbook. The author gives two functions from different dimensions of Euclidean space, and he precisely describes the image of arbitrary elements under these functions, but he does not state the arbitrary elements in the domains.
Here is the example from Chapter 5. Maybe someone has a different edition of the textbook.
Let $f: \mathbb{R}^{2} \to \mathbb{R}^{3}$ and $g: \mathbb{R}^{3} \to \mathbb{R}$ be defined by $f = (x,y,z)$ and $g = w$ where $$w = w(x,y,z) = xy + yz + xz$$ and $$x = x(s,t) = st, \quad y = y(s,t) = s\cos{t} \quad z = z(s,t) = s\sin{t}.$$
a.) Find the matrices that represent the linear transformations $(\mathrm{D}f)_{p}$ and $(\mathrm{D}g)_{q}$ where $p = (s_{\circ}, t_{\circ}) = (0,1)$ and $q = f(p)$.
b.) Use the Chain Rule to calculate the $1 \times 2$ matrix $[\partial{w}/\partial{s}, \partial{w}/\partial{t}]$ that represents $(\mathrm{D}(g\circ{f}))_{p}$.
c.) Substitute the functions $x = x(s,t)$, $y = y(s,t)$, and $z = z(s,t)$ directly into $w = w(x,y,z)$ in order to get $[\partial{w}/\partial{s}, \partial{w}/\partial{t}]$, verifying the answer in Part b.).
This gets a bit messy but we have the following... $$f:\mathbb{R}^2\rightarrow\mathbb{R}^3$$ $$f(s,t)=(x,y,z)=(st, s\cos(t), s\sin(t))$$
So the matrix which represents the total derivative of our map $f$ at the point $p=(s,t)$ would be...
$$(Df)_{(s,t)}=\begin{bmatrix} t &s \\ \cos(t)& -s\sin(t) \\ \sin(t)& s\cos(t) \end{bmatrix}$$
Additionally, we have the map; $$w:\mathbb{R}^3\rightarrow\mathbb{R}$$ $$w(x,y,z)=xy+yz+xz$$
Which yields;
$$(Dw)_{f(p)}=\begin{bmatrix} (y+z)& (x+z)& (y+x) \end{bmatrix}$$
So, by the chain rule, our map $D(w\circ f)_p$ would give us;
$$D(w\circ f)_p=D(w)_{f(p)}\circ D(f)_p=\begin{bmatrix} (y+z)& (x+z)& (y+x) \end{bmatrix}\begin{bmatrix} t &s \\ \cos(t)& -s\sin(t) \\ \sin(t)& s\cos(t) \end{bmatrix}$$
Letting $(s,t)=(1,0)$ gives us;
$$\begin{bmatrix} 1& 0& 1 \end{bmatrix}\begin{bmatrix} 0 &1 \\ 1& 0 \\ 0& 1 \end{bmatrix} = \begin{bmatrix} 0& 2 \end{bmatrix} $$
I'll leave it to you to confirm that;
$$\begin{bmatrix} \frac{\partial\omega}{\partial s}& \frac{\partial\omega}{\partial t} \end{bmatrix}_{(1,0)}$$
gives the same result.