What does the notation $f'(\mathbf{x_0})(\mathbf{x})$ and $h'(0,0)(0,\mathbf{w})$ mean in differential geometry book (Rudolph and Schmidt)?

Question

I am reading the book Differential Geometry and Mathematical Physics by Rudolph and Schmidt and on page 11 he gives a theorem called the 'Level Set Theorem.'

Let $U\subset \mathbb{R}^n$ be open, let $f:U\to \mathbb{R}^m$ be a differentiable mapping of class $C^k$ and let $\mathbf{c} \in \mathbb{R}^m$ be a regular value of $f$ such that the level set $M:=f^{-1}(\mathbf{c})$ is nonempty. Then, $M$ is a differentiable manifold of class $C^k$ and dimension $n-m$.

There are several things I am uncertain of notation-wise in the proof of this theorem.

1. Let $\mathbf{x_0} \in M$ and define $$ X_0 = \text{ker} f'(x_0) = \{\mathbf{x} \in \mathbb{R}^n : f'(\mathbf{x_0})(\mathbf{x}) = 0\} \subset \mathbb{R}^n. $$ What does the notation $f'(\mathbf{x_0})(\mathbf{x})$ mean? Is it a dot product..if so is this a standard way to indicate a dot product in differential geometry?
2. What is the particular derivative $f'(\mathbf{x})$? $f$ is defined as a map $f:\mathbb{R}^n \supset U \to \mathbb{R}^m$ so how would the derivative $f'(\mathbf{x})$ be expressed for the case of say, $3=m=2$?
3. Shortly after, he introduces a function $h:X_0 \to X_0^\perp$ given by $h(\mathbf{y},\mathbf{z}) = f(\mathbf{x_0} + \mathbf{y} + \mathbf{z}) - \mathbf{c}$ and says that for any $\mathbf{w} \in X_0^\perp$, $$ h'(0,0)(0,\mathbf{w}) = f'(\mathbf{x_0})(\mathbf{w}) $$ Again, he uses this strange notation except this time its even more awkward as he has $h'(0,0)(0,\mathbf{w})$ on the left hand side so I don't think this can be viewed as a dot product. So what is it?

Answer 1

Let's straighten out the confusion:

1. $f'(x_0)$ is a function in itself. This is the function that is linear and approximates $f$ best at the point $x_0$. As an example: If $$ f(x) = x^2$$ then at the point $3$ for example, $f'(3)$ is the linear function that approximates $x^2$ best at $3$. I promise you that if you calculate $$(3+h)^2 - 3^2$$ for $h=1$, you get $7 = 6h +$ something small. For $h = 0.1$, you get $0.601$, extremely small deviation from $6h = 0.6$. For smaller $h$, $x^2$ looks more and more like $6h = f'(3)(h)$ around $x=3$. It is in this sense we have that $$f'(3)(h) = 6h $$
2. If you have an explicit function $f: \mathbb{R^n} \to \mathbb{R^m}$ (which behaves nicely enough), you can get at this linear transformation by constructing a matrix where the first row contains all $n$ partial derivatives of the first function, second row contains all the partials of the second function, and so on, evaluating all the partials in a point $\bf{x_{0}}$$= (x_1 ... x_n)$, resulting in a $m \times n$-matrix. So in that case $f'(x_0)$ is a matrix.
3. First of all, note that $h$ cannot be defined the way you wrote it, since $h$ takes two arguments. In the book you referenced it states: $$h: X_0 \times X_0^\perp \to \mathbb{R}^m $$ and so $$h'(0,0)(0,w) = f'(x_0 + 0 + 0)(0+w) = f'(x_0)(w) $$ Here I have used that $w = 0 + w$ for $\mathbb{R^n}$ where $0 \in X_0$ and $w \in X_0^\perp$ which is exactly the element $(0,w)$ in the cartesian product.

Hope this helps.