For a function $f:X \to R$, where $X$ is a manifold. For any point $x \in X$, the derivative of $f$ at $x$ is $[Df(x)]:T_xX\to R$. And $T_xX$ is the tangent space of $X$ at point $x$. Why is the domain of $[Df(x)]$ $T_xX$?
At first, I read this in the book Vector Calculus, Linear Algebra and Differential Forms, a Unified Approach, 5e on page 311. I didn't quite understand, so I skipped.
Then I found it once again on page 350 in the margin note.
I can imagine this geographically, if we want to evaluate, by $[Df(x)]$, the influence of a small increment of $x$ on $f(x)$. Then we have to choose a very small $h$ to calculate $[Df(x)]h$, or otherwise $a+h$ will exceed X. And the $h$ is so small that it's tangent to $X$.
To persuade myself, I even came up with an example to help myself understand: the manifold is defined by $X=\{(x,y):x^2+y^2-1=0\}$. The function is defined on $X$ by $f:(x,y)\mapsto x^2+y^2$. Then I take the derivative of $f$, according to the definition of the derivative. We need to find a linear transformation $L$, which satisfies the following:
$$ \lim_{h\to 0}\frac{f(a+h)-f(a)-L(h)}{|h|}=0 $$
Because $a+h \in X$, so we have restriction $(1+h_1)^2+h_2^2-1=0$. Finally, I drew the conclusion that $L=0$, which complies with my imagination. So the domain of $[Df(x)]$ really seems $T_xX$. But how to prove it?