On the notion of differentiability of maps $\mathbb{R}^n \to \mathbb{R}^m$ and the definition of derivative This is probably a very basic question but I'm not seeing something. I recall that in my multivariable calculus course, the derivative of a $C^1$ map $f: \mathbb{R}^n \to \mathbb{R}^m $was defined to be the $m \times n$ matrix of continuous partial derivatives, often called the Jacobian matrix. I chanced upon this alternative definition, which is familiar-looking yet currently confusing to me:
For a smooth map $f: \mathbb{R}^n \to \mathbb{R}^m $ the differential of $f$ at $p \in \mathbb{R}^n$ (which I interpret as derivative evaluated at a point $p$) for a given $X \in \mathbb{R}^n$ is defined as 
$df_p(X) = \lim_{t \to 0} \frac{1}{t}(f(p + tX) - f(p)) = \frac{d}{dt} [f(p+tX)]_{t=0}$
I have a few concerns. First, I am not able to see how this is equal to $Df_p \dot X$ where $Df_p$ is the Jacobian matrix (of partial derivatives) evaluated at $p$. Second, why do we have this equality:
$\lim_{t \to 0} (f(p + tX) - f(p)) = \frac{d}{dt} [f(p+tX)]_{t=0}$ ? My main concern here is, since this is meant to be the "definition" of derivative, how is one supposed to know what $\frac{d}{dt}$ even is? And why, in any case, is this equality true? 
 A: The derivative $\frac{d}{dt}$ is being applied to the function with real domain $f \circ \gamma$, where $\gamma: I \to \mathbb{R}^n$ is the path $t \mapsto p+tX$. Thus, we "know" what $\frac{d}{dt}$ is, it is the standard component-wise derivative of functions $g: \mathbb{R} \to \mathbb{R}^m$. The equality is simply the definition.
To see how the differential of $f$ at $p$ is the Jacobian matrix (when we see the linear operator $df_p$ as a matrix in the canonical basis), note that $df_p(e_i)$ is by definition the $i$-th partial derivative (more precisely, the vector consisting of the $i$-th partial derivative of each component). Since a matrix representation of a linear map in a given basis has as the $l$-th collumn the image of the $l$-th basis element when represented in said basis, you get what you want.
Note, however, that only "having a well-defined differential", in your definition, is not the same as being (Fréchet-)differentiable. It is the same as being Gâteaux-differentiable, a weaker notion. (Your text has no problems with this, since it assumes that $f$ is smooth to talk about the differential, probably to sidestep these nuances. But since the title talks about differentiability, this observation is important).
