Derive the equation of first variation for a flow of a vector field. This is a problem from Susan Colley's Vector Calculus. I have trouble understanding the solution to it.
Problem: Derive the equation of first variation for a flow of a vector field. That is, if $\mathbf{F}$ is a vector field of class $C^1$ with flow $\phi$ of class $C^2$, show that 
$$\frac{\partial}{\partial t}D_{\mathbf{x}}\phi(\mathbf{x},t)=D\mathbf{F}(\phi(\mathbf{x},t))D_{\mathbf{x}}\phi(\mathbf{x},t).$$ 
Here the expression "$D_{\mathbf{x}}\phi(\mathbf{x},t)$" means to differentiate $\phi$ with respect to the variables $x_1,x_2,\ldots ,x_n,$ that is, by holding $t$ fixed.
Solution:
By definition of a flow of $\mathbf{F}$, we know that $\frac{\partial}{\partial t}\phi(\mathbf{x},t)=\mathbf{F}(\phi(\mathbf{x},t))$. So 
$$\frac{\partial}{\partial t}D_{\mathbf{x}}\phi(\mathbf{x},t)=D_\mathbf{x}(\frac{\partial}{\partial t}\phi(\mathbf{x},t))=D_{\mathbf{x}}\mathbf{F}(\phi(\mathbf{x},t)).$$
Now by the Chain Rule, 
$$D_{\mathbf{x}}\mathbf{F}(\phi(\mathbf{x},t))=D\mathbf{F}(\phi(\mathbf{x},t))D_{\mathbf{x}}\phi(\mathbf{x},t).$$
I don't understand the two equations shown in the solution. First, how can we interchange $\frac{\partial}{\partial t}$ and $D_{\mathbf{x}}$ in the first equation? 
Also, I don't understand how the Chain Rule leads to the second equation, since we're trying to get $D_\mathbf{x}\mathbf{F}$, not $D\mathbf{F}$. This seems to suggest that $D_\mathbf{x}\mathbf{F}$ is just the matrix of $D\mathbf{F}$ with the first $n$ columns. That is, the partials with respect to the variables $x_1,x_2,\ldots,x_n$. This result confuses me. How does this result make sense starting from the definition of the expression $D_{\mathbf{x}}$ given in the problem? 
I would greatly appreciate it if anyone could explain the above questions to me.
 A: N.B. I will be referring to the $\mathbf{4^{\mathrm{th}}}$ edition of the book. If you have another edition then the numbers of theorems, etc. might be different.
We are given a  function $\phi:\mathbb R^n\times (a,b)\to \mathbb R^n$, which is   of class $C^2$. Since it is $C^2$, theorem 4.5 of section 2.4 tells us that $$\frac{\partial^2\phi}{\partial t\partial x_i}=\frac{\partial ^2\phi}{\partial x_i\partial t},$$ for any $x_i$. Now we can simply write out $\dfrac{\partial}{\partial t}D_{\mathbf{x}}\phi(\mathbf{x},t)$ and use theorem 4,5:
\begin{align}\dfrac{\partial}{\partial t}D_{\mathbf{x}}\phi(\mathbf{x},t)=& \\
=&\dfrac{\partial}{\partial t}
\underbrace{\begin{pmatrix}\dfrac{\partial\phi_1}{\partial x_1} & \dfrac{\partial\phi_1}{\partial x_2}& \cdots &\dfrac{\partial\phi_1}{\partial x_n}\\
\dfrac{\partial\phi_2}{\partial x_1}& \dfrac{\partial\phi_2}{\partial x_2}&\cdots & \dfrac{\partial\phi_2}{\partial x_n}\\
\vdots & \vdots& \ddots & \vdots\\
\dfrac{\partial\phi_n}{\partial x_1}& \dfrac{\partial\phi_n}{\partial x_2}& \cdots & \dfrac{\partial\phi_n}{\partial x_n}\end{pmatrix}}_{\text{Make shure you understand why this  is $n\times n$.}}
=
\begin{pmatrix}\dfrac{\partial^2\phi_1}{\partial t\partial x_1} & \dfrac{\partial^2\phi_1}{\partial t\partial x_2}& \cdots &\dfrac{\partial^2\phi_1}{\partial t\partial x_n}\\
\dfrac{\partial^2\phi_2}{\partial t\partial x_1}& \dfrac{\partial^2\phi_2}{\partial t\partial x_2}&\cdots & \dfrac{\partial^2\phi_2}{\partial t\partial x_n}\\
\vdots & \vdots& \ddots & \vdots\\
\dfrac{\partial^2\phi_n}{\partial t\partial x_1}& \dfrac{\partial^2\phi_n}{\partial t\partial x_2}& \cdots & \dfrac{\partial^2\phi_n}{\partial t\partial x_n}\end{pmatrix}=\\
=& D_{\mathbf{x}}\frac{\partial \phi}{\partial t}=D_{\mathbf x}\mathbf{F}(\phi).
\end{align}
Above I wrote: "Make shure you understand that this is $n\times n$." You should also understand that $\phi (\mathbf x,t)=\big(\phi_1(\mathbf x,t),\ldots,\phi_n(\mathbf x,t)\big)$, i.e. is an element of $\mathbb R^n$.  
If you do, then it will quickly dawn on you that $D\mathbf F=D_{\mathbf x}\mathbf{F}$.
