Verfying partial derivatives for an ODE 
Let $\phi(t,x)$ : $[\mathbb{R}$ x $\mathbb{R^N}] \rightarrow \mathbb{R^N}$ have the following properties: for any $t\in\mathbb{R}$ and $x\in\mathbb{R^N}$,

*

*$\phi(t,x)$ is differentiable in both arguments and has a differentiable inverse.

*For any $s\in\mathbb{R}$, $\phi(t+s,x)=\phi(t,\phi(s,x))$.

Define T(t,x) = $\dfrac{\partial\phi(t,x)}{\partial{x}}$ and $f(x)=\dfrac{\partial\phi(t,x)}{\partial{t}}\big|_{t=0}$. Show that $\phi(t,x)$ and $T(t,x)$ satisfy:
a. $\dfrac{\partial\phi(t,x)}{\partial{t}} = f(\phi(t,x))$
b. $\dfrac{\partial{T(t,x)}}{\partial{t}} = Df(\phi(t,x))\hspace{0.5mm}T(t,x)$
where $Df$ is the Jacobian of f.

Here is my proposed solution.
For the first part, I need to show that $\dfrac{\partial\phi(t,x)}{\partial{t}} = f(\phi(t,x))$. Starting from the LHS,
\begin{equation}
\begin{split}
\dfrac{\partial\phi(t,x)}{\partial{t}}  & = \dfrac{\partial\phi(t+0,x)}{\partial{t}}  \\
 & = \dfrac{\partial\phi(t,\phi(0,x))}{\partial{t}} \\
 & = \dfrac{\partial\phi(t,\phi(t,x))}{\partial{t}}\big|_{t=0} \\
 & = f(\phi(t,x)) 
\end{split}
\end{equation}
I think that the second to last equality is justified since $\phi(t,x)$ is evaluated at $t=0$. Although, I'm not sure if I have overlooked something and made a mistake.
For the second part, I need to show that $\dfrac{\partial{T(t,x)}}{\partial{t}} = Df(\phi(t,x))\hspace{0.5mm}T(t,x)$. Starting from the LHS,
\begin{equation}
\begin{split}
\dfrac{\partial{T(t,x)}}{\partial{t}} & = \dfrac{\partial\phi(t,x)}{\partial{x}\hspace{0.5mm}\partial{t}}  \\
 & = \dfrac{f(\phi(t,x))}{\partial{x}} \\
 & = D{f(\phi(t,x))}\hspace{0.5mm}\dfrac{\partial\phi(t,x)}{\partial{x}} \\
 & = D{f(\phi(t,x))}\hspace{0.5mm}T(t,x)
\end{split}
\end{equation}
As $\phi\in{C^1}$, the Jacobian is defined by the third equality.
Is this approach correct? Please let me know if there are any better alternatives.
 A: For the first part, I recommend
\begin{align}
\frac{\partial \phi}{\partial t}(t, x)=&\ \frac{\partial}{\partial s}\phi(t+s, x)\Big|_{s=0}=\ \frac{\partial}{\partial s}\phi(s, \phi(t, x))\Big|_{s=0}\\
=&\ \frac{\partial \phi}{\partial t}(s, \phi(t, x))\Big|_{s=0} = f(\phi(t, x))
\end{align}
A: For the first part, we need to show that $\dfrac{\partial\phi(t,x)}{\partial{t}} = f(\phi(t,x))$. Observe that 


*

*$\phi$ depends on $(t,x)$.

*$t$ depends on $s$.


So, by the chain rule, we have the following observation (assuming t is fixed)
\begin{equation*}
\begin{split}
\dfrac{\partial\phi(t+s,x)}{\partial{s}}  & = \dfrac{\partial\phi(t+s,x)}{\partial{t}}\cdot\frac{d(t+s)}{ds}  \\
 & = \dfrac{\partial\phi(t+s,x)}{\partial{t}} \\
\end{split}
\end{equation*}
This implies that
$$\dfrac{\partial\phi(t+s,x)}{\partial{s}}\big|_{s=0}=\dfrac{\partial\phi(t,x)}{\partial{t}}$$
Therefore, we can apply the chain rule again to reduce the LHS to
\begin{equation*}
\begin{split}
\dfrac{\partial\phi(t,x)}{\partial{t}}  & = \dfrac{\partial\phi(t+s,x)}{\partial{s}}\big|_{s=0}  \\
 & = \dfrac{\partial\phi(s,\phi(t,x))}{\partial{s}}\big|_{s=0} \\
 & = \dfrac{\partial\phi(s,\phi(t,x))}{\partial{t}}\big|_{s=0}\cdot\frac{d}{ds}(s) \\
  & = \dfrac{\partial\phi(s,\phi(t,x))}{\partial{t}}\big|_{s=0} \\
 & = f(\phi(t,x)) 
\end{split}
\end{equation*}
For the second part, we need to show that $\dfrac{\partial{T(t,x)}}{\partial{t}} = Df(\phi(t,x))\hspace{0.5mm}T(t,x)$. Starting from the LHS, we can apply the chain rule to form
\begin{equation*}
\begin{split}
\dfrac{\partial{T(t,x)}}{\partial{t}} & = \dfrac{\partial\phi(t,x)}{\partial{x}\hspace{0.5mm}\partial{t}}  \\
 & = \dfrac{f(\phi(t,x))}{\partial{x}} \\
 & = D{f(\phi(t,x))}\hspace{0.5mm}\dfrac{\partial\phi(t,x)}{\partial{x}} \\
 & = D{f(\phi(t,x))}\hspace{0.5mm}T(t,x)
\end{split}
\end{equation*}
Remark: If $f\in{C^1}$ and is locally Lipschitz continuous, then $\frac{\partial{f(x_0)}}{x}$ is the Jacobian matrix at $x_0$. So, in our example, $Df(\phi(t,x))$ is the Jacobian matrix at $\phi(t,x)$.
