How to evaluate $\frac{d}{dt} \left\{\frac{\partial F}{\partial x_1}(tx)\right\}$? Let $F:\mathbb{R}^3\to\mathbb{R}$ be of class $C^\infty$. Consider
$$t\in\mathbb{R}^* \mapsto \frac{\partial F}{\partial x_1}(tx),$$
with $x\in\mathbb{R}^3$.
I am in trouble understanding what is
$$\frac{d}{dt} \left\{\frac{\partial F}{\partial x_1}(tx)\right\}.$$
I would say that it is something like
$$\frac{\partial^2 F}{\partial x_i \partial x_1}(tx)\cdot x,$$
but I am not confident at all.
Could someone please help me with that?
Thank you.
 A: The Leibniz notation could be sometimes very confusing, as it is in this case. If we write something like $\frac{d}{d s}H$ for some function $H$, then it means that the domain of $H$ is one-dimensional, say $\mathbb{R}$. However if we write $\frac{\partial}{\partial s}H$ it means that the domain of $H$ is multidimensional, say $\mathbb{R}^n$ for some $n>1$.
However sometimes we can write $\frac{d}{d \mathbf{x}}H$ to denote the Fréchet derivative of a function $H$ who domain is a vector space, but this notation is not used very much and in the context it is understood when $\mathbf{x}$ refers to a vector or an scalar.
Thus if you write $\frac{d}{d t}\left[\frac{\partial}{\partial x_1}F(t\mathbf{x})\right]$ then the function on the brackets just depends on $t$, this means that $t\mathbf{x}$ must depends only on $t$, that is, that $\mathbf{x}$ depends on $t$ also. Using the notation $\frac{\partial}{\partial x_k}$ together with $\mathbf{x}=(x_1,\ldots ,x_n)$ means that $\mathbf{x}$ is a function of $t$, otherwise the notation is misleading, so we can write
$$
\frac{\partial}{\partial x_1}F(t\mathbf{x})=\left(\frac{\partial}{\partial x_1}F\circ g\right)(t),\quad  g(t):=t\mathbf{x}(t)=(tx_1(t),\ldots,tx_n(t))=(g_1(t),\ldots,g_n(t))
$$
Therefore by the chain rule
$$
\begin{align*}
\frac{d}{d t}\left[\left(\frac{\partial}{\partial x_1}F\circ g\right)(t)\right]&=\left(\frac{d}{d \mathbf{x}}\frac{\partial}{\partial x_1}F\circ g\right)(t)\frac{d}{d t}g(t)\\
&=\sum_{k=1}^n\left(\frac{\partial^2}{\partial x_k\partial x_1}F\circ g\right)(t)g_k'(t)\\
&=\nabla \left(\frac{\partial}{\partial x_1}F\right)(t\mathbf{x}(t))\cdot (\mathbf{x}(t)+t\mathbf{x}'(t))\tag{*}
\end{align*}
$$
where $\frac{d}{d \mathbf{x}}$, as discussed before, represents the Fréchet (total) derivative. When $\mathbf{x}$ is constant, say $\mathbf{x}(t):=\mathbf{v}$ for some fixed vector $\mathbf{v}$ for every $t$, then the notation and the context must explicitly show it.
However if we would write $\frac{\partial}{\partial t}\left[\frac{\partial}{\partial x_1}F(t\mathbf{x})\right]$ then the notation means that $t$ is a coordinate of $\frac{\partial}{\partial x_1}F(t\mathbf{x})$ so it can be understood something like
$$
\frac{\partial}{\partial x_1}F(t\mathbf{x})=\left(\frac{\partial}{\partial x_1}F\circ h\right)(\mathbf{x},t),\; \text{ for }(\mathbf{x},t)=(x_1,\ldots,x_n ,t)\;\text{ and }\, h(\mathbf{x},t):=t\mathbf{x}
$$
Therefore, again using the chain rule, we have that
$$
\begin{align*}
\frac{\partial}{\partial t}\left[\left(\frac{\partial}{\partial x_1}F\circ h\right)(\mathbf{x},t)\right]&=\frac{d}{d(\mathbf{x},t) }\left[\left(\frac{\partial}{\partial x_1}F\circ h\right)(\mathbf{x},t)\right]\mathbf{e}_{n+1}\\
&=\left(\frac{d}{d \mathbf{x}}\frac{\partial}{\partial x_1}F\circ h\right)(\mathbf{x},t)\frac{d}{d (\mathbf{x},t)}h(\mathbf{x},t)\mathbf{e}_{n+1}\\
&=\left(\frac{d}{d \mathbf{x}}\frac{\partial}{\partial x_1}F\circ h\right)(\mathbf{x},t)\frac{\partial }{\partial  t}h(\mathbf{x},t)\\
&=\sum_{k=1}^n \left(\frac{\partial^2}{\partial x_k\partial x_1}F\circ h\right)(t)\frac{\partial}{\partial t}h_k(t)\\
&=\nabla \left(\frac{\partial}{\partial x_1}F\right)(t\mathbf{x})\cdot \mathbf{x}\tag{**}
\end{align*}
$$
where $\frac{d}{d (\mathbf{x},t)}$ means the Fréchet (total) derivative of the function, and $\mathbf{e}_{n+1}:=(0,0,\ldots,0 ,1)\in \mathbb{R}^{n+1}$, that is, is the unit-length vector in the direction of the last coordinate of $(\mathbf{x},t)$. Note the difference between (*) and (**).
A: Generally for a function $G:\mathbb{R}^n\longrightarrow \mathbb{R}$ we have
\begin{align*}\frac{dG}{dt} = \sum_{i=1}^n\frac{\partial G}{\partial x_i}\frac{dx_i}{dt}\end{align*}
so in your case we have
\begin{align*}
\frac{d}{dt}\frac{\partial F}{\partial x_1}(tx) = \sum_{i=1}^n\frac{\partial^2 F}{\partial x_i\partial x_1}\frac{d(tx_i)}{dt} = \sum_{i=1}^n\frac{\partial^2 F}{\partial x_i\partial x_1}x_i = \nabla(\frac{\partial F}{\partial x_1})(tx)\cdot x
\end{align*}
