Differentiation of scalar fields using tensor notation I'm learning tensor calculus to understand differential geometry. Please verify if I've understood how to employ Einstein's sum convention and index notation correctly.
Suppose that $\varphi := \varphi(x^1,\cdots,x^n)$ is a smooth scalar field, i.e. $\varphi: \mathbb{R}^n \to \mathbb{R}$ is a function that is infinitely differentiable.
The first derivative of $\varphi$ is calculated in this way, by using the chain rule:
$$\dfrac{d}{dt}\varphi = \varphi_i \dfrac{d}{dt}x^i$$
Where $i$ is a dummy variable and it's summed over from $1$ to $n$.
Here's my attempt to calculate the second derivative of $\varphi$:
$$\dfrac{d^2}{dt^2} \varphi = \dfrac{d}{dt}(\dfrac{d\varphi}{dt}) = (\varphi_i \dfrac{d}{dt}x^i)_j \dfrac{d}{dt}x^j = \left( \varphi_{ij}\dfrac{d}{dt}x^i+\underbrace{\varphi_i (\dfrac{d}{dt}x^i)_j}_{(*)} \right) \dfrac{d}{dt}x^j $$
But I believe we have $\varphi_i \left(\dfrac{d}{dt}x^i\right)_j= \varphi_i  \dfrac{d}{dt} \delta_{ij}$, therefore:
$$\dfrac{d^2\varphi}{dt^2}=\left( \varphi_{ij}\dfrac{d}{dt}x^i+\underbrace{\varphi_i \delta_{ij} \dfrac{d}{dt}}_{(*)} \right) \dfrac{d}{dt}x^j = \varphi_{ij}\dfrac{dx^i}{dt}\dfrac{dx^j}{dt}+\varphi_i \delta_{ij} \dfrac{d^2 x^j}{dt^2} = \varphi_{ij}\dfrac{dx^i}{dt}\dfrac{dx^j}{dt}+\varphi_j \dfrac{d^2 x^j}{dt^2}$$
The reason that I think $(\dfrac{d}{dt}x^i)_j = \dfrac{d}{dt}\delta_{ij}$ holds is this:
$$(\dfrac{d}{dt}x^i)_j = \dfrac{\partial}{\partial x^j}(\dfrac{d}{dt}x^i) = \frac{d}{dt}(\dfrac{\partial}{\partial x_j}x^i)= \dfrac{d}{dt} \delta_{ij}$$
I don't know why we can swap $\dfrac{\partial}{\partial x^j}$ and $\dfrac{d}{dt}$ though.
Are my calculations correct so far? (I want to calculate the third derivative of $\varphi$ as well, but I'll do it under my question as an answer because my question is getting too long).

EDIT:
I'm making the following assumptions in my calculations:


*

*$\dfrac{d}{dt}$ is a derivative operator and it acts to anything that is written right to it. Therefore $\dfrac{d}{dt} \space f = \dfrac{df}{dt}$ 

*I'm assuming that $\dfrac{\partial}{\partial x^j}\dfrac{d}{dt} = \dfrac{d}{dt}\dfrac{\partial}{\partial x^j}$
Using these two assumptions, we have:
$$(\dfrac{d}{dt}x^i)_j = \dfrac{d}{dt}(\dfrac{\partial}{\partial x^j} x^i) = \dfrac{d}{dt} \delta_{ij} = \delta_{ij} \dfrac{d}{dt}$$
The last equality holds because of linearity of $\dfrac{d}{dt}$. 
So, we have $(\dfrac{d}{dt}x^i)_j= \delta_{ij} \dfrac{d}{dt}$ as operators. Is that wrong?
 A: I do not fully understand your notation, when you write $(\varphi_i\frac{dx_i}{dt})_j$; using a more standard notation and  $\varphi=\varphi(x_1(t),\cdots,x_n(t))$ I get
$$\frac{d\varphi}{dt}=\sum_i \frac{d\varphi}{dx_i}\frac{dx_i}{dt} $$
and
$$\frac{d^2\varphi}{dt^2}= \sum_i \frac{d}{dt}\left(\frac{d\varphi}{dx_i}\right)\frac{dx_i}{dt}+\sum_i \frac{d\varphi}{dx_i}\frac{d^2x_i}{dt^2}=
\sum_i\sum_j \frac{d^2\varphi}{dx_idx_j}\frac{dx_j}{dt}\frac{dx_i}{dt}+\sum_i \frac{d\varphi}{dx_i}\frac{d^2x_i}{dt^2}.
$$
A consideration on the  notation $(\cdots)_j$ in the OP (add on).
In the OP $(\cdot)_j$ denotes $\frac{\partial}{\partial x_j}$. Then
$$(\frac{dx_i}{dt})_j =0$$
as $\frac{dx_i}{dt}$ does not depend on $x_j$, for all $i$' and $j$, but only on the parameter $t$ if $i\neq j$.
If $i=j$, then $g_j:=\frac{dx_j}{dt}$, with $g_j=g_j(x_j,t)$. We have
$$\frac{dg_j}{dt}=\frac{\partial g_j}{\partial x_j}\frac{dx_j}{dt}, $$
or $$\frac{\partial g_j}{\partial x_j}=\frac{dg_j}{dt}\frac{1}{\frac{dx_j}{dt}}=
\frac{d^2x_j}{dt^2}\frac{1}{\frac{dx_j}{dt}}$$
if $\frac{dx_j}{dt}\neq 0$ for all $t$. 
In summary
$$(\frac{dx_i}{dt})_j =\delta_{ij}\frac{d^2x_i}{dt^2}\frac{1}{\frac{dx_i}{dt}}. $$
