How to derive this inequality involving logarithms and Taylor series? Assume that it is given that,
$$|c^{-1/2}|z|^{(d-1)/4}e^{|z|/2} - t| \lesssim t \log^{-1/2}(t)\quad [1]$$
where $c>0$ and $d>0$ are some constants, $z:\mathbb{R}\to \mathbb{R}^d$ is a function of time $t\in \mathbb{R}.$ Next we also know that,
$$\left|\dot{v} + c\frac{z}{|z|}|z|^{-\frac{d-1}{2}} e^{-|z|}\right|\lesssim t^{-2}\log^{-1}(t).\quad [2]$$
Here $v$ is a function of time, $v:\mathbb{R}\to \mathbb{R}^{d}.$
The notation $X\lesssim Y$ means that there exists a fixed constant $C>0$ such that $X\leq C Y.$
I want to show that,
\begin{equation}
\left||\dot{v}| - t^{-2}\right|\lesssim t^{-2} \log^{-1}(t).\quad (*)
\end{equation}
I am first trying to show, 
$$\left|\dot{v} + \frac{z}{|z|}t^{-2}\right|\lesssim t^{-2}\log^{-1}(t).\quad (**)$$
To establish this estimate we must have something like this 
$$\left|c|z|^{-\frac{d-1}{2}}e^{-|z|}-t^{-2}\right|\lesssim t^{-2}\log^{-1}(t) \quad (***)$$
because then we can use triangle inequality to prove $(**).$ I tried to use $[1]$ to deduce $(***)$ however I am not able to get rid of the term $\log^{-1/2}(t).$ Thus, I would appreciate if someone could explain the following:


*

*Proof of $(**)\implies (*).$

*Proof of $[1]\implies (***).$
Edit:
This question follows from the remark 3.2.1 on page 17 of the article here.
 A: Let's start with some notations:
$$K(x):=c^{-1/2}x^{(d-1)/4}e^{x/2}\qquad w:=\dot{v}$$
I rephrase your question as simply:

Given $$|K(|z|)-t| \le \frac{C_1 t}{\sqrt{\ln(t)}}\quad(1)$$
$$\quad\left|w+\frac{z}{|z|}\frac1{K^2(|z|)}\right|\le\frac{C_2}{t^2\ln t}
\quad(2)$$
$$z\equiv z(t):\mathbb R\to\mathbb R^d \qquad w\equiv w(t):\mathbb R\to\mathbb R^d$$ 
  how to prove $$\left||w|-\frac1{t^2}\right|\le\frac{C_3}{t^2\ln t} \qquad(\star)$$ ?


To begin, let's simplify the inequalities.
$(\star)$:
$$\left||w|-\frac1{t^2}\right|\le\frac{C_3}{t^2\ln t} 
\quad\Leftrightarrow\quad \frac1{t^2}\left(1-\frac{C_3}{\ln t}\right)\le|w|\le\frac1{t^2}\left(1+\frac{C_3}{\ln t}\right)$$
In other words, we just want to obtain bounds on $|w|$.
Continue simplifying,
$(1)$:
$$|K(|z|)-t| \le \frac{C_1 t}{\sqrt{\ln(t)}}
\quad\Leftrightarrow\quad t\left(1-\frac{C_1}{\sqrt{\ln t}}\right)\le K(|z|)\le t\left(1+\frac{C_1}{\sqrt{\ln t}}\right)$$
$(2)$:
$$\left|w+\frac{z}{|z|}\frac1{K^2(|z|)}\right|\le\frac{C_2}{t^2\ln t}
\quad\Leftrightarrow\quad 
w\in\frac{C_2}{t^2\ln t}\cdot\mathcal B-\frac{z}{|z|}\frac1{K^2(|z|)}$$
where $\mathcal B$ is the closed unit ball in $\mathbb R^d$. Here, $k\mathcal B+\vec c$ represents the set created by multiplying every element in $\mathcal B$ by a scalar $k$ followed by adding $\vec c$ to every element in the 'scaled' $\mathcal B$.
Here, the strongest inequality on $|w|$ that can be deduced is
$$w\in\frac{C_2}{t^2\ln t}\cdot\mathcal B-\frac{z}{|z|}\frac1{K^2(|z|)}
\quad\implies\quad$$
$$\min_{\vec\beta\in \mathcal B}\left|\frac{C_2}{t^2\ln t}\vec\beta-\frac{z}{|z|}\frac1{K^2(|z|)}\right|
\le|w|\le
\max_{\vec\beta\in \mathcal B}\left|\frac{C_2}{t^2\ln t}\vec\beta-\frac{z}{|z|}\frac1{K^2(|z|)}\right| \qquad (\blacksquare)
$$
Obviously, the maximum is attained when $\vec\beta$ is parallel to $-\frac{z}{|z|}$, and minimum is reached when $\vec\beta$ is anti-parallel to $-\frac{z}{|z|}$ (of course $\left|\vec\beta\right|=1$ in both cases).
Together with
$$\frac{1}{t^2\left(1+\frac{C_1}{\sqrt {\ln t}}\right)^2}\le\frac1{K^2(|z|)}\le \frac{1}{t^2\left(1-\frac{C_1}{\sqrt {\ln t}}\right)^2}$$ 
we can obtain the minimum
$$\frac{1}{t^2\left(1+\frac{C_1}{\sqrt {\ln t}}\right)^2}-\frac{C_2}{t^2\ln t}\quad(\spadesuit)$$
and maximum
$$\frac{1}{t^2\left(1-\frac{C_1}{\sqrt {\ln t}}\right)^2}+\frac{C_2}{t^2\ln t}\quad(\heartsuit)$$

The bad news is: unfortunately these two bounds are weaker than the one we are attempting to prove.
To demonstrate this fact for the upper bound, suppose otherwise, i.e.
$$\begin{align}
\frac{1}{t^2\left(1-\frac{C_1}{\sqrt {\ln t}}\right)^2}+\frac{C_2}{t^2\ln t}&\le\frac1{t^2}\left(1+\frac{C_3}{\ln t}\right) \\ 
\frac{C_2-C_3}{\ln t}&\le 1-\left(1-\frac{C_1}{\sqrt {\ln t}}\right)^{-2}\\ 
\frac{C_2-C_3}{\ln t}&\le -\frac{C_1}{\sqrt{\ln t}-C_1}\cdot\frac{2\sqrt{\ln t}-C_1}{\sqrt{\ln t}-C_1}\\ 
C_3-C_2 &\ge \frac{C_1\ln t}{\sqrt{\ln t}-C_1}\cdot\frac{2\sqrt{\ln t}-C_1}{\sqrt{\ln t}-C_1}\\ 
\end{align}
$$
Taking $t\to\infty$, we require $C_3$ to be infinitely large, leading to a contradiction.
Similarly for the lower bound, if we suppose otherwise, i.e. 
$$\frac{1}{t^2\left(1+\frac{C_1}{\sqrt {\ln t}}\right)^2}-\frac{C_2}{t^2\ln t}\ge\frac1{t^2}\left(1-\frac{C_3}{\ln t}\right)$$ 
we finally get $$C_3-C_2\ge\frac{C_1\ln t}{\sqrt{\ln t}+C_1}\cdot\frac{2\sqrt{\ln t}+C_1}{\sqrt{\ln t}+C_1}$$ which clearly also contradicts the assumption.

As almost no approximations/loose bounds are applied above, I suspect the given inequalities are not sufficient to conclude $(\star)$. $(\spadesuit)$ and $(\heartsuit)$ are likely the best possible.
