Under what conditions is it true that if $f(x) \sim g(x)$ as $x \rightarrow x_0$, then $f'(x) \sim g'(x)$ as $x \rightarrow x_0$? So we say that two functions $f(x)$ and $g(x)$ of a real variable are asymptotic to each other as $x \rightarrow x_0$ ($x_0$ need not be finite) if
$$ \lim_{x \rightarrow x_0} \frac{f(x)}{g(x)} =1,$$
and in that case we write $f(x) \sim g(x)$ as $x \rightarrow x_0$. It is also not true generally that if $f(x) \sim g(x)$, then $f'(x) \sim g'(x)$. To see this, consider the cases $f(x) = x + \sin(x)$ and $g(x) = x$. In this cases it is true that $x + \sin(x) \sim x$ as $x \rightarrow \infty$, yet it is not true $1 + \cos(x) \sim 1$ as $x \rightarrow \infty$ (in fact the latter limit does not even exist).
My question is, under what conditions on $f(x)$ and $g(x)$ can we say that it is true that if $f(x) \sim g(x)$, then $f'(x) \sim g'(x)$ as $x \rightarrow x_0$.
 A: This is not a complete answer, but I fear the extra assumptions need to be too strong to be able to conclude something of this kind at this level of generality.
Indeed, if you know a bit of set theory, the asymptotic equivalence is an equivalence relation, and as such, it may be seen as a particular "equality" between functions, that tries to put together all of them in specific classes: like as if we take a great magnifying glass and we focus our attention really close to a particular point. In a nutshell, I may have a very monster function of which I may not be able to study its graph globally, so I try to use asymptotic equivalences to try to reduce, at least locally, the function to an easier one to be able to see its behaviour close to a point. For example, take $$f(x)=\sin(\log(1 + \sinh(e^{\arctan(x^3)}-1))) $$ which is asymptotically equivalent for $x \rightarrow 0$ to $$g(x)= x^3$$
But then, to be able to say their derivatives are also asymptotically equivalent for $x \rightarrow 0$, I think one must add a very restrictive assumption on the behavior of the functions in the neighborhood of zero, probably so much that the functions end up to be the same, at least in that neighborhood and at least for the first-order expansion around that specific point (but I do not have a proof of this, it is just my intuition).
Indeed, asymptotic calculus is not compatible with derivatives: your example is nice, as it is the following: $$f(x)=1+2x\sim_01\sim_0g(x)=1+x,$$ but $$f'(x)=2\nsim_0g'(x)=1$$
A: If we assume $f$ and $g$ are analytic near $0$ (replacing $x$ by $x - x_{0}$ if $x_{0}$ is finite), we may expand in power series
\begin{align*}
  f(x) &= a_{m}x^{m} + a_{m+1}x^{m+1} + \cdots
  = \sum_{k=m}^{\infty} a_{k}x^{k}, \\
  g(x) &= b_{n}x^{n} + b_{n+1}x^{n+1} + \cdots
  = \sum_{k=n}^{\infty} b_{k}x^{k},
\end{align*}
with $a_{m}$ and $b_{n}$ non-zero. Consequently, $f \sim g$ if and only if $m = n$ and $a_{m} = b_{m}$, i.e., the power series have the same first non-vanishing term.
Termwise differentiation gives
\begin{align*}
  f'(x) &= ma_{m}x^{m-1} + (m + 1)a_{m+1}x^{m} + \cdots
  = \sum_{k=m-1}^{\infty} (k + 1)a_{k+1}x^{k}, \\
  g'(x) &= mb_{m}x^{m-1} + (m + 1)b_{m+1}x^{m} + \cdots
  = \sum_{k=m-1}^{\infty} (k + 1)b_{k+1}x^{k}.
\end{align*}
If $m = 0$, i.e., $f(0) = g(0) \neq 0$, then $f' \sim g'$ if and only if $f$ and $g$ have the same first non-constant term, i.e., $f - f(0) \sim g - g(0)$.
If $m > 0$, i.e., $f$ and $g$ vanish to the same finite order at $0$, then $f' \sim g'$ is automatic since
\begin{align*}
  f'(x) &= ma_{m}x^{m-1}\bigl[1 + \frac{(m + 1)a_{m+1}}{ma_{m}} x + \cdots\bigr], \\
  g'(x) &= ma_{m}x^{m-1}\bigl[1 + \frac{(m + 1)b_{m+1}}{ma_{m}} x + \cdots\bigr],
\end{align*}
whose ratio has limit $1$ at $0$.
Modifications for the case $x_{0} = \infty$ are straightforward.
