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 $x \rightarrow x_0$.

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.
