Does $f(x)=O(\log x)\implies f'(x)=O(1/x)$? Does $f(x)=O(\log x)\implies f'(x)=O(1/x)$ ?
My attempt:
Since $f(x)=O(\log x)$, then there exists $C>0$ such that $f(x)<C\log x$ for sufficiently large $x$.
What I'm proposing is that $f'(x)=O(1/x)$, but this would mean there exists $C>0$ such that
$$f'(x)<\frac{C}{x},$$
and I don't think differentiation works in general over inequalities... in which case, can we say anything at all, perhaps with some assumptions on $f$ if necessary?
Update
In response to Surb's comment below, whilst differentiation does not generally hold over inequalities, integration does, e.g. suppose $f(x)=O(1/x)$, then $$\int_1^yf(x)dx=O(\log y).$$
Proof: if $f(x)=O(1/x)$, then there exists $C>0$ such that $$f(x)<\frac{C}{x}\implies \frac{C}{x}-f(x)>0,\tag{1}$$ for $x>x_0$ where $x_0$ is sufficiently large. Then
$$\int_1^y\left(\frac{C}{x}-f(x)\right)dx>0\implies C\log(y)-\int_1^yf(x)dx>0,$$
as a result of inequality (1). Hence, $$\int_1^yf(x)dx<C\log(y)\implies\int_1^yf(x)dx=O(\log y).$$
 A: It seems very difficult to obtain such conditions (now that I said it, I am sure that someone will prove me wrong, but anyway): Consider the functions $f_k(x)=\sin{(kx)}$ for some $k\in \mathbb N$. Then $f_k(x)=O(\log{x})$ for any $k\in \mathbb N$, however $$f'_k(x)=k\cos{(kx)}$$ which may take any value between $-k$ and $k$. 
A: What you want doesn't hold. Any function that behaves like $O(\log x)$ but has very small but increasingly violent oscillations will not be $O(x^{-1})$.
A: I will try to address the second part of your question. I can only show the $``\impliedby"$ direction instead under various conditions on $f, f', g, g'$. I hope I get the conditions right.
Suppose $f, g:(c, +\infty) \to \mathbb{R}$ are differentiable such that
$$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} g(x) = 0 \text{ or} \pm\infty$$
and $g, g' \ne 0$ on $(c, +\infty)$ and
$$\lim_{x \to +\infty} \frac{f'(x)}{g'(x)} = L$$
for some $L \in \mathbb{R}$. [Observe the last two conditions $\implies f' = \mathcal{O}(g')$ by the $\limsup$ characterization of big $\mathcal{O}$]
Then by L'Hôpital's rule,
$$\lim_{x \to +\infty} \frac{f(x)}{g(x)} = \lim_{x \to +\infty} \frac{f'(x)}{g'(x)} = L$$
Using the $\limsup$ characterization of big $\mathcal{O}$, 
since $$\limsup_{x \to +\infty} \left|\frac{f(x)}{g(x)}\right| = \lim_{x \to +\infty} \left|\frac{f(x)}{g(x)}\right| = \left|\lim_{x \to +\infty} \frac{f(x)}{g(x)}\right| = |L| \in \mathbb{R}$$
by continuity of absolute value,
we have $$f = \mathcal{O}(g)$$
A: No, this is not the case. Consider $f(x) = \log x + \sin x$, which is clearly $O(\log x)$, but $f'(x)$ is not $O(1/x)$.
If you $f$ to be monotonous, then consider the constinuous function where if $x$ is not in an interval $[10^n-1, 10^n]$, then it is constant, while if $x$ is in such an interval, it's increasing linearly from $n-1$ to $n$. You can even smooth out the corners at points of the form $10^n-1$ and $10^n$ such that the function is differentiable everywhere, and the derivative still attains a maximum value of $1$ arbitrarily far out on the number line.
A: This is not true. Consider $f(x)=\sin x$. Then with $C=1$, you can see that $f(x)=\mathcal{O}(\log x)$, but clearly, as $x\rightarrow \infty$, $\frac{1}{x}\rightarrow 0$, but $\cos x$ does not go to zero. 
In other words $f'(x)$ is not $\mathcal{O}(\frac{1}{x})$.
As to whether one can say anything at all about the order of $f'$,...No..not by just looking at $f$. One needs to look at $f'$ again. (Or in the terms you put it, any assumptions on f, cannot avoid stating something about $f'$)
A: Consider $f(x)=\sin(x^{2})$. Then $f(x)=O(1)$ so $f$ is certainly $O(\log(x))$. But
$$f'(x) = 2x\cos(x^{2}) = O(x)$$
which is certainly not $O(x^{-1})$. It is a general principle that bounded functions can oscillate very heavily (imagine any curve you like, perturbed by lots of small-amplitude but high-frequency wiggles). 
