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, $$\int_1^yf(x)dx<C\log(y)\implies\int_1^yf(x)dx=O(\log y).$$

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). 
