Let $f:\mathbb{R}^+\to\mathbb{R}^+$ and $g:\mathbb{R}^+\to\mathbb{R}^+$ be two strictly concave, strictly increasing, twice differentiable functions, such that $f(x)=O(g(x))$ as $x\to\infty$, i.e. there exists $M>0$ and $x_0$ such that $$f(x)\leq Mg(x)\qquad \forall x\geq x_0.$$ Is it true that $f'(x)=O(g'(x))$ as $x\to\infty$?
(this is an extensions of this question)
No. We first construct a function $h:[-1,\infty)\rightarrow\mathbb{R}$ as follows: For integers $n \in \{-1, 0, 1, 2, ...\}$ we want $h$ to be: $$ h(n) = \frac{1}{(n+2)!} \quad \forall n \in \{-1, 0, 1, 2, ...\} $$ These values are positive and strictly decreasing. So, we can interpolate them to get the full function $h(x)$ that is differentiable, positive, and satisfies $h'(x)<0$ for all $x \geq -1$.
Define $g:[-1,\infty)\rightarrow\mathbb{R}$ by $g(x) = \int_{-1}^{x} h(t)dt$. Then $g'(x)=h(x)>0$ and $g''(x)=h'(x)<0$ for all $x\geq -1$. So $g$ is increasing, twice differentiable, and strictly concave.
Now define $f:[0,\infty)\rightarrow\mathbb{R}$ by $f(x) = g(x-1)$. Then $$ f(x) = g(x-1) \leq g(x) \quad \forall x \in [0,\infty)$$ Furthermore, for integers $n \in \{1, 2, 3, ...\}$ we get: $$ \frac{f'(n)}{g'(n)} = \frac{g'(n-1)}{g'(n)} = \frac{1/(n+1)!}{1/(n+2)!}\rightarrow \infty $$ and so the conjecture $f'(x) = O(g'(x))$ fails.