Estimate for derivative of a function I'm self studying analytic number theory and I wonder how the proof goes for this problem: If $f (x) $ satisfies $f (x)=x^2+O (x) $, and $f $ is differentiable with nondecreasing derivative $f'(x) $ for sufficiently large $x $, then $f'(x)=2x+O (\sqrt{x}) $.
I have a proof but I did not use the fact that $f'$ is nondecreasing so I bet that my proof is wrong. Any help will be greatly appreciated.
It can be shown that for sufficiently large x, $f'(x)=2x+O (1) $. Basically, I just used the hypothesis and the limit definition of the derivative. So now, by dividing by $\sqrt {x} $, we obtain $\dfrac {f'(x)}{\sqrt{x}}=2\sqrt {x}+O \left(\dfrac {1}{\sqrt {x}}\right)=2\sqrt {x}+O (1) $. So multiplying by $\sqrt {x} $ we get the desired result. But I'm pretty sure that there is a flaw in my proof.
 A: It holds $$x^m = \mathcal{O}(x^n),\qquad m≤n, \qquad x→∞,$$
because of $$\frac{x^m}{x^n} = x^{m-n} →\begin{cases}0 &m≤n\\1 & m=n \\ ∞ &m≥n\end{cases}\quad \text{for } x→∞.$$
Therefore if $g(x)=\mathcal{O}(1)=\mathcal{O}(x^0)$ it follows $g(x)=\mathcal{O}(x^{\frac{1}{2}})$ for $x→∞$.
This should help you proof the desired result, with what you already have done so far.
The information $f'$ to be nondecreasing is an information, which is not necessary for the proof. IMO that can be best seen by the following argument, instead of a rigorous proof.  
If $f(x)=x^2+\mathcal{O}(x)$, then it is $f'(x)=x+\mathcal{O}(1)$. Let $f'(x)=h(x)+g(x)$, with $h(x)=x$ and $g$ being that function that behaves like $\mathcal{O}(1)$. 
This means
$$\frac{g(x)}{1}→c∈ℝ,\qquad x→∞.$$
And that means, that the governing part of $f'$ is $h(x)=x$, because $g$ can not grow as fast as $h$ does. So for sufficiently large $x$ the behaviour of $f'$ will be like $h$, and $h$ is nondecreasing.

Edit:
I somehow missed the last equation of your proof. What you did is correct, since you used what I stated in this answer, in your last "="-step.
