Show that if $f(x)=x^2+O(x)$, and $f$ is differentiable with non-decreasing derivative $f'(x)$, then $f'(x)=2x+O(\sqrt{x})$.
I know that if $f'$ is not non-decreasing, then the statement is not true. For example, consider the function $f(x)=x^2+\sin(x^2)$. I don't how to use the hypothesis that $f'$ is non-decreasing.
Let $a<1<\beta$ be real numbers whose value we will choose later. Since $f'$ is non-decreasing, we have the following inequality: $$\frac{1}{(1-\alpha) x}\int_{\alpha x}^x f'(y)\,dy\leq f'(x)\leq \frac{1}{(\beta-1)x}\int_{x}^{\beta x} f'(y)\,dy.$$ By the fundamental theorem of calculus, we get $$\frac{f(x)-f(\alpha x)}{(1-\alpha )x}\leq f'(x)\leq \frac{f(\beta x)-f(x)}{(\beta-1)x}.$$ Using the assumption $f(x)=x^2+O(x)$ to obtain $$(1+\alpha)x+O(\frac{1}{1-\alpha})\leq f'(x)\leq (\beta+1)x+O(\frac{1}{\beta-1}).$$ Now we choose $\alpha=1-\frac{1}{\sqrt{x}}$ and $\beta=1+\frac{1}{\sqrt{x}}$, then we get $$2x+O(\sqrt{x})\leq f'(x)\leq 2x+O(\sqrt{x}),$$ thus we have $f'(x)=2x+O(\sqrt{x})$.