Consider a real-valued function $f: \mathbb{R}\rightarrow \mathbb{R}$. Suppose we are said that the second derivative exists and is bounded in a neighbourhood of $x\in \mathbb{R}$. Does it imply that also the first derivative exists and is bounded in a neighbourhood of $x\in \mathbb{R}$?
[Math] Implications of bounded second derivative
derivativesfunctions
Best Answer
In order to have a second order derivative which exists, you need to assume that the first one also exists (because the second order derivative is the derivative of the first one). So you can not conclude that the first order derivative exists, you need to assume it.
Considering the boundedness, yes it is correct ; as $f'$ admits a derivative in your neighborhood, it is continuous, and therefore is bounded in a smaller neighborhood.
Note that I didn't use the assumption that $f''$ is bounded; it is a useless assumption in your question.
Therefore, if $f$ admits a second order derivative in a neighborhood of $x_0$, then $f'$ is bounded in a (possibly smaller) neighborhood $x_0$. (it makes more sense to give this statement at the first order...)