# The existence of the $n$th derivative at $c$ presumes the existence of the $(n-1)$st derivative in an interval containing $c$

The following is from Introduction to Real Analysis by Bartle.

If the derivative $f'(x)$ of a function $f$ exists at every point $x$ in an interval $I$ containing a point $c$, then we can consider the existence of the derivative of the function $f'$ at the point $c$. In case $f'$ has a derivative at the point $c$, we refer to the resulting number as the second derivative of $f$ at $c$ and we denote this number by $f''(c)$. In similar fashion we define the third derivative $f^{(3)}(c)$, and the nth derivative $f^{(n)}(c)$, whenever these derivatives exist. It is noted that the existence of the $n$th derivative at $c$ presumes the existence of the $(n-1)$st derivative in an interval containing $c$, but we do allow the possibility that $c$ might be an endpoint of such an interval.

My question is, does the existence of the $n$th derivative necessarily guarantee that the $(n-1)$st derivative exists in an interval containing $c$? I understand that $c$ must be an accumulation point in the domain of the $(n-1)$st derivative, but I don't see why the domain necessarily have to be an interval. For instance, $f(x)=x^2$ defined on ${\frac{1}{n}|n\in \mathbb{N}}\cup {0}$ has a derivative at $0$ but the domain is clearly not an interval.

You are looking for a function which is differentiable at precisely one point. Such functions exist; for example let $$f(x)=\begin{cases}x^2&;x\in\mathbb{Q}\\-x^2&;x\notin\mathbb{Q}.\end{cases}$$ Then, $f$ is differentialble only at $0$.