# Function is differentiable at a point so its differentiable in a region from the point

Given some function $$f: I \subseteq\mathbb R \rightarrow \mathbb R$$, Which is differentiable twice at some point $$a\in I$$. Can one say that there is a region around the point where the function is differentiable twice, without any other information?

so I assume that its true because if we look at the first derivative which is:

$$lim_{h\rightarrow0} \frac {f(a+h)-f(a)}h$$ then obliviously we can "take" h to be smaller as we want and then I can assume that if the limit exists then it exists at some region of that point a.

so I guess the same goes for the second derivative.

• How do you even define twice differentiable in a single point? If we have a function that is only differentiable in a single point, than the first derivative is only defined in this single point, and taking the derivative of that again is kind of useless. – Dirk Aug 14 at 11:29
• @Dirk I didn't defined the function to be differentiable only at a single point, I defined some function which can be differentiate twice in some point a. – Serlok Aug 14 at 11:32

## 1 Answer

Let $$f\colon\mathbb R\longrightarrow\mathbb R$$ a continuous function which is differentable nowhere, let $$F$$ be a primitive of $$f$$ and let $$g(x)=x^2F(x)$$. Then $$g$$ is differentiable: $$g'(x)=2xF(x)+x^2f(x)$$. And $$g'$$ is differentiable at $$0$$. But that's the only point of $$\mathbb R$$ at which $$g'$$ is differentiable.

• But because differentiable is defined by a limit, then if we can say the limit exists for 0, cant we say the limit exists in a region around zero? – Serlok Aug 14 at 11:45
• I am puzzled by your comment. Did I not prove that that does not occur? Short answer: no, you cannot say that. – José Carlos Santos Aug 14 at 11:48
• +1. Nice construction. – Kavi Rama Murthy Aug 14 at 11:48
• @KaviRamaMurthy Thank you. – José Carlos Santos Aug 14 at 11:50