# Can we say that, there is a neighborhood of $x_0$ such that, $f$ is differentiable in all points of neighborhood?

Let $f:\mathbb{R} \to \mathbb{R}$, and $f$ is differentiable in $x_0$.

Can we say that, there is a neighborhood of $x_0$ such that, $f$ is differentiable in all points of this neighborhood?

Which conditions say that this question is true?

• What have you tried already? How might one prove such a statement? What might a counter example look like? – Dan Robertson Jun 12 '16 at 23:53
• The first question is a duplicate. The second is interesting. Continuity of $f$ in a neighbourhood of $x_0$ is not sufficient considering $x^2 \cdot g(x)$ where $g(x)$ is the Weierstrass function. – MathematicsStudent1122 Jun 13 '16 at 0:31

No. Standard example: Let $f(x)=x^2$ for rational $x$, $f(x)=0$ for irrational $x$. Then $f$ is differentiable at $0$ but it is not even continuous at any other point.