Must the 1st derivative of phi be undetermined at V=infinity

[Happiness]

I think it is not true that a discontinuous $\nabla^2\psi$ implies a discontinuous $\nabla\psi$, because a continuous function can have a discontinuous derivative, eg. $y=|x|$.

Is it true that $\nabla\psi$ must always be undetermined at the boundary where $V=\infty$?

Attached below is the whole paragraph that the sentence appears in:

 

[DrClaude]

I think it is not true that a discontinuous $\nabla^2\phi$ implies a discontinuous $\nabla\phi$, because a continuous function can have a discontinuous derivative, eg. $y=|x|$.

That's not what it says. It says that the discontinuity of $\nabla^2\psi$ is the same as that of the potential, hence is a discontinuity of the second kind (infinite jump). Therefore the first derivative will be discontinuous.

 

[RUber]

While it is true that a discontinuous second derivative does not imply a discontinuous first derivative, it is true that a continuous function can not have an infinite derivative.

 

[Samy_A]

While it is true that a discontinuous second derivative does not imply a discontinuous first derivative, it is true that a continuous function can not have an infinite derivative.

Is there an implicit condition for this to be true?
Take $f(x)=x^{1/3}$. This function is continuous, but the first derivative in x=0 is infinite.

 

[RUber]

Thanks, Samy...I spoke too soon.