Having trouble understanding condition for Rolle's Theorem (russian translation)

$f$ having a finite derivative in $\xi$ means that it has a derivative in $\xi$ in the usual sense: $f'(\xi) = \lim_{x\to \xi} \dfrac{f(x) - f(\xi)}{x-\xi}$ is a real number.

$f$ having a definite sign infinite derivative means that $f'(\xi) = \lim_{x\to \xi} \dfrac{f(x) - f(\xi)}{x-\xi} = \infty$ or $f'(\xi) = \lim_{x\to \xi} \dfrac{f(x) - f(\xi)}{x-\xi} = -\infty$. This is usually denoted as an improper derivative at $\xi$. Geometrically it means that that the graph of $f$ has a vertical tangent at the point $(\xi,f(\xi))$. An example is $f(x) = \sqrt[3]{x}$. You have $f'(0) = \infty$.

For example $y=x^{1/3},$ a continuous function on all of $\mathbb R,$ is not differentiable at $0$ in the usual sense. But it has a "definite sign infinite derivative" of $+\infty$ at $0.$ I.e., both the left and right derivatives of $x^{1/3}$ at $0$ are $+\infty.$ Contrast this with $y=\sqrt {|x|},$ which has left and right infinite derivatives at $0$ of $-\infty,+\infty$ respectively.

I believe English definition you are referring to is the standard Rolle's theorem, while Russian is the generalized theorem.