Derived numbers of $-f(-x)$ I read (p. 321 here) that, if we respectively call $\Lambda_R,\Lambda_L,\lambda_R$ and $\lambda_L$ the right upper, left upper, right lower and left lower derived numbers of the function $f$ in a certain point $x$, and $\Lambda_R^\ast,\Lambda_L^\ast,\lambda_R^\ast$, $\lambda_L^\ast$ the corresponding numbers of fuction $f^\ast$ defined by $f^\ast(x)=-f(-x)$.
I am not familiar with the notations for derived numbers and I do not find much information on line, but I find it useful to take the liberty, for which I apologise, to introduce the notation $\Lambda_Rf(x)$, $\Lambda_Lf(x)$, $\lambda_Rf(x)$, $\lambda_Lf(x)$ for the respective derived numbers of $f$ in $x$. By using the definition of left and right limits and the fact that $\sup_{x\in S}f(x)=-\inf_{x\in S}-f(x)$ I think that, for the functions defined by $g(x):=f(-x)$ and $h(x):=-f(x)$, the following equalities are almost trivial:$$\Lambda_Rg(-x)=-\Lambda_Lf(x),\quad\Lambda_Lg(-x)=-\Lambda_Rf(x),\quad\lambda_Rg(-x)=-\lambda_Lf(x), \quad\lambda_Lg(-x)=-\lambda_Rf(x),$$
$$\Lambda_Rh(x)=-\lambda_Rf(x),\quad\Lambda_Lh(x)=-\lambda_Lf(x),\quad\lambda_Rh(x)=-\Lambda_Rf(x), \quad\lambda_Lh(x)=-\Lambda_Lf(x).$$
If those are correct, I would say that $\lambda_Lf^\ast(-x)=\Lambda_R f(x)$ and $\Lambda_Rf^\ast(-x)=\lambda_L f(x)$, i.e., in Kolmogorov-Fomin's notation, $\lambda_L\ast=\Lambda_R ,\Lambda_R^\ast=\lambda_L$, which is different from what Kolmogorov-Fomin's says: $\lambda_L\ast=\lambda_R ,\Lambda_R^\ast=\Lambda_L$. Therefore I strongly suspect that I am wrong, although the probability that a text contains an error is not $0$.
What do you think about it? Could anybody point out where my reasoning fails (if it does, which is likely)? I heartily thank you!
To add a graphical interpretation I insert the plots of a function $f(x)$ with the straight lines through $x_0$ with slopes equalt to the derived numbers in $x_0$ $\Lambda_R=\Lambda_{пр}$, $\lambda_R=\lambda_{пр}$, $\Lambda_L=\Lambda_{лев}$, $\lambda_L=\lambda_{лев}$, and of the function $-f(2x_0-x)$.

EDIT: Changed the point where the derived numbers of $f^\ast$ are taken from $x$ to $−x$, thanks to ald.li's remarks.
 A: It seems that the book actually wants to define $f^*$ as $f^*(x) := -f(2x_0-x)$ where $x_0$ is the point around which the derived numbers are subsequently taken. (I.e., mirror the function graph wrt lines $x = x_0$ and $y = 0$.) If so, then the claimed equalities $\lambda_L^* = \lambda_R$ and $\Lambda_R^* = \Lambda_L$ hold.
Indeed, note that $x\to x_0+$ is equivalent to $2x_0-x\to x_0-$ and
$$ \frac{f^*(x) - f^*(x_0)}{x-x_0} = \frac{-f(2x_0-x)-(-f(x_0))}{x-x_0}=
\frac{f(2x_0-x)-f(x_0)}{(2x_0-x)-x_0}.$$ Then 
$$\Lambda_R^* = \limsup_{x\to x_0+} \frac{f^*(x) - f^*(x_0)}{x-x_0} = \limsup_{2x_0-x \to x_0-}\frac{f(2x_0-x)-f(x_0)}{(2x_0-x)-x_0} = \Lambda_L.$$
For one way to imagine this let's say that $f$ is differentiable in the neighbourhood of $x_0$ (without $x_0$). Then $(f^*)'(x) = f'(2x_0-x)$ there (i.e. the graph of $(f^*)'$ is the mirror image of the graph of $f'$ wrt line $x=x_0$). It is clear now that $\limsup_{x \to x_0+} (f^*)'(x) = \limsup_{x\to x_0-}f'(x)$.
Note that the definition $f^*(x) := -f(-x)$ would not work as $f$ might not actually be defined at $-x$ or be differentiable and have a rather different derivative there. Take, e.g., $f = \int (x-2)^2 dx$.
