From the book of Kaplansky, A derivation of a ring $R$ is an additive mapping $r\rightarrow r'$ of $R$ ino itself satisfying $(ab)'=a'b+ab'$. We write $a'',a''',...,a^{(n)}$ for the successive derivatives.
Then, may I say, a derivation is only a differential function which satisfy additive endomorphism and Leibnitz's rule?
then, how if the derivative of an element of a ring does not exist? or how if the derivatives of an element of a ring is not in the ring itself?
notice Lie product, if a derivation $d$ in ring $R$ satisfy $d([x,y])=[x,y]$, it means $d$ satisfy additive endomorphism and $d(xy)=d(x)y+xd(y)$ for all $x,y \in R$ plus additional property which is $d[x,y]=d(xy-yx)=[x,y]=xy-yx$, right?
as usual, any comment and answer will be very appreciated!
A derivation of $R$ in itself is by definition a map $R\to R$ which, as you said, is a group homomorphism satisfying the Leibniz rule.
Note that such a map always exists (take the zero map for example). Existence of further non trivial derivations is I guess what you are asking about when you say "what if the derivative of an element of $R$ does not exist". But note that the term "derivative of an element" without mentioning the derivation (the map) does not make sense in general: $R$ is just an abstract ring.
This is a particular case of a more general notion, namely the notion of an $A$-derivation of an $A$-algebra $B$ into a $B$-module $M$. This is just an $A$-linear map which satisfies the Leibniz rule. In your case, $A=\mathbb{Z}$ and $B=M=R$.
This notion is closely related to the notion of Kähler differentials, which is an algebraic analog (and in fact a generalization) of the notion of differential used to define tangent spaces in smooth manifolds.