Free groups and derivative Per definition a derivative on a group $G$ is a mapping $D:G\rightarrow\mathbb{Z}G$ such that $D(gh)=D(g)+gD(h)$. Now my question:
uppose $G$ is a free group $F=F(X)$ with $X$ a finite set and suppose $D$ is any derivative. Why is $D$ completely determined by the values $D(x)$ (for $x\in X$) that it takes on the generators. Why holds the formula:$$D(w)=\sum_{x\in X}{a_x(w)D(x)},\ \ \ a_x(w)\in\mathbb{Z}F$$ and what are the coefficients? Here $\mathbb{Z}F$ is the groupring of $F$. Can someone help me with this question?
For more about a derivative defined above:


*

*$D(g^{-1})=-g^{-1}D(g)$ 

*$D(g^n)=\frac{g^n-1}{g-1}D(g)$


Because, $F$ is free on the set $X$, there are no relations among words (beyond those that make it a group:  associativity, identity, inverses).  You can prove your formula by induction on the length of the word $w$.
First, can you show that $D(1) = 0$, where $1 \in F$ is the identity?  Hint: $1 = 1 \cdot 1$.
While you're at it, why not observe that a word of length $1$ is just $w = x$ or $w = x^{-1}$ for $x \in F$.  Do you see why $D(w)$ is of the correct form?
Now, for the inductive step, any word can be written
$$
w = w_1 \cdots w_{n - 1} \cdot w_n = (w_1 \cdots w_{n - 1}) \cdot w_n.
$$
Use the Leibniz rule and the inductive hypothesis, and you've got your result.  Can you see how to finish it?
