TITLE: Help in understanding why the arithmetic derivative is well-defined.
QUESTION [18 upvotes]: I'm reading into the arithmetic derivative at the moment and I just don't get why it is well-defined. For reference, Ufnarovski tried explaining it here on page 2. I would like to understand the not explicitly stated induction. Thanks in advance!

REPLY [2 votes]: Unique prime factorization and the Leibniz rule are enough.  You can induct on the number of prime factors or on the total number of factors.  As an example, suppose $n=p^3q^2r$ with $p,q,r$ prime.  You can expand it to $n=pppqqr$ and use Leibniz to get $$n'=p'ppqqr+pp'pqqr+ppp'qqr+pppq'qr+pppqq'r+pppqqr'=3p^2q^2r+2p^3qr+p^3q^2$$
The point is that any intermediate decomposition of $n$ can be further decomposed into the unique set of prime factors.  
If you look at the formula for the arithmetic derivative in the article, you can just use strong induction on $n$.  Assume $n=pq$, with $p,q$ not assumed to be prime.  If you know the derivatives of $p,q$ are given by the formula because they are smaller than $n$, you can show the formula holds for $n$.