Let be a commutative ring and let . Recall that the derivative is defined to be . Prove that for all , , , and (for good measure) .
Let and . (For ease of exposition, pad or with zero terms so that they have the same nominal degree.)
We will now prove the ‘product rule’ in pieces. First, for monic monomials.
Lemma 0: If , then . Proof: .
Lemma 1: For all , we have . Proof: If , then as desired. Similarly for . Now .
Lemma 2: For all and , . Proof: We have as desired.
Lemma 3: For all , . Proof: We have as desired.
Now recall that if , then .
Lemma 4: For all , . Proof: We proceed by induction on . For and , the result is clear. (Note that .) Suppose the result holds for some . Now as desired.
Lemma 5: For all , . Proof: Let , , and , padding to make the degrees nominally . Then .
Lemma 6: For all , . Proof: We have as desired.