Below is my reply to this in an old sci.math post.
Zachary Turner wrote on 26 Jul 2002:
Let D = d/dx = derivative wrt x. Then
D[x^2] = D[x + x + ... + x (x times)]
= D[x] + D[x] + ... + D[x] (x times)
= 1 + 1 + ... + 1 (x times)
Notice that an obvious analogous fallacious argument proves both
D[x f(x)] = Df(x) (x times) = x Df(x)
D[x f(x)] = Dx (f(x) times) = f(x), via Dx = 1
vs. the correct result: their sum f(x) + x Df(x)
as given by the Leibniz product rule (= chain rule for times).
The error arises from overlooking the dependence upon x in both
arguments of the product x * f(x) when applying the chain rule.
The source of the error becomes clearer if we consider a
discrete analog. This will also eliminate any tangential
concerns on the meaning of "(x times)" for non-integer x.
Namely, we consider the shift operator S: n $\to$ n+1 on polynomials p(n)
with integer coef's, where S p(n) = p(n+1). Here's a similar fallacy
S[n^2] = S[n + n + ... + n (n times)]
= S[n] + S[n] + ... + S[n] (n times)
= n+1 + n+1 + ... + n+1 (n times)
But correct is $\rm\: S[n^2] = (n\!+\!1)^2.\:$ Here the "product rule" is
simply S[fg] = S[f] S[g], not S[f] g, as above.
The fallacy actually boils down to operator non-commutativity.
On the space of functions f(x), consider "x" as the linear
operator of multiplication by x, so x: f(x) $\to$ x f(x). Then
the linear operators D and x generate an operator algebra
of polynomials p(x,D) in NON-commutative indeterminates x,D
since we have
(Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so Dx = xD+1
(Sn)[f] = S[nf] = (n+1)S[f], so Sn = (n+1)S ≠ nS
This viewpoint reveals the error simply as mistakenly
assuming commutativity of the operators x,D or n,S.
Perhaps something to ponder on boring commutes !