Here's my explanation from 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)
= x
```

An obvious analogous fallacious argument proves both

vs. the correct result: their sum $\rm\:f(x) + x\, Df(x)\:$
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 $\rm\: x \ f(x)\:$ 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 $\rm\ S:\, n \to n+1\ $ on polynomials $\rm\:p(n)\:$ with integer coefficients, where $\rm\:S p(n) = p(n+1).\:$ Here is a similar fallacy

```
S[n^2] = S[n + n + ... + n (n times)]
= S[n] + S[n] + ... + S[n] (n times)
= 1+n + 1+n + ... + 1+n (n times)
= (1+n)n
```

But correct is $\rm\ S[n^2] = (n+1)^2.\:$ Here the "product rule" is
$\rm\ S[fg] = S[f]\, S[g],\ $ not $\rm\: S[f] g\:$ as above.

The fallacy actually boils down to operator *noncommutativity.*
On the space of functions $\rm\:f(x),\:$ consider "x" as the linear
operator of multiplication by x, so $\rm\ x:\, f(x) \to x f(x).\:$ Then
the linear operators $\rm\:D\:$ and $\rm\:x\:$ generate an operator algebra
of polynomials $\rm\:p(x,D)\:$ in NON-commutative indeterminates $\rm\:x,D\:$
since we have

```
(Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so Dx = xD + 1 ≠ xD
(Sn)[f] = S[nf] = (n+1)S[f], so Sn = (n+1)S ≠ nS
```

This view reveals the error as mistakenly
assuming commutativity of the operators $\rm\:x,D\:$ or $\rm\:n,S.$

Perhaps something to ponder on boring *commutes* !