RefacteThe product rule can be made into exact arithmetic. Replace a small displacement by a symbol whose square is zero. Then change that arithmetic to keep one more order, and the term that disappeared becomes a test of whether a motion can continue.
1. Start before taking a limit
A rectangle has sides x, y. Change them by u, v. Its area changes by the exact identity
The two strips contribute xv + yu; the corner contributes uv. When both displacements scale by a small parameter, the corner scales by its square. That is why it disappears from first-order variation.
We can encode that decision algebraically. Work over a field k and introduce a nonzero symbol ε with the rule ε² = 0. An element is a pair of coefficients, written a + bε. Equality means equality of both coefficients. Multiplication is
These are the dual numbers. The missing quadratic term is now a consequence of the chosen rule, not an estimate.
2. Let the arithmetic produce the rule
For a polynomial f, expand every monomial after substituting x+uε. Terms with two or more copies of ε vanish. Among the factors of xm, there are exactly m choices for the one displaced factor. Therefore
Multiplying two such expressions gives (fg)′ = f′g + fg′. Substituting the expression for g(x+uε) into f gives (f∘g)′(x) = f′(g(x))g′(x). Product and chain rules are consequences of multiplication and substitution in the testing algebra.
This calculation works in every characteristic. For example, in characteristic three, (x+uε)³ = x³: the formal derivative of x³ vanishes. It does not follow that the polynomial is constant. The test retains only first-order information. For smooth real functions, Taylor's remainder connects this algebraic calculation with the usual limiting derivative.
3. Turn a tangent direction into a substitution
Now consider the equation xy = 0 at the origin. It describes two crossing axes. Try the displacement x = aε, y = bε. Its product is zero because ε² = 0. Every pair (a,b) passes the first-order test.
Even (1,1) passes, although it points along neither axis. The test has not made an error: the equation has no linear term at the origin, so first order imposes no restriction. To ask whether the direction can continue, we need a test that retains the next term.
4. Restore the corner and try to repair it
Use k[ε]/(ε³), where ε² survives but ε³ vanishes. Allow arbitrary second-order corrections:
The product is xy = abε². Both correction coefficients disappear from this order: multiplying either correction by the other coordinate introduces at least ε³. A lift exists precisely when ab = 0. The mixed direction (1,1) cannot be repaired. Its corner is an obstruction.
Compare the smooth curve xy = 1 at (1,1). The first-order motion x = 1+tε, y = 1−tε has product 1−t²ε². This time a correction works:
The nonzero base coordinate lets the extra strip cancel the corner. At the node, both base coordinates are zero and that repair is unavailable.
5. What became elementary?
We changed the arithmetic used to probe an equation. One test makes first-order rules exact; a finer test distinguishes directions that can continue from those that cannot. The equation supplies the residual, and substitution tells us which corrections can affect it. A first-order description alone does not determine every higher-order obstruction.
Continue with the universal first-order construction, the node and its smoothing, or the complex that organises directions and obstruction targets. The separate oriented-boundary calculation explains cancellation by signs; it is a different operation from discarding quadratic terms.