Proof of the Product Rule

Why the Product Rule Needs a Proof

The Product Rule states how to differentiate a product when both factors vary. Its formula is familiar from the previous tutorial, but the derivative definition does not immediately produce that formula: the difference quotient contains changes in both factors at once. The proof must separate those changes without assuming that either factor stays fixed.

We will use two complementary approaches. First, an exact algebraic identity splits the product’s difference quotient into terms whose limits are known. Second, the first-order approximation characterization from earlier in this course shows how the linear approximations of the two factors combine, and how the remaining error becomes negligible. Both arguments rely on differentiability at the point, not on differentiability throughout the whole interval.

Proof by Splitting the Difference Quotient

For nonzero \(h\) small enough that \(a+h\in I\), start with the difference quotient of the product. Add and subtract \(f(a)g(a+h)\) in its numerator. This intermediate term is chosen so the resulting differences vary one factor at a time.

To check the identity, expand the numerator on the right: \(f(a+h)g(a+h)-f(a)g(a+h)+f(a)g(a+h)-f(a)g(a)\). The two middle terms cancel, leaving exactly the numerator on the left. Thus the equality is algebraic and holds for every such nonzero \(h\).

Proof of the Product Rule. Since \(f\) and \(g\) are differentiable at \(a\), the difference quotients satisfy \((f(a+h)-f(a))/h\to f'(a)\) and \((g(a+h)-g(a))/h\to g'(a)\) as \(h\to0\). Differentiability of \(g\) also implies continuity of \(g\) at \(a\), by the theorem that differentiability implies continuity. Hence \(g(a+h)\to g(a)\). Applying the product law for limits to the first term of the identity gives

The second term has a constant factor \(f(a)\), so

Both limits are finite. The sum law for limits therefore shows that the whole difference quotient tends to \(f'(a)g(a)+f(a)g'(a)\). By the definition of the derivative, \(fg\) is differentiable at \(a\) and has that derivative. This proves the Product Rule. \(\square\)

The continuity step is essential: the first difference quotient is multiplied by \(g(a+h)\), not by the fixed value \(g(a)\). Knowing only that the difference quotient tends to \(f'(a)\) would not, by itself, justify taking the limit of that product. Continuity supplies the limit of the other factor.

Proof Through First-Order Approximations

There is another way to track the same calculation. The First-Order Approximation Characterization says that differentiability at \(a\) gives an expansion into a constant term, a linear term, and a remainder that is small compared with \(h\). The following result spells out how two such expansions interact. It provides a useful technique beyond this particular derivative calculation: when multiplying approximations, one must account for every cross term and verify that the error terms remain negligible.

Proof. For \(h\ne0\), write \(r_f(h)=h\varepsilon_f(h)\) and \(r_g(h)=h\varepsilon_g(h)\), where \(\varepsilon_f(h)\to0\) and \(\varepsilon_g(h)\to0\). Multiplication gives

where the terms not included in \(AC+(AD+BC)h\) are

Dividing by \(h\) yields

The first two terms tend to zero. In the parentheses, \(B\) and \(D\) are fixed, while \(\varepsilon_f(h)\) and \(\varepsilon_g(h)\) tend to zero, so the parenthesized expression has a finite limit \(BD\). Multiplication by \(h\to0\) makes the final term tend to zero. Thus \(r(h)/h\to0\), proving the result. \(\square\)

For differentiable \(f\) and \(g\), the First-Order Approximation Characterization gives \(A=f(a)\), \(B=f'(a)\), \(C=g(a)\), and \(D=g'(a)\). The theorem then says that the product has constant term \(f(a)g(a)\), linear coefficient \(f'(a)g(a)+f(a)g'(a)\), and an error small compared with \(h\). By the same characterization, that linear coefficient is the derivative of the product. This recovers the Product Rule by following the first-order terms rather than taking limits of a split difference quotient.

Worked Applications of the Proof

What the Proof Does—and Does Not—Say

The proof succeeds because it turns a simultaneous change into two separate changes. The exact identity assigns one difference to \(f\) and the other to \(g\); in the limit, each change is weighted by the value of the other factor. The first-order approach expresses the same structure in another way: multiplying the linear approximations produces the two linear contributions, while products of changes contribute only to the smaller-order remainder.

A common error is to write \((fg)'(a)=f'(a)g'(a)\). The proof shows why this is not the formula: the terms arise from a change in one factor multiplied by the value of the other, not from multiplying the two changes. Another error is to use the theorem without checking differentiability of both factors at the point. The theorem provides a sufficient condition for the product to be differentiable; it does not say that both factors must be differentiable whenever their product is differentiable.

For example, \(u(x)=|x|\) is not differentiable at zero, but \(u(x)u(x)=x^2\) is differentiable there. This does not contradict the Product Rule, whose hypotheses are not met for the two factors \(u\) and \(u\). The distinction between a theorem’s sufficient hypotheses and necessary conditions is important throughout analysis.