Why a Product Needs Its Own Derivative Rule
A derivative describes how a function changes near a point. When a function is a product of two changing factors, both factors contribute to that change. The product rule records these two contributions precisely.
Rather than assume a rule for differentiating a product, we will derive it from the definition of the derivative. The key is to rewrite the change in a product so that each factor’s change appears separately. A limit then turns that rewrite into a derivative formula.
The Difference Quotient at a Point
A function \(F\) is differentiable at \(a\) if the following limit exists as a finite real number:
Here \(h\ne 0\) while the quotient is formed, and \(a+h\) must lie in the domain of \(F\). For the product rule, suppose \(f\) and \(g\) are both defined on an open interval containing \(a\), and each is differentiable at \(a\). Their product is the function \(F(x)=f(x)g(x)\), also defined near \(a\). Our task is to find the limit of its difference quotient.
The formula has two terms. In the first, the change in \(f\) is paired with the value of \(g\); in the second, the value of \(f\) is paired with the change in \(g\). We will see why both terms arise.
A Fact Needed for the Limit
The difference quotient for the product will contain \(f(a+h)\). To take its limit, we need to know that \(f(a+h)\) approaches \(f(a)\). Differentiability guarantees this, but it is useful to show the implication directly.
For \(h\ne0\), rewrite the difference \(f(a+h)-f(a)\) using the difference quotient:
As \(h\to0\), the first factor tends to \(0\). The expression in parentheses tends to the finite number \(f'(a)\), so it is bounded for all sufficiently small nonzero \(h\). Therefore the product tends to \(0\). This gives \(\lim_{h\to0}[f(a+h)-f(a)]=0\), which is exactly continuity at \(a\). The same argument applies to \(g\).
Deriving the Product Rule
Start with the difference quotient for \(F(x)=f(x)g(x)\). Add and subtract \(f(a+h)g(a)\) in its numerator. This changes nothing, because the added term is immediately subtracted, but it creates differences involving each factor separately.
The last equality is an algebraic factorization. In the first pair of terms, \(f(a+h)\) is common. In the second pair, \(g(a)\) is common. Each difference quotient is now one of the quotients whose limit defines a derivative.
Adding and subtracting \(f(a+h)g(a)\) rewrites the change in the product as a sum of two terms, one involving the change in \(g\) and one involving the change in \(f\).
By continuity, \(f(a+h)\to f(a)\). By differentiability, \(\frac{g(a+h)-g(a)}{h}\to g'(a)\). The first term therefore tends to \(f(a)g'(a)\).
The factor \(g(a)\) is constant with respect to \(h\), and \(\frac{f(a+h)-f(a)}{h}\to f'(a)\). The second term tends to \(g(a)f'(a)\).
Both limits exist and are finite, so the limit of their sum is the sum of their limits. Thus \(F'(a)=f(a)g'(a)+g(a)f'(a)\).
Reordering the two terms gives the standard statement:
This proves the rule at an arbitrary point \(a\) where both functions are differentiable. Consequently, wherever both are differentiable, their product has the derivative given by the same formula at each such point.
Worked Difference-Quotient Checks
Worked Example: A Quadratic Factor and a Linear Factor
Let \(f(x)=x^2\) and \(g(x)=x+1\). Compute the derivative of their product at \(a=2\) directly from the difference quotient, then compare it with the derived formula.
First, \(f(2)=4\) and \(g(2)=3\), so \(F(2)=12\). For \(h\ne0\),
Therefore,
Taking \(h\to0\) gives \(F'(2)=16\). Independently, \(f'(2)=4\) and \(g'(2)=1\), so the derived formula gives \(4\cdot3+4\cdot1=12+4=16\). As a check, \(F(x)=x^3+x^2\), whose derivative is \(3x^2+2x\); at \(x=2\), this is \(3\cdot4+4=16\), as required.
Worked Example: Two Factors That Both Change
Let \(f(x)=x\) and \(g(x)=x^2\), and find the derivative of \(F(x)=f(x)g(x)\) at \(a=1\) using the difference quotient.
At \(x=1\), \(F(1)=1\). For nonzero \(h\), expand the nearby product:
Thus,
The limit is \(F'(1)=3\). The formula gives \(f'(1)g(1)+f(1)g'(1)=1\cdot1+1\cdot2=3\). As a further check, \(F(x)=x^3\), so \(F'(x)=3x^2\) and \(F'(1)=3\). Both factors change with \(h\); the calculation retains contributions from each.
Worked Example: A Product That Simplifies Near the Point
Let \(f(x)=x^2\) and \(g(x)=1/x\), and consider \(a=2\). The functions are defined near \(2\), so the product difference quotient is valid for sufficiently small \(h\ne0\).
We have \(F(2)=2\). Also, \(2+h\ne0\) for \(h\) sufficiently close to zero, and
It follows that
so \(F'(2)=1\). To check the formula, \(f'(2)=4\), \(g(2)=1/2\), \(f(2)=4\), and \(g'(2)=-1/4\). Hence
The simplified product is \(F(x)=x\) near \(2\), whose derivative is \(1\). This agrees with both computations.
A Common Error: Keeping Only One Factor’s Change
A tempting but incorrect idea is that the change in a product is just the product of the two changes. For a concrete test, take \(f(x)=x\), \(g(x)=x\), and \(a=1\). Then \(f(1)=g(1)=1\), while \(f(1+h)-f(1)=h\) and \(g(1+h)-g(1)=h\). The incorrect calculation would treat the product’s difference quotient as
But \(F(x)=x^2\), so its derivative at \(1\) is \(2\), not \(0\). The error is that the product also changes when one factor changes while the other is at its original value. The exact decomposition from the proof makes the missing contributions visible:
In this example, dividing by \(h\) gives \(1+1+h\), which tends to \(2\). The final term contributes \(h\), which tends to zero, but the first two terms do not disappear. This is why the product rule has two terms rather than one.
What the Derivation Establishes
The proof relies on differentiability of both factors at the same point. That condition gives the two derivative limits and, through continuity, the limit of the factor \(f(a+h)\). The algebraic addition and subtraction alone do not establish the rule; the limit conditions are essential.
The result is local: it determines the derivative at \(a\) from the values and derivatives of the factors at \(a\). It does not require either factor to be nonzero. In particular, if one factor is zero at \(a\), the formula still applies as long as both factors are differentiable there.
The main idea to retain is the reasoning behind the formula. A product changes because either factor can change. Adding and subtracting an intermediate product isolates those changes; taking the limit turns them into the two derivative terms.
Check Your Understanding
Try answering these questions using the definition and the derivation.
- What assumptions on \(f\) and \(g\) are sufficient for the product rule at \(a\)?
- Why does differentiability of \(f\) at \(a\) imply that \(f(a+h)\to f(a)\)?
- Which term is added and subtracted in the product difference quotient, and how does it help?
- For \(f(x)=x\) and \(g(x)=x^2\) at \(a=1\), what are the two terms in the product-rule formula?
- Why does the product rule remain valid when one factor has value zero at \(a\)?
