Why Derive the Product Rule?
The derivative of a product is not obtained by simply multiplying the derivatives of its factors. The product rule has two terms, and the difference quotient explains why both are necessary. Its derivation also shows exactly which assumptions make the rule valid.
We will work at a single point \(a\). The functions do not need to be polynomials: the argument applies whenever both factors are differentiable at that point. The key algebraic move is to add and subtract a cross term so that the difference of products becomes a sum of two expressions involving familiar difference quotients.
Pointwise Derivatives and the Product Rule
A function \(f\) is differentiable at \(a\) when the following limit exists as a finite real number:
Here \(h\ne 0\) in the quotient, and the limit considers values of \(h\) approaching zero for which the function is defined. If \(f\) is differentiable at \(a\), then it is continuous there, so \(f(a+h)\to f(a)\) as \(h\to0\). We will use that fact in the derivation.
The rule is pointwise: differentiability of both factors at \(a\) is enough to conclude that the product is differentiable at \(a\). If both functions are differentiable throughout an interval, the same formula applies at every point in that interval.
The Difference Quotient for a Product
Start from the definition for \(p(x)=f(x)g(x)\). Its difference quotient at \(a\) is
The numerator is a difference of products. It does not yet display either derivative quotient. To separate it, add and subtract \(f(a+h)g(a)\), which is the cross term. Adding and subtracting the same quantity leaves the expression unchanged:
Factor each difference in this new sum. The first pair has the common factor \(f(a+h)\); the second pair has the common factor \(g(a)\). Dividing by \(h\) gives
Now the two difference quotients are recognizable: as \(h\to0\), the first quotient tends to \(g'(a)\), and the second tends to \(f'(a)\). The factor \(f(a+h)\) tends to \(f(a)\), while \(g(a)\) is constant with respect to \(h\). Taking the limit therefore gives the product rule.
Why the Limits Give the Formula
The details of the limit matter. Differentiability of \(g\) at \(a\) gives the finite limit \(g'(a)\) for its difference quotient. Differentiability of \(f\) gives the finite limit \(f'(a)\). Continuity of \(f\) at \(a\), which follows from its differentiability there, gives the limit of the changing factor \(f(a+h)\).
This uses the limit laws for sums and products: if two factors have finite limits, the limit of their product is the product of those limits, and the limit of a sum is the sum of its limits. Both derivative limits exist by the assumptions, and \(f(a+h)\) has a finite limit by continuity. Since the resulting limit exists and is finite, the product is differentiable at \(a\), as claimed.
Worked Examples: Following the Derivation
Worked Example: Derive the Product Derivative for \(x^2(x+1)\) at \(x=2\)
Let \(f(x)=x^2\), \(g(x)=x+1\), and \(a=2\). We will compute the difference quotient for the product using the cross term. First record the factor values and derivatives at the point: \(f(2)=4\), \(g(2)=3\), \(f'(2)=4\), and \(g'(2)=1\). Thus the product rule predicts \(4(1)+3(4)=16\).
The product at \(2+h\) is \((2+h)^2(3+h)\), and the product at \(2\) is \(4\cdot3=12\). Add and subtract the cross term \(f(2+h)g(2)=(2+h)^2\cdot3\):
Taking \(h\to0\) gives \(2^2+3(4)=4+12=16\). Check by expanding the product: \(x^2(x+1)=x^3+x^2\), whose derivative is \(3x^2+2x\). At \(x=2\), this is \(3(4)+4=16\), agreeing with the difference-quotient result.
Worked Example: Use a Zero Factor at \(x=3\)
Let \(f(x)=x-3\) and \(g(x)=x^2\), so \(p(x)=(x-3)x^2\). At \(a=3\), \(f(3)=0\), \(g(3)=9\), \(f'(3)=1\), and \(g'(3)=6\). The product rule predicts \(1(9)+0(6)=9\). The quotient derivation shows why the term involving \(g'(3)\) vanishes here.
For \(h\ne0\), \(f(3+h)=h\) and \(g(3+h)=(3+h)^2\). Add and subtract the cross term \(f(3+h)g(3)=9h\):
As \(h\to0\), \(h(6+h)\to0\), so the derivative is \(9\). This is also confirmed directly: \(p(x)=x^3-3x^2\), so \(p'(x)=3x^2-6x\), and \(p'(3)=27-18=9\).
Worked Example: Check the Two Terms at \(x=1\)
Take \(f(x)=2x+1\) and \(g(x)=x^2-1\), with \(a=1\). The values are \(f(1)=3\), \(g(1)=0\), \(f'(1)=2\), and \(g'(1)=2\). The product rule gives \(f'(1)g(1)+f(1)g'(1)=2(0)+3(2)=6\).
Using the cross term \(f(1+h)g(1)\), which is zero in this case, the quotient becomes
Its limit is \(3(2)=6\). For an independent check, expand \(p(x)=(2x+1)(x^2-1)=2x^3+x^2-2x-1\). Then \(p'(x)=6x^2+2x-2\), and \(p'(1)=6+2-2=6\), as required.
A Common Mistake: Leaving Out the Cross Term
A tempting but incorrect split is to replace the numerator \(f(a+h)g(a+h)-f(a)g(a)\) by \(f(a)[g(a+h)-g(a)]+g(a)[f(a+h)-f(a)]\). That expression uses only values at \(a\), and it omits the product of the two changes. A quick numerical check exposes the error.
Take \(f(x)=x\), \(g(x)=x\), \(a=1\), and \(h=1\). The actual product difference quotient is \((2\cdot2-1\cdot1)/1=3\). The incorrect split gives \(1(2-1)+1(2-1)=2\), not \(3\). The missing amount is the cross product of the changes: \((f(2)-f(1))(g(2)-g(1))=1\cdot1=1\).
The correction is to add and subtract a cross term before splitting. Using \(f(a+h)g(a)\), as in the proof, produces an exact identity for every allowed \(h\); it is not an approximation. In this example the corrected quotient is \(f(2)[g(2)-g(1)]+g(1)[f(2)-f(1)]=2(1)+1(1)=3\). In the limit, the first multiplier becomes \(f(a)\), yielding the established product rule.
Why This Derivation Matters
The product rule is more than a formula to memorize. The derivation identifies the roles played by algebra and calculus: adding and subtracting a carefully chosen quantity creates the two difference quotients, while differentiability and continuity determine their limits. This method also makes the assumptions visible. If either factor is not differentiable at \(a\), this proof cannot supply the required derivative limit; that does not by itself prove the product is not differentiable, but the theorem does not apply.
When using the result, keep the factors in their original roles long enough to identify their values and derivatives at the point. The two contributions are: the derivative of the first factor times the value of the second, and the value of the first times the derivative of the second. The derivation shows why neither contribution can generally be discarded.
Check Your Understanding
Answer each question using the difference-quotient derivation, not only the remembered formula.
- What quantity is added and subtracted to split the product difference quotient in the proof?
- Which two difference quotients appear after the cross-term decomposition, and what limits do they approach?
- Why does the factor \(f(a+h)\) approach \(f(a)\) as \(h\to0\) under the theorem's assumptions?
- For \(f(x)=x^2\) and \(g(x)=x-1\), calculate the product-rule derivative at \(x=1\), showing the factor values and derivatives.
- In the incorrect split, what part of the product change is omitted, and why can it matter before taking a limit?
