Why a Product Needs Its Own Rule
A polynomial may be written in standard form, such as \(x^2+3x+2\), or in factored form, such as \((x+1)(x+2)\). When a function is given as a product, differentiating each factor separately is not enough: both factors can change as \(x\) changes. The product rule accounts for both changes.
It is often useful to keep the original product factored while differentiating. This can avoid unnecessary expansion, preserve the structure of the function, and make it easier to substitute a particular input into the derivative.
The Product Rule
Let \(u\) and \(v\) be functions differentiable at \(x\). Their product \(u(x)v(x)\) is differentiable there, and its derivative is given by the product rule.
The order of the two terms is not important, but the pairing is: differentiate one factor while leaving the other unchanged, then add the term in which their roles are reversed. Do not differentiate both factors in the same term.
For algebraic functions in this tutorial, the factors will typically be polynomials or linear expressions. Polynomials are differentiable for every real \(x\), so the rule applies to their products everywhere on the real line. More generally, the rule applies at any input where both factors are differentiable.
A Justification from the Difference Quotient
The product rule follows from the definition of the derivative. This proof also explains why the derivative has two terms. Suppose \(u\) and \(v\) are differentiable at \(x\). For a nonzero increment \(h\), add and subtract \(u(x+h)v(x)\) in the numerator of the difference quotient:
As \(h\) approaches zero, differentiability of \(u\) and \(v\) implies that both are continuous at \(x\). Thus \(u(x+h)\) approaches \(u(x)\). The difference quotients for \(u\) and \(v\) approach \(u'(x)\) and \(v'(x)\), respectively. Taking the limit of the last line therefore gives
This is the derivative of \(u(x)v(x)\), so it proves the rule. The two terms correspond to the two ways the product changes: the first factor changes while the second is held at its current value, and the second factor changes while the first is held at its current value.
Applying the Rule to Algebraic Factors
A reliable procedure is to name the factors, differentiate each one, and then place them into the two terms of the rule. Keep the original factors visible until the derivative is assembled. This reduces the chance of losing a factor or applying the power rule to the wrong expression.
Write the product as \(u(x)v(x)\), naming each factor separately.
Find \(u'(x)\) and \(v'(x)\) using the rules for constants, powers, and sums.
Use \(u'v+uv'\), keeping the undifferentiated factor in each term.
Combine like terms if useful, or substitute the requested input into the derivative.
Worked Examples
Worked Example: A Quadratic Times a Linear Factor
Differentiate \(f(x)=(3x^2-2x+1)(x+4)\) without first expanding the product.
Set \(u(x)=3x^2-2x+1\) and \(v(x)=x+4\). Differentiate each factor:
Apply the product rule, \(f'=u'v+uv'\):
Check by expanding the original function: \(f(x)=3x^3+10x^2-7x+4\). Differentiating this polynomial gives \(9x^2+20x-7\), matching the product-rule result.
Worked Example: A Product of Two Polynomials
Find \(g'(x)\) for \(g(x)=(x^2-5)(2x^3+x)\), keeping the factors unexpanded when applying the rule.
Let \(u(x)=x^2-5\) and \(v(x)=2x^3+x\). Their derivatives are
Substitute into \(g'=u'v+uv'\), then multiply and collect terms:
Check: expanding the original product gives \(g(x)=2x^5+x^3-10x^3-5x=2x^5-9x^3-5x\). Its derivative is \(10x^4-27x^2-5\), as obtained above.
Worked Example: Evaluate a Product Derivative at an Input
Let \(h(x)=(2x-1)(x^2+3)\). Find \(h'(2)\).
Set \(u(x)=2x-1\), \(v(x)=x^2+3\); then \(u'(x)=2\) and \(v'(x)=2x\). The product rule gives
Now substitute \(x=2\):
Check by expansion: \(h(x)=2x^3-x^2+6x-3\), so \(h'(x)=6x^2-2x+6\), and \(h'(2)=24-4+6=26\). Both methods agree.
Keeping the Factors Straight
Each term in the product rule contains one differentiated factor and one unchanged factor. A useful way to check your setup before simplifying is to look at each term separately:
This structure is especially helpful when a factor has several terms. For instance, if one factor is \(x^2-5\), its derivative is \(2x\), but the other factor still appears unchanged in the first product-rule term. The product rule does not say to replace both factors by their derivatives.
There is no requirement to expand the original product before using the rule. In fact, keeping it factored during the first step makes the roles of the factors clear. Expansion may still be useful afterward to simplify the answer or to check it, as in the examples.
A Common Mistake
A frequent error is to differentiate both factors and multiply their derivatives. For \(p(x)=(x+2)(x^2+1)\), that incorrect approach gives
This omits the fact that either factor can change as \(x\) changes. Apply the two-term rule instead. With \(u=x+2\), \(u'=1\), \(v=x^2+1\), and \(v'=2x\),
Check by expanding \(p(x)=x^3+2x^2+x+2\). Its derivative is \(3x^2+4x+1\), not \(2x\). The correction is to include both product-rule terms, with one unchanged factor in each.
When This Rule Is Useful
The product rule is valuable whenever a function is naturally expressed as two changing factors. Factored form can reveal zeros, make evaluation convenient, or preserve a structure that would become less clear after expansion. For AP calculus problems, identifying the factors and writing the two terms accurately is often more important than doing extensive algebra.
Before accepting a result, check these points:
- Did you identify both factors correctly?
- Did you calculate each factor's derivative correctly?
- Does each product-rule term retain one unchanged factor?
- Did you add the two terms rather than multiply them together?
- If you expanded to check, does differentiating the expanded polynomial agree?
Check Your Understanding
For each question, keep track of which factor is differentiated in each term.
- Differentiate \(r(x)=(x^2+1)(x-3)\) using the product rule, without expanding first.
- For \(s(x)=(4x+1)(x^2-2x)\), write \(s'(x)\) as two product-rule terms before simplifying.
- Find the derivative of \(t(x)=(x^2-4)(3x+2)\) at \(x=1\).
- Explain why the expression \(u'(x)v'(x)\) is not the product rule for \(u(x)v(x)\).
- Check the derivative of \((x+1)(x^2+2)\) by comparing the product-rule result with the derivative obtained after expansion.
