Difficulty: Intermediate | Prerequisites: Differentiation rules (constant, power, constant multiple)
The product rule lets you differentiate functions that are the product of two (or more) simpler functions. Before this, you could only differentiate sums and individual terms. Most real-world models in business calculus involve products (revenue = price times quantity, for instance), so this rule comes up constantly. You should already be comfortable with the constant rule, power rule, and constant multiple rule before tackling this.
When a function is the product of two pieces, its derivative is the first piece times the derivative of the second, plus the second piece times the derivative of the first. The derivative of a product is not the product of the derivatives.
Product Rule
If w(x) = f(x) · g(x), then w'(x) = f'(x) · g(x) + f(x) · g'(x).
In simple terms, to differentiate a product of two functions, you take each function's turn being differentiated while the other stays put, then add the two results together.
Sum Rule (for contrast)
The derivative of a sum f(x) + g(x) is simply f'(x) + g'(x). No cross-terms needed.
Think of it as: addition is straightforward, but multiplication creates entanglement that the product rule untangles.
Limit Definition of the Derivative
f'(x) = lim (h → 0) [f(x + h) - f(x)] / h. The product rule is derived from applying this definition to f(x) · g(x) and using an algebraic trick (adding and subtracting a middle term) to separate the pieces.
Start with w(x) = f(x) · g(x) and apply the limit definition:
w'(x) = lim (h → 0) [f(x+h)g(x+h) - f(x)g(x)] / h
The trick is to add and subtract f(x+h)g(x) in the numerator:
= lim (h → 0) [f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)] / h
Group the terms:
= lim (h → 0) { f(x+h) · [g(x+h) - g(x)]/h + g(x) · [f(x+h) - f(x)]/h }
As h → 0, f(x+h) → f(x), the first bracket → g'(x), and the second bracket → f'(x). So:
w'(x) = f(x) · g'(x) + g(x) · f'(x)
You do not need to reproduce this derivation on an exam, but understanding it helps you see why the rule has two terms.
Identify f(x) and g(x), the two pieces being multiplied.
Find f'(x) and g'(x) separately, using the basic rules.
Plug into the formula: w'(x) = f'(x) · g(x) + f(x) · g'(x).
Simplify if the question asks for it.
Example 1: f(x) = (x - 1)(3x - 2)
Let the two pieces be (x - 1) and (3x - 2).
Derivatives: (1) and (3).
Product rule: f'(x) = (1)(3x - 2) + (3)(x - 1) = 3x - 2 + 3x - 3 = 6x - 5.
Example 2: f(x) = (2x + 1)(3x - 2)
Derivatives: (2) and (3).
Product rule: f'(x) = (2)(3x - 2) + (3)(2x + 1) = 6x - 4 + 6x + 3 = 12x - 1.
Example 3: f(x) = (2x + 1)(2x^2 - x - 1)
Derivatives: (2) and (4x - 1).
Product rule: f'(x) = (2)(2x^2 - x - 1) + (4x - 1)(2x + 1).
Expand and simplify from there.
Example 4: f(x) = 400(5 - x^2)(3x + 2)
Here the 400 is a constant multiple. Apply the constant multiple rule first, then the product rule to (5 - x^2)(3x + 2).
Derivatives of the two pieces: (-2x) and (3).
Product rule inside: (-2x)(3x + 2) + (3)(5 - x^2).
Full derivative: f'(x) = 400[(-2x)(3x + 2) + (3)(5 - x^2)].
If f(x) = (2x + 1) + (3x - 2), there is no product. This is just addition.
f'(x) = 2 + 3 = 5.
The product rule applies only when two expressions are multiplied together, not added.
\frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)Some students remember this as: "derivative of the first times the second, plus the first times the derivative of the second."
With three or more factors, extend the pattern: each factor takes its turn being differentiated while all the others stay put, and you sum the results.
Revenue often equals price times quantity, where both may vary with time or another variable. The product rule is how you find marginal revenue when neither factor is constant.
In finance, if the value of a portfolio is the number of shares times the share price, and both change over time, the product rule tells you the overall rate of change of the portfolio value.
Students often assume the derivative of f(x) · g(x) is simply f'(x) · g'(x). It is not. The product rule has two terms, and you cannot just multiply the individual derivatives.
Students sometimes apply the product rule when they should use the sum rule. If two expressions are added (not multiplied), differentiate each one separately and add the results. No cross-terms.
When a constant multiplies a product, students occasionally try to fold the constant into one of the factors and then lose track of it. Pull the constant out first, then apply the product rule to what remains.
Students forget to simplify after applying the product rule. On an exam, leaving the answer unsimplified may cost marks.
The product rule is tested directly on most calculus exams. Expect at least one question that requires it.
A common exam format gives you a function that looks like it might be a product, but is actually a sum (like the (2x + 1) + (3x - 2) example). Read the problem carefully for plus signs vs. multiplication.
Questions combining the constant multiple rule with the product rule (like the 400(5 - x^2)(3x + 2) example) test whether you can layer two rules in one problem.
True or false: The derivative of f(x) · g(x) is f'(x) · g'(x).
Answer: False. It is f'(x) · g(x) + f(x) · g'(x).
Fill in the blank: If f(x) = (x + 3)(2x - 1), the product rule gives f'(x) = (1)(___) + (2)(x + 3).
Answer: (2x - 1).
True or false: You need the product rule to differentiate f(x) = (x + 5) + (3x - 4).
Answer: False. This is a sum, not a product. Just differentiate each term: f'(x) = 1 + 3 = 4.
Fill in the blank: When differentiating 400 · f(x) · g(x), the 400 ___.
Answer: Stays outside as a constant multiple. Apply the product rule to f(x) · g(x), then multiply by 400.
True or false: The product rule produces exactly two terms (before simplifying).
Answer: True (for a product of two functions).
Q: Find f'(x) for f(x) = (3x + 4)(x^2 - 1).
A: f'(x) = (3)(x^2 - 1) + (2x)(3x + 4) = 3x^2 - 3 + 6x^2 + 8x = 9x^2 + 8x - 3.
Q: Find f'(x) for f(x) = (x^2 + 1)(x^3 - 2x).
A: f'(x) = (2x)(x^3 - 2x) + (3x^2 - 2)(x^2 + 1) = 2x^4 - 4x^2 + 3x^4 + 3x^2 - 2x^2 - 2 = 5x^4 - 3x^2 - 2.
Q: A company's revenue is R(x) = (50 + 2x)(1000 - 10x) dollars, where x is the number of price increases. Find R'(x) and evaluate at x = 5.
A: R'(x) = (2)(1000 - 10x) + (-10)(50 + 2x) = 2000 - 20x - 500 - 20x = 1500 - 40x. At x = 5: R'(5) = 1500 - 200 = 1300 dollars per price increase.
Q: Differentiate f(x) = (x - 1)(3x - 2) without the product rule, by expanding first. Verify it matches the product rule result.
A: Expanding: f(x) = 3x^2 - 2x - 3x + 2 = 3x^2 - 5x + 2. So f'(x) = 6x - 5. Product rule: f'(x) = (1)(3x - 2) + (3)(x - 1) = 3x - 2 + 3x - 3 = 6x - 5. They match.
The quotient rule, which you will meet next, is closely related to the product rule. In fact, the quotient rule can be derived from the product rule by rewriting f(x)/g(x) as f(x) · [g(x)]^(-1) and applying both the product rule and the chain rule.
The chain rule extends these ideas to composite functions, where one function is nested inside another. Combining the product rule with the chain rule lets you differentiate nearly any expression you will encounter in this course.
In business applications, marginal analysis relies heavily on the product rule whenever revenue, cost, or profit is expressed as a product of variable quantities.
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