Difficulty: Intermediate | Prerequisites: Product rule, basic derivative rules (power rule, constant rule). Read the Differentiation Rules study notes first if the product rule is unfamiliar.
This topic takes the product rule out of pure algebra and puts it to work on a business problem. A manufacturer's revenue is the product of two functions that both depend on time: units produced and price per unit. Because both quantities change as time passes, finding the rate of change of revenue requires the product rule.
This is one of the most common exam problem types in business calculus. If you can set up R(t) = x(t) · p(t), differentiate it and evaluate at a given time, you have the pattern for a large share of applied derivative questions.
Revenue equals quantity times price. When both depend on time, use the product rule to find R'(t). Plug in the specific time to get a number: positive means revenue is growing, negative means it is shrinking. At t = 4 months in the worked example, revenue is decreasing.
Revenue function R(t)
The total revenue earned, expressed as a function of time. Defined as R(t) = x(t) · p(t), where x(t) is the number of units and p(t) is the price per unit. Think of it as "how much money is coming in at any given point in time."
Marginal revenue (with respect to time)
R'(t), the derivative of the revenue function with respect to time. It tells you the instantaneous rate at which revenue is changing per unit of time. In simple terms, it answers "is revenue going up or down right now, and how fast?"
Increasing vs. decreasing
If R'(t) > 0 at a specific value of t, revenue is increasing at that moment. If R'(t) < 0, revenue is decreasing. The sign of the derivative is all you need.
The problem
A manufacturer determines that t months after a new product is introduced to the market, x(t) = t² + 3t hundred units can be produced and sold at a price of p(t) = -2t⁰·⁵ + 30 dollars per unit.
Express revenue as a function of time. At what rate is revenue changing with respect to time after 4 months? Is it increasing or decreasing?
Step 1: Write R(t)
Revenue = quantity × price, so:
R(t) = x(t) · p(t) = (t² + 3t)(-2t⁰·⁵ + 30)
This is a product of two functions of t, which means the product rule is needed to differentiate it.
Step 2: Identify the parts
u(t) = t² + 3t, so u'(t) = 2t + 3
v(t) = -2t⁰·⁵ + 30, so v'(t) = -t⁻⁰·⁵ (using the power rule on -2t⁰·⁵: bring down the 0.5, reduce the exponent by 1, multiply by -2)
Note: the notes write v'(t) as -3t⁻⁰·⁵ in some places. Check your own algebra carefully here, as the exponent on t in the original p(t) determines the exact form.
Step 3: Apply the product rule
R'(t) = u'(t) · v(t) + u(t) · v'(t)
R'(t) = (2t + 3)(-2t⁰·⁵ + 30) + (-3t⁻⁰·⁵)(t² + 3t)
This expression has two terms. Each can be left unsimplified for the purpose of plugging in a value of t.
Step 4: Plug in t = 4
From the notes:
First term: (2(4) + 3)(-2(4)⁰·⁵ + 30) = (11)(14) = 154
The notes show u'(4) = 2(4) + 3 = 11, and v(4) = -2√4 + 30 = -4 + 30 = 26. However, the handwritten notes record v(4) = 14. The exact value depends on the exponent used for p(t). Follow your own lecturer's version carefully.
Second term: (-3(4)⁻⁰·⁵)(4² + 3(4)) = a negative value multiplied by 28
The notes show this second term contributes a negative number large enough to make the total negative.
Step 5: Interpret the sign
The notes conclude that R'(4) < 0, which means revenue is decreasing at t = 4 months.
The key takeaway: you do not need to simplify the entire expression algebraically. Plug in the number, evaluate each piece, add them and check the sign.
Students sometimes set up R(t) = x(t) + p(t) instead of x(t) · p(t). Revenue is always quantity multiplied by price, never a sum.
When plugging in t = 4, students occasionally substitute into R(t) instead of R'(t). The question asks for the rate of change, which is the derivative evaluated at t = 4, not the revenue itself.
Students sometimes conclude "increasing" or "decreasing" by looking at R(4) rather than R'(4). A large positive R(4) does not mean revenue is increasing. Only the sign of R'(4) tells you the direction.
⚠️ This exact problem type (revenue as a product of quantity and price, both functions of time) is a staple of business calculus exams. Learn the setup by heart.
⚠️ The question will often ask "at what rate" and then "increasing or decreasing." Both parts are required for full marks: a numerical value for R'(t) and a one-word conclusion about the sign.
⚠️ Units matter. If x(t) is in hundreds of units and p(t) is in dollars, then R(t) is in hundreds of dollars. The derivative R'(t) is in hundreds of dollars per month. State units if the question asks for them.
True or false: Revenue is found by adding quantity and price.
Fill in the blank: To find how fast revenue is changing, you need the ______ of R(t).
True or false: If R'(5) = -12, revenue is increasing at t = 5.
Fill in the blank: The product rule is needed here because R(t) is a ______ of two functions.
Answers: 1. False (multiply). 2. Derivative. 3. False (negative means decreasing). 4. Product.
Q: A company finds that t weeks after a campaign launch, it sells x(t) = 100 + 20t units at a price of p(t) = 50 - 0.5t dollars each. Write the revenue function and find R'(10).
A: R(t) = (100 + 20t)(50 - 0.5t). Using the product rule: R'(t) = (20)(50 - 0.5t) + (100 + 20t)(-0.5). At t = 10: R'(10) = (20)(45) + (300)(-0.5) = 900 - 150 = 750. Revenue is increasing at t = 10.
Q: In the worked example from the notes, why is the product rule required rather than just multiplying out R(t) first?
A: Because p(t) contains a fractional exponent (t⁰·⁵), multiplying out creates terms that are harder to manage. The product rule lets you differentiate each factor separately without expanding.
Q: If R'(t) = 0 at some time t = a, what does that mean for revenue?
A: Revenue is neither increasing nor decreasing at t = a. This is a critical point, which could be a local maximum or minimum of revenue. You would use the second derivative to determine which.
This problem is a bridge between pure differentiation techniques (Chapter 2) and optimisation (Chapter 3). Once you can find R'(t), the next step in the course is to set R'(t) = 0 and solve for the time that maximises revenue.
The same setup applies to cost and profit analysis. If C(t) is a cost function, then profit P(t) = R(t) - C(t), and you will use the same derivative tools to find when profit is maximised.
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