Difficulty: Intermediate | Prerequisites: Basic derivative rules (constant rule, power rule, constant multiple rule), algebra of polynomial and rational functions.
Chapter 2 of Calculus for Business builds on the basic derivative rules you already know (power rule, constant rule) and introduces the tools you need when functions are multiplied together, divided, or nested inside one another. These are the product rule, quotient rule and chain rule (sometimes called the general power rule in this course). Without them, you cannot differentiate most real-world business functions, because revenue, cost and profit expressions almost always involve products or quotients of simpler functions.
You should already be comfortable taking derivatives of simple polynomials using the power rule. If you are not, revisit Chapter 1 before continuing.
When two functions are multiplied, use the product rule: f'g + fg'. When they are divided, use the quotient rule: (f'g - fg') / g². When a function sits inside another function, use the chain rule (general power rule): bring down the outer exponent, reduce it by one, then multiply by the derivative of the inside. These three rules, combined with the power rule, let you differentiate virtually any function you will see in a business calculus course.
Product rule
If f(x) = u(x) · v(x), then f'(x) = u'(x) · v(x) + u(x) · v'(x). In simple terms, differentiate the first and multiply by the second, then add the first multiplied by the derivative of the second.
Quotient rule
If f(x) = u(x) / v(x), then f'(x) = [u'(x) · v(x) - u(x) · v'(x)] / [v(x)]². Think of it as: "low d-high minus high d-low, over the square of what's below."
Chain rule (general power rule)
If f(x) = [g(x)]ⁿ, then f'(x) = n · [g(x)]ⁿ⁻¹ · g'(x). In simple terms, bring down the exponent, reduce it by one, then multiply by the derivative of everything inside the brackets.
Higher-order derivatives
The second derivative f''(x) is the derivative of the first derivative. The third derivative f'''(x) is the derivative of the second, and so on. Think of it as taking the derivative again on the result you already found.
Revenue function
R(t) = x(t) · p(t), where x(t) is quantity produced and p(t) is price per unit. Revenue is quantity times price, expressed as a function of whatever variable drives both.
The rule
If f(x) = u(x) · v(x), then f'(x) = u' · v + u · v'
Mnemonic: "first times derivative of second, plus second times derivative of first."
Worked example from the notes
Find f'(x) for f(x) = (1/5)(x⁵ - 2x³ + 1)(x - ½)
Identify u(x) = x⁵ - 2x³ + 1 and v(x) = x - ½
u'(x) = 5x⁴ - 6x²
v'(x) = 1 + x⁻²
Apply the product rule: f'(x) = (1/5)[(5x⁴ - 6x²)(x - ½) + (1 + x⁻²)(x⁵ - 2x³ + 1)]
Another worked example
Find f'(x) for f(x) = (x + 1/x)²
Rewrite as f(x) = (x + x⁻¹)² and expand, or use the chain rule combined with the product rule
f'(x) = (1 - x⁻²)(x + 1/x) + (1 - x⁻²)(x + 1/x)
This simplifies because both terms are identical: f'(x) = 2(1 - x⁻²)(x + x⁻¹)
The rule
If f(x) = u(x) / v(x), then f'(x) = [u'(x) · v(x) - u(x) · v'(x)] / [v(x)]²
The numerator order matters: it is "derivative of top times bottom, minus top times derivative of bottom." Swapping the subtraction gives the wrong sign.
Worked example 1
Find f'(x) for f(x) = (x² - 5x + 7) / (2x)
u = x² - 5x + 7, so u' = 2x - 5
v = 2x, so v' = 2
f'(x) = [(2x - 5)(2x) - (2)(x² - 5x + 7)] / (2x)²
Worked example 2
Find f'(x) for f(x) = (x + 1) / (x - 2)
u = x + 1, u' = 1
v = x - 2, v' = 1
f'(x) = [(1)(x - 2) - (1)(x + 1)] / (x - 2)²
Simplifies to f'(x) = -3 / (x - 2)²
Worked example 3
Find f'(x) for f(x) = (x² - 3x + 2) / (2x² + 5x - 1)
u = x² - 3x + 2, u' = 2x - 3
v = 2x² + 5x - 1, v' = 4x + 5
f'(x) = [(2x - 3)(2x² + 5x - 1) - (4x + 5)(x² - 3x + 2)] / (2x² + 5x - 1)²
The rule
If f(x) = [g(x)]ⁿ, then f'(x) = n · [g(x)]ⁿ⁻¹ · g'(x)
In words: bring down the power, reduce the power by one, multiply by the derivative of the inside function.
When to use it
Any time you see a function raised to a power, or a composite function where one function is plugged into another. Common triggers: expressions like (3x + 1)⁵, (x² + 1)¹⁰, or even something like √(x² + 1) which is (x² + 1)^(1/2).
Combining with the product rule
The notes show examples where the chain rule and product rule are used together. When a function is a product of two pieces and one or both involve a composite, apply the product rule first to set up u'v + uv', then use the chain rule within each derivative as needed.
Second derivative f''(x)
Take the derivative of f'(x) to get f''(x)
Example: f(x) = x², so f'(x) = 2x, so f''(x) = 2
Third derivative and beyond
Example: f(x) = 5x⁴ - 3x² - 3x + 7
f'(x) = 20x³ - 6x - 3
f''(x) = 60x² - 6
You could continue to f'''(x) = 120x
Why higher-order derivatives matter
The second derivative tells you about the rate of change of the rate of change. In business calculus, this is used to determine concavity and to find inflection points, which help identify whether a critical point is a maximum or a minimum.
Basic rules (review)
Constant rule: d/dx [c] = 0
Power rule: d/dx [xⁿ] = nxⁿ⁻¹
Constant multiple rule: d/dx [c · f(x)] = c · f'(x)
Chapter 2 rules
Product rule: d/dx [f(x) · g(x)] = f'(x) · g(x) + f(x) · g'(x)
Quotient rule: d/dx [f(x) / g(x)] = [f'(x) · g(x) - f(x) · g'(x)] / [g(x)]²
Chain rule (general power rule): d/dx [g(x)]ⁿ = n · [g(x)]ⁿ⁻¹ · g'(x)
The product rule is exactly what you use when revenue depends on two changing quantities (units sold and price per unit), both of which vary over time. The derivative of revenue tells a manufacturer whether revenue is growing or shrinking at a given moment, which informs production and pricing decisions.
The quotient rule appears whenever you need the rate of change of a ratio, such as average cost (total cost divided by quantity) or a financial ratio like profit margin.
Students often think the derivative of a product is simply the product of the derivatives: f' · g'. It is not. You must use f'g + fg'.
In the quotient rule, students frequently swap the subtraction order in the numerator. The correct order is "derivative of top times bottom minus top times derivative of bottom." Reversing it flips the sign of the entire answer.
Students sometimes forget to square the denominator in the quotient rule. The bottom function squared is always the denominator of the result.
When using the chain rule, students often forget to multiply by the derivative of the inside function. Leaving off g'(x) is the single most common chain rule error.
⚠️ You will almost certainly be asked to differentiate a product or quotient of two functions. Know both rules cold.
⚠️ Expect a word problem where revenue or cost is expressed as a product of two functions of time, and you are asked for the rate of change at a specific time. This is a product rule application with a numerical plug-in at the end.
⚠️ Higher-order derivative questions (find f''(x) or f'''(x)) are common. They are not conceptually hard, just require careful algebra on each successive derivative.
⚠️ "Increasing or decreasing" questions: after you find R'(t), plug in the given time. If R'(t) > 0, revenue is increasing. If R'(t) < 0, it is decreasing.
True or false: The derivative of f(x) · g(x) is f'(x) · g'(x).
Fill in the blank: In the quotient rule, the denominator of the result is always the original denominator function ______.
True or false: When using the chain rule on [g(x)]ⁿ, you must multiply by g'(x) at the end.
Fill in the blank: If R'(4) < 0, revenue at t = 4 is ______.
True or false: The second derivative is found by differentiating the original function twice in one step.
Answers: 1. False (use f'g + fg'). 2. Squared. 3. True. 4. Decreasing. 5. False (you differentiate the first derivative, not the original function again from scratch, though the result is the same).
Q: Find f'(x) for f(x) = (x² + 3x)(x³ - 2x + 5). Which rule do you use?
A: Product rule. Let u = x² + 3x and v = x³ - 2x + 5. Then u' = 2x + 3, v' = 3x² - 2. f'(x) = (2x + 3)(x³ - 2x + 5) + (x² + 3x)(3x² - 2).
Q: Find f'(x) for f(x) = (x + 1) / (x - 2).
A: Quotient rule. u = x + 1, v = x - 2, u' = 1, v' = 1. f'(x) = [(1)(x - 2) - (1)(x + 1)] / (x - 2)² = -3 / (x - 2)².
Q: A manufacturer finds that t months after launch, x(t) = t² + 3t hundred units can be produced, at a price of p(t) = -2t⁰·⁵ + 30 dollars per unit. Find R'(4) and state whether revenue is increasing or decreasing.
A: R(t) = x(t) · p(t) = (t² + 3t)(-2t⁰·⁵ + 30). Apply the product rule with u = t² + 3t, v = -2t⁰·⁵ + 30. At t = 4, R'(4) evaluates to a negative number, so revenue is decreasing.
Q: Find f''(x) for f(x) = 5x⁴ - 3x² - 3x + 7.
A: f'(x) = 20x³ - 6x - 3. f''(x) = 60x² - 6.
These differentiation rules are the foundation for Chapter 3 (applications of the derivative), where you will use first and second derivatives to find maxima, minima and inflection points of business functions. The revenue problem in these notes is a direct preview of optimisation problems.
The chain rule also connects to implicit differentiation and related rates, which appear later in the course. If you are comfortable with the chain rule now, those topics will be substantially easier.
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