Source: Calculus for Business, Florida International University, Chapter 2
Tags: product rule, chain rule, combined rules, related rates, cost function, calculus for business, FIU
Difficulty: Intermediate to Advanced | Prerequisites: Chain rule fundamentals (see Part 1 notes), product rule, quotient rule, basic cost/production functions.
Once you know the chain rule on its own, the next step is combining it with other differentiation rules. In practice, most functions you meet in business calculus are products or quotients of composite functions, so the chain rule rarely works in isolation. This set of notes covers the product rule combined with the chain rule, and introduces related rates: problems where two or more quantities change with time and you need to connect their rates of change. These are among the most commonly examined topics in the course.
When a function is a product of two composite pieces, you need both the product rule and the chain rule at the same time. For related rates, the chain rule lets you link how fast one quantity changes to how fast another changes, which is how you solve "at what rate is cost changing with time?" problems.
Product rule
If f(x) = u(x) · v(x), then f'(x) = u'(x) · v(x) + u(x) · v'(x). "First times derivative of second, plus second times derivative of first."
Think of it as: when two things are multiplied and both depend on x, you have to account for each one changing while the other is held steady.
Product rule combined with chain rule
When one or both factors in a product are themselves composite functions, you apply the product rule to split them, then use the chain rule to differentiate each factor.
In simple terms, this means: do the product rule first to set up the structure, then use the chain rule wherever you see a function inside a function.
Related rates
Problems where two or more quantities are connected by an equation, and you know the rate of change of one quantity and need to find the rate of change of another. The chain rule provides the link: dC/dt = (dC/dx) · (dx/dt).
Think of it as: if cost depends on how many units you produce, and the number of units depends on time, then the chain rule tells you how fast cost changes over time.
When f(x) = u(x) · v(x) and at least one of u or v is a composite function, the workflow is:
Apply the product rule: f'(x) = u'(x) · v(x) + u(x) · v'(x)
Use the chain rule to compute u'(x) and/or v'(x) wherever a composition appears.
Worked Example: f(x) = (x + 2)³ · (2x – 1)⁵
Let u = (x + 2)³ and v = (2x – 1)⁵
u'(x) = 3(x + 2)² · 1 = 3(x + 2)² (chain rule on the outer cube)
v'(x) = 5(2x – 1)⁴ · 2 = 10(2x – 1)⁴ (chain rule on the outer fifth power)
Product rule: f'(x) = 3(x + 2)² · (2x – 1)⁵ + (x + 2)³ · 10(2x – 1)⁴
Worked Example: f(x) = √((x² – 4)⁵ + 2x) = ((x² – 4)⁵ + 2x)^(1/2)
Outer function: u^(1/2), where u = (x² – 4)⁵ + 2x
The inner function itself contains another composition: (x² – 4)⁵
f'(x) = ½((x² – 4)⁵ + 2x)^(–1/2) · [5(x² – 4)⁴ · (2x) + 2]
The 5(x² – 4)⁴ · (2x) term comes from applying the chain rule to (x² – 4)⁵
The + 2 comes from d/dx(2x) = 2
This is the classic business calculus application of the chain rule.
Worked Example:
Cost function: C(x) = ⅓x² + 4x + 53 (dollars)
Production function: x(t) = 0.2t² + 0.03t (units produced after t hours)
Question: At what rate is cost changing with respect to time?
Solution, step by step:
Recognise this as dC/dt = (dC/dx) · (dx/dt)
dC/dx = ⅔x + 4
dx/dt = 0.4t + 0.03
dC/dt = (⅔x + 4)(0.4t + 0.03)
Substitute x = 0.2t² + 0.03t: dC/dt = (⅔(0.2t² + 0.03t) + 4)(0.4t + 0.03)
The final expression gives the rate of cost change (dollars per hour) at any time t.
Product Rule
d/dx [u(x) · v(x)] = u'(x) · v(x) + u(x) · v'(x)
Product Rule + Chain Rule (generalised power factors)
If f(x) = [g(x)]ᵐ · [h(x)]ⁿ, then:
f'(x) = m[g(x)]^(m–1) · g'(x) · [h(x)]ⁿ + [g(x)]ᵐ · n[h(x)]^(n–1) · h'(x)
Related Rates (chain rule in Leibniz form)
dC/dt = (dC/dx) · (dx/dt)
This generalises: for any dependent chain y depends on u depends on x, dy/dx = (dy/du) · (du/dx).
The related rates framework (dC/dt = dC/dx · dx/dt) is how businesses model how fast costs, revenues, or profits change over time when production levels are themselves changing. If a factory ramps up production on a curve, the chain rule tells management exactly how quickly their expenditure is climbing at any given hour, which feeds into shift-planning and budgeting decisions.
Students often confuse when to use the product rule versus the chain rule. The product rule is for f(x) = u(x) · v(x), where two separate functions are multiplied. The chain rule is for f(x) = f(g(x)), where one function is inside another. When both apply, do the product rule first to set up the terms, then chain rule inside each term.
In the worked example (x + 2)³ · (2x – 1)⁵, students sometimes try to expand everything before differentiating. This is unnecessary and error-prone. Keep the factors in their composed form and apply the rules directly.
For related rates, students frequently forget to substitute back for x in terms of t at the end. The question asks for a rate with respect to time, so the final answer should be entirely in terms of t (or evaluated at a specific t), not left with x in it.
Students sometimes apply the chain rule to only one factor in a product and forget that the other factor also needs the chain rule if it is composite.
⚠️ Product-of-composites problems are a staple of business calculus exams. If you see two bracketed expressions multiplied together, you almost certainly need both the product rule and the chain rule.
⚠️ Related rates problems ("at what rate is cost/revenue/profit changing with time?") appear on nearly every exam. Set up dC/dt = (dC/dx)(dx/dt) and show every step.
⚠️ Watch the algebra: after applying both rules, you will often have a long expression. Exams may ask you to simplify or factor out common terms. Practise combining like terms.
⚠️ If a problem gives you a specific time value (e.g., "after 2 hours"), remember to substitute t into dx/dt and into x(t), then into dC/dx, in the correct order.
True or False: When differentiating f(x) = (x + 1)² · (3x – 5)⁴, you only need the chain rule.
Answer: False. You need the product rule (two factors multiplied) and the chain rule (each factor is a composite).
Fill in the blank: dC/dt = (___) · (dx/dt).
Answer: dC/dx.
True or False: In a related rates problem, if C depends on x and x depends on t, the final answer for dC/dt should contain only the variable t (or a specific numerical value).
Answer: True.
Q: Find f'(x) for f(x) = (x² + 1)³ · (3x – 2)⁴.
A: Using the product rule with chain rule on each factor: f'(x) = 3(x² + 1)²(2x) · (3x – 2)⁴ + (x² + 1)³ · 4(3x – 2)³(3) = 6x(x² + 1)²(3x – 2)⁴ + 12(x² + 1)³(3x – 2)³.
Q: The revenue from selling x units is R(x) = 200x – 0.5x². During a promotional period, sales follow x(t) = 50 + 10t units after t days. Find dR/dt at t = 3.
A: dR/dx = 200 – x. dx/dt = 10. At t = 3, x = 50 + 30 = 80. So dR/dt = (200 – 80)(10) = 1200 dollars per day.
Q: Differentiate f(x) = √(x³ + 1) · (2x – 7)².
A: Let u = (x³ + 1)^(1/2) and v = (2x – 7)². Then u' = ½(x³ + 1)^(–1/2) · 3x² and v' = 2(2x – 7)(2) = 4(2x – 7). Product rule: f'(x) = [3x² / (2√(x³ + 1))] · (2x – 7)² + (x³ + 1)^(1/2) · 4(2x – 7).
Q: Why does the chain rule matter in related rates problems?
A: Because the quantity you care about (e.g., cost) depends on a variable (units produced) that itself depends on time. The chain rule links these dependencies: dC/dt = (dC/dx)(dx/dt). Without it, you cannot connect the rate you know to the rate you need.
Combining the product and chain rules prepares you for the quotient rule with composites (same idea, different formula) and for implicit differentiation, where you treat y as a function of x and apply the chain rule to every y-term. Related rates problems extend naturally into optimisation (Chapter 3), where you set a derivative equal to zero to find maximum or minimum values of cost, profit, or revenue functions.
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