Difficulty: Intermediate to Advanced. Prerequisites: Sections 2.1–2.5 (limits, continuity, basic derivatives).
Marginal analysis uses derivatives to approximate the cost, revenue or profit of producing one more unit, which is the core bridge between calculus and business decision-making. This set of sections also introduces differentiation of exponential and logarithmic functions, the product and quotient rules for products and fractions of functions, and the chain rule for composite functions. Together, these tools let you differentiate virtually any function you will encounter in this course.
Marginal cost function, C'(x)
The derivative of the total cost function. It approximates the cost of producing the (x + 1)th item. In simple terms, it tells you roughly how much one more unit will cost.
Marginal revenue function, R'(x)
The derivative of the total revenue function. It approximates the additional revenue from selling one more unit.
Marginal profit function, P'(x)
The derivative of the total profit function, where P(x) = R(x) − C(x).
Average cost, C̅(x)
Total cost divided by the number of units: C(x)/x. Its derivative, C̅'(x), is the marginal average cost.
Compound interest
Interest calculated on both the initial principal and accumulated interest. Formula: A = P(1 + r/m)^(mt), where m is the compounding frequency per year.
Continuous compounding
The limit of compound interest as the compounding frequency approaches infinity: A = Pe^(rt). Think of it as interest being added every instant.
Product rule
d/dx [F(x)·S(x)] = F'(x)·S(x) + F(x)·S'(x). In simple terms, differentiate the first and multiply by the second, then add the first times the derivative of the second.
Quotient rule
d/dx [T(x)/B(x)] = [B(x)·T'(x) − B'(x)·T(x)] / [B(x)]². A common mnemonic: "low d-high minus high d-low, over low squared."
Chain rule
For a composite function m(x) = E[I(x)], the derivative is m'(x) = E'[I(x)] · I'(x). Differentiate the outside function leaving the inside alone, then multiply by the derivative of the inside.
Logarithmic differentiation
A technique for complex products, quotients and powers: take ln of both sides, use log properties to simplify, then differentiate implicitly. Useful when the function has multiple factors with exponents.
C'(x) ≈ C(x + 1) − C(x). The marginal cost approximates the exact cost of the next unit.
To find the exact cost of the (n + 1)th item, compute C(n + 1) − C(n). To approximate it, compute C'(n). Both should be close for large n.
Revenue function: R(x) = x · p(x), where p(x) is the price-demand function. Profit: P(x) = R(x) − C(x).
Average cost: C̅(x) = C(x)/x. Marginal average cost: C̅'(x).
Key exam nuance: to approximate the cost of producing the 11th tire, evaluate C'(10), not C'(11). The marginal function at n approximates the (n + 1)th item.
Compound interest: A = P(1 + r/m)^(mt). m = 1 (annual), 12 (monthly), 365 (daily).
Continuous: A = Pe^(rt). To find doubling time, set 2P = Pe^(rt), cancel P, take ln: t = ln(2)/r.
Exponential decay uses the same formula with a negative rate: A = Pe^(rt) where r < 0. Find r from two data points, then predict.
d/dx [e^x] = e^x. The exponential function is its own derivative.
d/dx [e^(Cx)] = Ce^(Cx). The constant multiplier comes along.
d/dx [ln(x)] = 1/x.
Simplify before differentiating: ln(x⁴) = 4 ln(x), and ln(e^(4x+1)) = 4x + 1.
Product rule: d/dx [F·S] = F'·S + F·S'. Apply when two functions are multiplied and you cannot easily expand them.
Quotient rule: d/dx [T/B] = [B·T' − B'·T] / B². Apply when one function is divided by another.
Example: f(x) = x/(4x − 8). T = x, B = 4x − 8. f'(x) = [(4x − 8)(1) − (4)(x)] / (4x − 8)² = −8/(4x − 8)².
To find where the tangent line is horizontal, set the numerator of f'(x) = 0 (denominator must not be zero there).
General power rule: if y = [u(x)]^n, then y' = n[u(x)]^(n−1) · u'(x). Identify the inner function u(x) and its derivative u'(x).
Chain rule for exponentials: d/dx [e^(g(x))] = e^(g(x)) · g'(x).
Chain rule for logarithms: d/dx [ln(g(x))] = g'(x)/g(x).
Example: y = (4x + 1)⁵. Inner: u = 4x + 1, u' = 4. y' = 5(4x + 1)⁴ · 4 = 20(4x + 1)⁴.
Example: y = e^(3x² − 1). y' = e^(3x² − 1) · 6x.
Combinations: use the product or quotient rule on the outer structure, and the chain rule on each factor.
For functions like f(x) = e^(4x)(3x + 1)² / √(5x² − 2):
Take ln of both sides: ln f(x) = 4x + 2 ln(3x + 1) − (1/2) ln(5x² − 2).
Differentiate both sides: f'(x)/f(x) = 4 + 6/(3x + 1) − 5x/(5x² − 2).
Multiply both sides by f(x) to isolate f'(x).
C'(x) \approx C(x+1) - C(x) \quad\text{(marginal cost approximation)}A = P\left(1 + \frac{r}{m}\right)^{mt} \quad\text{(compound interest)}A = Pe^{rt} \quad\text{(continuous compounding)}\frac{d}{dx}[e^x] = e^x \qquad \frac{d}{dx}[e^{Cx}] = Ce^{Cx} \qquad \frac{d}{dx}[\ln x] = \frac{1}{x}\text{Product: } (FS)' = F'S + FS'\text{Quotient: } \left(\frac{T}{B}\right)' = \frac{BT' - B'T}{B^2}\text{Chain: } \frac{d}{dx}[E(I(x))] = E'(I(x)) \cdot I'(x)\text{General power: } \frac{d}{dx}[u^n] = nu^{n-1} \cdot u'Log properties (used in logarithmic differentiation):
Rule | Formula |
|---|---|
Product | ln(uv) = ln(u) + ln(v) |
Quotient | ln(u/v) = ln(u) − ln(v) |
Power | ln(u^r) = r·ln(u) |
Students often evaluate C'(11) to approximate the 11th item. The correct input is C'(10), because the marginal function at n approximates the (n + 1)th item.
Students sometimes apply the product rule when they could simply expand the expression first. If you can multiply out, do so; it is usually faster.
Students forget to multiply by the inner derivative when using the chain rule. If you differentiate (4x + 1)⁵ and get 5(4x + 1)⁴ without the extra factor of 4, you have missed the chain rule.
Students confuse d/dx [e^(3x)] = 3e^(3x) with d/dx [e^x] = e^x. The constant in the exponent must come out front.
⚠️ Marginal analysis word problems (set up cost, revenue, profit functions and interpret the derivative) are a staple of business calculus exams.
⚠️ Continuous compounding and doubling-time problems appear frequently. Know A = Pe^(rt) and how to solve for t using natural log.
⚠️ The quotient rule is tested by asking you to find where the tangent line is horizontal (set the numerator of f'(x) = 0).
⚠️ Chain rule problems often combine with the product or quotient rule. Expect nested functions.
True or False: C'(100) approximates the cost of producing the 100th item. False. It approximates the 101st item.
Fill in the blank: d/dx [ln(x)] = ________. 1/x.
True or False: d/dx [e^(5x)] = e^(5x). False. It is 5e^(5x).
What rule do you use to differentiate f(x) = x³/(x + 1)? Quotient rule.
Fill in the blank: Doubling time under continuous compounding is t = ________. ln(2)/r.
Q: A company has C(x) = 3x + 5000 and price p(x) = x + 2. Find the marginal profit function.
A: R(x) = x·p(x) = x² + 2x. P(x) = R(x) − C(x) = x² + 2x − 3x − 5000 = x² − x − 5000. P'(x) = 2x − 1.
Q: Differentiate f(x) = (x² − 4x + 1)(x² − 1) using the product rule.
A: F = x² − 4x + 1, S = x² − 1. F' = 2x − 4, S' = 2x. f'(x) = (2x − 4)(x² − 1) + (x² − 4x + 1)(2x).
Q: Differentiate y = (2x − 1)³ / e^(5x).
A: Use the quotient rule. T = (2x − 1)³, B = e^(5x). T' = 3(2x − 1)² · 2 = 6(2x − 1)² (chain rule). B' = 5e^(5x). y' = [e^(5x) · 6(2x − 1)² − 5e^(5x) · (2x − 1)³] / [e^(5x)]².
Q: How long to double $200 at 3% compounded continuously?
A: 2(200) = 200e^(0.03t). Divide: 2 = e^(0.03t). ln(2) = 0.03t. t = ln(2)/0.03 ≈ 23.1 years.
Marginal analysis connects directly to optimisation (Sections 4.5–4.6), where you set P'(x) = 0 to find the production level that maximises profit. The chain rule is essential for implicit differentiation and related rates in more advanced courses. Compound interest and exponential growth/decay reappear in integral calculus (Section 5.1) and in differential equations.
Marginal cost, marginal revenue, marginal profit, average cost, marginal analysis, compound interest, continuous compounding, doubling time, exponential growth, exponential decay, product rule, quotient rule, chain rule, general power rule, composite function, inner function, outer function, logarithmic differentiation, e^x, ln(x), MAC 2233, business calculus, FIU