Difficulty: Intermediate | Prerequisites: Differentiation Rules notes
This is where calculus meets business. The derivative is no longer an abstract slope; it becomes the marginal cost of producing one more unit, the rate at which a population grows, or the percentage at which cost is changing. These problems follow a pattern: you are given a function (cost, revenue, profit, population), you take its derivative, and you interpret the result in context. If you can differentiate confidently, the applied part is mostly reading comprehension.
Marginal cost/revenue/profit is the derivative of the respective function, evaluated at a given production level. The "exact" cost of the nth unit is C(n) - C(n-1). The "estimated" cost uses the derivative at a nearby point. Percentage rate of change is [f'(x)/f(x)] x 100. Chain rule problems link two related rates together.
Marginal cost, C'(x)
The derivative of the cost function. It estimates the cost of producing one additional unit when x units have already been made. Think of it as: "How much more will it cost to make one more?"
Marginal revenue, R'(x)
The derivative of the revenue function. Estimates the additional revenue from selling one more unit.
Marginal profit, P'(x)
The derivative of the profit function. Since P(x) = R(x) - C(x), marginal profit = marginal revenue - marginal cost.
Exact cost of the nth unit
C(n) - C(n-1). The actual difference in total cost between producing n units and n-1 units.
Estimated cost of the nth unit
C'(a) where a is a value near n (typically the midpoint n - 0.5, or the value n-1). Uses the derivative as an approximation.
Percentage rate of change
[f'(x) / f(x)] x 100. Tells you the rate of change as a percentage of the current value. In simple terms, this means "by what percent is the function growing (or shrinking) at this moment?"
Cost of producing a fractional unit more
C'(x) times the fraction. For example, the cost of making half a unit more when 3 are produced: C'(3) x 0.5.
Exact: C(n) - C(n-1). Plug n and n-1 into the cost function and subtract.
Estimated: Take the derivative C'(x) and evaluate at a point near n. The course uses the midpoint: C'(n - 0.5).
Example: C(x) = x² + 2x + 1000. Exact cost of the 3rd unit: C(3) - C(2) = 1015 - 1008 = $7. Estimated cost: C'(x) = 2x + 2, so C'(2.5) = 2(2.5) + 2 = 7. (Here exact and estimate happen to match.)
Multiply the marginal cost by the fraction.
Example: How much does it cost to make 1/2 unit more when 3 have been produced? C'(3) = 2(3) + 2 = 8. Cost of half a unit: 8 x 0.5 = $4.
Formula: [C'(x) / C(x)] x 100
Example: At x = 3, C'(3) = 8 and C(3) = 1015. Percentage rate = (8/1015) x 100 = 0.79%.
Exact: P(n) - P(n-1).
Estimated: P'(a) where a is near n.
Example: P(x) = x² + 10x + 200. Exact profit from the 4th unit: P(4) - P(3) = 256 - 239 = $17. Estimated: P'(x) = 2x + 10, P'(3.5) = 2(3.5) + 10 = 17.
Given P(x) = x² + 5x + 100 (population x months from now).
"By how much did the population increase during the 3rd month?" means P(3) - P(2) = 124 - 114 = 10 persons. This is the exact increase.
"At what rate is the population changing at 5 months?" means find P'(5). P'(x) = 2x + 5, so P'(5) = 15 persons/month.
"At what percentage rate?" means [P'(5)/P(5)] x 100 = (15/150) x 100 = 10%.
Q(k) = 600k^(1/2) where k is capital in units of $1000.
Q'(k) = 300k^(-1/2).
Current capital is $100,000, so k = 100. Q'(100) = 300/10 = 30.
Estimate the increase from raising capital by $400: since $400 = 0.4 units of $1000, the increase is approximately Q'(100) x 0.4 = 30 x 0.4 = 12 units.
If cost C depends on units x, and units x depend on time t, then the rate of change of cost with respect to time is: dC/dt = C'(x) · x'(t). This is the chain rule applied to real-world quantities.
Example: C(x) = x² + 2x + 200 and x(t) = 2t² + t. At t = 1: x(1) = 2(1)² + 1 = 3 (so 3 notepads are produced). C'(x) = 2x + 2, so C'(3) = 8. x'(t) = 4t + 1, so x'(1) = 5. Rate of cost change with respect to time: 8 x 5 = 40 dollars per hour.
Same principle: if x(t) = t² + 2t, then x'(t) = 2t + 2. At t = 2: x'(2) = 6 units per hour. Combined with C'(x), you get the cost change per hour.
Students confuse "exact cost of the nth unit" with "estimated cost." Exact means C(n) - C(n-1), two full evaluations. Estimated means using the derivative C'(a) at a nearby point.
When estimating the cost of the nth unit, students plug in x = n into C'(x). The course convention is to use the midpoint: C'(n - 0.5). Pay attention to how your professor phrases it.
For percentage rate of change, students sometimes forget to multiply by 100 or divide by the function value. The formula is [C'(x)/C(x)] x 100, not just C'(x).
In chain rule applied problems (dC/dt), students often forget the second factor. dC/dt is not just C'(x); it is C'(x) times x'(t). Both pieces are required.
⚠️ "Find the exact cost of the nth unit" vs "estimate the cost" are different exam questions with different methods. Know which is which.
⚠️ Percentage rate of change appears on nearly every applied test. Memorise [f'(x)/f(x)] x 100.
⚠️ Chain rule word problems (cost depends on units, units depend on time) are a favourite test item. The answer is always two derivatives multiplied together.
⚠️ Population growth problems test the same skills: P(n) - P(n-1) for exact, P'(x) for rate, [P'(x)/P(x)] x 100 for percentage rate.
⚠️ Read the question carefully for units. "Dollars per unit" vs "dollars per hour" vs "percentage" each require a different calculation.
Fill in the blank: The exact cost of the 5th unit is C() - C().
C(5) - C(4).
True or false: Marginal cost C'(x) gives the exact cost of the next unit.
False. It gives an estimate. The exact cost is C(x+1) - C(x).
Fill in the blank: Percentage rate of change = [___ / ___] x 100.
f'(x) / f(x).
True or false: If C depends on x and x depends on t, then dC/dt = C'(x) + x'(t).
False. It is C'(x) times x'(t) (multiplication, not addition).
Q: C(x) = x² + 2x + 1000. What is the exact cost of manufacturing the 3rd unit?
A: C(3) - C(2) = (9 + 6 + 1000) - (4 + 4 + 1000) = 1015 - 1008 = $7.
Q: Using the same cost function, estimate the cost of the 3rd unit.
A: C'(x) = 2x + 2. Evaluate at x = 2.5: C'(2.5) = 7. Estimated cost is $7.
Q: P(x) = x² + 5x + 100 represents population x months from now. At what percentage rate is the population changing at 5 months?
A: P'(x) = 2x + 5, so P'(5) = 15. P(5) = 25 + 25 + 100 = 150. Percentage rate = (15/150) x 100 = 10%.
Q: C(x) = x² + 2x + 200 and x(t) = 2t² + t. At what rate is manufacturing cost changing with respect to time after 1 hour?
A: At t = 1, x(1) = 3. C'(3) = 8. x'(1) = 5. dC/dt = 8 x 5 = 40 dollars per hour.
Q: Q(k) = 600k^(1/2), k in units of $1000. Current capital is $100,000. Estimate the increase in output from a $400 increase in capital.
A: k = 100. Q'(k) = 300k^(-1/2), so Q'(100) = 30. $400 = 0.4 units of $1000. Increase = 30 x 0.4 = 12 units.
Marginal analysis is the business application of derivatives, so it depends on all the differentiation rules from the previous unit. Optimisation (next topic) builds on this: to maximise profit or minimise cost, you set the marginal function to zero and solve. Elasticity of demand uses a similar ratio structure. Integration (later in the course) reverses the process: given a marginal cost function, you integrate to recover the total cost function.
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