Difficulty: Intermediate | Prerequisites: Limits and Continuity (Sections 2.1–2.3), basic algebra
The derivative is the central idea of differential calculus: it measures how fast a function's output changes relative to its input at a single instant. In a business context, this translates directly to marginal cost, marginal revenue, and marginal profit, the quantities that tell a firm whether producing one more unit helps or hurts. This chapter moves from the formal limit definition of the derivative through the shortcut rules (power rule, basic differentiation) and into applications: finding tangent lines, computing rates of change, and interpreting results in business language.
The derivative f'(x) gives the instantaneous rate of change of f at any point x. You can find it from the limit definition or from shortcut rules (power rule). Once you have f'(x), you can write tangent-line equations, find where the function is flat, and interpret rates of change in revenue, cost, and profit problems.
Derivative (f'(x) or dy/dx)
The instantaneous rate of change of a function at a given point. Geometrically, it is the slope of the tangent line to the curve at that point. Think of it as: the speedometer reading at one exact moment, not the average speed over a trip.
Definition of the derivative (limit definition)
f'(x) = lim (h→0) [f(x+h) – f(x)] / h. This is the formal way to compute f'(x) from first principles. Some exam problems require this method specifically.
Tangent line
The straight line that just touches a curve at a single point and has the same slope as the curve at that point. Its equation in slope-intercept form is y = f'(a)(x – a) + f(a), where a is the x-value of the point of tangency.
Horizontal tangent line
A tangent line with slope zero. Occurs wherever f'(x) = 0. These points are candidates for local maxima and minima.
Average rate of change
The slope of the secant line between two points: [f(b) – f(a)] / (b – a). In business, this is the average change in revenue (or cost, or profit) per unit over an interval.
Instantaneous rate of change
The derivative evaluated at a specific point: f'(a). This is the rate of change at exactly that one production level, not over a range.
Marginal cost / marginal revenue / marginal profit
The derivative of the cost, revenue, or profit function, respectively. Marginal cost C'(x) approximates the cost of producing one additional unit beyond x. Same logic for revenue and profit.
Marginal average profit
The derivative of the average profit function. Average profit = P(x)/x, and the marginal average profit is d/dx [P(x)/x].
Profit function
P(x) = R(x) – C(x). Revenue minus cost. The marginal profit P'(x) = R'(x) – C'(x).
Some problems require this method and will give zero credit for shortcut rules. The process:
Write f(x+h): replace every x in f(x) with (x+h).
Form [f(x+h) – f(x)] / h.
Simplify the numerator (expand, combine like terms).
Cancel h from numerator and denominator.
Take the limit as h → 0.
Example: f(x) = 6/(x–2).
f(x+h) = 6/((x+h)–2).
[f(x+h) – f(x)]/h = [6/((x+h)–2) – 6/(x–2)] / h.
Common denominator for the numerator: [6(x–2) – 6((x+h)–2)] / [((x+h)–2)(x–2) · h].
Simplify numerator: 6x – 12 – 6x – 6h + 12 = –6h.
Cancel h: –6 / [((x+h)–2)(x–2)].
Take limit as h→0: f'(x) = –6/(x–2)².
If f(x) = x^n, then f'(x) = n·x^(n–1).
Constant multiple rule: d/dx [c·f(x)] = c·f'(x).
Sum/difference rule: d/dx [f(x) ± g(x)] = f'(x) ± g'(x).
Rewrite roots and reciprocals as powers before differentiating.
√x = x^(1/2), so d/dx[√x] = (1/2)x^(–1/2).
1/x = x^(–1), so d/dx[1/x] = –x^(–2) = –1/x².
⁴√(x⁵) = x^(5/4), so d/dx = (5/4)x^(1/4).
7/x = 7x^(–1), so d/dx = –7x^(–2) = –7/x².
Find f'(x) using differentiation rules.
Evaluate f'(a) to get the slope m at x = a.
Evaluate f(a) to get the y-coordinate of the point.
Write the equation: y = m(x – a) + f(a), then simplify to slope-intercept form y = mx + b.
Example: f(x) = 4x^(5/4) – x³ – 7/x at x = 1.
f'(x) = 5x^(1/4) – 3x² + 7x^(–2).
f'(1) = 5 – 3 + 7 = 9. Slope = 9.
f(1) = 4 – 1 – 7 = –4. Point is (1, –4).
Tangent line: y = 9(x – 1) – 4 = 9x – 13.
Set f'(x) = 0 and solve for x. These are the x-values where the curve flattens out.
Example: f(x) = 4x³ + 6x² – 3. f'(x) = 12x² + 12x = 12x(x + 1) = 0. Horizontal tangents at x = 0 and x = –1.
Revenue at a production level: substitute x into R(x).
Average change in revenue from x = a to x = b: [R(b) – R(a)] / (b – a).
Instantaneous rate of change of revenue at x = a: R'(a).
Interpretation matters: state the units and what the number means in context.
Example: R(x) = 200x – 0.035x² – 500, at x = 380.
R(380) = 200(380) – 0.035(380)² – 500 = 76,000 – 5,054 – 500 = $70,446.
R'(x) = 200 – 0.07x. R'(380) = 200 – 26.6 = $173.40 per stroller.
Interpretation: When 380 strollers are produced, revenue is $70,446 and is increasing at a rate of approximately $173.40 per additional stroller.
Profit function: P(x) = R(x) – C(x).
Marginal profit: P'(x) = R'(x) – C'(x).
Average profit: P̄(x) = P(x)/x.
Marginal average profit: P̄'(x) = d/dx [P(x)/x] (use the quotient rule).
To estimate the profit from selling the (n+1)th unit, evaluate P'(n). This gives an approximation, not the exact profit from that single unit.
f'(x) = lim (h→0) [f(x+h) – f(x)] / h
d/dx [x^n] = n · x^(n–1), for any real number n
y = f'(a)(x – a) + f(a)
Revenue: R(x) = x · p(x), where p(x) is the price-demand function
Profit: P(x) = R(x) – C(x)
Average cost: C̄(x) = C(x)/x
Marginal cost: C'(x)
Marginal revenue: R'(x)
Marginal profit: P'(x) = R'(x) – C'(x)
Students often leave the limit notation on after substituting in the limit definition. Once you have cancelled h and taken the limit, the lim symbol goes away.
Students often forget to rewrite radicals and reciprocals as fractional/negative powers before applying the power rule. You cannot differentiate √x directly; rewrite as x^(1/2) first.
Students confuse average rate of change (secant slope over an interval) with instantaneous rate of change (derivative at a point). The exam will ask for both, and they are different calculations.
Students often forget to write an interpretation sentence when the problem asks for one. State the value, the units, and what it means in context.
⚠️ If a problem says "use the definition of the derivative," the power rule will score zero.
⚠️ Tangent line problems require both the derivative (for slope) and the original function (for the point). Show both calculations clearly separated.
⚠️ Word problems will ask you to interpret your numerical answer in a sentence. Include units (dollars, dollars per unit, etc.).
⚠️ Marginal analysis: P'(300) estimates the profit from the 301st unit, not the 300th.
True or False: The derivative of 7/x is –7/x². (True. Rewrite as 7x^(–1), apply power rule: –7x^(–2) = –7/x².)
Fill in the blank: A horizontal tangent line occurs where f'(x) = __________. (0)
True or False: The average rate of change of R(x) from x = 100 to x = 200 equals R'(100). (False. Average rate of change is [R(200)–R(100)]/100, not R'(100).)
Fill in the blank: P'(500) estimates the profit from the sale of the __________th unit. (501st)
True or False: To find the tangent line at x = a you need both f(a) and f'(a). (True.)
Q: Use the definition of the derivative to find f'(x) for f(x) = 6/(x–2).
A: f'(x) = lim (h→0) [6/((x+h)–2) – 6/(x–2)] / h. Common denominator, simplify numerator to –6h, cancel h, take limit: f'(x) = –6/(x–2)².
Q: Find the x-values where the tangent line to f(x) = 4x³ + 6x² – 3 is horizontal.
A: f'(x) = 12x² + 12x = 12x(x+1) = 0. Solutions: x = 0 and x = –1.
Q: The revenue from selling x strollers is R(x) = 200x – 0.035x² – 500. Find the average change in revenue from 300 to 380 strollers.
A: R(380) = 70,446. R(300) = 56,350. Average change = (70,446 – 56,350)/(380 – 300) = 14,096/80 = $176.20 per stroller.
Q: For C(x) = 3x – 400√x + 3000 and R(x) = 16x, find the profit function and estimate the profit from the 301st unit.
A: P(x) = R(x) – C(x) = 16x – (3x – 400√x + 3000) = 13x + 400√x – 3000. P'(x) = 13 + 200/√x. P'(300) = 13 + 200/√300 ≈ 13 + 11.547 ≈ $24.55. The estimated profit from the 301st unit is approximately $24.55.
The derivative is used everywhere from this point forward. In Sections 3.2–3.4, you will differentiate exponential and logarithmic functions using the chain rule, product rule, and quotient rule. In Sections 4.1–4.6, the first and second derivatives drive the graphing strategy (increasing/decreasing, concavity, extrema). Marginal analysis reappears whenever a business application problem involves optimisation.
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