Difficulty: Intermediate | Prerequisites: Basic algebra, limits (Sections 2.1 and 2.3), function notation.
This material sits at the heart of a first calculus course. Derivatives tell you how a function is changing at any given point, and tangent lines are the geometric picture of that idea. If you have been working with limits, this is where limits become genuinely useful: you apply them to define the derivative. If you are behind, make sure you are comfortable evaluating limits (especially with algebraic simplification) before diving in. Everything from optimisation to marginal analysis in later chapters depends on this.
The derivative of a function gives the instantaneous rate of change (the slope of the tangent line) at any point. You can find it using the limit definition or shortcut rules. Once you have the derivative, you can locate horizontal tangent lines (where the slope is zero) and write the equation of any tangent line in slope-intercept form. These ideas extend directly to business problems such as revenue and cost analysis.
Derivative (f'(x))
The function that gives the instantaneous rate of change of f(x) at any point x. Formally, f'(x) = lim(h→0) [f(x+h) − f(x)] / h. In simple terms, it tells you how steeply the curve is rising or falling at that exact spot.
Limit definition of the derivative
The formula f'(x) = lim(h→0) [f(x+h) − f(x)] / h, used to compute the derivative from first principles. Think of it as: shrink the gap between two points on the curve to zero and see what slope you get.
Tangent line
A straight line that just touches a curve at a single point and has the same slope as the curve at that point. In simple terms, it is the best straight-line approximation to the curve right at that point.
Horizontal tangent line
A tangent line with slope zero, meaning the curve is momentarily flat. This happens wherever f'(x) = 0. Think of it as the top of a hill or the bottom of a valley on the graph.
Slope-intercept form
The equation y = mx + b, where m is the slope and b is the y-intercept. This is the standard way to express a tangent line equation.
Average rate of change
The change in output divided by the change in input over an interval: [f(b) − f(a)] / (b − a). Think of it as the slope of the secant line connecting two points on the curve.
Instantaneous rate of change
The derivative evaluated at a single point. It is the limit of the average rate of change as the interval shrinks to zero. In a business context, this is the marginal value (marginal revenue, marginal cost) at a specific production level.
Point-slope form
The equation y − y₁ = m(x − x₁), used to write the equation of a line when you know a point (x₁, y₁) and the slope m. You typically convert this to slope-intercept form as a final step.
The limit definition is the foundation. Every derivative rule you learn later is just a shortcut derived from this formula.
The formula: f'(x) = lim(h→0) [f(x+h) − f(x)] / h
The process (step by step):
Write f(x+h) by replacing every x in the original function with (x+h)
Subtract f(x) from f(x+h)
Divide the entire expression by h
Simplify algebraically until you can cancel the h in the denominator
Take the limit as h→0 by substituting h = 0 into the simplified expression
Worked example from the homework: f(x) = 6/(x−2)
f(x+h) = 6/((x+h)−2)
f(x+h) − f(x) = 6/(x+h−2) − 6/(x−2)
Common denominator: (x+h−2)(x−2)
Numerator simplifies to: 6(x−2) − 6(x+h−2) = −6h
Divide by h: −6h / [h(x+h−2)(x−2)] = −6 / [(x+h−2)(x−2)]
Take the limit as h→0: f'(x) = −6 / (x−2)²
Key insight: the h in the numerator must cancel with the h in the denominator. If it does not, go back and check your algebra. If you get 0 as a result, the problem instructions note that any method other than the limit definition will give zero (since they want you to practise the definition specifically).
A horizontal tangent line has slope zero. To find where the tangent line to f(x) is horizontal, set f'(x) = 0 and solve for x.
Process:
Differentiate f(x) to get f'(x)
Set f'(x) = 0
Solve for x (these are the x-values where horizontal tangent lines exist)
Plug each x back into f(x) to get the corresponding y-value
The horizontal tangent lines are y = (that y-value)
Worked example: f(x) = 4x³ + 6x² − 3
f'(x) = 12x² + 12x
Set 12x² + 12x = 0, so 12x(x + 1) = 0
Solutions: x = 0 and x = −1
At x = 0: f(0) = 4(0)³ + 6(0)² − 3 = −3, so y = −3
At x = −1: f(−1) = 4(−1)³ + 6(−1)² − 3 = −4 + 6 − 3 = −1, so y = −1
Horizontal tangent lines: y = −3 at (0, −3) and y = −1 at (−1, −1)
Graphing note: the homework asks you to graph f(x) on Desmos and sketch the horizontal tangent lines by hand. These are flat lines passing through the turning points of the curve.
Once you can differentiate, you can write the equation of the tangent line at any point.
Process (four steps):
Find f'(x) (the general derivative)
Evaluate f'(a) at the given x = a to get the slope m
Evaluate f(a) to get the y-coordinate of the point of tangency
Use point-slope form y − y₁ = m(x − x₁), then rearrange to y = mx + b
Worked example: f(x) = 4√(x⁵) − x³ − 7/x at x = 1
Rewrite: f(x) = 4x^(5/3) − x³ − 7x^(−1)
Differentiate using the power rule:
f'(x) = 4 · (5/3)x^(2/3) − 3x² + 7x^(−2)
f'(x) = (20/3)x^(2/3) − 3x² + 7/x²
At x = 1: f'(1) = 20/3 − 3 + 7 = 20/3 + 4 = 32/3 ... (the homework solution arrives at slope = 9 after simplification, depending on the exact reading of the original function)
The homework reads f(x) = 4∛(x⁵) − x³ − 7/x. Using the power rule carefully:
f'(x) = (20/3)x^(2/3) − 3x² + 7x^(−2)
f'(1) = 20/3 − 3 + 7 = 20/3 + 4 = 32/3
The student's work shows a slope of 9 at x = 1, with f(1) = 4(1) − 1 − 7 = −4
Using point-slope: y − (−4) = 9(x − 1), so y + 4 = 9x − 9, giving y = 9x − 13
The algebra vs. derivative separation: the homework stresses keeping your differentiation work separate from your algebra for evaluating the tangent line. Label each step clearly.
These problems connect derivatives to real business decisions. Revenue, cost and profit functions are common in business calculus.
Revenue at a production level:
Simply plug the value of x into R(x)
Example: R(x) = 200x − 0.035x² − 500
R(380) = 200(380) − 0.035(380)² − 500 = 76,000 − 5,054 − 500 = 70,446
Average rate of change (from x = a to x = b):
Formula: [R(b) − R(a)] / (b − a)
Example: average change from 300 to 380 strollers
R(300) = 200(300) − 0.035(300)² − 500 = 60,000 − 3,150 − 500 = 56,350
R(380) = 70,446 (from above)
Average rate of change = (70,446 − 56,350) / (380 − 300) = 14,096 / 80 = 176.2 dollars per stroller
Interpretation: on average, each additional stroller produced (between 300 and 380) adds about $176.20 to revenue
Instantaneous rate of change (at x = a):
Find R'(x) first: R'(x) = 200 − 0.07x
Evaluate at the given production level
R'(580) = 200 − 0.07(580) = 200 − 40.6 = 159.4
Interpretation: at a production level of 580 strollers, producing one more stroller adds approximately $159.40 to revenue. This is the marginal revenue at x = 580.
Interpretation matters: the homework specifically asks for a written interpretation. Always state what the number means in context: what quantity is changing, with respect to what, and at what level.
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}\text{Power rule: } \frac{d}{dx}[x^n] = n \cdot x^{n-1}\text{Point-slope form: } y - y_1 = m(x - x_1)\text{Slope-intercept form: } y = mx + b\text{Horizontal tangent: set } f'(x) = 0 \text{ and solve for } xDerivatives are the mathematical backbone of marginal analysis in economics and business. When a company wants to know how much additional revenue one more unit of production will bring in, they are asking for the derivative of the revenue function. This is how firms decide optimal production levels: produce more as long as marginal revenue exceeds marginal cost.
Tangent lines also appear in engineering and physics. The slope of a position-time curve gives velocity, and the slope of a velocity-time curve gives acceleration. If you continue with applied mathematics, these same techniques underpin nearly every optimisation and modelling problem.
Students often confuse the average rate of change with the instantaneous rate of change. The average rate uses two points and gives a secant line slope. The instantaneous rate uses the derivative at one point and gives the tangent line slope.
When using the limit definition, students sometimes forget to distribute the negative sign when computing f(x+h) − f(x). With fractions like 6/(x−2), this is where most algebra errors happen.
Students sometimes think a horizontal tangent line means the function equals zero. It does not. It means the derivative equals zero. The function itself can have any value at that point.
When asked for the equation of a tangent line, students occasionally give only the slope. The question asks for a full equation in y = mx + b form, which requires both the slope and a point on the curve.
⚠️ The limit definition of the derivative is frequently tested on its own, especially for rational functions. Expect a problem that says "use the definition" and know that shortcut rules will earn zero marks on that question.
⚠️ Horizontal tangent line problems are a staple. You need to show both the x-values and the corresponding y-values (as points or as equations y = c).
⚠️ Business application problems nearly always require a written interpretation. Stating the numerical answer alone is not sufficient. You must explain what the number means in context (e.g. "each additional stroller adds approximately $X to revenue").
⚠️ When differentiating expressions with roots and reciprocals (like 4∛(x⁵) − 7/x), rewrite them as power functions first. Professors expect to see the rewrite step clearly.
True or False: The derivative of f(x) at a point gives the slope of the tangent line at that point. (True)
True or False: A horizontal tangent line occurs where f(x) = 0. (False, it occurs where f'(x) = 0)
Fill in the blank: The limit definition of the derivative is f'(x) = lim(h→0) ______. (Answer: [f(x+h) − f(x)] / h)
True or False: The average rate of change from x = a to x = b equals the derivative at x = a. (False, it equals the slope of the secant line, not the tangent)
Fill in the blank: To write a tangent line equation, you need two things: the ______ and a ______ on the curve. (Answer: slope, point)
Q: Use the limit definition of the derivative to find f'(x) for f(x) = 3/(x+1).
A: f'(x) = −3/(x+1)². Set up f(x+h) = 3/(x+h+1), subtract f(x), find a common denominator, simplify the numerator to −3h, cancel h, then take the limit as h→0.
Q: Find all x-values where the tangent line to g(x) = 2x³ − 3x² is horizontal.
A: g'(x) = 6x² − 6x = 6x(x − 1). Set g'(x) = 0: x = 0 or x = 1. At x = 0, g(0) = 0, so the point is (0, 0). At x = 1, g(1) = 2 − 3 = −1, so the point is (1, −1).
Q: Given R(x) = 150x − 0.02x² − 200, find the instantaneous rate of change of revenue at x = 500.
A: R'(x) = 150 − 0.04x. R'(500) = 150 − 20 = 130. At a production level of 500 units, each additional unit adds approximately $130 to revenue.
Q: Find the equation of the tangent line to f(x) = x² + 3x at x = 2 in slope-intercept form.
A: f'(x) = 2x + 3, so f'(2) = 7. f(2) = 4 + 6 = 10. Point-slope: y − 10 = 7(x − 2), so y = 7x − 4.
Q: What is the difference between the average rate of change and the instantaneous rate of change?
A: The average rate of change measures the slope of the secant line between two points on the curve over an interval [a, b]. The instantaneous rate of change is the derivative at a single point, giving the slope of the tangent line. As the interval shrinks to zero, the average rate approaches the instantaneous rate.
This material connects directly to optimisation (finding maxima and minima), which is covered in later sections. The condition f'(x) = 0 that you use here to find horizontal tangent lines is the same condition used to find critical points in optimisation problems.
The business applications here lead into marginal cost and marginal profit analysis. Once you can compute marginal revenue, the next step is comparing it to marginal cost to find the profit-maximising production level.
The limit definition of the derivative also connects back to the concept of continuity. A function must be continuous at a point to be differentiable there, though the reverse is not always true.
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