Difficulty: Introductory | Prerequisites: Basic limits (Section 2.3), function notation, algebraic manipulation of polynomials.
Section 2.4 introduces the derivative, the single most important concept in all of calculus. Up to this point you have been computing limits as a technique. Now limits become the engine behind a new tool: the derivative, which measures how fast a function's output changes when its input changes. In a business calculus course, this is the foundation for marginal cost, marginal revenue, and optimisation problems you will meet in later chapters. You should already be comfortable evaluating limits and expanding algebraic expressions such as (x + h) squared.
The derivative of a function is defined as a limit and tells you the slope of the tangent line at any point on the curve. You can find the derivative using the limit definition, then plug in a specific x-value to get the slope there, and from that write the equation of the tangent line using point-slope form.
Derivative
The instantaneous rate of change of a function at a given point, defined formally as the limit of the difference quotient as h approaches 0. In simple terms, it tells you how quickly the output of a function is changing at a specific input value.
Limit definition of the derivative
The formula dy/dx = lim(h→0) [f(x+h) − f(x)] / h. Think of it as zooming in on two points on a curve until they are infinitely close together, so the secant line becomes the tangent line.
Tangent line
A straight line that touches a curve at exactly one point (locally) and has the same slope as the curve at that point. In simple terms, if you zoomed in far enough on a smooth curve, it would look like this line.
Difference quotient
The expression [f(x+h) − f(x)] / h, which gives the average rate of change (slope of the secant line) between x and x+h. The derivative is what this becomes when h shrinks to zero.
Point-slope form
The equation y − y₁ = m(x − x₁), used to write the equation of a line when you know one point (x₁, y₁) and the slope m. This is the standard way to write a tangent line equation once you have the derivative value and the point of tangency.
The derivative is defined as:
dy/dx = lim(h→0) [f(x+h) − f(x)] / h
This formula works by computing the slope between two points on f(x), at x and at x+h, and then letting the gap h shrink to zero.
The result is a new function, f′(x), that outputs the slope of the original function at any x you choose.
At any point on a curve, the derivative gives the slope of the tangent line to the curve at that point.
A tangent line touches the curve at one point and matches the curve's direction there.
So if f′(a) = 5, the tangent line at x = a rises with slope 5.
Step 1: Apply the limit definition
Compute f(x+h): substitute (x+h) into the function.
f(x+h) = 3(x+h)² − 2(x+h) + 1
Expand: 3(x² + 2xh + h²) − 2x − 2h + 1
Simplify: 3x² + 6xh + 3h² − 2x − 2h + 1
Form the difference quotient: [f(x+h) − f(x)] / h
Numerator: (3x² + 6xh + 3h² − 2x − 2h + 1) − (3x² − 2x + 1)
Cancel matching terms: 6xh + 3h² − 2h
Factor out h: h(6x + 3h − 2)
Divide by h: 6x + 3h − 2
Take the limit as h→0: f′(x) = 6x − 2
Step 2: Find the tangent line at x = −2
Find the y-coordinate: f(−2) = 3(4) − 2(−2) + 1 = 12 + 4 + 1 = 17. So the point is (−2, 17).
Find the slope: f′(−2) = 6(−2) − 2 = −12 − 2 = −14.
Write the tangent line using point-slope form:
y − 17 = −14(x − (−2))
y − 17 = −14(x + 2)
y = −14x − 28 + 17
y = −14x − 11
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}y - y_1 = m(x - x_1)The first formula is the limit definition of the derivative. The second is point-slope form, used to write the equation of a tangent line once you know the slope (from the derivative) and the point of tangency (from the original function).
In business, the derivative at a point is a marginal quantity. If C(x) is the cost of producing x units, then C′(x) tells you the approximate cost of making one more unit. Similarly, R′(x) gives marginal revenue. Tangent lines let you build linear approximations, which are used constantly in forecasting: "if we increase production by 10 units from here, cost rises by roughly 10 × C′(x)."
Students often forget to subtract the entire original function f(x) when forming the difference quotient. Every term in f(x) must be subtracted, including constants. Missing a sign here is the most common algebra error in this section.
The derivative f′(x) is a function, not a single number. It only becomes a number when you evaluate it at a specific x-value. Writing "f′(x) = −14" as the general derivative is wrong; −14 is f′(−2).
A tangent line is not the same as a secant line. The secant crosses the curve at two points. The tangent touches at one point and matches the curve's slope there. The derivative gives the tangent slope, not the secant slope.
Students sometimes plug the x-value into the derivative before finishing the limit process. Complete the algebra, cancel h, and take the limit first. Only then substitute your specific x-value.
⚠️ The limit definition of the derivative is almost certainly on your exam. Be prepared to apply it from scratch to a polynomial, not just state the formula.
⚠️ Expect a question that asks for both the derivative and the tangent line equation at a given point. This is a two-part process: derivative first (slope), then point-slope form.
⚠️ Show every algebraic step when using the limit definition. Partial credit depends on the expansion, the cancellation, and the limit being clearly taken. Jumping straight to the power rule (Section 2.5) will not earn credit here.
⚠️ Know the vocabulary: "instantaneous rate of change" = the derivative. If the exam says "find the instantaneous rate of change," it means find f′(x) or f′(a).
True or False: The derivative of a function at a point gives the slope of the secant line at that point.
Answer: False. It gives the slope of the tangent line.
Fill in the blank: f′(x) = lim(h→0) ____
Answer: [f(x+h) − f(x)] / h
True or False: If f(x) = 3x² − 2x + 1, then f′(x) = 6x − 2.
Answer: True.
Fill in the blank: To write the equation of a tangent line, you need two things: the ____ and a ____ on the curve.
Answer: slope (from the derivative), point (from the original function).
True or False: When using the limit definition, you should substitute your specific x-value before cancelling h.
Answer: False. Cancel h and take the limit first, then substitute.
Q: Use the limit definition to find the derivative of g(x) = x² + 4x.
A: g(x+h) = (x+h)² + 4(x+h) = x² + 2xh + h² + 4x + 4h. The difference quotient becomes (2xh + h² + 4h)/h = 2x + h + 4. Taking the limit as h→0 gives g′(x) = 2x + 4.
Q: For f(x) = 3x² − 2x + 1, find the equation of the tangent line at x = 1.
A: f(1) = 3 − 2 + 1 = 2, so the point is (1, 2). f′(x) = 6x − 2, so f′(1) = 4. The tangent line is y − 2 = 4(x − 1), which simplifies to y = 4x − 2.
Q: What does the derivative f′(a) represent geometrically?
A: It represents the slope of the line tangent to the graph of f at the point (a, f(a)).
Q: If the tangent line to y = f(x) at x = 3 has a slope of 0, what does that tell you about the graph at that point?
A: The graph has a horizontal tangent at x = 3. The function is neither increasing nor decreasing at that instant; this often indicates a local maximum or minimum.
This section builds directly on limits (Section 2.3). Every derivative is a limit, so weak limit skills will follow you here. In Section 2.5, you will learn shortcut rules (the power rule, constant rule, sum rule) that let you skip the long limit process for standard functions, but those rules are derived from this definition.
Later in the course, the derivative is used to find where functions increase or decrease (first derivative test), where they curve upward or downward (concavity and the second derivative), and where profit, cost, or revenue are optimised. The tangent line itself reappears as the tool behind linear approximation and differentials.
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