Difficulty: Intermediate | Prerequisites: Ch. 2.1–2.3 (limits, continuity, average rate of change)
The derivative is the central idea in differential calculus. Everything that follows in this course, from marginal cost to optimisation, depends on understanding what a derivative is and how to compute one from the definition. If you are comfortable evaluating limits and simplifying algebraic expressions, you are ready for this chapter. If limits still feel shaky, revisit Ch. 2.1 first.
The derivative of a function at a point gives the instantaneous rate of change at that point. You find it by writing the difference quotient, simplifying, and then taking the limit as h approaches 0. The result is usually a new function that tells you the slope of the original function at any x value.
Derivative
The limit of the difference quotient as h approaches 0. Written f′(x). It gives the instantaneous rate of change of f(x) at a given point.
Think of it as: the exact slope of the curve at one specific x value, rather than an average slope between two points.
Difference quotient
The expression [f(x + h) – f(x)] / h, where h ≠ 0. This measures the average rate of change of f over the interval from x to x + h.
In simple terms, this is the slope formula (rise over run) applied to a curve, with the two points separated by a distance h.
Average rate of change
For a function f, the average rate of change from a to a + h is [f(a + h) – f(a)] / h, where h ≠ 0.
Think of it as: the slope of the straight line (secant line) connecting two points on the curve.
Average velocity
Distance covered divided by elapsed time: v = d / t. A special case of average rate of change where the function describes position over time.
In simple terms, this means how fast something moved on average, ignoring speed-ups and slow-downs along the way.
Instantaneous rate of change
The rate of change at a single moment, obtained by taking the limit of the average rate of change as the interval shrinks to zero. This is what the derivative gives you.
Think of it as: the speedometer reading at one exact instant, as opposed to the trip-average speed.
The derivative from the definition (sometimes called "the long way" or "using first principles") follows four steps every time. Learn this sequence cold; exam problems will ask you to apply it to different function types.
Step 1: Find f(x + h)
Take the original function and replace every x with (x + h).
Expand and simplify fully.
Example: if f(x) = 4x – x², then f(x + h) = 4(x + h) – (x + h)² = 4x + 4h – x² – 2xh – h².
Step 2: Find f(x + h) – f(x)
Subtract the original f(x) from the result in Step 1.
Group and cancel terms.
Example: (4x + 4h – x² – 2xh – h²) – (4x – x²) = 4h – 2xh – h².
Step 3: Find [f(x + h) – f(x)] / h
Divide the entire numerator from Step 2 by h.
Factor h from the numerator so it cancels with the denominator.
Example: (4h – 2xh – h²) / h = h(4 – 2x – h) / h = 4 – 2x – h.
Step 4: Take the limit as h → 0
Let h = 0 in the simplified expression from Step 3.
The result is f′(x), the derivative.
Example: lim(h→0) (4 – 2x – h) = 4 – 2x.
So f′(x) = 4 – 2x.
The answer is normally another equation (a function of x), not a single number. You plug specific x values into f′(x) when you need the slope at a particular point.
The four-step process works on every function, but the algebra in Steps 1–3 looks different depending on the function type. Three common types from this chapter:
Rational functions (e.g. f(x) = 1/x)
Step 1: f(x + h) = 1/(x + h).
Step 2: f(x + h) – f(x) = 1/(x + h) – 1/x. Find a common denominator: = [x – (x + h)] / [x(x + h)] = –h / [x(x + h)].
Step 3: Divide by h: [–h / x(x + h)] / h = –1 / [x(x + h)], where h ≠ 0.
Step 4: lim(h→0) –1 / [x(x + h)] = –1/x², where x ≠ 0.
Result: f′(x) = –1/x².
The key algebra move here is finding the common denominator in Step 2. This is where most students trip up.
Square root functions (e.g. f(x) = √x + 2)
Step 1: f(x + h) = √(x + h) + 2.
Step 2: f(x + h) – f(x) = √(x + h) + 2 – (√x + 2) = √(x + h) – √x.
Step 3: Divide by h: [√(x + h) – √x] / h. Multiply numerator and denominator by the conjugate [√(x + h) + √x] to rationalise: = [(x + h) – x] / [h(√(x + h) + √x)] = 1 / [√(x + h) + √x].
Step 4: lim(h→0) 1 / [√(x + h) + √x] = 1 / (2√x).
Result: f′(x) = 1 / (2√x).
The critical technique here is multiplying by the conjugate. Without it, you cannot cancel the h from the denominator.
Applied example: sales analysis
Suppose total sales are modelled by S(t) = √t + 2.
S(25) = √25 + 2 = 5 + 2 = 7 (total sales at t = 25).
Since S(t) has the same form as √x + 2 above, S′(t) = 1 / (2√t).
S′(25) = 1 / (2√25) = 1 / 10 = 0.1.
Interpretation: at t = 25, sales are increasing at a rate of 0.1 units per time period.
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}This is the definition of the derivative. Every derivative problem in this chapter starts here.
\text{Average rate of change} = \frac{f(a+h) - f(a)}{h}, \quad h \neq 0\text{Average velocity} = \frac{d}{t} = \frac{\text{distance covered}}{\text{elapsed time}}Results worth memorising from the worked examples:
If f(x) = 4x – x², then f′(x) = 4 – 2x.
If f(x) = 1/x, then f′(x) = –1/x².
If f(x) = √x + 2, then f′(x) = 1/(2√x).
In business calculus, the derivative tells you marginal quantities: marginal cost, marginal revenue, marginal profit. If C(x) is total cost, then C′(x) is the approximate cost of producing one more unit. This is why the derivative matters for business students: it converts "how much in total" into "how much per additional unit."
The sales analysis example (S(t) = √t + 2) shows exactly this. S′(25) = 0.1 means that at t = 25, each additional time period adds roughly 0.1 units of sales. The rate is slowing down (compare S′(1) = 0.5 to S′(25) = 0.1), which tells a manager that growth is decelerating.
Students often confuse the average rate of change with the derivative. The average rate of change uses a finite interval (h is some number). The derivative takes the limit as h goes to 0. They are related, but they are not the same thing.
Forgetting to distribute the negative sign in Step 2. When you subtract f(x) from f(x + h), every term of f(x) gets subtracted, not just the first one. This is the single most common algebraic error.
Thinking the derivative is always a number. For most functions, f′(x) is itself a function. You only get a single number when you evaluate f′ at a specific x value.
With square roots, students try to simplify [√(x + h) – √x] / h by "cancelling" the square roots. You cannot do this. Multiply by the conjugate instead.
⚠️ "Find the derivative using the definition" is a standard exam instruction. Shortcut rules (power rule, etc.) come in later chapters, but exams on Ch. 2.4 require the four-step limit process.
⚠️ Show every step. Partial credit on these problems comes from showing the algebra in Steps 1–3 clearly, even if you make a small error.
⚠️ Rational and square root derivatives are favourite exam choices because they test algebraic technique (common denominators, conjugate multiplication) alongside the limit concept.
⚠️ Applied problems (like the sales analysis) may ask you to interpret the derivative in context. Be ready to write a sentence: "At t = 25, sales are increasing at a rate of 0.1 units per period."
True or False: The derivative f′(x) gives the average rate of change of f.
Fill in the blank: The first step in finding a derivative from the definition is to find f(______).
True or False: When finding the derivative of f(x) = √x, you should multiply by the conjugate to simplify.
Fill in the blank: f′(x) = lim as h → ______ of [f(x + h) – f(x)] / h.
True or False: The derivative of a function is always a single number.
Answers: 1. False (it gives the instantaneous rate of change). 2. x + h. 3. True. 4. 0. 5. False (it is usually a function of x).
Q: Using the four-step process, find the derivative of f(x) = 3x – x².
A: f(x + h) = 3(x + h) – (x + h)² = 3x + 3h – x² – 2xh – h². Then f(x + h) – f(x) = 3h – 2xh – h². Divide by h: 3 – 2x – h. Take the limit as h → 0: f′(x) = 3 – 2x.
Q: Find f′(x) for f(x) = 1/x using the definition.
A: f(x + h) – f(x) = 1/(x + h) – 1/x = –h/[x(x + h)]. Divide by h: –1/[x(x + h)]. Limit as h → 0: f′(x) = –1/x².
Q: If S(t) = √t + 2 and S′(t) = 1/(2√t), what is the rate of change of sales at t = 4?
A: S′(4) = 1/(2√4) = 1/4 = 0.25 units per time period.
Q: Explain in one sentence what f′(10) = 3 means if f(x) represents total revenue in thousands of dollars and x is the number of units sold.
A: At 10 units sold, revenue is increasing at approximately $3,000 per additional unit.
Q: Why can you not simply substitute h = 0 into the difference quotient before simplifying?
A: Substituting h = 0 directly gives 0/0, which is undefined. You must simplify first to cancel the h from the denominator, then take the limit.
This material connects directly to Ch. 2.5 and beyond, where shortcut rules (power rule, product rule, quotient rule) replace the four-step process. Those rules are derived from the limit definition taught here, so understanding the process now makes the shortcut rules meaningful rather than arbitrary.
The derivative also connects to Ch. 3 (applications), where you will use f′(x) to find marginal cost, marginal revenue, and to solve optimisation problems (finding maximum profit, minimum cost). Every applied problem in that chapter depends on being able to compute and interpret a derivative.
derivative, difference quotient, limit definition of derivative, first principles, instantaneous rate of change, average rate of change, f prime, f′(x), slope of tangent line, marginal analysis, rational function derivative, square root derivative, conjugate method, four-step derivative process, calculus for business, FIU calculus, Ch. 2.4