Difficulty: Introductory
Prerequisites: Basic algebra (expanding brackets, factoring, simplifying fractions). Familiarity with function notation f(x). No prior calculus required.
Big picture: This is the foundational concept of differential calculus. Before any shortcut rules (power rule, product rule, chain rule), you need to understand where derivatives come from. The limit definition of the derivative gives you the instantaneous rate of change of a function at any point, built up from the idea of average rate of change over a shrinking interval. Everything else in the course builds on this. If you can do these problems mechanically and understand what the result means, the later chapters will make far more sense.
The derivative f'(x) measures how fast f(x) is changing at any given x. You find it by taking the limit of [f(x + h) - f(x)] / h as h approaches 0. The algebra is mostly expanding, cancelling, factoring out h, then substituting h = 0. Once you have f'(x), you can tell whether the original function is increasing (f'(x) > 0) or decreasing (f'(x) < 0) at any point.
Derivative (f'(x))
The instantaneous rate of change of a function f at x. Formally, f'(x) = lim(h→0) [f(x + h) - f(x)] / h, provided that limit exists.
In simple terms, it tells you how quickly the output of the function is changing at a specific input value. If f is a profit function, f'(x) is the rate profit changes per unit.
Difference quotient
The expression [f(x + h) - f(x)] / h before you take the limit. This is the average rate of change of f over the interval from x to x + h.
Think of it as the slope of a line connecting two points on the curve. As h shrinks to 0, that line becomes the tangent line, and the difference quotient becomes the derivative.
Limit (as h approaches 0)
The value that an expression approaches as h gets arbitrarily close to 0. You cannot simply plug in h = 0 directly (that would give 0/0), so you simplify first, cancel the h, then substitute.
In simple terms, you are asking: "What does this expression settle down to as h gets closer and closer to zero?"
Differentiable
A function is differentiable at a point x = c if the derivative f'(c) exists (the limit above produces a finite number). If differentiable, the function has a well-defined tangent line at that point.
Increasing function
A function is increasing at x = c if f'(c) > 0. The output is getting larger as x moves to the right.
Think of it as: the curve is going uphill at that point.
Decreasing function
A function is decreasing at x = c if f'(c) < 0. The output is getting smaller as x moves to the right.
Think of it as: the curve is going downhill at that point.
Every problem using the limit definition follows the same pattern:
Write out the formula: f'(x) = lim(h→0) [f(x + h) - f(x)] / h
Substitute f(x + h) and f(x) into the formula, then expand and simplify the numerator
Factor h out of every term in the numerator, then cancel it with the h in the denominator
Substitute h = 0 into what remains
The algebra in step 2 is where most of the work happens. The rest is mechanical.
Write the definition: f'(x) = lim(h→0) [16(x + h)² - 16x²] / h
Expand: 16(x² + 2xh + h²) = 16x² + 32xh + 16h²
Subtract 16x²: numerator becomes 32xh + 16h²
Factor out h: h(32x + 16h)
Cancel h with the denominator: lim(h→0) (32x + 16h)
Substitute h = 0: 32x + 16(0) = 32x
Result: f'(x) = 32x
Write the definition: f'(x) = lim(h→0) [5(x + h) - 3 - (5x - 3)] / h
Expand: 5x + 5h - 3 - 5x + 3 = 5h
The numerator simplifies to 5h immediately
Cancel h: lim(h→0) 5
The limit of a constant is the constant itself
Result: f'(x) = 5
Notice the derivative of a linear function is just its slope. This will always be the case.
Write the definition: f'(x) = lim(h→0) [2 - 7(x + h) - (2 - 7x)] / h
Expand: 2 - 7x - 7h - 2 + 7x = -7h
Cancel h: lim(h→0) -7
Result: f'(x) = -7
Again, the derivative equals the slope of the original linear function.
Write the definition: f'(x) = lim(h→0) [(x + h)² - 1 - (x² - 1)] / h
Expand: x² + 2xh + h² - 1 - x² + 1 = 2xh + h²
Factor out h: h(2x + h)
Cancel h: lim(h→0) (2x + h)
Substitute h = 0: 2x
Result: f'(x) = 2x
Quadratic f(x) = ax² + bx + c: The derivative is always linear, f'(x) = 2ax + b
Linear f(x) = mx + b: The derivative is the constant slope, f'(x) = m
Constant f(x) = c: The derivative is 0 (no change)
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}This is the limit definition of the derivative. It is also called the derivative from first principles.
The expression inside the limit, before taking h to 0, is the difference quotient:
\frac{f(x+h) - f(x)}{h}For increasing and decreasing behaviour at a point x = c:
f'(c) > 0 means f is increasing at c
f'(c) < 0 means f is decreasing at c
f'(c) = 0 means f has a horizontal tangent at c (could be a maximum, minimum, or neither)
Derivatives appear constantly in business calculus as marginal functions. "Marginal" just means "the derivative of": marginal cost is the derivative of the cost function, marginal revenue is the derivative of revenue, and marginal profit is the derivative of the profit function.
Setup: You own a small manufacturing firm. When x thousand units are produced and sold, profit is p(x) = -400x² + 6800x - 12,000 dollars. At what rate is profit changing when 9,000 units are sold? Is profit increasing or decreasing?
Step 1: Find p'(x) using the limit definition.
p'(x) = lim(h→0) [-400(x + h)² + 6800(x + h) - 12000 - (-400x² + 6800x - 12000)] / h
Expand -400(x + h)²: -400x² - 800xh - 400h²
Combine all terms: -400x² - 800xh - 400h² + 6800x + 6800h - 12000 + 400x² - 6800x + 12000
Simplify: -800xh - 400h² + 6800h
Factor out h: h(-800x - 400h + 6800)
Cancel h: lim(h→0) (-800x - 400h + 6800)
Substitute h = 0: -800x + 6800
Result: p'(x) = -800x + 6800
Step 2: Evaluate at x = 9 (since x is in thousands, 9,000 units means x = 9).
p'(9) = -800(9) + 6800 = -7200 + 6800 = -400
Step 3: Interpret. The rate of change of profit at 9,000 units is -400 dollars per thousand units. Because p'(9) = -400 < 0, profit is decreasing at this production level. Producing more units beyond 9,000 will reduce profit.
This is the kind of analysis businesses use to find optimal production levels. The profit-maximising output is where p'(x) = 0.
Students often substitute h = 0 too early. You cannot plug h = 0 into the difference quotient before simplifying, because that gives 0/0, which is undefined. You must simplify the expression algebraically (expand, cancel the h) and only then set h = 0.
Students forget to distribute the negative sign. When computing f(x + h) - f(x), the subtraction applies to the entire f(x), not just the first term. Missing this, especially with functions like p(x) = -400x² + 6800x - 12000, causes sign errors that cascade through the rest of the problem.
Students confuse the derivative with the original function. The derivative f'(x) tells you the rate of change, not the value of f(x) itself. If p'(9) = -400, that does not mean the profit at 9,000 units is -400 dollars. It means profit is changing at a rate of -400 dollars per thousand units at that production level.
Students assume f'(x) < 0 means the function is negative. A negative derivative means the function is decreasing (going down), not that the function value is below zero. A function can be positive and decreasing at the same time.
⚠️ The limit definition of the derivative is almost certainly on the exam. Expect at least one problem where you must find f'(x) using the definition, not shortcut rules.
⚠️ Business application problems (marginal profit, marginal cost, marginal revenue) are a staple in business calculus exams. You need to find the derivative, evaluate it at a given point, and interpret the sign (increasing or decreasing) in context.
⚠️ Be ready to state whether a function is increasing or decreasing at a specific point using the sign of f'(c). This is a quick marks question that trips up students who confuse the value of f with the value of f'.
⚠️ Show every step. In limit-definition problems, partial credit is given for the setup and algebraic steps. Skipping to the answer is risky.
True or false: To find f'(x) using the limit definition, you can substitute h = 0 directly into [f(x + h) - f(x)] / h without simplifying first. False. That gives 0/0.
Fill in the blank: The derivative of a linear function f(x) = mx + b is always ____. m (the slope).
True or false: If f'(5) = -3, then f is increasing at x = 5. False. A negative derivative means f is decreasing.
Fill in the blank: In the difference quotient, the key algebraic step after expanding is to factor ____ out of the numerator so it cancels with the denominator. h.
True or false: If p'(x) = -800x + 6800 and you evaluate at x = 9, the result p'(9) = -400 tells you the profit at 9,000 units is -400 dollars. False. It tells you profit is changing at a rate of -400 dollars per thousand units.
Q: Use the limit definition to find f'(x) for f(x) = 3x² + 2x.
A: Set up f'(x) = lim(h→0) [3(x + h)² + 2(x + h) - 3x² - 2x] / h. Expand: 3x² + 6xh + 3h² + 2x + 2h - 3x² - 2x = 6xh + 3h² + 2h. Factor: h(6x + 3h + 2). Cancel h: lim(h→0) (6x + 3h + 2) = 6x + 2. So f'(x) = 6x + 2.
Q: Use the limit definition to find f'(x) for f(x) = 4x - 9.
A: f'(x) = lim(h→0) [4(x + h) - 9 - (4x - 9)] / h = lim(h→0) 4h / h = lim(h→0) 4 = 4. The derivative of a linear function is its slope.
Q: A company's revenue is R(x) = -200x² + 5000x dollars, where x is in thousands of units. Find R'(x) using the limit definition, then determine whether revenue is increasing or decreasing at x = 15.
A: Using the limit definition and expanding: R'(x) = -400x + 5000. At x = 15: R'(15) = -400(15) + 5000 = -6000 + 5000 = -1000. Since R'(15) = -1000 < 0, revenue is decreasing at 15,000 units.
Q: If f'(c) = 0, is the function increasing, decreasing, or neither at x = c?
A: Neither increasing nor decreasing at that exact point. f'(c) = 0 means the tangent line is horizontal there. The point could be a local maximum, a local minimum, or an inflection point. Further analysis is needed to determine which.
Q: Explain in your own words what the difference quotient [f(x + h) - f(x)] / h represents geometrically.
A: It represents the slope of the secant line connecting the points (x, f(x)) and (x + h, f(x + h)) on the graph. As h approaches 0, the secant line approaches the tangent line, and the difference quotient approaches the derivative.
This connects to derivative shortcut rules (power rule, product rule, quotient rule) because those rules are derived from this limit definition. Once you understand why the derivative of x² is 2x from first principles, the power rule (d/dx of xⁿ = nxⁿ⁻¹) becomes a natural generalisation rather than an arbitrary formula to memorise.
This also connects directly to optimisation problems later in the course. Finding where f'(x) = 0 (the production level that maximises profit, the price that maximises revenue) relies on computing and interpreting derivatives. The business problem here is an early version of those optimisation questions.
The concept of increasing and decreasing functions connects to curve sketching and the first derivative test, which you will use to classify critical points as maxima or minima.
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