Difficulty: Intermediate. Prerequisites: Sections 2.1–2.3 (limits and continuity).
The derivative measures how fast a function is changing at a single point. It is defined as a limit of the difference quotient and gives you the slope of the tangent line at any point on a graph. Shortcut rules (power rule, constant rule, sum rule) let you differentiate most polynomial and root functions without grinding through the limit definition every time.
Average rate of change
The change in f(x) divided by the change in x between two points: (f(x₂) − f(x₁)) / (x₂ − x₁). Think of it as the slope of the straight line connecting two points on the graph.
Secant line
The straight line passing through two points on a curve. Its slope equals the average rate of change between those two points.
Instantaneous rate of change
The rate of change at a single point, obtained by taking the limit of the average rate of change as the two points collapse into one. In simple terms, it is the speed of the function at one exact moment.
Tangent line
The line that just touches the curve at a single point, matching the curve's direction there. Its slope is the derivative at that point.
Derivative
The function f'(x) = lim(h→0) [f(x + h) − f(x)] / h. It gives the slope of the tangent line at every point where it exists. Also written as df/dx, y', or dy/dx.
Power rule
If f(x) = x^n, then f'(x) = n·x^(n−1) for any real n ≠ 0. Think of it as: bring the exponent down, then reduce the exponent by one.
Constant rule
The derivative of a constant is zero. If f(x) = c, then f'(x) = 0.
Sum and difference rule
The derivative of a sum (or difference) is the sum (or difference) of the derivatives: (u ± v)' = u' ± v'.
Differentiation
The process of finding the derivative of a function. "Differentiate f(x)" and "find f'(x)" mean the same thing.
The average rate of change between x = a and x = a + h is the difference quotient: [f(a + h) − f(a)] / h. This is the slope of the secant line.
As h → 0, the secant line rotates toward the tangent line. The limiting slope is the instantaneous rate of change, i.e. the derivative.
Example: P(x) = −x² + 20x (profit from Chucky dolls, x in thousands). Average change from x = 10 to x = 12: [P(12) − P(10)] / (12 − 10) = [96 − 100] / 2 = −2 thousand dollars per thousand units.
f'(x) = lim(h→0) [f(x + h) − f(x)] / h.
The four-step process (used in exam "show your work" questions):
Write out f(x + h) by substituting (x + h) for every x.
Compute f(x + h) − f(x) and simplify.
Divide by h and cancel.
Take the limit as h → 0.
Example: f(x) = x² − x + 5. Then f(x + h) = (x + h)² − (x + h) + 5 = x² + 2xh + h² − x − h + 5. Subtract f(x): 2xh + h² − h. Divide by h: 2x + h − 1. Limit as h → 0: f'(x) = 2x − 1.
Example: f(x) = 3/(x + 1). The algebra involves combining fractions with a common denominator, then cancelling h. Result: f'(x) = −3/(x + 1)².
This problem type shows up on nearly every exam.
Step 1: Find f'(x).
Step 2: Evaluate m = f'(a) to get the slope at x = a.
Step 3: Evaluate f(a) to get the point (a, f(a)).
Step 4: Plug into point-slope form: y − f(a) = m(x − a). Convert to slope-intercept form if asked.
Example: f(x) = x² − x + 5 at x = 1. f'(x) = 2x − 1, so m = f'(1) = 1. f(1) = 5. Point-slope: y − 5 = 1(x − 1), giving y = x + 4.
A horizontal tangent line has slope zero, so set f'(x) = 0 and solve for x.
Example: f(x) = x³ − 3x² + 1. f'(x) = 3x² − 6x = 3x(x − 2). Setting f'(x) = 0 gives x = 0 and x = 2. Horizontal tangent lines at y = f(0) = 1 and y = f(2) = −3.
Power rule: d/dx [x^n] = n·x^(n−1). Works for fractions and negatives too. Rewrite roots and reciprocals first: √x = x^(1/2), 1/x³ = x^(−3).
Constant rule: d/dx [c] = 0.
Sum/difference rule: d/dx [u(x) ± v(x)] = u'(x) ± v'(x).
Before differentiating, always simplify first. Expand products like (2x − 3)(x² − 4x) and split fractions like (7x − 6x⁵ + 2)/x² into separate terms.
Example: f(x) = −2x⁴ − x^(−5) + 4x^(7/3). Then f'(x) = −8x³ + 5x^(−6) + (28/3)x^(4/3).
\text{Average rate of change} = \frac{f(a+h) - f(a)}{h}f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\text{Power rule: } \frac{d}{dx}[x^n] = nx^{n-1} \quad (n \neq 0)\text{Constant rule: } \frac{d}{dx}[c] = 0\text{Sum/difference: } \frac{d}{dx}[u \pm v] = u' \pm v'\text{Tangent line: } y - f(a) = f'(a)(x - a)Algebra refresher for rewriting before differentiation:
Original form | Rewritten form | Why |
|---|---|---|
1/x^n | x^(−n) | Negative exponent |
ⁿ√(x^m) | x^(m/n) | Fractional exponent |
(product of two brackets) | Expand, then differentiate each term | No product rule needed yet |
Students often think the derivative at a point is the same as the function value at that point. f'(a) is the slope of the tangent line at x = a, not the height of the graph.
Students sometimes skip rewriting before differentiating. You cannot apply the power rule to 1/x³ as written; you must rewrite it as x^(−3) first.
Students forget that the derivative of a constant is zero, and try to apply the power rule to a plain number like 31.
Students confuse average rate of change (secant line, two points needed) with instantaneous rate of change (tangent line, one point, found via derivative).
⚠️ Finding the equation of the tangent line at a given point appears on nearly every exam in this course. Practise until it is automatic.
⚠️ The limit-definition derivation (the four-step process) is tested directly. You will be asked to differentiate a function using the definition, not shortcut rules.
⚠️ Horizontal tangent lines (set f'(x) = 0) come up in the context of optimisation and curve sketching later in the course.
⚠️ Rewriting radicals and reciprocals as power-rule-friendly forms (fractional and negative exponents) is tested constantly.
True or False: The derivative of 7x⁰ is 0. False. 7x⁰ = 7·1 = 7, and the derivative of the constant 7 is 0. But 7x⁰ itself is already 7, so d/dx [7] = 0 is correct. (The statement is actually true. The derivative is 0.)
Fill in the blank: The slope of the tangent line to f(x) at x = a is ________. f'(a).
True or False: To find horizontal tangent lines, set f(x) = 0. False. Set f'(x) = 0.
Differentiate: f(x) = 3x⁵ − 2x + 9. f'(x) = 15x⁴ − 2.
True or False: The average rate of change and the instantaneous rate of change are the same thing. False. Average uses two points; instantaneous uses one (via a limit).
Q: Use the definition of the derivative to find f'(x) for f(x) = x² − x + 5.
A: f(x + h) = (x + h)² − (x + h) + 5 = x² + 2xh + h² − x − h + 5. Then f(x + h) − f(x) = 2xh + h² − h. Dividing by h gives 2x + h − 1. Taking the limit as h → 0: f'(x) = 2x − 1.
Q: Find the equation of the tangent line to f(x) = 3√(x³) − 2x + 1 at x = 1.
A: Rewrite: f(x) = 3x^(3/2) − 2x + 1. Derivative: f'(x) = (9/2)x^(1/2) − 2. At x = 1: f'(1) = 9/2 − 2 = 5/2. f(1) = 3 − 2 + 1 = 2. Tangent line: y − 2 = (5/2)(x − 1), or y = (5/2)x − 1/2.
Q: Find the x-values where f(x) = x³ − 3x² + 1 has horizontal tangent lines.
A: f'(x) = 3x² − 6x = 3x(x − 2) = 0. So x = 0 and x = 2.
Q: Differentiate y = (7x − 6x⁵ + 2)/x².
A: Split: y = 7x^(−1) − 6x³ + 2x^(−2). Then y' = −7x^(−2) − 18x² − 4x^(−3).
The derivative definition uses limits (Sections 2.1–2.3). Marginal analysis in business applications (Section 2.7) is a direct application of the derivative: the marginal cost function is the derivative of the total cost function. The product rule, quotient rule (Section 3.2–3.3) and chain rule (Section 3.4) extend differentiation to more complex expressions. The first and second derivative tests (Sections 4.1–4.2) use the derivative's sign to determine where a function increases, decreases, or has local extrema.
Derivative, differentiation, difference quotient, limit definition, average rate of change, instantaneous rate of change, secant line, tangent line, slope of tangent, power rule, constant rule, sum rule, difference rule, horizontal tangent, f'(x), dy/dx, df/dx, MAC 2233, business calculus, FIU