What is AP Calculus AB/BC unit 2?
Differentiation is the process of finding instantaneous rates of change. Unit 2 starts by connecting the average rate of change formula (f(b) - f(a))/(b - a) to the derivative through a limiting process, then develops the rules that make computing derivatives practical for polynomials, trig functions, exponentials, and logarithms.
The derivative f'(x) is the limit of the difference quotient as h approaches 0. Unit 2 gives you the rules to compute that limit efficiently without returning to the definition every time: the Power Rule, constant and sum rules, Product Rule, Quotient Rule, and the standard derivatives of sin x, cos x, e^x, ln x, tan x, cot x, sec x, and csc x.
From average to instantaneous
The average rate of change over [a, b] is the slope of the secant line. As the interval shrinks to zero, that slope becomes the derivative at a point: f'(a) = lim(h to 0) (f(a+h) - f(a))/h. Topics 2.1-2.3 build this idea from tables, graphs, and algebra.
Differentiability and continuity
Topic 2.4 establishes a one-way relationship: differentiable implies continuous, but not the reverse. Corners (like |x| at 0), cusps, vertical tangents, and discontinuities are all points where the derivative fails to exist.
Differentiation rules
Topics 2.5-2.10 give you the computational toolkit: Power Rule for x^r, constant and sum rules for polynomials, derivatives of sin x, cos x, e^x, and ln x, the Product Rule (fg)' = f'g + fg', the Quotient Rule (f/g)' = (gf' - fg')/g^2, and the four remaining trig derivatives derived from those rules.
Why the limit definition matters even after you learn the rulesEvery shortcut in Unit 2 is a consequence of the limit definition. On the AP exam you may be asked to evaluate a limit by recognizing it as a derivative in disguise, for example lim(h to 0) (e^(x+h) - e^x)/h = e^x. Understanding the definition lets you work in both directions: from function to derivative and from limit expression back to a known derivative value.
Unit 2 review notes
2.1
Defining the Derivative from Average to Instantaneous Rate of Change
The average rate of change of f on [a, b] is (f(b) - f(a))/(b - a), which equals the slope of the secant line through those two points. The instantaneous rate of change at x = a is the limit of that quotient as the interval collapses: f'(a) = lim(h to 0) (f(a+h) - f(a))/h. The derivative function f'(x) extends this to every point in the domain. The slope of the tangent line at x = a equals f'(a), and the tangent line equation uses point-slope form: y - f(a) = f'(a)(x - a).
- Average rate of change: (f(b) - f(a))/(b - a); slope of the secant line between (a, f(a)) and (b, f(b)).
- Limit definition of the derivative: f'(a) = lim(h to 0) (f(a+h) - f(a))/h, or equivalently lim(x to a) (f(x) - f(a))/(x - a).
- Derivative function: f'(x) = lim(h to 0) (f(x+h) - f(x))/h; gives the instantaneous rate of change at every x in the domain.
- Derivative notation: f'(x), y', and dy/dx all represent the same derivative; choose notation to match the problem context.
- Tangent line: The line through (a, f(a)) with slope f'(a); equation is y - f(a) = f'(a)(x - a).
Given f(x) = x^2, use the limit definition to show f'(3) = 6, then write the equation of the tangent line at x = 3.
| Concept | Formula | Geometric meaning |
|---|
| Average rate of change | (f(b)-f(a))/(b-a) | Slope of secant line |
| Instantaneous rate of change | lim(h to 0)(f(a+h)-f(a))/h | Slope of tangent line |
| Derivative function | lim(h to 0)(f(x+h)-f(x))/h | Slope function for all x |
2.3
Estimating Derivatives from Tables and Graphs
When you cannot compute a derivative algebraically, estimate it using nearby data. From a table, use the symmetric difference quotient (f(a+h) - f(a-h))/(2h) when interior values are available, or a one-sided quotient at endpoints. From a graph, estimate the slope of the tangent line by reading rise over run between two close points. Always include units when the context provides them.
- Symmetric difference quotient: (f(a+h) - f(a-h))/(2h); generally the most accurate table-based estimate for f'(a) at interior points.
- One-sided estimate: Use (f(a+h) - f(a))/h or (f(a) - f(a-h))/h when only one neighboring value is available, such as at a table endpoint.
- Graphical estimation: Read two nearby points on the curve and compute slope; the closer the points, the better the approximation of the tangent slope.
A table gives f(2) = 5, f(3) = 8, f(4) = 13. Estimate f'(3) using the symmetric difference quotient.
| Method | Formula | Best used when |
|---|
| Symmetric difference quotient | (f(a+h)-f(a-h))/(2h) | Interior table point with equal spacing |
| Forward difference quotient | (f(a+h)-f(a))/h | Left endpoint or only right neighbor available |
| Backward difference quotient | (f(a)-f(a-h))/h | Right endpoint or only left neighbor available |
2.4
Differentiability and Continuity
Differentiability implies continuity: if f is differentiable at x = a, it must be continuous there. The converse is false. A function can be continuous at a point but still fail to be differentiable. The four main failure cases are corners, cusps, vertical tangents, and discontinuities. To justify non-differentiability on the AP exam, name the specific reason rather than just identifying the point.
- Differentiable implies continuous: If f'(a) exists, then f is continuous at a. Contrapositive: if f is discontinuous at a, then f is not differentiable at a.
- Corner: Left-hand and right-hand difference quotient limits both exist but are not equal; example: f(x) = |x| at x = 0.
- Vertical tangent: The difference quotient limit is infinite; example: f(x) = x^(1/3) at x = 0, where the slope grows without bound.
- Cusp: One-sided difference quotient limits are infinite with opposite signs; the graph has a sharp point with infinite steepness.
- Discontinuity: Any jump, removable, or infinite discontinuity at a point prevents differentiability there.
Explain why f(x) = |x - 2| is continuous but not differentiable at x = 2. Compute the left-hand and right-hand difference quotient limits to support your answer.
| Failure type | Example | One-sided limits of difference quotient |
|---|
| Corner | |x| at x=0 | Left: -1, Right: +1 (unequal, both finite) |
| Vertical tangent | x^(1/3) at x=0 | Both approach infinity |
| Discontinuity | Piecewise jump at x=a | At least one limit does not match function value |
2.5
Power Rule and Algebraic Derivative Rules
The Power Rule states d/dx(x^r) = r*x^(r-1) for any real number r. It applies to integer, negative, and fractional exponents, so rewrite radicals and denominators as powers before differentiating. The constant rule, sum rule, difference rule, and constant multiple rule let you differentiate any polynomial term by term. The derivative of a constant is zero; a constant factor pulls out front.
- Power Rule: d/dx(x^r) = r*x^(r-1); works for all real r, including r = 1/2 (square root) and r = -1 (reciprocal).
- Constant rule: d/dx(c) = 0 for any constant c.
- Constant multiple rule: d/dx(c*f(x)) = c*f'(x); the constant factor is preserved.
- Sum and difference rules: d/dx(f(x) +/- g(x)) = f'(x) +/- g'(x); differentiate term by term.
- Rewriting before differentiating: Convert sqrt(x) to x^(1/2) and 1/x^2 to x^(-2) before applying the Power Rule to avoid errors.
Find the derivative of f(x) = 3x^4 - 5x^(1/2) + 7x^(-1) + 9 using the Power Rule and algebraic rules.
| Rule | Formula | Example |
|---|
| Power Rule | d/dx(x^r) = r*x^(r-1) | d/dx(x^5) = 5x^4 |
| Constant Multiple | d/dx(c*f) = c*f' | d/dx(4x^3) = 12x^2 |
| Sum/Difference | d/dx(f +/- g) = f' +/- g' | d/dx(x^2 + x) = 2x + 1 |
| Constant | d/dx(c) = 0 | d/dx(7) = 0 |
2.7
Derivatives of sin x, cos x, e^x, and ln x
Four essential derivatives must be memorized: d/dx(sin x) = cos x, d/dx(cos x) = -sin x, d/dx(e^x) = e^x, and d/dx(ln x) = 1/x. The negative sign on the cosine derivative is the most common source of errors. The self-derivative property of e^x is unique among functions. These can be combined with sum, difference, and constant multiple rules. A secondary skill from this topic is recognizing a limit expression as a derivative definition and evaluating it using a known derivative.
- d/dx(sin x) = cos x: Derived from the limit definition using the sine addition formula and the special limits lim(sin h/h) = 1 and lim((1-cos h)/h) = 0.
- d/dx(cos x) = -sin x: The negative sign is essential; forgetting it is one of the most common errors on trig derivative problems.
- d/dx(e^x) = e^x: The exponential function e^x is its own derivative; this property defines e and makes it central to differential equations.
- d/dx(ln x) = 1/x: Valid for x > 0; derived using the inverse function relationship between e^x and ln x.
- Limit-as-derivative recognition: If a limit matches the form lim(h to 0)(f(a+h)-f(a))/h, evaluate it as f'(a) using a known derivative instead of computing the limit directly.
Evaluate lim(h to 0) (sin(pi/6 + h) - sin(pi/6))/h by recognizing it as a derivative, then confirm by differentiating sin x directly.
| Function | Derivative | Key note |
|---|
| sin x | cos x | Positive; use radian measure |
| cos x | -sin x | Negative sign required |
| e^x | e^x | Self-derivative; unique to e |
| ln x | 1/x | Domain x > 0 |
2.8
Product Rule and Quotient Rule
When two differentiable functions are multiplied, use the Product Rule: d/dx(f*g) = f'*g + f*g'. When one differentiable function is divided by another, use the Quotient Rule: d/dx(f/g) = (g*f' - f*g')/g^2. The subtraction order in the Quotient Rule is fixed; reversing it changes the sign of the answer. Neither rule allows you to simply multiply or divide the individual derivatives.
- Product Rule: d/dx(f*g) = f'*g + f*g'; first times derivative of second, plus second times derivative of first.
- Quotient Rule: d/dx(f/g) = (g*f' - f*g')/g^2; denominator times derivative of numerator minus numerator times derivative of denominator, all over denominator squared.
- Order matters in Quotient Rule: g*f' - f*g' is not the same as f*g' - g*f'; always put the denominator function first in the numerator of the formula.
- When to use each rule: Product Rule for expressions written as a product; Quotient Rule for explicit fractions. Rewriting a quotient as a product with a negative exponent and using the Product Rule is an alternative.
Differentiate h(x) = x^2 * sin x using the Product Rule, then differentiate k(x) = e^x / (x^2 + 1) using the Quotient Rule.
| Rule | Formula | Common error |
|---|
| Product Rule | (fg)' = f'g + fg' | Multiplying f' and g' instead of using the formula |
| Quotient Rule | (f/g)' = (gf' - fg')/g^2 | Reversing subtraction order or forgetting to square the denominator |
2.10
Derivatives of tan x, cot x, sec x, and csc x
The four remaining trig derivatives follow from rewriting each function as a ratio of sin x and cos x and applying the Quotient Rule. The results are: d/dx(tan x) = sec^2 x, d/dx(cot x) = -csc^2 x, d/dx(sec x) = sec x * tan x, d/dx(csc x) = -csc x * cot x. The negative signs on cot x and csc x derivatives are the most frequently missed details. All four require radian measure.
- d/dx(tan x) = sec^2 x: Derived by applying the Quotient Rule to sin x / cos x; uses the identity sin^2 x + cos^2 x = 1 to simplify.
- d/dx(cot x) = -csc^2 x: Derived from cos x / sin x via the Quotient Rule; the negative sign is required.
- d/dx(sec x) = sec x * tan x: Derived from 1/cos x using the reciprocal or Quotient Rule.
- d/dx(csc x) = -csc x * cot x: Derived from 1/sin x; the negative sign is required and often forgotten.
- Derivation strategy: Rewrite tan x = sin x/cos x, cot x = cos x/sin x, sec x = 1/cos x, csc x = 1/sin x, then apply the Quotient Rule to derive each formula rather than memorizing without understanding.
Derive d/dx(tan x) = sec^2 x by applying the Quotient Rule to sin x / cos x. Then state d/dx(sec x) and d/dx(csc x) from memory.
| Function | Rewrite | Derivative |
|---|
| tan x | sin x / cos x | sec^2 x |
| cot x | cos x / sin x | -csc^2 x |
| sec x | 1 / cos x | sec x * tan x |
| csc x | 1 / sin x | -csc x * cot x |
Practice AP Calculus AB/BC unit 2 questions
Try AP-style multiple-choice questions and written prompts after you review the notes.
QuestionThe function h(x)=x2+4ex has derivative h′(x)=(x2+4)2(x2+4)ex−2xex by the quotient rule. A student constructs the second-degree Taylor polynomial P2(x) centered at x=0 to approximate h(0.4). It is known that h(0)=0.25, h′(0)=0.25, h′′(0)=−0.1875, and ∣h′′′(x)∣≤1.2 for all x∈[0,0.4]. Which statement correctly describes how the error bound constrains the actual value of h(0.4)?
The actual value h(0.4) is guaranteed to lie within 0.0128 of P2(0.4), so ∣h(0.4)−P2(0.4)∣≤0.0128.
The actual value h(0.4) is guaranteed to lie within 0.0192 of P2(0.4) because the second derivative bound is ∣−0.1875∣=0.1875.
The actual value h(0.4) is guaranteed to lie within 0.0064 of P2(0.4) because the error bound uses only the fourth power of the distance.
The actual value cannot be bounded because the quotient rule produces a derivative that changes with x, making error bounds impossible to establish.
QuestionConsider g(x)=4(3x3−2x). Which rule should be applied first to find g′(x)?
Constant Multiple Rule — because 4 multiplies the entire expression (3x3−2x)
Power Rule — because you must differentiate 3x3 and −2x individually before factoring out the constant 4
Constant Multiple Rule — because 4 is a constant, but you should apply it AFTER using the Sum Rule on the terms inside the parentheses
Constant Multiple Rule — because the coefficient 3 in 3x3 is a constant that multiplies a power of x
1. A cup of coffee is poured at time t=0 minutes. The temperature of the coffee, in degrees Celsius, is modeled by the function T defined by T(t)=20+65e−0.12t for 0≤t≤20. A graph of T on the interval 0≤t≤20 is shown in Figure 1.
Figure 1. Graph of T(t)=20+65e^{-0.12t} for 0 ≤ t ≤ 20 (t in minutes, T in °C).
2. A bacterial culture is observed for the first 6 hours after a nutrient is added. The number of bacteria, in thousands, at time t hours is modeled by the function B given by B(t)=4+2e0.5t−ln(t+1) for 0≤t≤6.
t (hours) | B(t) (thousands of bacteria) |
|---|
1 | 4+2e0.5−ln2 |
1.1 | 4+2e0.55−ln2.1 |
1.2 | 4+2e0.6−ln2.2 |
1.3 | 4+2e0.65−ln2.3 |