AP Calculus BC, AP exams, 4–7%

Unit 2: Differentiation: Definition and Fundamental Properties

Unit 2 defines the derivative through difference quotients and establishes the basic differentiation rules. Differentiability and its relationship to continuity, along with derivatives of the standard elementary functions, are covered. The rules built here are used without comment in every later unit.

Position: 2 of 10 units
Published weighting: 4–7% in the official framework
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Calculus BC. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Defining the derivative: Average rate of change, the difference quotient, and the derivative as its limit.
Derivative as a function: Estimating derivative values from graphs and tables and sketching derivative graphs.
Differentiability and continuity: Why differentiability implies continuity and where differentiability fails.
Basic derivative rules: Power, constant multiple, sum, and difference rules on polynomial and radical functions.
Derivatives of transcendental functions: Sine, cosine, exponential, and natural logarithm derivatives.
Product and quotient rules: Applying both rules and recognizing when simplification first is faster.

Study tips

How to practice Differentiation Basics.

Section-specific guidance that follows from how this material is usually tested.

Keep the limit definition fresh; it appears in multiple-choice questions phrased as a limit to be recognized.
Drill the elementary derivatives to automaticity so attention can go to setup in later units.
Practice moving between a function graph and its derivative graph in both directions.

Where this fits

Unit 2: Differentiation: Definition and Fundamental Properties is 2 of 10 units in the AP Calculus BC guide, and it carries a published weighting of 4–7% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus BC guide Official exam information

Sample Differentiation Basics questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
If f(x) = 4√x + 3x for x > 0, what is f'(x)?
1. A2/√x + 3
2. B1/(2√x) + 3
3. C2√x + 3
4. D4/√x + 3
Show answer
Answer: A. 2/√x + 3
Rewrite 4√x as 4x^(1/2); the power rule gives 4(1/2)x^(-1/2) = 2/√x, and the derivative of 3x is 3. Choosing 1/(2√x) + 3 forgets to carry the coefficient 4 through the power rule.
Key insight: Convert every radical to a fractional exponent before applying the power rule.
Question 2Foundational
What is the derivative of f(x) = e^x cos(x)?
1. Ae^x cos(x) - e^x sin(x)
2. Be^x cos(x) + e^x sin(x)
3. C-e^x sin(x)
4. D-e^x sin(x) cos(x)
Show answer
Answer: A. e^x cos(x) - e^x sin(x)
The product rule gives e^x cos(x) + e^x times the derivative of cos(x), and that derivative is -sin(x), so the second term is -e^x sin(x). Adding instead of subtracting misses the sign in the derivative of cosine, and multiplying the two derivatives together ignores the product rule entirely.
Key insight: Product rule first, then substitute the derivative of each factor with its correct sign.
Question 3Intermediate
What is the value of lim h→0 of (√(4 + h) - 2)/h?
1. A1/4
2. B1/2
3. C2
4. D0
Show answer
Answer: A. 1/4
This is the difference quotient for f(x) = √x at x = 4, so the limit is f'(4). Since f'(x) = 1/(2√x), the value is 1/(2 times 2) = 1/4. Answering 1/2 evaluates 1/(2√x) as though √4 were 1.
Key insight: Recognizing a difference quotient converts a hard limit into a routine derivative.
Question 4Intermediate
What is the equation of the line tangent to y = x^2 - 4x at the point where x = 1?
1. Ay = -2x - 1
2. By = -2x - 3
3. Cy = -2x + 1
4. Dy = 2x - 5
Show answer
Answer: A. y = -2x - 1
At x = 1 the point is (1, -3) because 1 - 4 = -3, and the slope is y'(1) = 2(1) - 4 = -2. Point-slope form gives y + 3 = -2(x - 1), which simplifies to y = -2x - 1. Writing y = -2x - 3 substitutes the y-value as the intercept without adjusting for the shift.
Key insight: A tangent line needs both the point on the curve and the derivative evaluated there.
Question 5Advanced
For f(x) = x|x|, what is f'(0)?
1. A0
2. B1
3. C-1
4. Df'(0) does not exist
Show answer
Answer: A. 0
The difference quotient is (h|h| - 0)/h = |h|, which approaches 0 from both sides, so f'(0) = 0. The function equals x^2 for x ≥ 0 and -x^2 for x < 0, and both pieces have slope 0 at the origin, so unlike |x| this function is differentiable there.
Key insight: An absolute value does not automatically create a corner; test the difference quotient itself.

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Unit 2: Differentiation: Definition and Fundamental Properties