Drill 1 · Math · Definition of the Derivative
AP Calculus AB: Definition of the Derivative (Drill 1) is a Math practice drill covering Definition of the Derivative. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice using the limit definition of the derivative and interpreting derivatives as instantaneous rates of change. These AP Calculus AB skills appear on both the multiple-choice and free-response sections.
Question 1. If \( f(x) = x^2 + 3x \), which of the following correctly expresses \( f'(x) \) using the limit definition of the derivative?
Explanation: Choice A is correct. The limit definition of the derivative is \( f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} \). Substituting \( f(x) = x^2 + 3x \) gives exactly the expression in Choice A. Choice B is incorrect because it omits the \( -f(x) \) term in the numerator, the student forgot to subtract \( x^2 + 3x \). Choice C is incorrect because it omits division by \( h \) entirely; this expression equals 0 after taking the limit, not the derivative. Choice D is incorrect because it takes \( x \to 0 \) rather than \( h \to 0 \); the variable approaching 0 must be the increment \( h \), not \( x \).
Question 2. Using the limit definition of the derivative, what is \( f'(2) \) if \( f(x) = 3x^2 \)?
Explanation: Choice B is correct. We compute \( f'(2) = \lim_{h \to 0} \dfrac{3(2+h)^2 - 3(4)}{h} = \lim_{h \to 0} \dfrac{12h + 3h^2}{h} = \lim_{h \to 0}(12 + 3h) = 12 \). Choice A is incorrect because the student obtained \( f'(x) = 6x \) but evaluated it at \( x = 1 \) instead of \( x = 2 \). Choice C is incorrect because the student dropped the coefficient 3 when differentiating, obtaining \( 2x \) evaluated at \( x = 2 \) without the factor. Choice D is incorrect because the student evaluated \( f(2) = 3(4) = 12 \) and then multiplied by the exponent 2, confusing the function value with power rule mechanics rather than computing the derivative.
Question 3. The table below shows selected values of a function \( g \).
| x | 1 | 3 | 5 | 7 |
|---|---|---|---|---|
| g(x) | 4 | 10 | 18 | 24 |
Explanation: Choice B is correct. The best estimate for \( g'(5) \) uses the symmetric difference quotient: \( g'(5) \approx \dfrac{g(7) - g(3)}{7 - 3} = \dfrac{24 - 10}{4} = 3.5 \). This uses the points on either side of \( x = 5 \) for a centered, more accurate estimate. Choice A is incorrect because the student used only the one-sided difference quotient \( \dfrac{g(7)-g(5)}{7-5} = \dfrac{6}{2} = 3 \) rather than the symmetric quotient. Choice C is incorrect because the student reported \( g(5) = 18 \), confusing the function value with the derivative. Choice D is incorrect because the student correctly computed the centered numerator \( g(7) - g(3) = 14 \) but divided by 2 instead of 4, treating the denominator as the half-interval width rather than the full distance between \( x = 3 \) and \( x = 7 \).
Question 4. A company's revenue (in thousands of dollars) \( t \) months after launch is modeled by a differentiable function \( R(t) \). It is given that \( R(6) = 85 \) and \( R'(6) = 4.2 \). Which of the following best interprets \( R'(6) = 4.2 \)?
Explanation: Choice B is correct. \( R'(6) \) is the instantaneous rate of change of revenue at \( t = 6 \). Since \( R \) is measured in thousands of dollars and \( t \) in months, \( R'(6) = 4.2 \) means revenue is growing at 4.2 thousand dollars ($4,200) per month at that instant. Choice A is incorrect because it confuses \( R'(6) \) with \( R(6) \); the revenue at 6 months is $85,000, not $4,200. Choice C is incorrect because it confuses instantaneous rate of change with average rate of change; the average monthly revenue would be \( \dfrac{R(6)}{6} \), not \( R'(6) \). Choice D is incorrect because it misreads the derivative as a future revenue prediction rather than a current instantaneous rate.
Question 5. Which of the following limits equals \( f'(5) \) if \( f(x) = \sqrt{x} \)?
Explanation: Choice A is correct. By definition, \( f'(5) = \lim_{h \to 0} \dfrac{f(5+h) - f(5)}{h} = \lim_{h \to 0} \dfrac{\sqrt{5+h} - \sqrt{5}}{h} \). Choice B is incorrect because it omits the \( -\sqrt{5} \) term, the student forgot to subtract \( f(5) \) from the numerator. Choice C is incorrect because it divides by the constant 5 rather than by the increment \( x - 5 \); the correct alternate form is \( \lim_{x \to 5} \dfrac{\sqrt{x} - \sqrt{5}}{x - 5} \). Choice D is incorrect because the argument of the square root in the numerator is \( h \) instead of \( 5 + h \), so the expression does not evaluate \( f \) at \( 5 + h \).