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AP Calculus BC Quiz

AP Calculus BC Quiz: Estimating Derivatives Of A Function

Practice Estimating Derivatives Of A Function in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

The height of a plant, H(t)H(t)H(t), in centimeters, is a differentiable function of time ttt in days. Measurements are recorded as follows: H(10)=20H(10) = 20H(10)=20, H(12)=24H(12) = 24H(12)=24, and H(15)=33H(15)=33H(15)=33.

Which of the following is the best estimate of the growth rate H′(12)H'(12)H′(12) in centimeters per day?

Select an answer to continue

What this quiz covers

This quiz focuses on Estimating Derivatives Of A Function, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The height of a plant, H(t)H(t)H(t), in centimeters, is a differentiable function of time ttt in days. Measurements are recorded as follows: H(10)=20H(10) = 20H(10)=20, H(12)=24H(12) = 24H(12)=24, and H(15)=33H(15)=33H(15)=33.

Which of the following is the best estimate of the growth rate H′(12)H'(12)H′(12) in centimeters per day?

  1. 2.02.02.0
  2. 2.52.52.5 (correct answer)
  3. 2.62.62.6
  4. 3.03.03.0

Explanation: To get the best estimate for H′(12)H'(12)H′(12), we should average the rates of change from the intervals on either side of t=12t=12t=12. The backward difference (rate on [10,12][10, 12][10,12]) is H(12)−H(10)12−10=24−202=2 \frac{H(12)-H(10)}{12-10} = \frac{24-20}{2} = 212−10H(12)−H(10)​=224−20​=2. The forward difference (rate on [12,15][12, 15][12,15]) is H(15)−H(12)15−12=33−243=3 \frac{H(15)-H(12)}{15-12} = \frac{33-24}{3} = 315−12H(15)−H(12)​=333−24​=3. The best estimate is the average of these two rates: 2+32=2.5 \frac{2+3}{2} = 2.522+3​=2.5 cm/day.

Question 2

The temperature TTT, in degrees Celsius, of a chemical reaction is a differentiable function of time ttt, in minutes. At time t=5t=5t=5 minutes, the temperature is 40∘C40^{\circ}C40∘C. At time t=5.5t=5.5t=5.5 minutes, the temperature is 48∘C48^{\circ}C48∘C.

Based on this information, which of the following is the best estimate for the rate of change of the temperature, in degrees Celsius per minute, at time t=5t=5t=5 minutes?

  1. 888
  2. 161616 (correct answer)
  3. 1.451.451.45
  4. 444444

Explanation: The rate of change of the temperature at t=5t=5t=5 can be estimated by the average rate of change over the interval [5,5.5][5, 5.5][5,5.5]. T′(5)≈T(5.5)−T(5)5.5−5=48−400.5=80.5=16T'(5) \approx \frac{T(5.5) - T(5)}{5.5 - 5} = \frac{48 - 40}{0.5} = \frac{8}{0.5} = 16T′(5)≈5.5−5T(5.5)−T(5)​=0.548−40​=0.58​=16 degrees Celsius per minute.

Question 3

The depth of water in a reservoir, D(t)D(t)D(t), is a differentiable function of time ttt. On a certain day, at t=12t=12t=12 hours, the depth is 50 feet. Two hours prior, the depth was 50.8 feet, and two hours later, the depth is 49.6 feet.

What is the best estimate for D′(12)D'(12)D′(12), the rate of change of the depth at t=12t=12t=12 hours, in feet per hour?

  1. −0.2-0.2−0.2
  2. −0.3-0.3−0.3 (correct answer)
  3. −0.4-0.4−0.4
  4. 0.30.30.3

Explanation: We are given D(10)=50.8D(10) = 50.8D(10)=50.8, D(12)=50D(12) = 50D(12)=50, and D(14)=49.6D(14) = 49.6D(14)=49.6. The best estimate for D′(12)D'(12)D′(12) uses the symmetric interval around t=12t=12t=12, which is [10,14][10, 14][10,14]. D′(12)≈D(14)−D(10)14−10=49.6−50.84=−1.24=−0.3D'(12) \approx \frac{D(14) - D(10)}{14 - 10} = \frac{49.6 - 50.8}{4} = \frac{-1.2}{4} = -0.3D′(12)≈14−10D(14)−D(10)​=449.6−50.8​=4−1.2​=−0.3 feet per hour.

Question 4

The derivative of a function fff at x=cx=cx=c, denoted f′(c)f'(c)f′(c), is best approximated by the slope of a secant line through two points on the graph of fff. For the most accurate approximation, the two points should be:

  1. chosen far apart to capture the overall trend of the function.
  2. chosen symmetrically about x=cx=cx=c and as close to ccc as possible. (correct answer)
  3. chosen such that one of the points has an x-coordinate of ccc.
  4. located at local maximum or minimum values near x=cx=cx=c.

Explanation: The derivative is the limit of the slope of the secant line as the interval between the points approaches zero. Therefore, the best approximation is obtained when the points are very close to x=cx=cx=c. A symmetric interval, using x=c−hx=c-hx=c−h and x=c+hx=c+hx=c+h, often provides a better approximation than a one-sided interval.

Question 5

The velocity of a particle, v(t)v(t)v(t), in meters per second, is a differentiable function of time ttt in seconds. At t=2t=2t=2 seconds, the velocity is v(2)=10v(2) = 10v(2)=10 m/s. At t=2.1t=2.1t=2.1 seconds, the velocity is v(2.1)=10.5v(2.1)=10.5v(2.1)=10.5 m/s.

Based on this information, what is a possible estimate for the acceleration of the particle at t=2t=2t=2 seconds, in m/s2^22?

  1. 0.250.250.25
  2. 0.50.50.5
  3. 555 (correct answer)
  4. 10.2510.2510.25

Explanation: Acceleration is the derivative of velocity, so we need to estimate a(2)=v′(2)a(2) = v'(2)a(2)=v′(2). We can use the average rate of change of velocity over the interval [2,2.1][2, 2.1][2,2.1]. a(2)≈v(2.1)−v(2)2.1−2=10.5−100.1=0.50.1=5a(2) \approx \frac{v(2.1) - v(2)}{2.1 - 2} = \frac{10.5 - 10}{0.1} = \frac{0.5}{0.1} = 5a(2)≈2.1−2v(2.1)−v(2)​=0.110.5−10​=0.10.5​=5 m/s2^22.

Question 6

A function fff is differentiable for all real numbers. If f(2.9)=5.8f(2.9) = 5.8f(2.9)=5.8 and f(3.1)=6.2f(3.1) = 6.2f(3.1)=6.2, which of the following is the best estimate for f′(3)f'(3)f′(3)?

  1. 0.50.50.5
  2. 222 (correct answer)
  3. 0.20.20.2
  4. 666

Explanation: The derivative at a point can be estimated by the slope of the secant line through two nearby points. The best estimate for f′(3)f'(3)f′(3) using the given values is the average rate of change over the interval [2.9,3.1][2.9, 3.1][2.9,3.1]. f′(3)≈f(3.1)−f(2.9)3.1−2.9=6.2−5.80.2=0.40.2=2f'(3) \approx \frac{f(3.1) - f(2.9)}{3.1 - 2.9} = \frac{6.2 - 5.8}{0.2} = \frac{0.4}{0.2} = 2f′(3)≈3.1−2.9f(3.1)−f(2.9)​=0.26.2−5.8​=0.20.4​=2.

Question 7

A tank is being filled with water. The volume V(t)V(t)V(t) of water in the tank, in gallons, is a differentiable function of time ttt in hours. At t=2t=2t=2 hours, the volume is 300 gallons. A half-hour earlier, at t=1.5t=1.5t=1.5 hours, the volume was 280 gallons.

Based on this information, what is the best estimate for the rate at which the volume is changing at t=2t=2t=2 hours, in gallons per hour?

  1. 101010
  2. 202020
  3. 404040 (correct answer)
  4. 0.0250.0250.025

Explanation: The rate of change of the volume at t=2t=2t=2 can be estimated by the average rate of change over the interval [1.5,2][1.5, 2][1.5,2]. V′(2)≈V(2)−V(1.5)2−1.5=300−2800.5=200.5=40V'(2) \approx \frac{V(2) - V(1.5)}{2 - 1.5} = \frac{300 - 280}{0.5} = \frac{20}{0.5} = 40V′(2)≈2−1.5V(2)−V(1.5)​=0.5300−280​=0.520​=40 gallons per hour.

Question 8

A function g(x)g(x)g(x) is differentiable. Selected values of g(x)g(x)g(x) are g(1.9)=4.2g(1.9) = 4.2g(1.9)=4.2, g(2.0)=4.5g(2.0) = 4.5g(2.0)=4.5, g(2.1)=4.9g(2.1) = 4.9g(2.1)=4.9, and g(2.2)=5.4g(2.2) = 5.4g(2.2)=5.4.

Which of the following is the best estimate for g′(2)g'(2)g′(2)?

  1. 3.03.03.0
  2. 3.53.53.5 (correct answer)
  3. 4.04.04.0
  4. 4.54.54.5

Explanation: To find the best estimate for g′(2)g'(2)g′(2), we should use the smallest symmetric interval around x=2x=2x=2. The values at x=1.9x=1.9x=1.9 and x=2.1x=2.1x=2.1 provide this interval. g′(2)≈g(2.1)−g(1.9)2.1−1.9=4.9−4.20.2=0.70.2=3.5g'(2) \approx \frac{g(2.1) - g(1.9)}{2.1 - 1.9} = \frac{4.9 - 4.2}{0.2} = \frac{0.7}{0.2} = 3.5g′(2)≈2.1−1.9g(2.1)−g(1.9)​=0.24.9−4.2​=0.20.7​=3.5.

Question 9

Let f(x)=x2cos⁡(x)f(x) = x^2 \cos(\sqrt{x})f(x)=x2cos(x​). It is given that f(3.99)≈6.431f(3.99) \approx 6.431f(3.99)≈6.431 and f(4.01)≈6.277f(4.01) \approx 6.277f(4.01)≈6.277.

Which of the following is the best approximation for f′(4)f'(4)f′(4)?

  1. −7.7-7.7−7.7 (correct answer)
  2. 6.3546.3546.354
  3. 0.020.020.02
  4. −0.154-0.154−0.154

Explanation: The derivative f′(4)f'(4)f′(4) can be approximated by the slope of the secant line through the points at x=3.99x=3.99x=3.99 and x=4.01x=4.01x=4.01. f′(4)≈f(4.01)−f(3.99)4.01−3.99=6.277−6.4310.02=−0.1540.02=−7.7f'(4) \approx \frac{f(4.01) - f(3.99)}{4.01 - 3.99} = \frac{6.277 - 6.431}{0.02} = \frac{-0.154}{0.02} = -7.7f′(4)≈4.01−3.99f(4.01)−f(3.99)​=0.026.277−6.431​=0.02−0.154​=−7.7.

Question 10

For a differentiable function h(t)h(t)h(t), the following values are known: h(4.8)=10.2h(4.8) = 10.2h(4.8)=10.2, h(5.0)=10.8h(5.0) = 10.8h(5.0)=10.8, and h(5.2)=11.6h(5.2) = 11.6h(5.2)=11.6.

Which of the following is the best estimate for h′(5.0)h'(5.0)h′(5.0)?

  1. 3.03.03.0
  2. 3.53.53.5 (correct answer)
  3. 4.04.04.0
  4. 0.70.70.7

Explanation: The best estimate for h′(5.0)h'(5.0)h′(5.0) can be found by using the symmetric difference quotient over the smallest symmetric interval [4.8,5.2][4.8, 5.2][4.8,5.2]. This is equivalent to averaging the backward and forward difference quotients. h′(5.0)≈h(5.2)−h(4.8)5.2−4.8=11.6−10.20.4=1.40.4=3.5h'(5.0) \approx \frac{h(5.2) - h(4.8)}{5.2 - 4.8} = \frac{11.6 - 10.2}{0.4} = \frac{1.4}{0.4} = 3.5h′(5.0)≈5.2−4.8h(5.2)−h(4.8)​=0.411.6−10.2​=0.41.4​=3.5.

Question 11

The population of a town, P(t)P(t)P(t), in thousands, is a differentiable function of time ttt in years since the beginning of 2010. It is known that P(8)=150P(8) = 150P(8)=150 and P(10)=162P(10) = 162P(10)=162.

Which of the following is the best estimate for the rate at which the population was growing at the beginning of 2019 (t=9t=9t=9), in thousands of people per year?

  1. 666 (correct answer)
  2. 121212
  3. 156156156
  4. 1.21.21.2

Explanation: The rate of population growth at t=9t=9t=9 is P′(9)P'(9)P′(9). We can estimate this value using the average rate of change over the interval [8,10][8, 10][8,10], for which t=9t=9t=9 is the midpoint. P′(9)≈P(10)−P(8)10−8=162−1502=122=6P'(9) \approx \frac{P(10) - P(8)}{10 - 8} = \frac{162 - 150}{2} = \frac{12}{2} = 6P′(9)≈10−8P(10)−P(8)​=2162−150​=212​=6. This is 6 thousand people per year.

Question 12

Let fff be a differentiable function with f(3.4)=7.1f(3.4) = 7.1f(3.4)=7.1 and f(3.8)=6.5f(3.8) = 6.5f(3.8)=6.5.

Which of the following is the best approximation for f′(3.6)f'(3.6)f′(3.6)?

  1. −0.6-0.6−0.6
  2. −1.5-1.5−1.5 (correct answer)
  3. 0.40.40.4
  4. −0.16-0.16−0.16

Explanation: The point x=3.6x=3.6x=3.6 is the midpoint of the interval [3.4,3.8][3.4, 3.8][3.4,3.8]. The best estimate for the derivative at the midpoint of an interval is the average rate of change over that interval. f′(3.6)≈f(3.8)−f(3.4)3.8−3.4=6.5−7.10.4=−0.60.4=−1.5f'(3.6) \approx \frac{f(3.8) - f(3.4)}{3.8 - 3.4} = \frac{6.5 - 7.1}{0.4} = \frac{-0.6}{0.4} = -1.5f′(3.6)≈3.8−3.4f(3.8)−f(3.4)​=0.46.5−7.1​=0.4−0.6​=−1.5.

Question 13

A differentiable function ggg has values g(4.98)=1.25g(4.98) = 1.25g(4.98)=1.25 and g(5.02)=1.35g(5.02) = 1.35g(5.02)=1.35.

Using the secant line between x=4.98x=4.98x=4.98 and x=5.02x=5.02x=5.02 as an approximation to the tangent line at x=5x=5x=5, what is the estimated value of g′(5)g'(5)g′(5)?

  1. 0.100.100.10
  2. 0.400.400.40
  3. 1.301.301.30
  4. 2.502.502.50 (correct answer)

Explanation: The slope of the tangent line at x=5x=5x=5, which is g′(5)g'(5)g′(5), is approximated by the slope of the secant line through the nearby points at x=4.98x=4.98x=4.98 and x=5.02x=5.02x=5.02. g′(5)≈g(5.02)−g(4.98)5.02−4.98=1.35−1.250.04=0.100.04=2.5g'(5) \approx \frac{g(5.02) - g(4.98)}{5.02 - 4.98} = \frac{1.35 - 1.25}{0.04} = \frac{0.10}{0.04} = 2.5g′(5)≈5.02−4.98g(5.02)−g(4.98)​=0.041.35−1.25​=0.040.10​=2.5.

Question 14

Let f(x)f(x)f(x) be a differentiable function. The expression D(h)=f(c+h)−f(c−h)2hD(h) = \frac{f(c+h) - f(c-h)}{2h}D(h)=2hf(c+h)−f(c−h)​ is used to approximate f′(c)f'(c)f′(c).

If f(x)=x3f(x) = x^3f(x)=x3, which of the following is the value of D(h)D(h)D(h) used to approximate f′(c)f'(c)f′(c)?

  1. 3c23c^23c2
  2. 3c2+h23c^2 + h^23c2+h2 (correct answer)
  3. 3c2+3ch+h23c^2+3ch+h^23c2+3ch+h2
  4. 6c2+2h26c^2+2h^26c2+2h2

Explanation: We substitute f(x)=x3f(x)=x^3f(x)=x3 into the expression for D(h)D(h)D(h): D(h)=(c+h)3−(c−h)32hD(h) = \frac{(c+h)^3 - (c-h)^3}{2h}D(h)=2h(c+h)3−(c−h)3​ =(c3+3c2h+3ch2+h3)−(c3−3c2h+3ch2−h3)2h= \frac{(c^3+3c^2h+3ch^2+h^3) - (c^3-3c^2h+3ch^2-h^3)}{2h}=2h(c3+3c2h+3ch2+h3)−(c3−3c2h+3ch2−h3)​ =6c2h+2h32h= \frac{6c^2h + 2h^3}{2h}=2h6c2h+2h3​ =3c2+h2= 3c^2 + h^2=3c2+h2. As h→0h \to 0h→0, this expression approaches 3c23c^23c2, which is f′(c)f'(c)f′(c).

Question 15

To estimate the value of f′(2)f'(2)f′(2) for the function f(x)=x3−xf(x) = x^3 - xf(x)=x3−x, an analyst uses the slope of the secant line connecting the points on the graph of fff at x=1x=1x=1 and x=3x=3x=3. What is the value of this estimate?

  1. 242424
  2. 121212 (correct answer)
  3. 111111
  4. 888

Explanation: First, find the function values at x=1x=1x=1 and x=3x=3x=3. f(1)=13−1=0f(1) = 1^3 - 1 = 0f(1)=13−1=0. f(3)=33−3=27−3=24f(3) = 3^3 - 3 = 27 - 3 = 24f(3)=33−3=27−3=24. The slope of the secant line is the average rate of change between these points: msec=f(3)−f(1)3−1=24−02=12m_{sec} = \frac{f(3) - f(1)}{3 - 1} = \frac{24 - 0}{2} = 12msec​=3−1f(3)−f(1)​=224−0​=12. This value is an estimate for f′(2)f'(2)f′(2).

Question 16

The quantity Q=g(a+0.01)−g(a)0.01Q = \frac{g(a+0.01) - g(a)}{0.01}Q=0.01g(a+0.01)−g(a)​ is calculated for a differentiable function ggg. Which of the following statements about QQQ is the most accurate?

  1. QQQ is an approximation for the value of g(a)g(a)g(a).
  2. QQQ is the exact value of the average rate of change of ggg on [a,a+0.01][a, a+0.01][a,a+0.01]. (correct answer)
  3. QQQ is the exact value of the instantaneous rate of change g′(a)g'(a)g′(a).
  4. QQQ is an approximation for the average rate of change of ggg on [a,a+0.01][a, a+0.01][a,a+0.01].

Explanation: The expression g(b)−g(a)b−a\frac{g(b) - g(a)}{b - a}b−ag(b)−g(a)​ defines the exact average rate of change of the function ggg over the interval [a,b][a, b][a,b]. The given quantity QQQ matches this form with b=a+0.01b=a+0.01b=a+0.01. This value is often used to approximate the instantaneous rate of change g′(a)g'(a)g′(a), but the quantity itself is the exact average rate of change.

Question 17

Let kkk be a differentiable function. For a small positive value hhh, it is known that k(3)−k(3−h)≈2hk(3) - k(3-h) \approx 2hk(3)−k(3−h)≈2h. Which of the following is the best estimate for k′(3)k'(3)k′(3)?

  1. 2h2h2h
  2. hhh
  3. −2-2−2
  4. 222 (correct answer)

Explanation: The derivative k′(3)k'(3)k′(3) can be estimated using a difference quotient. The given information k(3)−k(3−h)≈2hk(3) - k(3-h) \approx 2hk(3)−k(3−h)≈2h can be rearranged by dividing by hhh: k(3)−k(3−h)h≈2 \frac{k(3) - k(3-h)}{h} \approx 2hk(3)−k(3−h)​≈2. The left side is the backward difference quotient k(3)−k(3−h)3−(3−h)\frac{k(3) - k(3-h)}{3 - (3-h)}3−(3−h)k(3)−k(3−h)​, which is an approximation for k′(3)k'(3)k′(3). Therefore, the best estimate for k′(3)k'(3)k′(3) is 2.

Question 18

The amount of a certain drug in a patient's bloodstream, A(t)A(t)A(t), in milligrams (mg), is a differentiable function of time ttt in hours. At t=2t=2t=2 hours, A(2)=15A(2) = 15A(2)=15 mg. At t=2.5t=2.5t=2.5 hours, A(2.5)=12A(2.5) = 12A(2.5)=12 mg.

Using this data, what is the approximate rate of change of the amount of the drug in the bloodstream at t=2.25t=2.25t=2.25 hours, in mg per hour?

  1. −6-6−6 (correct answer)
  2. −3-3−3
  3. 333
  4. −1.2-1.2−1.2

Explanation: The rate of change at t=2.25t=2.25t=2.25 can be estimated by the average rate of change over the interval [2,2.5][2, 2.5][2,2.5], since t=2.25t=2.25t=2.25 is the midpoint of this interval. A′(2.25)≈A(2.5)−A(2)2.5−2=12−150.5=−30.5=−6A'(2.25) \approx \frac{A(2.5) - A(2)}{2.5 - 2} = \frac{12 - 15}{0.5} = \frac{-3}{0.5} = -6A′(2.25)≈2.5−2A(2.5)−A(2)​=0.512−15​=0.5−3​=−6 mg per hour.

Question 19

Let fff be a differentiable function. Given f(x0−Δx)=y1f(x_0 - \Delta x) = y_1f(x0​−Δx)=y1​ and f(x0+Δx)=y2f(x_0 + \Delta x) = y_2f(x0​+Δx)=y2​, where Δx>0\Delta x > 0Δx>0. Which expression represents the best estimate for f′(x0)f'(x_0)f′(x0​)?

  1. y2−y12Δx\frac{y_2 - y_1}{2 \Delta x}2Δxy2​−y1​​ (correct answer)
  2. y2−y1Δx\frac{y_2 - y_1}{\Delta x}Δxy2​−y1​​
  3. y2+y12\frac{y_2 + y_1}{2}2y2​+y1​​
  4. y2−y1x0\frac{y_2 - y_1}{x_0}x0​y2​−y1​​

Explanation: The best estimate for the derivative at a point x0x_0x0​ is the slope of the secant line over a small symmetric interval around x0x_0x0​. The interval is [x0−Δx,x0+Δx][x_0 - \Delta x, x_0 + \Delta x][x0​−Δx,x0​+Δx]. The change in yyy is y2−y1y_2 - y_1y2​−y1​. The change in xxx is (x0+Δx)−(x0−Δx)=2Δx(x_0 + \Delta x) - (x_0 - \Delta x) = 2 \Delta x(x0​+Δx)−(x0​−Δx)=2Δx. Therefore, the slope is y2−y12Δx\frac{y_2 - y_1}{2 \Delta x}2Δxy2​−y1​​.

Question 20

Let fff be a differentiable function. The average rate of change of fff on the interval [c,c+h][c, c+h][c,c+h] is given by the expression 4−2h4 - 2h4−2h. Which of the following is the value of f′(c)f'(c)f′(c)?

  1. 444 (correct answer)
  2. 4−2h4-2h4−2h
  3. 222
  4. It cannot be determined from the information given.

Explanation: The instantaneous rate of change, f′(c)f'(c)f′(c), is the limit of the average rate of change as the interval length hhh approaches 0. We are given that the average rate of change on [c,c+h][c, c+h][c,c+h] is f(c+h)−f(c)h=4−2h \frac{f(c+h)-f(c)}{h} = 4-2hhf(c+h)−f(c)​=4−2h. Taking the limit as h→0h \to 0h→0 gives: f′(c)=lim⁡h→0(4−2h)=4−2(0)=4f'(c) = \lim_{h \to 0} (4 - 2h) = 4 - 2(0) = 4f′(c)=limh→0​(4−2h)=4−2(0)=4.