AP Calculus BC Quiz: Estimating Derivatives Of A Function

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Estimating Derivatives Of A FunctionQuestion 1 of 20

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

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

2.0

2.5

2.6

3.0

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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.

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.

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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.

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Question 1

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

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

$2.0$
$2.5$ (correct answer)
$2.6$
$3.0$

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

Question 2

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

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 $t=5$ minutes?

$8$
$16$ (correct answer)
$1.45$
$44$

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

Question 3

The depth of water in a reservoir, $D(t)$ , is a differentiable function of time $t$ . On a certain day, at $t=12$ 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 $D'(12)$ , the rate of change of the depth at $t=12$ hours, in feet per hour?

$-0.2$
$-0.3$ (correct answer)
$-0.4$
$0.3$

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

Question 4

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

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

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 $x=c$ . A symmetric interval, using $x=c-h$ and $x=c+h$ , often provides a better approximation than a one-sided interval.

Question 5

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

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

$0.25$
$0.5$
$5$ (correct answer)
$10.25$

Explanation: Acceleration is the derivative of velocity, so we need to estimate $a(2) = v'(2)$ . We can use the average rate of change of velocity over the interval $[2, 2.1]$ . $a(2) \approx \frac{v(2.1) - v(2)}{2.1 - 2} = \frac{10.5 - 10}{0.1} = \frac{0.5}{0.1} = 5$ m/s $^2$ .

Question 6

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

$10$
$20$
$40$ (correct answer)
$0.025$

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

Question 7

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

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

$3.0$
$3.5$ (correct answer)
$4.0$
$4.5$

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

Question 8

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

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

$-7.7$ (correct answer)
$6.354$
$0.02$
$-0.154$

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

Question 9

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

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

$3.0$
$3.5$ (correct answer)
$4.0$
$0.7$

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

Question 10

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

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

$6$ (correct answer)
$12$
$156$
$1.2$

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

Question 11

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

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

$-0.6$
$-1.5$ (correct answer)
$0.4$
$-0.16$

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

Question 12

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

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

$0.10$
$0.40$
$1.30$
$2.50$ (correct answer)

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

Question 13

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

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

$3c^2$
$3c^2 + h^2$ (correct answer)
$3c^2+3ch+h^2$
$6c^2+2h^2$

Explanation: We substitute $f(x)=x^3$ into the expression for $D(h)$ : $D(h) = \frac{(c+h)^3 - (c-h)^3}{2h}$ $= \frac{(c^3+3c^2h+3ch^2+h^3) - (c^3-3c^2h+3ch^2-h^3)}{2h}$ $= \frac{6c^2h + 2h^3}{2h}$ $= 3c^2 + h^2$ . As $h \to 0$ , this expression approaches $3c^2$ , which is $f'(c)$ .

Question 14

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

$24$
$12$ (correct answer)
$11$
$8$

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

Question 15

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

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

$-6$ (correct answer)
$-3$
$3$
$-1.2$

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

Question 16

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

$0.5$
$2$ (correct answer)
$0.2$
$6$

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

Question 17

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

$Q$ is an approximation for the value of $g(a)$ .
$Q$ is the exact value of the average rate of change of $g$ on $[a, a+0.01]$ . (correct answer)
$Q$ is the exact value of the instantaneous rate of change $g'(a)$ .
$Q$ is an approximation for the average rate of change of $g$ on $[a, a+0.01]$ .

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

Question 18

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

$2h$
$h$
$-2$
$2$ (correct answer)

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

Question 19

Let $f$ be a differentiable function. Given $f(x_0 - \Delta x) = y_1$ and $f(x_0 + \Delta x) = y_2$ , where $\Delta x > 0$ . Which expression represents the best estimate for $f'(x_0)$ ?

$\frac{y_2 - y_1}{2 \Delta x}$ (correct answer)
$\frac{y_2 - y_1}{\Delta x}$
$\frac{y_2 + y_1}{2}$
$\frac{y_2 - y_1}{x_0}$

Explanation: The best estimate for the derivative at a point $x_0$ is the slope of the secant line over a small symmetric interval around $x_0$ . The interval is $[x_0 - \Delta x, x_0 + \Delta x]$ . The change in $y$ is $y_2 - y_1$ . The change in $x$ is $(x_0 + \Delta x) - (x_0 - \Delta x) = 2 \Delta x$ . Therefore, the slope is $\frac{y_2 - y_1}{2 \Delta x}$ .

Question 20

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

$4$ (correct answer)
$4-2h$
$2$
It cannot be determined from the information given.

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