Business Calculus Quiz: Derivative As Instantaneous Rate

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Derivative As Instantaneous RateQuestion 1 of 14

The total cost in dollars to produce $x$ units of a commodity is $C(x) = 0.5x^2 + 20x + 1800$ . What is the instantaneous rate of change of the average cost with respect to the number of units produced when $x = 30$ ?

$-1.50 per unit per unit

$2.50 per unit per unit

$50.00 per unit

$95.00 per unit

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Business Calculus Quiz

Business Calculus Quiz: Derivative As Instantaneous Rate

Practice Derivative As Instantaneous Rate in Business Calculus 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 Derivative As Instantaneous Rate, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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

$-1.50 per unit per unit (correct answer)
$2.50 per unit per unit
$50.00 per unit
$95.00 per unit

Explanation: First, find the average cost function,

\bar{C}(x) = C(x)/x

\bar{C}(x) = \frac{0.5x^2 + 20x + 1800}{x} = 0.5x + 20 + 1800x^{-1}

. Next, find the derivative of the average cost function,

\bar{C}'(x)

\bar{C}'(x) = 0.5 - 1800x^{-2} = 0.5 - \frac{1800}{x^2}

. Finally, evaluate this derivative at

x = 30

\bar{C}'(30) = 0.5 - \frac{1800}{30^2} = 0.5 - \frac{1800}{900} = 0.5 - 2 = -1.5

. This means the average cost is decreasing at a rate of $1.50 per unit per unit.

Question 2

A manufacturer's total cost to produce $x$ items is given by $C(x) = 40 \ln(x+1) + 500$ . What is the marginal cost when production is at 19 items? The marginal cost is the instantaneous rate of change of the total cost.

$2.00 (correct answer)
$26.50
$119.83
$619.83

Explanation: The marginal cost is the derivative of the cost function,

C'(x)

. The derivative of

40 \ln(x+1) + 500

C'(x) = 40 \cdot \frac{1}{x+1} \cdot 1 + 0 = \frac{40}{x+1}

. To find the marginal cost at a production level of 19 items, we evaluate

C'(19)

C'(19) = \frac{40}{19+1} = \frac{40}{20} = 2

. Therefore, the marginal cost is $2.00 per item.

Question 3

The price-demand function for a particular brand of tablet is given by $p(x) = 1200 - 1.5x$ , where $p$ is the wholesale price in dollars at which $x$ thousand tablets can be sold. Find the instantaneous rate of change of revenue with respect to the number of tablets sold when sales are at 100,000 tablets.

$1050 per thousand tablets
$900 per thousand tablets (correct answer)
$105,000,000 per thousand tablets
$1,050,000 per thousand tablets

Explanation: First, define the revenue function,

R(x) = x \cdot p(x)

. Note that

x

is in thousands of tablets. So,

R(x) = x(1200 - 1.5x) = 1200x - 1.5x^2

. The instantaneous rate of change of revenue is the derivative,

R'(x) = 1200 - 3x

. The problem asks for this rate when sales are 100,000 tablets, which corresponds to

x=100

. Evaluate

R'(100) = 1200 - 3(100) = 1200 - 300 = 900

. The units are dollars per thousand tablets.

Question 4

The value $V$ (in dollars) of a piece of industrial equipment after $t$ years is modeled by the function $V(t) = \frac{50000}{t+1} + 2000$ . At what instantaneous rate is the value of the equipment changing at the end of the fourth year ( $t=4$ )?

Decreasing by $2,000 per year. (correct answer)
Decreasing by $10,000 per year.
The value is $12,000.
Increasing by $2,000 per year.

Explanation: The instantaneous rate of change is the derivative,

V'(t)

. First, rewrite

V(t) = 50000(t+1)^{-1} + 2000

. Using the power rule and chain rule,

V'(t) = -50000(t+1)^{-2} \cdot (1) = -\frac{50000}{(t+1)^2}

. Now, evaluate this at

t=4

V'(4) = -\frac{50000}{(4+1)^2} = -\frac{50000}{25} = -2000

. The negative sign indicates the value is decreasing. Thus, the value is decreasing at a rate of $2,000 per year.

Question 5

A population model predicts that the number of bacteria in a culture is $N(t) = \frac{5000}{1 + 49e^{-0.8t}}$ , where $t$ is time in hours. What is the instantaneous rate of population growth when the population reaches exactly $2500$ bacteria?

$800$ bacteria per hour
$1000$ bacteria per hour (correct answer)
$1200$ bacteria per hour
$1600$ bacteria per hour

Explanation: When

N(t) = 2500

2500 = \frac{5000}{1 + 49e^{-0.8t}}

, so

1 + 49e^{-0.8t} = 2

, giving

e^{-0.8t} = \frac{1}{49}

. Using the quotient rule:

N'(t) = \frac{5000 \cdot 49 \cdot 0.8 \cdot e^{-0.8t}}{(1 + 49e^{-0.8t})^2} = \frac{196000e^{-0.8t}}{(1 + 49e^{-0.8t})^2}

. Substituting

e^{-0.8t} = \frac{1}{49}

N'(t) = \frac{196000 \cdot \frac{1}{49}}{(1 + 1)^2} = \frac{4000}{4} = 1000

. Choice A forgets the factor of 0.8. Choice C uses incorrect denominator. Choice D doubles the correct answer.

Question 6

A drug concentration in the bloodstream is modeled by $C(t) = \frac{15t}{t^2 + 4}$ mg/L, where $t$ is hours after administration. What is the instantaneous rate of concentration change when the concentration is at its maximum value?

$0$ mg/L per hour (correct answer)
$1.875$ mg/L per hour
$3.75$ mg/L per hour
$-1.875$ mg/L per hour

Explanation: To find the maximum, we need

C'(t) = 0

. Using quotient rule:

C'(t) = \frac{15(t^2 + 4) - 15t(2t)}{(t^2 + 4)^2} = \frac{15t^2 + 60 - 30t^2}{(t^2 + 4)^2} = \frac{60 - 15t^2}{(t^2 + 4)^2}

. Setting

C'(t) = 0

60 - 15t^2 = 0

, so

t^2 = 4

, giving

t = 2

(positive time). At the maximum, by definition,

C'(t) = 0

, so the instantaneous rate of change is

0

mg/L per hour. Choice B gives the maximum concentration value. Choice C doubles that value. Choice D gives the derivative just after the maximum.

Question 7

Six months after the launch of a new software product, a company's analysts report the following data about its monthly revenue function, $R(t)$ , where $t$ is in months: $R(6) = 200,000$ , $R'(6) = 15,000$ , and $R''(6) = -1,200$ .

Based on the data in the passage, which statement best describes the revenue situation at $t=6$ months?

Revenue is increasing, and the rate of revenue growth is also increasing.
Revenue is increasing, but the rate of revenue growth is decreasing. (correct answer)
Revenue has reached its maximum and is about to start decreasing.
Revenue is decreasing, but the rate of decrease is slowing.

Explanation:

R(6) = 200,000

is the revenue at 6 months.

R'(6) = 15,000

represents the instantaneous rate of change of revenue. Since

R'(6)

is positive, revenue is increasing.

R''(6) = -1,200

represents the rate of change of the rate of change. Since

R''(6)

is negative, the rate of growth is decreasing (the revenue curve is concave down). Therefore, at 6 months, revenue is still increasing, but its rate of growth is slowing down.

Question 8

The profit $P(x)$ in dollars from producing $x$ units of a particular product is given by the function $P(x) = -0.01x^2 + 80x - 15000$ . At what production level $x$ is the profit increasing at an instantaneous rate of $40 per unit?

200 units
2,000 units (correct answer)
4,000 units
6,000 units

Explanation: The instantaneous rate of change of profit is given by the derivative of the profit function,

P'(x)

. First, find the derivative:

P'(x) = -0.02x + 80

. The question asks for the production level

x

where this rate is $40 per unit. Set $P'(x) = 40

and solve for

:

-0.02x + 80 = 40

. Subtracting 80 from both sides gives

-0.02x = -40

. Dividing by

-0.02

gives

x = 2000$.

Question 9

A company finds that its monthly revenue, in thousands of dollars, can be modeled by $R(a) = -2a^2 + 80a + 50$ , where $a$ is the amount spent on advertising, in thousands of dollars. The company's current advertising budget is $15 thousand. Which of the following is the best interpretation of $R'(15)$?

The total revenue is $700,000 when advertising is $15,000.
An advertising budget of $20,000 is required to maximize the monthly revenue.
When spending $15,000 on advertising, each additional dollar spent is expected to generate approximately $50 in revenue.
When spending $15,000 on advertising, each additional dollar spent is expected to generate approximately $20 in revenue. (correct answer)

Explanation: When you see a revenue function and need to interpret its derivative at a specific point, you're working with marginal revenue – the rate of change of revenue with respect to the input variable. To find

R'(15)

, first calculate the derivative:

R'(a) = -4a + 80

. Then evaluate at

a = 15

R'(15) = -4(15) + 80 = -60 + 80 = 20

. This means that when the advertising budget is $15,000, the marginal revenue is $20 per additional dollar spent on advertising. Since the revenue function is in thousands of dollars and the advertising variable is also in thousands of dollars, $

R'(15) = 20

represents 20 thousand dollars of additional revenue per thousand dollars of additional advertising. This translates to

20 of additional revenue per additional dollar of advertising. Choice A incorrectly interprets the derivative as the actual revenue value – that would be

R(15)

, not

R'(15)

. Choice B attempts to find where revenue is maximized, which would require setting

R'(a) = 0

and solving for

a

, giving

a = 20

. However, this isn't what

R'(15)

represents. Choice C uses the wrong value – it might come from misreading the derivative calculation or confusing it with another part of the revenue function. Remember that derivatives represent rates of change, not absolute values. When interpreting marginal revenue, always pay attention to the units: the derivative tells you how much the output changes per unit change in the input at that specific point.

Question 10

The number of users for a new mobile app is modeled by the logistic function $P(t) = \frac{20000}{1 + 4e^{-0.5t}}$ , where $t$ is the number of weeks since launch. At what instantaneous rate is the user base growing at the moment when $t = 2 \ln(4)$ weeks?

1,600 users per week
10,000 users per week
5,000 users per week
2,500 users per week (correct answer)

Explanation: When you encounter a logistic growth model and need to find an instantaneous rate of change, you're looking for the derivative of the function at a specific point. The logistic function models situations where growth starts slowly, accelerates, then levels off as it approaches a carrying capacity. To find the growth rate at

t = 2\ln(4)

, you need to calculate

P'(t)

and evaluate it at this point. Using the quotient rule or chain rule, the derivative of

P(t) = \frac{20000}{1 + 4e^{-0.5t}}

is:

P'(t) = \frac{20000 \cdot 4 \cdot 0.5 \cdot e^{-0.5t}}{(1 + 4e^{-0.5t})^2} = \frac{40000e^{-0.5t}}{(1 + 4e^{-0.5t})^2}

t = 2\ln(4)

, we have

e^{-0.5 \cdot 2\ln(4)} = e^{-\ln(4)} = \frac{1}{4}

. Substituting:

P'(2\ln(4)) = \frac{40000 \cdot \frac{1}{4}}{(1 + 4 \cdot \frac{1}{4})^2} = \frac{10000}{(1 + 1)^2} = \frac{10000}{4} = 2500

Answer D (2,500 users per week) is correct. Answer B (10,000) represents the numerator before dividing by the denominator squared. Answer A (1,600) might come from calculation errors with the exponential evaluation. Answer C (5,000) results from incorrectly computing the denominator as 2 instead of 4. Remember: For logistic growth problems, carefully evaluate exponential expressions at the given point, and don't forget to square the denominator when applying the derivative formula.

Question 11

The concentration $C$ of a drug in a patient's bloodstream, in milligrams per liter, $t$ hours after injection is given by $C(t) = \frac{10t}{t^2 + 1}$ . The drug is considered effective as long as its concentration is increasing. For how many hours is the drug's concentration increasing?

For all $t > 0$
For 0.5 hours
For 1 hour (correct answer)
For 2 hours

Explanation: The concentration is increasing when its rate of change,

C'(t)

, is positive. First, find the derivative using the quotient rule:

C'(t) = \frac{(10)(t^2 + 1) - (10t)(2t)}{(t^2 + 1)^2} = \frac{10t^2 + 10 - 20t^2}{(t^2 + 1)^2} = \frac{10 - 10t^2}{(t^2 + 1)^2}

. To find where the concentration is increasing, set

C'(t) > 0

. Since the denominator

(t^2 + 1)^2

is always positive, we only need to solve

10 - 10t^2 > 0

. This simplifies to

10 > 10t^2

, or

1 > t^2

. Since time

t

must be non-negative, this inequality holds for

0 \le t < 1

. Thus, the concentration is increasing for 1 hour.

Question 12

The temperature $T$ of a chemical reaction follows $T(t) = 20 + 45e^{-0.3t}$ degrees Celsius, where $t$ is time in minutes. At what rate is the temperature changing when the temperature has decreased to exactly $35°C$ for the first time?

$-4.5$ degrees Celsius per minute (correct answer)
$-6.75$ degrees Celsius per minute
$-9.0$ degrees Celsius per minute
$-13.5$ degrees Celsius per minute

Explanation: First, find when

T(t) = 35

35 = 20 + 45e^{-0.3t}

, so

15 = 45e^{-0.3t}

, giving

e^{-0.3t} = 1/3

. The derivative is

T'(t) = 45(-0.3)e^{-0.3t} = -13.5e^{-0.3t}

. When

e^{-0.3t} = 1/3

, we have

T'(t) = -13.5(1/3) = -4.5

degrees per minute. Choice B uses

e^{-0.3t} = 1/2

. Choice C uses the coefficient without the exponential factor. Choice D uses the full derivative coefficient.

Question 13

The number of units $N$ a new employee can assemble on day $t$ of their employment is modeled by $N(t) = \frac{150t}{t+3}$ . What is the instantaneous rate of change of the employee's daily assembly output on day $t=2$ ?

60 units per day
30 units per day
18 units per day (correct answer)
-18 units per day

Explanation: When you encounter a question asking for "instantaneous rate of change," you're being asked to find the derivative of the function at a specific point. The derivative measures how quickly the output is changing at any given moment. To find the instantaneous rate of change of

N(t) = \frac{150t}{t+3}

t=2

, you need to use the quotient rule since this is a fraction where both numerator and denominator contain the variable

t

. The quotient rule states:

\frac{d}{dt}\left[\frac{f(t)}{g(t)}\right] = \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2}

Here,

f(t) = 150t

f'(t) = 150

, and

g(t) = t+3

g'(t) = 1

. Applying the quotient rule:

N'(t) = \frac{150(t+3) - 150t(1)}{(t+3)^2} = \frac{150t + 450 - 150t}{(t+3)^2} = \frac{450}{(t+3)^2}

t=2

N'(2) = \frac{450}{(2+3)^2} = \frac{450}{25} = 18

units per day. Choice A (60) likely comes from incorrectly evaluating the original function at

t=2

rather than finding its derivative. Choice B (30) might result from computational errors in the quotient rule. Choice D (-18) has the correct magnitude but wrong sign, suggesting confusion about whether the function is increasing or decreasing. Remember: "instantaneous rate of change" always means derivative. Master the quotient rule—it's essential for rational functions in business calculus.

Question 14

The number of units sold of a product is given by $S(p) = \frac{2000}{p + 5}$ , where $p$ is the price in dollars. If the price is increasing at a rate of $2 per week, at what rate are sales changing when the price is $15?

$-5$ units per week
$-10$ units per week (correct answer)
$-20$ units per week
$-25$ units per week

Explanation: This requires the chain rule since we want

\frac{dS}{dt}

given

\frac{dp}{dt} = 2

. We have

\frac{dS}{dt} = \frac{dS}{dp} \cdot \frac{dp}{dt}

. First,

\frac{dS}{dp} = \frac{d}{dp}\left(\frac{2000}{p + 5}\right) = \frac{-2000}{(p + 5)^2}

. When

p = 15

\frac{dS}{dp} = \frac{-2000}{(15 + 5)^2} = \frac{-2000}{400} = -5

. Therefore,

\frac{dS}{dt} = (-5)(2) = -10

units per week. Choice A forgets to multiply by

\frac{dp}{dt}

. Choice C uses the wrong denominator calculation. Choice D uses

p = 15

directly instead of

p + 5 = 20