Business Calculus Quiz: Chain Rule

15 questions · exam conditions

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Chain RuleQuestion 1 of 15

A company's manufacturing cost, $C$ , in thousands of dollars, is a function of the number of units produced, $q$ , according to $C(q) = \sqrt{q^2 + 567}$ . The number of units produced is a function of time, $t$ , in hours, given by $q(t) = t^3$ . What is the rate of change of the manufacturing cost with respect to time, in thousands of dollars per hour, when $t=3$ ?

$27.00

$0.75

$20.25

$3.38

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Business Calculus Quiz

Business Calculus Quiz: Chain Rule

Practice Chain Rule 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 Chain Rule, 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

$27.00
$0.75
$20.25 (correct answer)
$3.38

Explanation: The problem asks for

dC/dt

t=3

. Using the chain rule,

dC/dt = (dC/dq) \cdot (dq/dt)

. First, find the derivatives:

C'(q) = \frac{1}{2}(q^2+567)^{-1/2}(2q) = \frac{q}{\sqrt{q^2+567}}

and

q'(t) = 3t^2

. Next, evaluate the functions at

t=3

. When

t=3

, the production level is

q(3) = 3^3 = 27

units. The rate of production is

q'(3) = 3(3^2) = 27

units/hour. The marginal cost at

q=27

C'(27) = \frac{27}{\sqrt{27^2 + 567}} = \frac{27}{\sqrt{729+567}} = \frac{27}{\sqrt{1296}} = \frac{27}{36} = 0.75

thousand dollars/unit. Finally,

dC/dt|_{t=3} = C'(27) \cdot q'(3) = 0.75 \cdot 27 = 20.25

thousand dollars/hour.

Question 2

The profit $P(x)$ , in hundreds of dollars, from a concert is modeled by $P(x) = 200\ln(S(x))$ , where $S(x) = 0.5x^2 + 10$ is an audience satisfaction index and $x$ is the number of tickets sold in thousands. Find the marginal profit, $P'(x)$ , when 10,000 tickets are sold.

$100/3$ (correct answer)
$10/3$
$20$
$200\ln(60)$

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

P'(x)

. We use the chain rule:

P'(x) = 200 \cdot \frac{1}{S(x)} \cdot S'(x)

. The derivative of the inner function is

S'(x) = 2(0.5)x = x

. Substituting this in gives

P'(x) = 200 \cdot \frac{1}{0.5x^2 + 10} \cdot x = \frac{200x}{0.5x^2 + 10}

. We need to evaluate this at

x=10

(since

x

is in thousands).

P'(10) = \frac{200(10)}{0.5(10)^2 + 10} = \frac{2000}{0.5(100) + 10} = \frac{2000}{50 + 10} = \frac{2000}{60} = \frac{100}{3}

Question 3

The value of a particular asset is given by the function $V(x) = (x^2+3)\sqrt{4x-3}$ , where $x$ is the number of years since its acquisition. Find the rate at which the asset's value is changing after 3 years.

$20$
$8$
$18$
$26$ (correct answer)

Explanation: We need to find

V'(3)

. The function is a product

V(x) = u(x)v(x)

where

u(x)=x^2+3

and

v(x)=(4x-3)^{1/2}

. We use the product rule,

V'(x) = u'(x)v(x) + u(x)v'(x)

. The derivatives are

u'(x)=2x

and, using the chain rule,

v'(x) = \frac{1}{2}(4x-3)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x-3}}

. So,

V'(x) = (2x)\sqrt{4x-3} + (x^2+3)\frac{2}{\sqrt{4x-3}}

. Now, we evaluate at

x=3

V'(3) = (2\cdot3)\sqrt{4\cdot3-3} + (3^2+3)\frac{2}{\sqrt{4\cdot3-3}} = 6\sqrt{9} + (12)\frac{2}{\sqrt{9}} = 6(3) + \frac{24}{3} = 18 + 8 = 26

Question 4

The population of a city t years after 2020 is modeled by $P(t) = (t^2 + 100)^{3/2}$ . Find the rate at which the population growth is accelerating in 2030.

$30\sqrt{2}$
$15\sqrt{2}$
$300\sqrt{2}$
$45\sqrt{2}$ (correct answer)

Explanation: The rate of acceleration is the second derivative,

P''(t)

. First, find the first derivative,

P'(t)

, using the chain rule:

P'(t) = \frac{3}{2}(t^2+100)^{1/2} \cdot (2t) = 3t(t^2+100)^{1/2}

. Next, find the second derivative,

P''(t)

, using the product rule and the chain rule:

P''(t) = (3)(t^2+100)^{1/2} + (3t) \left[ \frac{1}{2}(t^2+100)^{-1/2} \cdot (2t) \right]

. This simplifies to

P''(t) = 3\sqrt{t^2+100} + \frac{3t^2}{\sqrt{t^2+100}}

. The year 2030 corresponds to

t=10

. Evaluating at

t=10

P''(10) = 3\sqrt{10^2+100} + \frac{3(10^2)}{\sqrt{10^2+100}} = 3\sqrt{200} + \frac{300}{\sqrt{200}} = 3(10\sqrt{2}) + \frac{300}{10\sqrt{2}} = 30\sqrt{2} + \frac{30}{\sqrt{2}}

. Rationalizing the second term gives

30\sqrt{2} + \frac{30\sqrt{2}}{2} = 30\sqrt{2} + 15\sqrt{2} = 45\sqrt{2}

Question 5

A factory's daily production cost is $C(n)$ dollars, where $n$ is the number of units produced. The number of units produced $t$ hours into the workday is given by $n(t)$ . The derivative $C'(n)$ represents the marginal cost per unit, and $n'(t)$ is the production rate in units per hour. What is the practical interpretation of the composite derivative $(C \circ n)'(3)$ ?

The total cost of production during the first three hours of the workday.
The marginal cost of production, in dollars per unit, at the production level reached after three hours.
The rate at which production costs are changing, in dollars per hour, three hours into the workday. (correct answer)
The change in the marginal cost with respect to time, three hours into the workday.

Explanation: By the chain rule,

(C \circ n)'(t) = \frac{d}{dt}C(n(t)) = C'(n(t)) \cdot n'(t)

. Let's analyze the units.

C'(n)

has units of dollars per unit.

n'(t)

has units of units per hour. Their product,

C'(n(t)) \cdot n'(t)

, has units of (dollars/unit)

\times

(units/hour) = dollars/hour. This represents the rate of change of cost with respect to time. Evaluating at

t=3

gives this rate at the specific moment three hours into the workday.

Question 6

The number of subscribers $S$ to a streaming service, in millions, t months after launch is modeled by $S(t) = (3t^2 + 4)^{3/2}$ . What is the rate of growth of subscribers 2 months after launch, in millions per month?

6
64
144
72 (correct answer)

Explanation: The rate of growth is the derivative,

S'(t)

. We use the chain rule:

S'(t) = \frac{3}{2}(3t^2 + 4)^{(3/2 - 1)} \cdot \frac{d}{dt}(3t^2+4)

. This gives

S'(t) = \frac{3}{2}(3t^2 + 4)^{1/2} \cdot (6t)

. Simplifying,

S'(t) = 9t\sqrt{3t^2+4}

. We need to evaluate this at

t=2

S'(2) = 9(2)\sqrt{3(2^2)+4} = 18\sqrt{3(4)+4} = 18\sqrt{12+4} = 18\sqrt{16} = 18(4) = 72

. So the growth rate is 72 million subscribers per month.

Question 7

If $y = e^{x \ln(x^2 + 1)}$ , then $\frac{dy}{dx}$ equals:

$e^{x \ln(x^2 + 1)} \left(\ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}\right)$ (correct answer)
$e^{x \ln(x^2 + 1)} \left(\ln(x^2 + 1) + \frac{x^2}{x^2 + 1}\right)$
$e^{x \ln(x^2 + 1)} \left(\ln(x^2 + 1) + \frac{2x}{x^2 + 1}\right)$
$e^{x \ln(x^2 + 1)} \left(\frac{2x^2}{x^2 + 1}\right)$

Explanation: Using chain rule on

y = e^{u}

where

u = x\ln(x^2 + 1)

\frac{dy}{dx} = e^u \cdot \frac{du}{dx}

. For

\frac{du}{dx}

, use product rule:

\frac{du}{dx} = 1 \cdot \ln(x^2 + 1) + x \cdot \frac{2x}{x^2 + 1} = \ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}

. Choice B has wrong coefficient in the fraction. Choice C has

2x

instead of

2x^2

in numerator. Choice D omits the

\ln(x^2 + 1)

term completely.

Question 8

The weekly revenue $R$ from selling a new software is given by $R(x) = 5000e^{-0.01x}$ dollars, where $x$ is the number of units sold. The number of units sold depends on the advertising budget $a$ , in thousands of dollars, according to $x(a) = a^2 + 9a$ . Find the rate of change of revenue with respect to the advertising budget when the budget is $1,000.

$-550e^{-0.1}$ (correct answer)
$550e^{-0.1}$
$-50e^{-0.1}$
$-550e^{-0.01}$

Explanation: We need to find

dR/da

a=1

(since

a

is in thousands of dollars). By the chain rule,

dR/da = (dR/dx) \cdot (dx/da)

. First, find the derivatives:

dR/dx = 5000e^{-0.01x} \cdot (-0.01) = -50e^{-0.01x}

and

dx/da = 2a+9

. When

a=1

, the number of units sold is

x(1) = 1^2 + 9(1) = 10

. The rate of change of sales is

x'(1) = 2(1)+9=11

. Now evaluate

dR/dx

x=10

dR/dx|_{x=10} = -50e^{-0.01(10)} = -50e^{-0.1}

. Finally, multiply the rates:

dR/da|_{a=1} = (-50e^{-0.1}) \cdot (11) = -550e^{-0.1}

Question 9

A company finds that its daily cost $C$ is a function of the natural logarithm of its production output $x$ , given by the formula $C(x) = (\ln(x^3 + 1))^2$ . Find the marginal cost $C'(x)$ when production output is $x=1$ .

$6\ln(2)$
$3/4$
$3\ln(2)$ (correct answer)
$\ln(2)$

Explanation: To find the marginal cost

C'(x)

, we must differentiate

C(x)

using the chain rule twice. Let

u = \ln(v)

and

v = x^3+1

, so

C=u^2

. The derivative is

C'(x) = \frac{dC}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}

. The component derivatives are

\frac{dC}{du} = 2u = 2\ln(x^3+1)

\frac{du}{dv} = \frac{1}{v} = \frac{1}{x^3+1}

, and

\frac{dv}{dx} = 3x^2

. Combining these gives

C'(x) = 2\ln(x^3+1) \cdot \frac{1}{x^3+1} \cdot 3x^2 = \frac{6x^2 \ln(x^3+1)}{x^3+1}

. Evaluating at

x=1

C'(1) = \frac{6(1)^2 \ln(1^3+1)}{1^3+1} = \frac{6\ln(2)}{2} = 3\ln(2)

Question 10

Let $h(x) = \sin(\cos(x^3))$ . Which expression represents $h'(x)$ ?

$\cos(\cos(x^3)) \cdot \sin(x^3) \cdot 3x^2$
$\cos(\cos(x^3)) \cdot (-\sin(x^3)) \cdot 3x^2$ (correct answer)
$\cos(\sin(x^3)) \cdot (-\sin(x^3)) \cdot 3x^2$
$\cos(\cos(x^3)) \cdot (-\sin(x^3)) \cdot x^2$

Explanation: Using chain rule with three layers:

h'(x) = \cos(\cos(x^3)) \cdot \frac{d}{dx}[\cos(x^3)] \cdot \frac{d}{dx}[x^3] = \cos(\cos(x^3)) \cdot (-\sin(x^3)) \cdot 3x^2

. Choice A forgets the negative sign when differentiating cosine. Choice C incorrectly switches sine and cosine in the outer function. Choice D has the wrong coefficient for

x^2

Question 11

Let $h(x) = f(e^{x^2})$ , where $f$ is a differentiable function. If it is known that $f'(e^4) = 10$ , what is the value of $h'(2)$ ?

$20e^4$
$40e^4$ (correct answer)
$10e^2$
$80e^2$

Explanation: This question tests the chain rule with composite functions, which is essential when dealing with nested business functions like compound interest or layered cost structures. To find

h'(2)

, you need to apply the chain rule to

h(x) = f(e^{x^2})

. The chain rule states that if you have a composition

f(g(x))

, then the derivative is

f'(g(x)) \cdot g'(x)

. Here, the outer function is

f

and the inner function is

g(x) = e^{x^2}

. So

h'(x) = f'(e^{x^2}) \cdot \frac{d}{dx}[e^{x^2}]

. First, find the derivative of the inner function:

\frac{d}{dx}[e^{x^2}] = e^{x^2} \cdot 2x

(using the chain rule again). Therefore:

h'(x) = f'(e^{x^2}) \cdot e^{x^2} \cdot 2x

x = 2

h'(2) = f'(e^{2^2}) \cdot e^{2^2} \cdot 2(2) = f'(e^4) \cdot e^4 \cdot 4

Since

f'(e^4) = 10

h'(2) = 10 \cdot e^4 \cdot 4 = 40e^4

Choice A (

20e^4

) results from forgetting to apply the chain rule to

e^{x^2}

, missing the

2x

factor. Choice C (

10e^2

) incorrectly uses

e^2

instead of

e^4

and omits the

2x

factor. Choice D (

80e^2

) has the wrong exponential term but accidentally gets close by doubling somewhere. Strategy tip: When seeing nested functions, identify each layer clearly and apply the chain rule systematically. Always multiply by the derivative of each inner function as you work outward.

Question 12

A particle moves along a path where its position is given by $s(t) = \tan^{-1}(t^3 - 2t)$ . What is the particle's velocity when $t = 1$ ?

$\frac{1}{2}$ (correct answer)
$\frac{1}{4}$
$\frac{3}{2}$
$1$

Explanation: Using chain rule:

s'(t) = \frac{1}{1 + (t^3 - 2t)^2} \cdot (3t^2 - 2)

. At

t = 1

: the inner function

t^3 - 2t = 1 - 2 = -1

, and its derivative is

3(1)^2 - 2 = 1

. So

s'(1) = \frac{1}{1 + (-1)^2} \cdot 1 = \frac{1}{2}

. Choice B incorrectly calculates

1 + (-1)^2 = 4

. Choice C uses the wrong derivative of the inner function. Choice D forgets to evaluate the denominator correctly.

Question 13

Let $f(x) = \ln(\sin(e^x))$ . Which of the following represents $f'(x)$ ?

$\cos(e^x) \cdot e^x$
$\frac{\cos(e^x)}{\sin(e^x)}$
$\frac{e^x}{\sin(e^x)}$
$\frac{\cos(e^x) \cdot e^x}{\sin(e^x)}$ (correct answer)

Explanation: When you encounter a composite function like

f(x) = \ln(\sin(e^x))

, you need to apply the chain rule systematically. This function has three layers: the natural logarithm on the outside, sine in the middle, and an exponential function on the inside. To find

f'(x)

, work from the outside in. Start with the derivative of

\ln(u)

, which is

\frac{1}{u}

. Here,

u = \sin(e^x)

, so the first step gives us

\frac{1}{\sin(e^x)}

. Next, multiply by the derivative of the middle layer:

\sin(e^x)

. The derivative of

\sin(v)

\cos(v)

, where

v = e^x

, giving us

\cos(e^x)

. Finally, multiply by the derivative of the innermost function

e^x

, which is

e^x

. Combining all parts:

f'(x) = \frac{1}{\sin(e^x)} \cdot \cos(e^x) \cdot e^x = \frac{\cos(e^x) \cdot e^x}{\sin(e^x)}

Choice A (

\cos(e^x) \cdot e^x

) represents only the derivative of

\sin(e^x)

but ignores the outer logarithm layer. Choice B (

\frac{\cos(e^x)}{\sin(e^x)}

) correctly handles the logarithm and sine layers but forgets to differentiate

e^x

. Choice C (

\frac{e^x}{\sin(e^x)}

) incorrectly treats the sine function as if it weren't there during differentiation. Remember: with nested functions, the chain rule requires you to differentiate every layer and multiply all those derivatives together. Missing any layer leads to an incomplete answer.

Question 14

A company's market share $M$ (as a percentage) $t$ years from now is given by $M(t) = \frac{100t}{(t^2+9)^{3/2}}$ . At what rate will the company's market share be changing in 4 years?

$2.336\%$ per year
$-0.736\%$ per year (correct answer)
$0.608\%$ per year
$0.736\%$ per year

Explanation: We need to find

M'(4)

. Using the quotient rule,

M'(t) = \frac{100 \cdot (t^2+9)^{3/2} - 100t \cdot \frac{d}{dt}(t^2+9)^{3/2}}{((t^2+9)^{3/2})^2}

. The derivative of the denominator requires the chain rule:

\frac{d}{dt}(t^2+9)^{3/2} = \frac{3}{2}(t^2+9)^{1/2}(2t) = 3t(t^2+9)^{1/2}

. Substituting this in:

M'(t) = \frac{100(t^2+9)^{3/2} - 100t(3t(t^2+9)^{1/2})}{(t^2+9)^3}

. Factor out

100(t^2+9)^{1/2}

from the numerator:

M'(t) = \frac{100(t^2+9)^{1/2}[(t^2+9)-3t^2]}{(t^2+9)^3} = \frac{100(9-2t^2)}{(t^2+9)^{5/2}}

. Now evaluate at

t=4

M'(4) = \frac{100(9-2(4^2))}{(4^2+9)^{5/2}} = \frac{100(9-32)}{(16+9)^{5/2}} = \frac{100(-23)}{25^{5/2}} = \frac{-2300}{(\sqrt{25})^5} = \frac{-2300}{5^5} = \frac{-2300}{3125} = -0.736

Question 15

A company's cost function is $C(q) = 50\sqrt{q^2 + 4q + 13}$ dollars for producing $q$ units. What is the marginal cost when $q = 3$ units?

$\frac{50(q + 2)}{\sqrt{q^2 + 4q + 13}}$ evaluated at $q = 3$
$\frac{50(2q + 4)}{\sqrt{q^2 + 4q + 13}}$ evaluated at $q = 3$
$\frac{25(2q + 4)}{\sqrt{q^2 + 4q + 13}}$ evaluated at $q = 3$ (correct answer)
$\frac{100(q + 2)}{\sqrt{q^2 + 4q + 13}}$ evaluated at $q = 3$

Explanation: Using chain rule:

C'(q) = 50 \cdot \frac{1}{2\sqrt{q^2 + 4q + 13}} \cdot (2q + 4) = \frac{25(2q + 4)}{\sqrt{q^2 + 4q + 13}}

. At

q = 3

C'(3) = \frac{25(10)}{\sqrt{34}} = \frac{250}{\sqrt{34}}

. Choice A incorrectly factors the derivative of the inner function. Choice B forgets to apply the

\frac{1}{2}

factor from differentiating the square root. Choice D doubles the coefficient incorrectly.