Business Calculus Quiz: Power Product And Quotient Rules

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Power Product And Quotient RulesQuestion 1 of 14

Find the derivative, $\frac{dy}{dx}$ , for the function $y = \frac{2x^2 + x - 1}{x^2 + 3}$ .

\frac{-x^2 + 14x + 3}{(x^2+3)^2}

\frac{x^2 - 14x - 3}{(x^2+3)^2}

\frac{4x+1}{2x}

\frac{-x^2 + 14x + 3}{x^2+3}

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Business Calculus Quiz

Business Calculus Quiz: Power Product And Quotient Rules

Practice Power Product And Quotient Rules 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 Power Product And Quotient Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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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

Find the derivative, $\frac{dy}{dx}$ , for the function $y = \frac{2x^2 + x - 1}{x^2 + 3}$ .

$\frac{-x^2 + 14x + 3}{(x^2+3)^2}$ (correct answer)
$\frac{x^2 - 14x - 3}{(x^2+3)^2}$
$\frac{4x+1}{2x}$
$\frac{-x^2 + 14x + 3}{x^2+3}$

Explanation: Use the quotient rule,

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

. Let

u(x) = 2x^2 + x - 1

, so

u'(x) = 4x + 1

. Let

v(x) = x^2 + 3

, so

v'(x) = 2x

. Plugging into the formula:

\frac{dy}{dx} = \frac{(4x+1)(x^2+3) - (2x^2+x-1)(2x)}{(x^2+3)^2}

Expand the numerator: Numerator

= (4x^3 + 12x + x^2 + 3) - (4x^3 + 2x^2 - 2x)

Numerator

= 4x^3 + x^2 + 12x + 3 - 4x^3 - 2x^2 + 2x

Numerator

= -x^2 + 14x + 3

So, the derivative is

\frac{-x^2 + 14x + 3}{(x^2+3)^2}

. (B) This results from reversing the terms in the numerator of the quotient rule, calculating

uv' - u'v

instead of

u'v - uv'

. (C) This is the result of the common mistake of taking the derivative of the numerator divided by the derivative of the denominator,

u'/v'

. (D) This results from correctly calculating the numerator but forgetting to square the denominator in the quotient rule.

Question 2

A manufacturing company models its cost function as $C(q) = \frac{q^3}{(q^2 + 4)^2}$ where $q$ is the quantity produced. The marginal cost function $C'(q)$ simplifies to:

$\frac{q^2(12 - q^2)}{(q^2 + 4)^3}$ (correct answer)
$\frac{q^2(8 - q^2)}{(q^2 + 4)^3}$
$\frac{q^2(16 - q^2)}{(q^2 + 4)^3}$
$\frac{q^2(10 - q^2)}{(q^2 + 4)^3}$

Explanation: Using the quotient rule with

u = q^3

and

v = (q^2 + 4)^2

u' = 3q^2

and

v' = 2(q^2 + 4) \cdot 2q = 4q(q^2 + 4)

. So

C'(q) = \frac{3q^2(q^2 + 4)^2 - q^3 \cdot 4q(q^2 + 4)}{(q^2 + 4)^4} = \frac{q^2(q^2 + 4)[3(q^2 + 4) - 4q^2]}{(q^2 + 4)^4} = \frac{q^2(3q^2 + 12 - 4q^2)}{(q^2 + 4)^3} = \frac{q^2(12 - q^2)}{(q^2 + 4)^3}

. Choice B results from miscalculating

3(4) = 8

instead of 12. Choice C comes from incorrectly using

4 \cdot 4 = 16

. Choice D results from arithmetic errors in the simplification.

Question 3

The price-demand function for a product is $p(x) = \frac{4000}{0.5x + 8}$ , where $x$ is the number of units sold. The revenue is given by $R(x) = x \cdot p(x)$ . Which of the following is the marginal revenue function, $R'(x)$ ?

$\frac{-2000}{(0.5x + 8)^2}$
$\frac{32000}{(0.5x + 8)^2}$ (correct answer)
$\frac{4000x + 32000}{(0.5x + 8)^2}$
$\frac{32000}{0.5x+8}$

Explanation: First, define the revenue function:

R(x) = x \cdot p(x) = x \cdot \frac{4000}{0.5x + 8} = \frac{4000x}{0.5x + 8}

. To find the marginal revenue,

R'(x)

, we use the quotient rule with

u(x) = 4000x

and

v(x) = 0.5x + 8

. The derivatives are

u'(x) = 4000

and

v'(x) = 0.5

R'(x) = \frac{u'v - uv'}{v^2} = \frac{(4000)(0.5x+8) - (4000x)(0.5)}{(0.5x+8)^2}

Now, simplify the numerator:

R'(x) = \frac{2000x + 32000 - 2000x}{(0.5x+8)^2}

R'(x) = \frac{32000}{(0.5x+8)^2}

. (A) This is the derivative of the price function,

p'(x)

, not the revenue function. This represents the marginal price. (C) This results from incorrectly adding the terms in the numerator of the quotient rule (

u'v + uv'

) instead of subtracting them. (D) This results from correctly finding the numerator but forgetting to square the denominator in the quotient rule formula.

Question 4

What is the derivative of the function $f(x) = \frac{x^3 - 8}{x^2 + 2x + 4}$ ?

$\frac{x^4 + 4x^3 + 12x^2 + 16x + 16}{(x^2 + 2x + 4)^2}$
$x-2$
$\frac{3x^2}{2x+2}$
$1$ (correct answer)

Explanation: This problem can be solved by first simplifying the function

f(x)

. The numerator,

x^3 - 8

, is a difference of cubes, which factors as

x^3 - a^3 = (x-a)(x^2 + ax + a^2)

. For

x^3 - 8 = x^3 - 2^3

, the factorization is

(x-2)(x^2 + 2x + 4)

. So, the function is

f(x) = \frac{(x-2)(x^2 + 2x + 4)}{x^2 + 2x + 4}

. The term

(x^2 + 2x + 4)

cancels out (it is never zero for real

x

), leaving

f(x) = x - 2

. The derivative of this simplified function is

f'(x) = 1

. (A) This is the correct, unsimplified result of applying the quotient rule to the original function. The numerator and denominator are identical, so the expression simplifies to 1. A student who correctly applies the quotient rule but fails to simplify would arrive at this complex expression. (B) This is the simplified function

f(x)

, not its derivative

f'(x)

. (C) This is the result of the common mistake of taking the derivative of the numerator divided by the derivative of the denominator.

Question 5

Given the function $h(t) = \frac{4\sqrt{t}}{t^{1/3} + 3}$ , determine the value of $h'(1)$ .

$6$
$\frac{7}{12}$
$\frac{5}{12}$ (correct answer)
$\frac{11}{12}$

Explanation: To find

h'(t)

, use the quotient rule. First, write the function with exponents:

h(t) = \frac{4t^{1/2}}{t^{1/3} + 3}

. Let

u(t) = 4t^{1/2}

and

v(t) = t^{1/3} + 3

. Then

u'(t) = 4(\frac{1}{2})t^{-1/2} = 2t^{-1/2}

. And

v'(t) = \frac{1}{3}t^{-2/3}

. Apply the quotient rule:

h'(t) = \frac{u'v - uv'}{v^2} = \frac{(2t^{-1/2})(t^{1/3}+3) - (4t^{1/2})(\frac{1}{3}t^{-2/3})}{(t^{1/3}+3)^2}

. Evaluate at

t=1

h'(1) = \frac{(2(1)^{-1/2})(1^{1/3}+3) - (4(1)^{1/2})(\frac{1}{3}(1)^{-2/3})}{(1^{1/3}+3)^2}

h'(1) = \frac{(2)(1+3) - (4)(\frac{1}{3})}{(1+3)^2}

h'(1) = \frac{8 - \frac{4}{3}}{16} = \frac{\frac{24}{3} - \frac{4}{3}}{16} = \frac{\frac{20}{3}}{16} = \frac{20}{48} = \frac{5}{12}

. (A) This result comes from the mistake of calculating

u'(1)/v'(1)

, which would be

2 / (1/3) = 6

. (B) This results from using a plus sign instead of a minus sign in the numerator of the quotient rule, giving

\frac{8 + 4/3}{16} = \frac{28/3}{16} = \frac{28}{48} = \frac{7}{12}

. (D) This results from an error in the power rule, treating the derivative of

\sqrt{t}

as 1 instead of

\frac{1}{2}t^{-1/2}

. This makes

u'=4

, leading to

\frac{4(4) - 4(1/3)}{16} = \frac{16 - 4/3}{16} = \frac{44/3}{16} = \frac{44}{48} = \frac{11}{12}

Question 6

What is the y-intercept of the line tangent to the graph of $f(x) = \frac{x-3}{x+1}$ at the point where $x=1$ ?

$-1$
$1$
$-2$ (correct answer)
$0$

Explanation: This is a multi-step problem. First, find the point of tangency. At

x=1

, the y-coordinate is

f(1) = \frac{1-3}{1+1} = \frac{-2}{2} = -1

. The point is

(1, -1)

. Next, find the slope of the tangent line by finding the derivative

f'(x)

using the quotient rule. Let

u=x-3

and

v=x+1

. Then

u'=1

and

v'=1

f'(x) = \frac{u'v - uv'}{v^2} = \frac{(1)(x+1) - (x-3)(1)}{(x+1)^2} = \frac{x+1 - x+3}{(x+1)^2} = \frac{4}{(x+1)^2}

. Evaluate the slope at

x=1

m = f'(1) = \frac{4}{(1+1)^2} = \frac{4}{4} = 1

. Now, use the point-slope form of a line,

y - y_1 = m(x - x_1)

, with point

(1, -1)

and slope

m=1

y - (-1) = 1(x - 1)

y + 1 = x - 1

y = x - 2

. The y-intercept is the value of

y

when

x=0

. In this case,

y = 0 - 2 = -2

. (A) This is the y-coordinate of the point of tangency, not the y-intercept of the tangent line. (B) This is the slope of the tangent line, not the y-intercept. (D) This would be the y-intercept if there were a sign error in the quotient rule calculation, leading to a slope of

m=-1

and an equation of

y=-x

Question 7

Two differentiable functions, $f(x)$ and $g(x)$ , have the following properties: $f(3)=4$ , $f'(3)=-2$ , $g(3)=8$ , and $g'(3)=6$ . If $h(x) = \frac{f(x)}{g(x)}$ , what is the value of $h'(3)$ ?

$\frac{1}{8}$
$-5$
$\frac{5}{8}$
$-\frac{5}{8}$ (correct answer)

Explanation: To find

h'(x)

, we apply the quotient rule:

h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

. We need to evaluate this derivative at

x=3

using the given values:

h'(3) = \frac{f'(3)g(3) - f(3)g'(3)}{[g(3)]^2}

Substitute the given values into the formula:

h'(3) = \frac{(-2)(8) - (4)(6)}{(8)^2}

h'(3) = \frac{-16 - 24}{64}

h'(3) = \frac{-40}{64}

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, 8:

h'(3) = -\frac{5}{8}

. (A) This is the result of incorrectly applying the product rule formula to the numerator,

f'(3)g(3) + f(3)g'(3) = (-2)(8) + (4)(6) = -16+24=8

, and then dividing by the denominator,

8/64 = 1/8

. (B) This results from forgetting to square the denominator in the quotient rule:

\frac{-40}{8} = -5

. (C) This is the result of reversing the terms in the numerator of the quotient rule:

\frac{f(x)g'(x) - f'(x)g(x)}{[g(x)]^2} = \frac{(4)(6) - (-2)(8)}{64} = \frac{24+16}{64} = \frac{40}{64} = \frac{5}{8}

Question 8

Consider the function $f(x) = \frac{x^2 \sqrt{x+3}}{x-1}$ . What is the value of the derivative $f'(-2)$ ?

$\frac{2}{9}$ (correct answer)
$-\frac{2}{9}$
$\frac{10}{9}$
$\frac{13}{18}$

Explanation: This requires using the quotient rule, and the derivative of the numerator will require the product rule. Let the numerator be

u(x) = x^2 (x+3)^{1/2}

and the denominator be

v(x) = x-1

. First, find

u'(x)

using the product rule:

u'(x) = (2x)(x+3)^{1/2} + (x^2)(\frac{1}{2}(x+3)^{-1/2})

. Now, apply the quotient rule:

f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}

v'(x) = 1

. It is easiest to evaluate the pieces at

x=-2

before substituting into the full formula:

u(-2) = (-2)^2 \sqrt{-2+3} = 4\sqrt{1} = 4

v(-2) = -2-1 = -3

u'(-2) = (2(-2))\sqrt{-2+3} + ((-2)^2)(\frac{1}{2}(-2+3)^{-1/2}) = (-4)(1) + (4)(\frac{1}{2}(1)^{-1/2}) = -4 + 2 = -2

v'(-2) = 1

. Now substitute these values into the quotient rule formula:

f'(-2) = \frac{u'(-2)v(-2) - u(-2)v'(-2)}{[v(-2)]^2} = \frac{(-2)(-3) - (4)(1)}{(-3)^2} = \frac{6 - 4}{9} = \frac{2}{9}

. (B) This results from reversing the terms in the numerator of the quotient rule:

\frac{u(-2)v'(-2) - u'(-2)v(-2)}{[v(-2)]^2} = \frac{4 - 6}{9} = -\frac{2}{9}

. (C) This results from using a plus sign instead of a minus sign in the numerator of the quotient rule:

\frac{6 + 4}{9} = \frac{10}{9}

. (D) This results from an error in the product rule for

u'(x)

, such as adding the derivatives instead of applying the rule correctly.

Question 9

Let $h(x) = (x^3 + 2x) g(x)$ . Given that $g(2) = -1$ and $g'(2) = 3$ , find the value of $h'(2)$ .

$22$ (correct answer)
$42$
$30$
$-50$

Explanation: To find

h'(x)

, we use the product rule. Let

f(x) = x^3 + 2x

. Then

h(x) = f(x)g(x)

. The product rule states that

h'(x) = f'(x)g(x) + f(x)g'(x)

. First, we need to find

f(x)

and its derivative

f'(x) = 3x^2 + 2

. Next, we evaluate

f(x)

and

f'(x)

x=2

f(2) = (2)^3 + 2(2) = 8 + 4 = 12

f'(2) = 3(2)^2 + 2 = 3(4) + 2 = 12 + 2 = 14

. Now, substitute all the given and calculated values into the product rule formula at

x=2

h'(2) = f'(2)g(2) + f(2)g'(2)

h'(2) = (14)(-1) + (12)(3)

h'(2) = -14 + 36 = 22

. (B) This is the result of the common misconception that the derivative of a product is the product of the derivatives,

f'(2)g'(2) = (14)(3) = 42

. (C) This result comes from an incorrect formula, such as

f(2)g(2) + f'(2)g'(2) = (12)(-1) + (14)(3) = -12 + 42 = 30

. (D) This could result from confusing the product rule with the quotient rule, using a minus sign:

f'(2)g(2) - f(2)g'(2) = (14)(-1) - (12)(3) = -14 - 36 = -50

Question 10

The demand for a product is given by $p(q) = (q^2 + 100)(1 + \frac{200}{q})$ for $q > 0$ . The revenue is $R(q) = q \cdot p(q)$ . Find the marginal revenue at a production level of $q=10$ .

$20$
$12,400$
$42,000$
$4,400$ (correct answer)

Explanation: First, find the revenue function

R(q) = q \cdot p(q)

R(q) = q(q^2 + 100)(1 + \frac{200}{q})

. It's easiest to simplify this expression before differentiating:

R(q) = (q^3 + 100q)(1 + \frac{200}{q})

R(q) = (q^3 + 100q)(1 + 200q^{-1})

R(q) = q^3(1+200q^{-1}) + 100q(1+200q^{-1})

R(q) = q^3 + 200q^2 + 100q + 20000

. Now, find the marginal revenue,

R'(q)

, by differentiating this polynomial:

R'(q) = 3q^2 + 400q + 100

. Finally, evaluate at

q=10

R'(10) = 3(10)^2 + 400(10) + 100

R'(10) = 3(100) + 4000 + 100 = 300 + 4000 + 100 = 4400

. (A) This is the value of the marginal price,

p'(10)

, not the marginal revenue. (B) This result is obtained from a sign error when differentiating

200/q

, treating its derivative as

200q^{-2}

instead of

-200q^{-2}

when applying the product rule to the unsimplified form. (C) This is the total revenue

R(10)

, not the marginal revenue

R'(10)

Question 11

Let the function $f(x)$ be defined as the product $f(x) = (x^3 - 4x)\sqrt{x}$ . What is the value of $f'(4)$ ?

$88$
$11$
$100$ (correct answer)
$76$

Explanation: To find the derivative,

f'(x)

, one can use the product rule. First, rewrite the function as

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

. Let

u(x) = x^3 - 4x

and

v(x) = x^{1/2}

. Then

u'(x) = 3x^2 - 4

and

v'(x) = \frac{1}{2}x^{-1/2}

. By the product rule,

f'(x) = u'(x)v(x) + u(x)v'(x) = (3x^2 - 4)x^{1/2} + (x^3 - 4x)(\frac{1}{2}x^{-1/2})

. Now, evaluate

f'(x)

x=4

f'(4) = (3(4^2) - 4)\sqrt{4} + (4^3 - 4(4))(\frac{1}{2}(4)^{-1/2})

f'(4) = (3(16) - 4)(2) + (64 - 16)(\frac{1}{2} \cdot \frac{1}{2})

f'(4) = (48 - 4)(2) + (48)(\frac{1}{4})

f'(4) = 44(2) + 12 = 88 + 12 = 100

. Alternatively, one could first expand

f(x) = x^{7/2} - 4x^{3/2}

. Then

f'(x) = \frac{7}{2}x^{5/2} - 4(\frac{3}{2})x^{1/2} = \frac{7}{2}x^{5/2} - 6x^{1/2}

. At

x=4

f'(4) = \frac{7}{2}(4)^{5/2} - 6(4)^{1/2} = \frac{7}{2}(32) - 6(2) = 7(16) - 12 = 112 - 12 = 100

. (A) This is the value of the first term,

u'(4)v(4)

, in the product rule calculation. (B) This result is obtained by incorrectly applying the product rule as

f' = u'v'

, which would be

(3x^2-4)(\frac{1}{2}x^{-1/2})

. At

x=4

, this gives

(44)(1/4) = 11

. (D) This results from a calculation error, such as subtracting the two terms in the final step instead of adding them (

88 - 12 = 76

Question 12

For which value(s) of $x$ does the graph of the function $f(x) = \frac{x^2 + 12}{2x + 1}$ have a horizontal tangent line?

$x=3$ only
$x=-4$ only
$x=3$ and $x=-4$ (correct answer)
$x=-3$ and $x=4$

Explanation: A horizontal tangent line occurs where the derivative of the function is equal to zero. First, we must find the derivative,

f'(x)

, using the quotient rule. Let

u(x) = x^2 + 12

and

v(x) = 2x + 1

. Then

u'(x) = 2x

and

v'(x) = 2

f'(x) = \frac{u'v - uv'}{v^2} = \frac{(2x)(2x+1) - (x^2+12)(2)}{(2x+1)^2}

. Set the derivative equal to zero. A fraction is zero only when its numerator is zero.

(2x)(2x+1) - 2(x^2+12) = 0

4x^2 + 2x - 2x^2 - 24 = 0

2x^2 + 2x - 24 = 0

Divide the entire equation by 2 to simplify:

x^2 + x - 12 = 0

Factor the quadratic equation:

(x+4)(x-3) = 0

The solutions are

x = -4

and

x = 3

. At these two x-values, the function has a horizontal tangent line. (A) This is an incomplete answer; it identifies only one of the correct values. (B) This is an incomplete answer; it identifies only one of the correct values. (D) This results from a sign error when factoring the quadratic equation

x^2 + x - 12 = 0

(x-4)(x+3)=0

Question 13

The function $F(x) = (x^2 + 1)^3 \cdot (2x - 3)^{-2}$ represents a financial model. Which of the following expressions is equivalent to $\frac{F'(x)}{F(x)}$ ?

$\frac{6x}{x^2 + 1} - \frac{4}{2x - 3}$ (correct answer)
$\frac{6x}{x^2 + 1} + \frac{4}{2x - 3}$
$\frac{3x}{x^2 + 1} - \frac{2}{2x - 3}$
$\frac{12x}{x^2 + 1} - \frac{8}{2x - 3}$

Explanation: This question asks for the logarithmic derivative

\frac{F'(x)}{F(x)}

. Taking the natural log of both sides:

\ln F(x) = 3\ln(x^2 + 1) - 2\ln(2x - 3)

. Differentiating both sides:

\frac{F'(x)}{F(x)} = 3 \cdot \frac{2x}{x^2 + 1} - 2 \cdot \frac{2}{2x - 3} = \frac{6x}{x^2 + 1} - \frac{4}{2x - 3}

. Choice B has the wrong sign on the second term. Choice C has incorrect coefficients (3 instead of 6, and 2 instead of 4). Choice D has incorrect coefficients (12 instead of 6, and 8 instead of 4). The logarithmic derivative method is much more efficient than using the product rule directly for this type of problem.

Question 14

A company's revenue function is $R(x) = x^3 \cdot \sqrt{2x + 1}$ where $x$ represents the number of units sold (in thousands). What is $R'(x)$ ?

$\frac{x^2(7x + 3)}{\sqrt{2x + 1}}$ (correct answer)
$\frac{x^2(8x + 3)}{\sqrt{2x + 1}}$
$\frac{x^2(6x + 3)}{\sqrt{2x + 1}}$
$\frac{x^2(5x + 3)}{\sqrt{2x + 1}}$

Explanation: Using the product rule with

u = x^3

and

v = (2x + 1)^{1/2}

u' = 3x^2

and

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

. So

R'(x) = 3x^2(2x + 1)^{1/2} + x^3(2x + 1)^{-1/2} = \frac{3x^2(2x + 1) + x^3}{\sqrt{2x + 1}} = \frac{6x^3 + 3x^2 + x^3}{\sqrt{2x + 1}} = \frac{x^2(7x + 3)}{\sqrt{2x + 1}}

. Choice B results from incorrectly computing

6x^3 + x^3 = 8x^3

. Choice C comes from forgetting the coefficient 2 in the chain rule. Choice D results from computational errors in combining terms.