AP Calculus BC Quiz: The Quotient Rule

What this quiz covers

This quiz focuses on The Quotient Rule, 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.

Question 1

1. $\dfrac{(2t+3)(t-1)-(t^2+3t)}{(t-1)^2}$ (correct answer)
2. $\dfrac{(t-1)(t^2+3t)-(2t+3)}{(t-1)^2}$
3. $\dfrac{(2t+3)(t-1)+(t^2+3t)}{(t-1)^2}$
4. $\dfrac{(2t+3)(t-1)-(t^2+3t)}{t-1}$
5. $\dfrac{(t^2+3t)-(2t+3)(t-1)}{(t-1)^2}$

Explanation: This problem requires the quotient rule to find the derivative of R(t) = (t² + 3t)/(t - 1). The quotient rule states that if f(x) = g(x)/h(x), then f'(x) = [g'(x)·h(x) - g(x)·h'(x)]/[h(x)]². Here, g(t) = t² + 3t with g'(t) = 2t + 3, and h(t) = t - 1 with h'(t) = 1. Applying the formula: R'(t) = [(2t + 3)(t - 1) - (t² + 3t)(1)]/(t - 1)². Choice B incorrectly reverses the order in the numerator, which would give a different result. Remember: in the quotient rule, it's always "derivative of top times bottom minus top times derivative of bottom, all over bottom squared."

Question 2

1. $\dfrac{\left(\dfrac{1}{x}\right)(x+2)-\ln x}{(x+2)^2}$ (correct answer)
2. $\dfrac{\left(\dfrac{1}{x}\right)(x+2)+\ln x}{(x+2)^2}$
3. $\dfrac{\ln x-\left(\dfrac{1}{x}\right)(x+2)}{(x+2)^2}$
4. $\dfrac{\left(\dfrac{1}{x}\right)(x+2)-\ln x}{x+2}$
5. $\dfrac{\left(\dfrac{1}{x}\right)(x+2)-\ln x}{(x+2)}$

Explanation: To find g'(x) for g(x) = $\dfrac{\ln x}{x + 2}$, we use the quotient rule with numerator $\ln x$ (derivative: $\dfrac{1}{x}$) and denominator $x + 2$ (derivative: 1). Applying the quotient rule formula: g'(x) = $\dfrac{ \left( \dfrac{1}{x} \right) (x + 2) - (\ln x)(1) }{ (x + 2)^2 }$. This simplifies to $\dfrac{ \left( \dfrac{1}{x} \right) (x + 2) - \ln x }{ (x + 2)^2 }$. Choice B incorrectly adds instead of subtracts $\ln x$ in the numerator, violating the quotient rule's subtraction requirement. When using the quotient rule, always maintain the correct order and operation: it's subtraction, not addition, between the two products in the numerator.

Question 3

1. $\dfrac{\sec^2 x\cdot x-\tan x}{x^2}$ (correct answer)
2. $\dfrac{\sec^2 x\cdot x+\tan x}{x^2}$
3. $\dfrac{\tan x-\sec^2 x\cdot x}{x^2}$
4. $\dfrac{\sec^2 x\cdot x-\tan x}{x}$
5. $\dfrac{\sec^2 x\cdot x-\tan x}{(\sqrt{x})^2}$

Explanation: Finding w'(x) for w(x) = tan x/x requires the quotient rule with numerator tan x (derivative: sec² x) and denominator x (derivative: 1). Applying the quotient rule formula: w'(x) = [(sec² x)·x - (tan x)·1]/x² = (x sec² x - tan x)/x². Choice C incorrectly reverses the order of subtraction in the numerator, which would change the sign and give an incorrect result. The quotient rule's mnemonic "low d-high minus high d-low, all over low squared" helps remember the correct order, where maintaining the subtraction order is essential for the correct derivative.

Question 4

1. $\dfrac{\left(\dfrac{1}{2\sqrt{x}}\right)(x-3)-(\sqrt{x}+1)(1)}{(x-3)^2}$ (correct answer)
2. $\dfrac{(\sqrt{x}+1)(1)-\left(\dfrac{1}{2\sqrt{x}}\right)(x-3)}{(x-3)^2}$
3. $\dfrac{\left(\dfrac{1}{2\sqrt{x}}\right)(x-3)+(\sqrt{x}+1)(1)}{(x-3)^2}$
4. $\dfrac{\left(\dfrac{1}{2\sqrt{x}}\right)(x-3)-(\sqrt{x}+1)(1)}{x-3}$
5. $\dfrac{1}{2\sqrt{x}(x-3)^2}$

Explanation: This problem requires the quotient rule to find the derivative of the rate function F(x) = (√x + 1)/(x - 3). The quotient rule states that if f(x) = num(x)/den(x), then f'(x) = [num'(x) den(x) - num(x) den'(x)] / [den(x)]². Here, num(x) = √x + 1 with num'(x) = 1/(2√x), and den(x) = x - 3 with den'(x) = 1. Applying the rule gives [(1/(2√x))(x - 3) - (√x + 1)·1] / (x - 3)², matching the correct choice. A tempting distractor like choice D fails by omitting the square on the denominator, which is a key part of the quotient rule. Remember, a transferable strategy for the quotient rule is to always write 'low d-high minus high d-low over low squared' to keep the order and signs correct.

Question 5

1. $\dfrac{5(x^2+1)+(5x-2)(2x)}{(x^2+1)^2}$
2. $\dfrac{5(x^2+1)-(5x-2)(2x)}{(x^2+1)^2}$ (correct answer)
3. $\dfrac{(5x-2)(2x)-5(x^2+1)}{(x^2+1)^2}$
4. $\dfrac{5(x^2+1)-(5x-2)(2x)}{x^2+1}$
5. $\dfrac{5x-2}{(x^2+1)^2}$

Explanation: This problem requires the quotient rule to find the derivative of the average cost function A(x) = (5x - 2)/(x² + 1). The quotient rule states that if f(x) = num(x)/den(x), then f'(x) = [num'(x) den(x) - num(x) den'(x)] / [den(x)]². Here, num(x) = 5x - 2 with num'(x) = 5, and den(x) = x² + 1 with den'(x) = 2x. Applying the rule gives [5(x² + 1) - (5x - 2)(2x)] / (x² + 1)², matching the correct choice. A tempting distractor like choice A fails because it uses a plus sign instead of a minus in the numerator, reversing the subtraction required by the rule. Remember, a transferable strategy for the quotient rule is to always write 'low d-high minus high d-low over low squared' to keep the order and signs correct.

Question 6

1. $\dfrac{(1)(x^2-9)-(x-2)(2x)}{(x^2-9)^2}$ (correct answer)
2. $\dfrac{(1)(x^2-9)+(x-2)(2x)}{(x^2-9)^2}$
3. $\dfrac{(x-2)(2x)-(1)(x^2-9)}{(x^2-9)^2}$
4. $\dfrac{(1)(x^2-9)-(x-2)(2x)}{(x^2-9)}$
5. $\dfrac{(1)(x^2-9)-(x-2)(2x)}{(x^2-9)^3}$

Explanation: The simplified score s(x) = (x - 2) / (x² - 9) is differentiated using the quotient rule. Let u = x - 2 with u' = 1 and v = x² - 9 with v' = 2x. Applying the rule results in [1 (x² - 9) - (x - 2) 2x] / (x² - 9)², matching choice A. The numerator simplifies to x² - 9 - 2x(x - 2) for further analysis if needed. Choice C incorrectly subtracts in reverse, yielding the opposite sign. Always compute the denominator squared accurately to avoid power errors in quotient rule problems.

Question 7

1. $\dfrac{3x^2(x^2+4)-(x^3-1)(2x)}{(x^2+4)^2}$ (correct answer)
2. $\dfrac{3x^2(x^2+4)+(x^3-1)(2x)}{(x^2+4)^2}$
3. $\dfrac{(x^3-1)(x^2+4)-3x^2(2x)}{(x^2+4)^2}$
4. $\dfrac{3x^2(x^2+4)-(x^3-1)(2x)}{x^2+4}$
5. $\dfrac{(x^3-1)(2x)-3x^2(x^2+4)}{(x^2+4)^2}$

Explanation: To differentiate g(x) = (x³ - 1)/(x² + 4), we use the quotient rule. The quotient rule formula is [f'(x)·g(x) - f(x)·g'(x)]/[g(x)]² where f is the numerator and g is the denominator. The numerator (x³ - 1) has derivative 3x², and the denominator (x² + 4) has derivative 2x. Substituting: g'(x) = [3x²(x² + 4) - (x³ - 1)(2x)]/(x² + 4)². Choice E reverses the order of subtraction in the numerator, which is incorrect—the quotient rule specifically requires numerator' × denominator minus numerator × denominator'. Always maintain the correct order: (top)' × (bottom) - (top) × (bottom)'.

Question 8

1. $\dfrac{e^t(t+2)-e^t}{(t+2)^2}$ (correct answer)
2. $\dfrac{e^t(t+2)+e^t}{(t+2)^2}$
3. $\dfrac{e^t(t+2)-e^t}{t+2}$
4. $\dfrac{(t+2)-e^t}{(t+2)^2}$
5. $\dfrac{e^t-(t+2)}{(t+2)^2}$

Explanation: Finding C'(t) for C(t) = eᵗ/(t + 2) requires the quotient rule. The quotient rule states that for f/g, the derivative is (f'g - fg')/g². Here, the numerator eᵗ has derivative eᵗ, and the denominator (t + 2) has derivative 1. Applying the formula: C'(t) = [eᵗ(t + 2) - eᵗ(1)]/(t + 2)² = [eᵗ(t + 2) - eᵗ]/(t + 2)². Choice C shows the common mistake of forgetting to square the denominator in the quotient rule. To avoid errors, always write the squared denominator first, then work on the numerator calculation.

Question 9

1. $\dfrac{2x\cdot x^3-(x^2+1)\cdot 3x^2}{(x^3)^2}$ (correct answer)
2. $\dfrac{2x\cdot x^3+(x^2+1)\cdot 3x^2}{(x^3)^2}$
3. $\dfrac{2x\cdot x^3-(x^2+1)\cdot 3x^2}{x^3}$
4. $\dfrac{(x^2+1)\cdot x^3-2x\cdot 3x^2}{(x^3)^2}$
5. $\dfrac{(x^2+1)\cdot 3x^2-2x\cdot x^3}{(x^3)^2}$

Explanation: Finding m'(x) for m(x) = (x² + 1)/x³ requires the quotient rule. The quotient rule states (u/v)' = (u'v - uv')/v². The numerator (x² + 1) has derivative 2x, and the denominator x³ has derivative 3x². Applying: m'(x) = [2x·x³ - (x² + 1)·3x²]/(x³)² = [2x⁴ - 3x⁴ - 3x²]/x⁶ = [-x⁴ - 3x²]/x⁶. Choice C forgets to square the denominator, showing x³ instead of (x³)² = x⁶. When using the quotient rule, always remember to square the entire denominator expression in your final answer.

Question 10

1. $\dfrac{(2x+4)\sqrt{x}-(x^2+4x)\left(\dfrac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2}$ (correct answer)
2. $\dfrac{(2x+4)\sqrt{x}+(x^2+4x)\left(\dfrac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2}$
3. $\dfrac{(x^2+4x)\left(\dfrac{1}{2\sqrt{x}}\right)-(2x+4)\sqrt{x}}{(\sqrt{x})^2}$
4. $\dfrac{(2x+4)\sqrt{x}-(x^2+4x)\left(\dfrac{1}{2\sqrt{x}}\right)}{\sqrt{x}}$
5. $\dfrac{(2x+4)\sqrt{x}-(x^2+4x)\left(\dfrac{1}{2\sqrt{x}}\right)}{(\sqrt{x})}$

Explanation: To differentiate $f(x)=\frac{x^2+4x}{\sqrt{x}}$, we use the quotient rule. The quotient rule formula is $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}$. With $u(x)=x^2+4x$ giving $u'(x)=2x+4$, and $v(x)=\sqrt{x}$ giving $v'(x)=\frac{1}{2\sqrt{x}}$, we get: $f'(x)=\frac{(2x+4)\sqrt{x}-(x^2+4x)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2}$. Choice D forgets to square the denominator, writing just $\sqrt{x}$ instead of $(\sqrt{x})^2=x$. The quotient rule requires squaring the entire denominator function—this ensures dimensional consistency and the correct rate of change.

Question 11

1. $\dfrac{(2t+3)(t-1)-(t^2+3t)}{(t-1)^2}$ (correct answer)
2. $\dfrac{(2t+3)(t-1)+(t^2+3t)}{(t-1)^2}$
3. $\dfrac{(t^2+3t)(t-1)-(2t+3)}{(t-1)^2}$
4. $\dfrac{(2t+3)(t-1)-(t^2+3t)}{t-1}$
5. $\dfrac{(t^2+3t)-(2t+3)(t-1)}{(t-1)^2}$

Explanation: This problem requires the quotient rule to find the derivative of R(t) = (t² + 3t)/(t - 1). The quotient rule states that if f(x) = g(x)/h(x), then f'(x) = [g'(x)·h(x) - g(x)·h'(x)]/[h(x)]². Here, g(t) = t² + 3t with g'(t) = 2t + 3, and h(t) = t - 1 with h'(t) = 1. Applying the formula: R'(t) = [(2t + 3)(t - 1) - (t² + 3t)(1)]/(t - 1)². Choice D incorrectly omits squaring the denominator, a common error when rushing through the quotient rule. Remember: the quotient rule always produces a squared denominator—write it first to avoid this mistake.

Question 12

1. $\dfrac{1\cdot \cos x-x(-\sin x)}{(\cos x)^2}$ (correct answer)
2. $\dfrac{1\cdot \cos x-x\sin x}{(\cos x)^2}$
3. $\dfrac{1\cdot \cos x-x(-\sin x)}{\cos x}$
4. $\dfrac{\cos x-1\cdot x(-\sin x)}{(\cos x)^2}$
5. $\dfrac{x(-\sin x)-1\cdot \cos x}{(\cos x)^2}$

Explanation: To differentiate q(x) = x/cos x, we use the quotient rule. The quotient rule formula is (fg' - gf')/g² where f is the numerator and g is the denominator. The numerator x has derivative 1, and the denominator cos x has derivative -sin x. Applying: q'(x) = [1·cos x - x(-sin x)]/(cos x)² = [cos x + x sin x]/(cos x)². Choice B incorrectly writes -sin x as just sin x, forgetting the negative sign in the derivative of cosine. Remember that d/dx(cos x) = -sin x; the negative sign is crucial for correct differentiation.

Question 13

1. $\dfrac{x^2e^x-e^x(2x)}{x^4}$ (correct answer)
2. $\dfrac{x^2e^x+e^x(2x)}{x^4}$
3. $\dfrac{e^x(2x)-x^2e^x}{x^4}$
4. $\dfrac{x^2e^x-e^x(2x)}{x^2}$
5. $\dfrac{x^2e^x-e^x(2x)}{x^3}$

Explanation: This problem requires the quotient rule to differentiate σ(x) = e^x/x². The quotient rule states that for f(x) = g(x)/h(x), we have f'(x) = [h(x)·g'(x) - g(x)·h'(x)]/[h(x)]². Here, g(x) = e^x with g'(x) = e^x, and h(x) = x² with h'(x) = 2x. Applying the quotient rule: σ'(x) = [x²·e^x - e^x·(2x)]/[x²]² = [x²e^x - 2xe^x]/x⁴ = e^x(x² - 2x)/x⁴ = e^x(x - 2)/x³. Choice E incorrectly uses x³ instead of x⁴ in the denominator, forgetting that (x²)² = x⁴. When squaring the denominator in the quotient rule, remember to double the exponent.

Question 14

1. $\dfrac{(x^2+1)\cos x-\sin x(2x)}{(x^2+1)^2}$ (correct answer)
2. $\dfrac{(x^2+1)\cos x+\sin x(2x)}{(x^2+1)^2}$
3. $\dfrac{\cos x(x^2+1)-\sin x(2x)}{x^2+1}$
4. $\dfrac{\sin x(2x)-(x^2+1)\cos x}{(x^2+1)^2}$
5. $\dfrac{(x^2+1)\cos x-\sin x(2x)}{(x^2+1)}$

Explanation: This problem requires the quotient rule to find R'(x) for R(x) = sin x/(x² + 1). The quotient rule formula is: if f(x) = g(x)/h(x), then f'(x) = [h(x)·g'(x) - g(x)·h'(x)]/[h(x)]². For this function, g(x) = sin x with g'(x) = cos x, and h(x) = x² + 1 with h'(x) = 2x. Substituting into the quotient rule: R'(x) = [(x² + 1)·cos x - sin x·(2x)]/[(x² + 1)²]. Choice B incorrectly adds instead of subtracts in the numerator, violating the quotient rule's requirement to subtract g(x)·h'(x). To remember the quotient rule correctly, think: 'lo·d(hi) - hi·d(lo), all over lo²' where 'lo' is the denominator and 'hi' is the numerator.

Question 15

1. $\dfrac{\left(\dfrac{1}{x}\right)(x^3)-(\ln x)(3x^2)}{(x^3)^2}$ (correct answer)
2. $\dfrac{\left(\dfrac{1}{x}\right)(x^3)+(\ln x)(3x^2)}{(x^3)^2}$
3. $\dfrac{(\ln x)(3x^2)-\left(\dfrac{1}{x}\right)(x^3)}{(x^3)^2}$
4. $\dfrac{\left(\dfrac{1}{x}\right)(x^3)-(\ln x)(3x^2)}{x^3}$
5. $\dfrac{\left(\dfrac{1}{x}\right)(x^3)-(\ln x)(3x^2)}{(x^3)^3}$

Explanation: The quotient rule is needed to find q'(x) for q(x) = ln x / x³ where x > 0. Identify u = ln x with u' = 1/x and v = x³ with v' = 3x². Substituting into the rule produces [(1/x) x³ - (ln x) 3x²] / (x³)², as shown in choice A. This simplifies to (x² - 3x² ln x) / x^6 if desired. Choice C reverses the terms in the numerator, changing the sign and thus the derivative's value. A useful strategy is to write out the quotient rule formula explicitly before substituting to ensure correct application every time.

Question 16

1. $\dfrac{(3x^2)\sqrt{x}-(x^3-4)\left(\dfrac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2}$ (correct answer)
2. $\dfrac{(3x^2)\sqrt{x}+(x^3-4)\left(\dfrac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2}$
3. $\dfrac{(x^3-4)\sqrt{x}-(3x^2)\left(\dfrac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2}$
4. $\dfrac{(3x^2)\sqrt{x}-(x^3-4)\left(\dfrac{1}{2\sqrt{x}}\right)}{\sqrt{x}}$
5. $\dfrac{(3x^2)\sqrt{x}-(x^3-4)\left(\dfrac{1}{2\sqrt{x}}\right)}{\sqrt{x^2}}$

Explanation: To find P'(x) for P(x) = (x³ - 4)/√x, we apply the quotient rule with numerator f(x) = x³ - 4 and denominator g(x) = √x = x^(1/2). The derivative of the numerator is f'(x) = 3x², and the derivative of the denominator is g'(x) = (1/2)x^(-1/2) = 1/(2√x). Using the quotient rule formula: P'(x) = [(3x²)(√x) - (x³ - 4)(1/(2√x))]/(√x)². Note that (√x)² = x, not √(x²) as in choice E. Choice D incorrectly omits squaring the denominator, which is essential in the quotient rule. Remember: always square the original denominator in the quotient rule result.

Question 17

1. $\dfrac{(1)(1+t^2)-t(2t)}{(1+t^2)^2}$ (correct answer)
2. $\dfrac{(1)(1+t^2)+t(2t)}{(1+t^2)^2}$
3. $\dfrac{t(2t)-(1)(1+t^2)}{(1+t^2)^2}$
4. $\dfrac{(1)(1+t^2)-t(2t)}{1+t^2}$
5. $\dfrac{(1)(1+t^2)-t(2t)}{(1+t^2)}$

Explanation: Finding h'(t) for h(t) = t/(1 + t²) requires the quotient rule with numerator t (derivative: 1) and denominator 1 + t² (derivative: 2t). The quotient rule formula gives: h'(t) = [(1)(1 + t²) - t(2t)]/(1 + t²)². This simplifies to [1 + t² - 2t²]/(1 + t²)² = (1 - t²)/(1 + t²)². Choice D incorrectly omits squaring the denominator, while choice E has the same error (since (1 + t²) without exponent means (1 + t²)¹). Remember that the quotient rule always requires squaring the original denominator in the final result.

Question 18

1. $\dfrac{(2x)\cos x-(x^2+1)(-\sin x)}{(\cos x)^2}$ (correct answer)
2. $\dfrac{(2x)\cos x+(x^2+1)(-\sin x)}{(\cos x)^2}$
3. $\dfrac{(x^2+1)(-\sin x)-(2x)\cos x}{(\cos x)^2}$
4. $\dfrac{(2x)\cos x-(x^2+1)(-\sin x)}{\cos x}$
5. $\dfrac{(2x)\cos x-(x^2+1)(-\sin x)}{\cos(x^2)}$

Explanation: To find m'(x) for m(x) = (x² + 1)/cos x, we apply the quotient rule with numerator x² + 1 (derivative: 2x) and denominator cos x (derivative: -sin x). Using the quotient rule: m'(x) = [(2x)(cos x) - (x² + 1)(-sin x)]/(cos x)². Note the double negative becomes positive: (2x cos x + (x² + 1)sin x)/(cos x)². Choice C incorrectly reverses the order of subtraction, which would give a different result. The quotient rule maintains a specific order: "top' × bottom - top × bottom'" where careful attention to signs is crucial, especially with trigonometric derivatives.

Question 19

1. $\dfrac{(2x-5)(x^2+4)-(x^2-5x)(2x)}{(x^2+4)^2}$ (correct answer)
2. $\dfrac{(2x-5)(x^2+4)+(x^2-5x)(2x)}{(x^2+4)^2}$
3. $\dfrac{(x^2-5x)(2x)-(2x-5)(x^2+4)}{(x^2+4)^2}$
4. $\dfrac{(2x-5)(x^2+4)-(x^2-5x)(2x)}{x^2+4}$
5. $\dfrac{(2x-5)(x^2+4)-(x^2-5x)(2x)}{(x^2+4)}$

Explanation: To find s'(x) for s(x) = (x² - 5x)/(x² + 4), we apply the quotient rule with numerator x² - 5x (derivative: 2x - 5) and denominator x² + 4 (derivative: 2x). The quotient rule gives: s'(x) = [(2x - 5)(x² + 4) - (x² - 5x)(2x)]/(x² + 4)². This can be expanded and simplified, but the form shown in choice A is correct. Choice D incorrectly omits squaring the denominator, a common error. Remember that the quotient rule always results in the original denominator squared in the final expression, regardless of what the original denominator was.

Question 20

1. $\dfrac{\cos t\,(t^2+1)-\sin t\,(2t)}{(t^2+1)^2}$ (correct answer)
2. $\dfrac{\sin t\,(2t)-\cos t\,(t^2+1)}{(t^2+1)^2}$
3. $\dfrac{\cos t\,(t^2+1)+\sin t\,(2t)}{(t^2+1)^2}$
4. $\dfrac{\cos t\,(t^2+1)-\sin t\,(2t)}{t^2+1}$
5. $\dfrac{\cos t}{t^2+1}$

Explanation: This problem requires the quotient rule to find the derivative of the signal function S(t) = sin t / (t² + 1). The quotient rule states that if f(t) = num(t)/den(t), then f'(t) = [num'(t) den(t) - num(t) den'(t)] / [den(t)]². Here, num(t) = sin t with num'(t) = cos t, and den(t) = t² + 1 with den'(t) = 2t. Applying the rule gives [cos t (t² + 1) - sin t (2t)] / (t² + 1)², which is correct. A tempting distractor like choice B fails because it reverses the subtraction, starting with the second term instead of the first as per the rule. Remember, a transferable strategy for the quotient rule is to always write 'low d-high minus high d-low over low squared' to keep the order and signs correct.