AP Calculus BC Quiz: The Product Rule

What this quiz covers

This quiz focuses on The Product 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. $g'(x)=\frac{1}{1+x^2}(x^3-2)$
2. $g'(x)=\arctan(x)(3x^2)$
3. $g'(x)=\frac{1}{1+x^2}(3x^2)$
4. $g'(x)=\frac{x^3-2}{1+x^2}+3x^2\arctan(x)$ (correct answer)
5. $g'(x)=\arctan(x)\cdot\frac{x^3-2}{1+x^2}$

Explanation: This problem requires the product rule to differentiate g(x) = arctan(x)(x³ - 2). The product rule states that for f(x) = u(x)·v(x), we have f'(x) = u'(x)·v(x) + u(x)·v'(x). Here, u(x) = arctan(x) with u'(x) = 1/(1 + x²), and v(x) = x³ - 2 with v'(x) = 3x². Applying the product rule: g'(x) = (1/(1 + x²))·(x³ - 2) + arctan(x)·3x² = (x³ - 2)/(1 + x²) + 3x²arctan(x). Choice A shows only the first term, missing the crucial second product of arctan(x) and 3x². The product rule always yields two terms—both must be included to obtain the correct derivative.

Question 2

1. $(12r^2)\arctan r$
2. $(4r^3)\cdot \dfrac{1}{1+r^2}$
3. $(12r^2)\arctan r+(4r^3)\dfrac{1}{1+r^2}$ (correct answer)
4. $(12r^2)\dfrac{1}{1+r^2}$
5. $(4r^3)(12r^2)\dfrac{1}{1+r^2}$

Explanation: This problem requires the product rule to find the derivative of a volume function given as a product of two functions. To apply the product rule, identify u(r) = 4r³ with u'(r) = 12r², and v(r) = arctan r with v'(r) = 1/(1 + r²). Then, the derivative V'(r) is u'(r)v(r) + u(r)v'(r), which expands to 12r² arctan r + 4r³ [1/(1 + r²)]. This matches choice C, providing the complete derivative. A tempting distractor like choice A omits the second term, failing to account for the derivative of the arctan function. Always remember to differentiate both parts of the product and add the results for accurate application of the product rule.

Question 3

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

Explanation: This problem requires the product rule to find the derivative of a temperature model given as a product of two functions. To apply the product rule, identify u(x) = $x^2 - 2x$ with u'(x) = $2x - 2$, and v(x) = $e^{3x}$ with v'(x) = $3e^{3x}$. Then, the derivative T'(x) is $u'(x)v(x) + u(x)v'(x)$, which expands to $(2x - 2)e^{3x} + (x^2 - 2x)(3e^{3x})$. This matches choice A, providing the complete derivative. A tempting distractor like choice B omits the second term, failing to include the derivative of the exponential function. Always remember to differentiate both parts of the product and add the results for accurate application of the product rule.

Question 4

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

Explanation: This problem requires the product rule to differentiate A(x) = sin x · (x³ - 1). The product rule states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). Here, u(x) = sin x with u'(x) = cos x, and v(x) = x³ - 1 with v'(x) = 3x². Applying the product rule: A'(x) = (cos x)(x³ - 1) + (sin x)(3x²) = cos x(x³ - 1) + sin x(3x²). A common mistake is to compute only cos x(3x²) (choice B), which incorrectly pairs the derivative of sin x with the derivative of (x³ - 1). The product rule pairs each function with the derivative of the other—not derivative with derivative.

Question 5

1. $e^t\cos t-e^t\sin t$ (correct answer)
2. $e^t(-\sin t)$
3. $e^t\cos t$
4. $e^t\sin t-e^t\cos t$
5. $e^t\cos t\sin t$

Explanation: This problem requires the product rule to differentiate T(t) = eᵗcos t. The product rule states that if f(t) = u(t)v(t), then f'(t) = u'(t)v(t) + u(t)v'(t). Here, u(t) = eᵗ with u'(t) = eᵗ, and v(t) = cos t with v'(t) = -sin t. Applying the product rule: T'(t) = (eᵗ)(cos t) + (eᵗ)(-sin t) = eᵗcos t - eᵗsin t. A common mistake is to forget the first term and write only eᵗ(-sin t) (choice B), which comes from differentiating just the cosine factor. The product rule requires both parts: keep track of differentiating each factor while holding the other constant.

Question 6

1. $\tan t+(t-1)\sec^2 t$ (correct answer)
2. $(t-1)\sec^2 t$
3. $\sec^2 t$
4. $\tan t\sec^2 t$
5. $\tan t+(t-1)\tan t$

Explanation: This problem requires the product rule to differentiate y(t) = (t - 1)tan t. The product rule states that if f(t) = u(t)v(t), then f'(t) = u'(t)v(t) + u(t)v'(t). Here, u(t) = t - 1 with u'(t) = 1, and v(t) = tan t with v'(t) = sec²t. Applying the product rule: dy/dt = (1)(tan t) + (t - 1)(sec²t) = tan t + (t - 1)sec²t. A common error is to write only (t - 1)sec²t (choice B), forgetting the term that comes from differentiating the first factor. Remember the product rule pattern: derivative of first times second, plus first times derivative of second—both terms are essential.

Question 7

1. $D'(p)=-p^2+(10-p)2p$ (correct answer)
2. $D'(p)=(10-p)2p$
3. $D'(p)=-p^2$
4. $D'(p)=-1\cdot 2p$
5. $D'(p)=(10-p)p^2$

Explanation: For $D(p)=(10-p)(p^2)$, we apply the product rule with $u=10-p$ and $v=p^2$. We have $u'=-1$ and $v'=2p$. Using the product rule: $D'(p)=(-1)(p^2)+(10-p)(2p)=-p^2+(10-p)2p$. Choice B shows only $(10-p)2p$, which is just the second term and misses the $-p^2$ from the first term. The product rule formula $(uv)'=u'v+uv'$ requires both terms; omitting either one gives an incomplete derivative.

Question 8

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

Explanation: This problem requires the product rule to find the derivative of a model given as a product of two functions. To apply the product rule, identify u(x) = x + 1 with u'(x) = 1, and v(x) = ln(x²) with v'(x) = 2/x. Then, the derivative M'(x) is u'(x)v(x) + u(x)v'(x), which expands to ln(x²) + (x + 1)(2/x). This matches choice A, providing the complete derivative. A tempting distractor like choice B omits the logarithmic term, failing to account for the derivative of the linear function. Always remember to differentiate both parts of the product and add the results for accurate application of the product rule.

Question 9

1. $T'(t)=\dfrac{1}{2\sqrt{t}}(t^2+1)+\sqrt{t}(2t)$ (correct answer)
2. $T'(t)=\dfrac{1}{2\sqrt{t}}(2t)$
3. $T'(t)=\dfrac{1}{2\sqrt{t}}+2t$
4. $T'(t)=\sqrt{t}(2t)$
5. $T'(t)=\dfrac{t^2+1}{2t}$

Explanation: This problem requires the product rule to find the derivative of the temperature model given as a product. Let u(t) = √t with u'(t) = 1/(2√t), and v(t) = t² + 1 with v'(t) = 2t. Then, T'(t) = u'(t)v(t) + u(t)v'(t) = [1/(2√t)](t² + 1) + √t (2t), which is choice A. This expression fully captures the differentiation of both factors. Choice D fails as a distractor by omitting the first term and only differentiating the second factor. A transferable strategy is to rewrite functions in exponential form if needed, but always apply the product rule systematically to both parts.

Question 10

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

Explanation: This problem requires the product rule to differentiate the cost function as a product. Set u(x) = $x^2 - 4x$ with u'(x) = $2x - 4$, and v(x) = $\sqrt{x}$ with v'(x) = $\frac{1}{2\sqrt{x}}$. Thus, C'(x) = u'(x)v(x) + u(x)v'(x) = $(2x - 4)\sqrt{x} + (x^2 - 4x)(\frac{1}{2\sqrt{x}})$, matching choice A. This correctly handles the square root derivative. Choice D is a distractor as it omits the second term, only using the polynomial's derivative. A useful strategy is to express roots as exponents to ease differentiation, but always include both product rule terms.

Question 11

1. $s'(t)=3t^2\cos t$
2. $s'(t)=3t^2\sin t+t^3\cos t$ (correct answer)
3. $s'(t)=t^3\cos t$
4. $s'(t)=3t^2\sin t$
5. $s'(t)=3t^2\sin t\cos t$

Explanation: This problem asks us to differentiate $s(t)=t^3\sin t$ using the product rule. The product rule states $(fg)'=f'g+fg'$, where $f(t)=t^3$ and $g(t)=\sin t$. We find $f'(t)=3t^2$ and $g'(t)=\cos t$. Applying the product rule: $s'(t)=(3t^2)(\sin t)+(t^3)(\cos t)=3t^2\sin t+t^3\cos t$. Choice A incorrectly replaces $\sin t$ with $\cos t$ in the first term, showing $3t^2\cos t$ instead of $3t^2\sin t$. When using the product rule, keep the original functions intact in their respective terms—don't accidentally differentiate a function twice.

Question 12

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

Explanation: This problem requires the product rule to differentiate F(x) = (x² + 3)e^(-x). The product rule states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). Here, u(x) = x² + 3 with u'(x) = 2x, and v(x) = e^(-x) with v'(x) = -e^(-x). Applying the product rule: F'(x) = (2x)(e^(-x)) + (x² + 3)(-e^(-x)) = 2xe^(-x) - (x² + 3)e^(-x). A common error is to compute only (x² + 3)(-e^(-x)) (choice C), forgetting the term from differentiating the first factor. The product rule requires both contributions: always calculate derivative of first times second, plus first times derivative of second.

Question 13

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

Explanation: This problem requires the product rule to differentiate c(t) = ln(t + 1) · (t² - 4t). The product rule states that if f(t) = u(t)v(t), then f'(t) = u'(t)v(t) + u(t)v'(t). Here, u(t) = ln(t + 1) with u'(t) = 1/(t + 1), and v(t) = t² - 4t with v'(t) = 2t - 4. Applying the product rule: c'(t) = (1/(t + 1))(t² - 4t) + (ln(t + 1))(2t - 4) = (t² - 4t)/(t + 1) + ln(t + 1)(2t - 4). A common mistake is to write only ln(t + 1)(2t - 4) (choice D), omitting the first term entirely. The product rule generates two terms—never drop either one when differentiating a product of functions.

Question 14

1. $V'(t)=2t\sqrt{t}+(t^2-1)\dfrac{1}{2\sqrt{t}}$ (correct answer)
2. $V'(t)=2t\sqrt{t}$
3. $V'(t)=(t^2-1)\dfrac{1}{2\sqrt{t}}$
4. $V'(t)=2t\cdot\dfrac{1}{2\sqrt{t}}$
5. $V'(t)=(2t)(t^2-1)+\sqrt{t}$

Explanation: This problem requires the product rule to differentiate V(t) = (t² - 1)√t. The product rule tells us that for h(t) = f(t)·g(t), we have h'(t) = f'(t)·g(t) + f(t)·g'(t). Here, f(t) = t² - 1 with f'(t) = 2t, and g(t) = √t = t^(1/2) with g'(t) = (1/2)t^(-1/2) = 1/(2√t). Applying the product rule: V'(t) = (2t)(√t) + (t² - 1)(1/(2√t)) = 2t√t + (t² - 1)/(2√t). Choice B shows only 2t√t, which is just the first term and ignores the second term. Always remember: the product rule gives you the sum of two terms, never just one.

Question 15

1. $C'(x)=(2x+3)\ln x+(x^2+3x)\cdot\dfrac{1}{x}$ (correct answer)
2. $C'(x)=(2x+3)\ln x$
3. $C'(x)=(x^2+3x)\cdot\dfrac{1}{x}$
4. $C'(x)=(2x+3)\cdot\dfrac{1}{x}$
5. $C'(x)=(2x+3)\ln x\cdot\dfrac{1}{x}$

Explanation: This problem requires the product rule to differentiate C(x) = (x² + 3x)ln x. The product rule states that if f(x) = u(x)·v(x), then f'(x) = u'(x)·v(x) + u(x)·v'(x). Here, u(x) = x² + 3x with u'(x) = 2x + 3, and v(x) = ln x with v'(x) = 1/x. Applying the product rule: C'(x) = (2x + 3)·ln x + (x² + 3x)·(1/x). Choice B shows only the first term (2x + 3)ln x, omitting the second term where we multiply (x² + 3x) by 1/x. The product rule requires both terms—always include the "first times derivative of second" term even when it involves fractions.

Question 16

1. $T'(x)=3\cos(2x)+(3x+2)(-2\sin(2x))$ (correct answer)
2. $T'(x)=3\cos(2x)$
3. $T'(x)=(3x+2)(-2\sin(2x))$
4. $T'(x)=3(-2\sin(2x))+(3x+2)\cos(2x)$
5. $T'(x)=3\cos(2x)+(3x+2)\sin(2x)$

Explanation: This problem requires the product rule to differentiate T(x) = (3x + 2)cos(2x). The product rule states that if f(x) = u(x)·v(x), then f'(x) = u'(x)·v(x) + u(x)·v'(x). Here, u(x) = 3x + 2 with u'(x) = 3, and v(x) = cos(2x) with v'(x) = -2sin(2x) by the chain rule. Applying the product rule: T'(x) = 3·cos(2x) + (3x + 2)·(-2sin(2x)). Choice E incorrectly uses sin(2x) instead of -2sin(2x) for the derivative of cos(2x), forgetting both the negative sign and the chain rule factor of 2. When applying the product rule with composite functions, carefully use the chain rule for each derivative.

Question 17

1. $R'(x)=\ln(x)+\dfrac{x+1}{x}$ (correct answer)
2. $R'(x)=\dfrac{1}{x}$
3. $R'(x)=(x+1)\cdot\dfrac{1}{x}$
4. $R'(x)=\ln(x)$
5. $R'(x)=\dfrac{\ln(x)}{x+1}$

Explanation: To find $R'(x)$ for $R(x)=(x+1)\ln(x)$, we apply the product rule. The product rule tells us that $(uv)'=u'v+uv'$ where $u=x+1$ and $v=\ln(x)$. We have $u'=1$ and $v'=\frac{1}{x}$. Substituting into the product rule: $R'(x)=(1)(\ln(x))+(x+1)(\frac{1}{x})=\ln(x)+\frac{x+1}{x}$. Choice C shows only the second term $(x+1)\cdot\frac{1}{x}$, missing the crucial first term $\ln(x)$. Always remember to include both parts of the product rule: the derivative of the first function times the second, plus the first function times the derivative of the second.

Question 18

1. $s'(t)=\cos(t)(t^2-4)+\sin(t)(2t)$ (correct answer)
2. $s'(t)=\cos(t)(2t)$
3. $s'(t)=\cos(t)(t^2-4)$
4. $s'(t)=\sin(t)(2t)$
5. $s'(t)=\cos(t)\sin(t)(2t)(t^2-4)$

Explanation: This problem requires the product rule to differentiate $s(t) = \sin(t)(t^2 - 4)$. The product rule tells us that if $f(t) = u(t) \cdot v(t)$, then $f'(t) = u'(t) \cdot v(t) + u(t) \cdot v'(t)$. Here, $u(t) = \sin(t)$ with $u'(t) = \cos(t)$, and $v(t) = t^2 - 4$ with $v'(t) = 2t$. Applying the product rule: $s'(t) = \cos(t) \cdot (t^2 - 4) + \sin(t) \cdot 2t = \cos(t)(t^2 - 4) + \sin(t)(2t)$. Choice C incorrectly uses only the first term, omitting the crucial second part where we multiply $\sin(t)$ by the derivative of $(t^2 - 4)$. The product rule always yields a sum of two products—both parts are essential for the correct derivative.

Question 19

1. $\Pi'(x)=(2x+3)\ln(2x)+(x^2+3x)\cdot\frac{1}{x}$ (correct answer)
2. $\Pi'(x)=(2x+3)\ln(2x)$
3. $\Pi'(x)=(x^2+3x)\cdot\frac{1}{2x}$
4. $\Pi'(x)=(2x+3)\cdot\frac{1}{x}$
5. $\Pi'(x)=(2x+3)\ln(2x)+(x^2+3x)\cdot\frac{1}{2x}$

Explanation: This problem requires the product rule to differentiate $\Pi(x) = (x^2 + 3x)\ln(2x)$. The product rule states that if $f(x) = u(x) \cdot v(x)$, then $f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$. Here, $u(x) = x^2 + 3x$ with $u'(x) = 2x + 3$, and $v(x) = \ln(2x)$ with $v'(x) = \frac{1}{2x} \cdot 2 = \frac{1}{x}$ by the chain rule. Applying the product rule: $\Pi'(x) = (2x + 3) \cdot \ln(2x) + (x^2 + 3x) \cdot \frac{1}{x} = (2x + 3)\ln(2x) + \frac{x^2 + 3x}{x}$. Choice E incorrectly computes the derivative of $\ln(2x)$ as $\frac{1}{2x}$, missing that $\frac{d}{dx}[\ln(2x)] = \frac{1}{2x} \cdot 2 = \frac{1}{x}$. When differentiating logarithms of linear functions, remember to apply the chain rule correctly.

Question 20

1. $4t^3\sqrt{t}$
2. $t^4\cdot \frac{1}{2\sqrt{t}}$
3. $4t^3\sqrt{t}+t^4\cdot \frac{1}{2\sqrt{t}}$ (correct answer)
4. $4t^3\cdot \frac{1}{2\sqrt{t}}$
5. $\frac{9}{2}t^{7/2}$

Explanation: This problem requires the product rule to differentiate P(t) = t⁴√t = t⁴ · t^(1/2). The product rule states that if f(t) = u(t)v(t), then f'(t) = u'(t)v(t) + u(t)v'(t). Here, u(t) = t⁴ with u'(t) = 4t³, and v(t) = √t = t^(1/2) with v'(t) = (1/2)t^(-1/2) = 1/(2√t). Applying the product rule: P'(t) = (4t³)(√t) + (t⁴)(1/(2√t)) = 4t³√t + t⁴/(2√t). A tempting shortcut is to combine t⁴√t = t^(9/2) and differentiate directly to get (9/2)t^(7/2) (choice E), but the question asks us to use the product rule form. When applying the product rule, maintain the structure of two separate terms rather than simplifying first.