Calculus 1 Quiz: Product Rule
20 questions · exam conditions
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Product RuleQuestion 1 of 20

Let h(x)=f(x)g(x)h(x) = f(x)g(x). If f(2)=3f(2) = 3, f′(2)=−1f'(2) = -1, g(2)=5g(2) = 5, and g′(2)=4g'(2) = 4, what is the value of h′(2)h'(2)?

−4-4
77
1717
−17-17
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Calculus 1 Quiz

Calculus 1 Quiz: Product Rule

Practice Product Rule in Calculus 1 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 Product Rule, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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

Let h(x)=f(x)g(x)h(x) = f(x)g(x). If f(2)=3f(2) = 3, f′(2)=−1f'(2) = -1, g(2)=5g(2) = 5, and g′(2)=4g'(2) = 4, what is the value of h′(2)h'(2)?

  1. −4-4
  2. 77 (correct answer)
  3. 1717
  4. −17-17
Explanation: The product rule states that h′(x)=f′(x)g(x)+f(x)g′(x)h'(x) = f'(x)g(x) + f(x)g'(x). To find h′(2)h'(2), we substitute the given values: h′(2)=f′(2)g(2)+f(2)g′(2)=(−1)(5)+(3)(4)=−5+12=7h'(2) = f'(2)g(2) + f(2)g'(2) = (-1)(5) + (3)(4) = -5 + 12 = 7.
  • A is incorrect because it results from the common mistake of multiplying the derivatives: f′(2)g′(2)=(−1)(4)=−4f'(2)g'(2) = (-1)(4) = -4.
  • C is incorrect because it incorrectly uses the structure of the quotient rule: f(2)g′(2)−f′(2)g(2)=(3)(4)−(−1)(5)=12+5=17f(2)g'(2) - f'(2)g(2) = (3)(4) - (-1)(5) = 12 + 5 = 17.
  • D is incorrect because it incorrectly uses the structure of the quotient rule with a sign error: f′(2)g(2)−f(2)g′(2)=(−1)(5)−(3)(4)=−5−12=−17f'(2)g(2) - f(2)g'(2) = (-1)(5) - (3)(4) = -5 - 12 = -17.

Question 2

Let kk be a constant. If h(x)=(kx3)f(x)h(x) = (kx^3) f(x), and we know f(1)=2f(1)=2 and f′(1)=−1f'(1)=-1, for what value of kk is h′(1)=10h'(1)=10?

  1. k=2k=2 (correct answer)
  2. k=−5k=-5
  3. k=−10k=-10
  4. k=1k=1
Explanation: First, find the derivative of h(x)h(x) using the product rule. Let u=kx3u = kx^3 and v=f(x)v = f(x). Then u′=3kx2u' = 3kx^2 and v′=f′(x)v' = f'(x). So, h′(x)=(3kx2)f(x)+(kx3)f′(x)h'(x) = (3kx^2)f(x) + (kx^3)f'(x). Now, evaluate at x=1x=1: h′(1)=(3k(1)2)f(1)+(k(1)3)f′(1)=3kf(1)+kf′(1)h'(1) = (3k(1)^2)f(1) + (k(1)^3)f'(1) = 3kf(1) + kf'(1). Substitute the given values: h′(1)=3k(2)+k(−1)=6k−k=5kh'(1) = 3k(2) + k(-1) = 6k - k = 5k. We are given that h′(1)=10h'(1) = 10, so we set 5k=105k = 10 and solve for kk, which gives k=2k=2.
  • B is incorrect and would result from an error such as h′(x)=k⋅f′(x)⋅3x2h'(x) = k \cdot f'(x) \cdot 3x^2 , giving h′(1)=k(−1)(3)=−3k=10h'(1) = k(-1)(3) = -3k = 10, or a different error leading to −2k=10-2k = 10.
  • C is incorrect and would result from neglecting the first term of the product rule: h′(x)=kx3f′(x)h'(x) = kx^3 f'(x), leading to h′(1)=kf′(1)=−k=10h'(1) = kf'(1) = -k = 10, so k=−10k = -10.
  • D is incorrect and would be chosen if a student made a significant calculation error.

Question 3

Find the slope of the tangent line to the curve x2y+y3=10x^2 y + y^3 = 10 at the point (1,2)(1, 2).

  1. −413-\frac{4}{13} (correct answer)
  2. 413\frac{4}{13}
  3. −213-\frac{2}{13}
  4. −411-\frac{4}{11}
Explanation: We use implicit differentiation. Differentiating both sides with respect to xx, the term x2yx^2 y requires the product rule: ddx(x2y)=(2x)y+x2dydx\frac{d}{dx}(x^2 y) = (2x)y + x^2 \frac{dy}{dx}. The full differentiated equation is 2xy+x2dydx+3y2dydx=02xy + x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0. Now, solve for dydx\frac{dy}{dx}: dydx(x2+3y2)=−2xy\frac{dy}{dx}(x^2 + 3y^2) = -2xy, so dydx=−2xyx2+3y2\frac{dy}{dx} = \frac{-2xy}{x^2 + 3y^2}. Finally, substitute the point (1,2)(1, 2) into this expression for the slope: dydx=−2(1)(2)(1)2+3(2)2=−41+3(4)=−41+12=−413\frac{dy}{dx} = \frac{-2(1)(2)}{(1)^2 + 3(2)^2} = \frac{-4}{1 + 3(4)} = \frac{-4}{1+12} = -\frac{4}{13}.
  • B is incorrect due to a sign error in the final calculation.
  • C is incorrect because it omits the yy factor in the numerator term −2xy-2xy.
  • D is incorrect due to a calculation error in the denominator: 1+3(22)=1+10=111 + 3(2^2) = 1+10=11 instead of 1+12=131+12=13.

Question 4

For the function f(x)=xln⁡(x)f(x) = x \ln(x), on which interval is the first derivative, f′(x)f'(x), positive?

  1. (0,∞)(0, \infty)
  2. (0,e−1)(0, e^{-1})
  3. (e,∞)(e, \infty)
  4. (e−1,∞)(e^{-1}, \infty) (correct answer)
Explanation: First, note that the domain of f(x)=xln⁡(x)f(x) = x \ln(x) is x>0x>0. Next, find the derivative using the product rule: f′(x)=(1)ln⁡(x)+x(1x)=ln⁡(x)+1f'(x) = (1)\ln(x) + x(\frac{1}{x}) = \ln(x) + 1. We want to find where f′(x)>0f'(x) > 0, so we solve the inequality ln⁡(x)+1>0\ln(x) + 1 > 0. This gives ln⁡(x)>−1\ln(x) > -1. To solve for xx, we exponentiate both sides: eln⁡(x)>e−1e^{\ln(x)} > e^{-1}, which simplifies to x>e−1x > e^{-1}. The interval is therefore (e−1,∞)(e^{-1}, \infty).
  • A is incorrect. For xx values between 0 and e−1e^{-1}, ln⁡(x)\ln(x) is less than -1, so ln⁡(x)+1\ln(x)+1 is negative.
  • B is incorrect. This is the interval where f′(x)<0f'(x) < 0.
  • C is incorrect. This results from solving the inequality ln⁡(x)>1\ln(x) > 1 instead of ln⁡(x)>−1\ln(x) > -1.

Question 5

What is the equation of the line tangent to the curve y=x2cos⁡(x)y = x^2 \cos(x) at x=πx = \pi?

  1. y=−π2y = -\pi^2
  2. y=−2πx+π2y = -2\pi x + \pi^2 (correct answer)
  3. y=−2πx−π2y = -2\pi x - \pi^2
  4. y=−2πx+3π2y = -2\pi x + 3\pi^2
Explanation: First, find the y-coordinate of the point of tangency: y(π)=π2cos⁡(π)=π2(−1)=−π2y(\pi) = \pi^2 \cos(\pi) = \pi^2(-1) = -\pi^2. The point is (π,−π2)(\pi, -\pi^2). Next, find the derivative using the product rule: y′=ddx(x2)cos⁡(x)+x2ddx(cos⁡(x))=2xcos⁡(x)−x2sin⁡(x)y' = \frac{d}{dx}(x^2)\cos(x) + x^2\frac{d}{dx}(\cos(x)) = 2x\cos(x) - x^2\sin(x). Evaluate the derivative at x=πx=\pi to find the slope of the tangent line: y′(π)=2πcos⁡(π)−π2sin⁡(π)=2π(−1)−π2(0)=−2πy'(\pi) = 2\pi\cos(\pi) - \pi^2\sin(\pi) = 2\pi(-1) - \pi^2(0) = -2\pi. Finally, use the point-slope form: y−y1=m(x−x1)y - y_1 = m(x - x_1), which gives y−(−π2)=−2π(x−π)y - (-\pi^2) = -2\pi(x - \pi). Simplifying, we get y+π2=−2πx+2π2y + \pi^2 = -2\pi x + 2\pi^2, so y=−2πx+π2y = -2\pi x + \pi^2.
  • A is incorrect because it uses a slope of 0, which would happen if the derivative was incorrectly calculated as y′=2x(−sin⁡(x))y' = 2x(-\sin(x)).
  • C is incorrect due to a sign error when substituting the y-coordinate into the point-slope formula.
  • D is incorrect due to an algebraic error in the final step of isolating y: adding π2\pi^2 to both sides instead of subtracting.

Question 6

For which values of xx does the function f(x)=x2e−xf(x) = x^2 e^{-x} have a horizontal tangent line?

  1. x=0x=0 only
  2. x=2x=2 only
  3. x=0,x=2x=0, x=2 (correct answer)
  4. x=0,x=−2x=0, x=-2
Explanation: A horizontal tangent line occurs where the derivative is zero. First, find the derivative using the product rule: f′(x)=(ddxx2)e−x+x2(ddxe−x)=2xe−x+x2(−e−x)f'(x) = (\frac{d}{dx}x^2)e^{-x} + x^2(\frac{d}{dx}e^{-x}) = 2xe^{-x} + x^2(-e^{-x}). Factor the expression: f′(x)=xe−x(2−x)f'(x) = xe^{-x}(2-x). Set the derivative equal to zero: xe−x(2−x)=0xe^{-x}(2-x) = 0. Since e−xe^{-x} is never zero, we solve x=0x=0 or 2−x=02-x=0. This gives the solutions x=0x=0 and x=2x=2.
  • A is incorrect because it is an incomplete solution, missing the second value.
  • B is incorrect because it is an incomplete solution, missing the first value.
  • D is incorrect due to a sign error when solving the equation 2−x=02-x=0.

Question 7

Which of the following is the derivative of h(x)=xexcos⁡(x)h(x) = x e^x \cos(x)?

  1. ex(cos⁡(x)+xcos⁡(x)−xsin⁡(x))e^x(\cos(x) + x\cos(x) - x\sin(x)) (correct answer)
  2. ex(−sin⁡(x))e^x(-\sin(x))
  3. ex(cos⁡(x)−xsin⁡(x))e^x(\cos(x) - x\sin(x))
  4. ex(1+cos⁡(x)−sin⁡(x))e^x(1 + \cos(x) - \sin(x))
Explanation: To find the derivative of a product of three functions, we can apply the Product Rule twice. Let's group f(x)=(xex)f(x) = (xe^x) and g(x)=cos⁡(x)g(x) = \cos(x). h′(x)=f′(x)g(x)+f(x)g′(x)h'(x) = f'(x)g(x) + f(x)g'(x). First, find f′(x)f'(x) using the Product Rule: f′(x)=(1)ex+x(ex)=ex+xexf'(x) = (1)e^x + x(e^x) = e^x + xe^x. Now substitute this and g′(x)=−sin⁡(x)g'(x) = -\sin(x) into the main formula: h′(x)=(ex+xex)cos⁡(x)+(xex)(−sin⁡(x))h'(x) = (e^x + xe^x)\cos(x) + (xe^x)(-\sin(x)) h′(x)=excos⁡(x)+xexcos⁡(x)−xexsin⁡(x)h'(x) = e^x\cos(x) + xe^x\cos(x) - xe^x\sin(x) Factoring out exe^x gives: h′(x)=ex(cos⁡(x)+xcos⁡(x)−xsin⁡(x))h'(x) = e^x(\cos(x) + x\cos(x) - x\sin(x)).

Question 8

The electrical power PP (in Watts) in a circuit is given by P(t)=V(t)I(t)P(t) = V(t)I(t), where VV is the voltage (in Volts) and II is the current (in Amperes) at time tt (in seconds).

At a certain moment, the voltage is 120 V and is increasing at a rate of 2 V/s. At the same moment, the current is 10 A and is decreasing at a rate of 0.5 A/s. What is the rate of change of the power at that moment, in Watts per second?

  1. −60-60 W/s
  2. −40-40 W/s (correct answer)
  3. 2020 W/s
  4. 8080 W/s
Explanation: We are asked to find the rate of change of power, dPdt\frac{dP}{dt}. The relationship is P=VIP = VI. We differentiate with respect to time tt using the Product Rule: dPdt=d(VI)dt=dVdtI+VdIdt\frac{dP}{dt} = \frac{d(VI)}{dt} = \frac{dV}{dt}I + V\frac{dI}{dt}. The given values are: V=120V = 120 V I=10I = 10 A dVdt=2\frac{dV}{dt} = 2 V/s (increasing) dIdt=−0.5\frac{dI}{dt} = -0.5 A/s (decreasing) Substitute these values into the derivative equation: dPdt=(2)(10)+(120)(−0.5)\frac{dP}{dt} = (2)(10) + (120)(-0.5) dPdt=20−60=−40\frac{dP}{dt} = 20 - 60 = -40 W/s.

Question 9

If f(x)=sin⁡(2x)cos⁡(x)f(x) = \sin(2x)\cos(x), what is the value of f′(π6)f'(\frac{\pi}{6})?

  1. 00
  2. −12-\frac{1}{2}
  3. 334\frac{3\sqrt{3}}{4}
  4. 34\frac{\sqrt{3}}{4} (correct answer)
Explanation: We use the Product Rule, f′(x)=u′v+uv′f'(x) = u'v + uv', combined with the Chain Rule for the term sin⁡(2x)\sin(2x). Let u(x)=sin⁡(2x)u(x) = \sin(2x) and v(x)=cos⁡(x)v(x) = \cos(x). u′(x)=cos⁡(2x)⋅2=2cos⁡(2x)u'(x) = \cos(2x) \cdot 2 = 2\cos(2x). v′(x)=−sin⁡(x)v'(x) = -\sin(x). So, f′(x)=(2cos⁡(2x))(cos⁡(x))+(sin⁡(2x))(−sin⁡(x))=2cos⁡(2x)cos⁡(x)−sin⁡(2x)sin⁡(x)f'(x) = (2\cos(2x))(\cos(x)) + (\sin(2x))(-\sin(x)) = 2\cos(2x)\cos(x) - \sin(2x)\sin(x). Now we evaluate at x=π6x = \frac{\pi}{6}: f′(π6)=2cos⁡(2⋅π6)cos⁡(π6)−sin⁡(2⋅π6)sin⁡(π6)f'(\frac{\pi}{6}) = 2\cos(2 \cdot \frac{\pi}{6})\cos(\frac{\pi}{6}) - \sin(2 \cdot \frac{\pi}{6})\sin(\frac{\pi}{6}) =2cos⁡(π3)cos⁡(π6)−sin⁡(π3)sin⁡(π6)= 2\cos(\frac{\pi}{3})\cos(\frac{\pi}{6}) - \sin(\frac{\pi}{3})\sin(\frac{\pi}{6}) Using the values cos⁡(π3)=12\cos(\frac{\pi}{3})=\frac{1}{2}, cos⁡(π6)=32\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}, sin⁡(π3)=32\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}, and sin⁡(π6)=12\sin(\frac{\pi}{6})=\frac{1}{2}: =2(12)(32)−(32)(12)= 2(\frac{1}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{3}}{2})(\frac{1}{2}) =32−34=234−34=34= \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4}.

Question 10

Let ff be a differentiable function. Let k(x)=(3x2)f(x)k(x) = (3x^2)f(x).

If f(1)=2f(1) = 2 and f′(1)=−4f'(1) = -4, what is the value of k′(1)k'(1)?

  1. 00 (correct answer)
  2. −24-24
  3. −12-12
  4. 1212
Explanation: To find k′(x)k'(x), we use the Product Rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. Let u(x)=3x2u(x) = 3x^2 and v(x)=f(x)v(x) = f(x). Then u′(x)=6xu'(x) = 6x and v′(x)=f′(x)v'(x) = f'(x). Applying the rule: k′(x)=(6x)f(x)+(3x2)f′(x)k'(x) = (6x)f(x) + (3x^2)f'(x). Now, substitute x=1x=1 and the given values for f(1)f(1) and f′(1)f'(1): k′(1)=(6⋅1)f(1)+(3⋅12)f′(1)k'(1) = (6 \cdot 1)f(1) + (3 \cdot 1^2)f'(1) k′(1)=6(2)+3(−4)k'(1) = 6(2) + 3(-4) k′(1)=12−12=0k'(1) = 12 - 12 = 0.

Question 11

Let p(x)=f(x)g(x)p(x) = f(x)g(x), where ff and gg are differentiable functions.

Which of the following sets of conditions is sufficient to guarantee that p′(c)>0p'(c) > 0 for some number cc?

  1. f(c)>0,g(c)>0,f′(c)<0,g′(c)>0f(c) > 0, g(c) > 0, f'(c) < 0, g'(c) > 0
  2. f(c)<0,g(c)>0,f′(c)<0,g′(c)>0f(c) < 0, g(c) > 0, f'(c) < 0, g'(c) > 0
  3. f(c)<0,g(c)<0,f′(c)<0,g′(c)<0f(c) < 0, g(c) < 0, f'(c) < 0, g'(c) < 0 (correct answer)
  4. f(c)>0,g(c)<0,f′(c)>0,g′(c)<0f(c) > 0, g(c) < 0, f'(c) > 0, g'(c) < 0
Explanation: The derivative of p(x)p(x) is given by the Product Rule: p′(x)=f′(x)g(x)+f(x)g′(x)p'(x) = f'(x)g(x) + f(x)g'(x). We need to find the conditions that guarantee p′(c)>0p'(c) > 0. Let's analyze each choice: A) p′(c)=f′(c)g(c)+f(c)g′(c)=(negative)(positive)+(positive)(positive)=(negative)+(positive)p'(c) = f'(c)g(c) + f(c)g'(c) = (\text{negative})(\text{positive}) + (\text{positive})(\text{positive}) = (\text{negative}) + (\text{positive}). The sign is indeterminate. B) p′(c)=f′(c)g(c)+f(c)g′(c)=(negative)(positive)+(negative)(positive)=(negative)+(negative)p'(c) = f'(c)g(c) + f(c)g'(c) = (\text{negative})(\text{positive}) + (\text{negative})(\text{positive}) = (\text{negative}) + (\text{negative}). This is always negative. C) p′(c)=f′(c)g(c)+f(c)g′(c)=(negative)(negative)+(negative)(negative)=(positive)+(positive)p'(c) = f'(c)g(c) + f(c)g'(c) = (\text{negative})(\text{negative}) + (\text{negative})(\text{negative}) = (\text{positive}) + (\text{positive}). This is always positive. D) p′(c)=f′(c)g(c)+f(c)g′(c)=(positive)(negative)+(positive)(negative)=(negative)+(negative)p'(c) = f'(c)g(c) + f(c)g'(c) = (\text{positive})(\text{negative}) + (\text{positive})(\text{negative}) = (\text{negative}) + (\text{negative}). This is always negative.

Question 12

Let h(x)=exf(x)h(x) = \frac{e^x}{f(x)}. The function f(x)f(x) is defined as f(x)=xg(x)f(x) = x g(x).

If g(2)=1g(2) = 1 and g′(2)=4g'(2) = 4, what is the value of h′(2)h'(2)?

  1. −e22-\frac{e^2}{2}
  2. −7e24-\frac{7e^2}{4} (correct answer)
  3. −3e22-\frac{3e^2}{2}
  4. 11e24\frac{11e^2}{4}
Explanation: This problem tests your ability to apply both the quotient rule and product rule in combination. When you see a function that's a quotient where both numerator and denominator need differentiation, the quotient rule is your go-to tool. To find h′(2)h'(2), you need to differentiate h(x)=exf(x)h(x) = \frac{e^x}{f(x)} using the quotient rule: h′(x)=f(x)⋅(ex)′−ex⋅f′(x)[f(x)]2=f(x)⋅ex−ex⋅f′(x)[f(x)]2h'(x) = \frac{f(x) \cdot (e^x)' - e^x \cdot f'(x)}{[f(x)]^2} = \frac{f(x) \cdot e^x - e^x \cdot f'(x)}{[f(x)]^2} Since f(x)=xg(x)f(x) = x g(x), you need f′(x)f'(x) using the product rule: f′(x)=1⋅g(x)+x⋅g′(x)=g(x)+xg′(x)f'(x) = 1 \cdot g(x) + x \cdot g'(x) = g(x) + x g'(x) At x=2x = 2: f(2)=2⋅g(2)=2⋅1=2f(2) = 2 \cdot g(2) = 2 \cdot 1 = 2 and f′(2)=g(2)+2⋅g′(2)=1+2⋅4=9f'(2) = g(2) + 2 \cdot g'(2) = 1 + 2 \cdot 4 = 9 Therefore: h′(2)=2⋅e2−e2⋅922=2e2−9e24=−7e24h'(2) = \frac{2 \cdot e^2 - e^2 \cdot 9}{2^2} = \frac{2e^2 - 9e^2}{4} = \frac{-7e^2}{4} Answer B is correct. Answer A represents forgetting the product rule when differentiating f(x)f(x), using only f′(2)=g(2)=1f'(2) = g(2) = 1. Answer C likely comes from sign errors or incorrectly applying the quotient rule formula. Answer D results from reversing the quotient rule (putting ex⋅f′(x)−f(x)⋅exe^x \cdot f'(x) - f(x) \cdot e^x in the numerator instead). Remember: quotient rule problems often involve multiple differentiation rules. Work systematically through each component before substituting values, and double-check the quotient rule formula—the order matters for the sign.

Question 13

Let yy be a function of xx such that x3y2=8x^3 y^2 = 8.

What is the value of dydx\frac{dy}{dx} at the point (2,2)(2, \sqrt{2})?

  1. 324\frac{3\sqrt{2}}{4}
  2. 00
  3. −324-\frac{3\sqrt{2}}{4} (correct answer)
  4. −22-\frac{\sqrt{2}}{2}
Explanation: We use implicit differentiation with respect to xx. The left side, x3y2x^3y^2, requires the Product Rule. ddx(x3y2)=ddx(8)\frac{d}{dx}(x^3 y^2) = \frac{d}{dx}(8) (ddxx3)y2+x3(ddxy2)=0(\frac{d}{dx}x^3)y^2 + x^3(\frac{d}{dx}y^2) = 0 (3x2)y2+x3(2ydydx)=0(3x^2)y^2 + x^3(2y \frac{dy}{dx}) = 0. Now, solve for dydx\frac{dy}{dx}: 2x3ydydx=−3x2y22x^3y \frac{dy}{dx} = -3x^2y^2 dydx=−3x2y22x3y=−3y2x\frac{dy}{dx} = \frac{-3x^2y^2}{2x^3y} = \frac{-3y}{2x}. Finally, substitute the coordinates of the point (2,2)(2, \sqrt{2}) into this expression: dydx=−3(2)2(2)=−324\frac{dy}{dx} = \frac{-3(\sqrt{2})}{2(2)} = -\frac{3\sqrt{2}}{4}.

Question 14

A student is asked to find the derivative of h(x)=(x2+1)(2x−1)h(x) = (x^2+1)(2x-1). Their work is shown below. Step 1: Let f(x)=x2+1f(x) = x^2+1 and g(x)=2x−1g(x)=2x-1. Step 2: Find derivatives: f′(x)=2xf'(x)=2x and g′(x)=2g'(x)=2. Step 3: Apply product rule: h′(x)=f′(x)g′(x)h'(x) = f'(x)g'(x). Step 4: h′(x)=(2x)(2)=4xh'(x) = (2x)(2) = 4x.

In which step did the student first make an error?

  1. Step 1
  2. Step 2
  3. Step 3 (correct answer)
  4. Step 4
Explanation: The student makes their first error in Step 3. The product rule is h′(x)=f′(x)g(x)+f(x)g′(x)h'(x) = f'(x)g(x) + f(x)g'(x), not h′(x)=f′(x)g′(x)h'(x) = f'(x)g'(x). This is a very common misconception. Steps 1 and 2 are correct identifications and differentiations of the component functions. Step 4 is a correct calculation based on the incorrect formula from Step 3, so the error originates in Step 3.
  • A is incorrect because Step 1 correctly identifies the functions in the product.
  • B is incorrect because Step 2 correctly finds the derivatives of the functions from Step 1.
  • D is incorrect because while the final answer is wrong, the error was made in the previous step's formula, not in the calculation in Step 4.

Question 15

Let f(x)f(x) and g(x)g(x) be differentiable functions. Let p(x)=f(x)g(x)p(x) = f(x)g(x) and q(x)=f(x)g(x)q(x) = \frac{f(x)}{g(x)}. If f(a)=0f(a) = 0, g(a)≠0g(a) \neq 0, and f′(a)≠0f'(a) \neq 0, what is the relationship between p′(a)p'(a) and q′(a)q'(a)?

  1. p′(a)=q′(a)p'(a) = q'(a)
  2. p′(a)=−q′(a)p'(a) = -q'(a)
  3. p′(a)=[g(a)]2q′(a)p'(a) = [g(a)]^2 q'(a) (correct answer)
  4. p′(a)=g(a)q′(a)p'(a) = g(a) q'(a)
Explanation: First, find an expression for p′(a)p'(a) using the product rule: p′(x)=f′(x)g(x)+f(x)g′(x)p'(x) = f'(x)g(x) + f(x)g'(x). At x=ax=a, p′(a)=f′(a)g(a)+f(a)g′(a)p'(a) = f'(a)g(a) + f(a)g'(a). Since f(a)=0f(a)=0, this simplifies to p′(a)=f′(a)g(a)p'(a) = f'(a)g(a). Next, find an expression for q′(a)q'(a) using the quotient rule: q′(x)=f′(x)g(x)−f(x)g′(x)[g(x)]2q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}. At x=ax=a, q′(a)=f′(a)g(a)−f(a)g′(a)[g(a)]2q'(a) = \frac{f'(a)g(a) - f(a)g'(a)}{[g(a)]^2}. Since f(a)=0f(a)=0, this simplifies to q′(a)=f′(a)g(a)[g(a)]2q'(a) = \frac{f'(a)g(a)}{[g(a)]^2}. From this, we see that f′(a)g(a)=[g(a)]2q′(a)f'(a)g(a) = [g(a)]^2 q'(a). Since p′(a)=f′(a)g(a)p'(a) = f'(a)g(a), we can conclude that p′(a)=[g(a)]2q′(a)p'(a) = [g(a)]^2 q'(a).
  • A is incorrect. This would only be true if [g(a)]2=1[g(a)]^2 = 1.
  • B is incorrect as there is no basis for a negative relationship.
  • D is incorrect as it is missing a factor of g(a)g(a).

Question 16

The volume VV of a cylinder is changing over time. The volume is given by V(t)=π[r(t)]2h(t)V(t) = \pi [r(t)]^2 h(t). At a moment when the radius rr is 2 inches and the height hh is 5 inches, the radius is increasing at 0.1 in/s and the volume is constant. What is the rate of change of the height, h′(t)h'(t)?

  1. -0.5 in/s (correct answer)
  2. -0.25 in/s
  3. 0 in/s
  4. 0.25 in/s
Explanation: Differentiate V=πr2hV = \pi r^2 h with respect to tt using the product rule on r2r^2 and hh. V′(t)=π[(ddtr2)h+r2h′]V'(t) = \pi [(\frac{d}{dt}r^2)h + r^2 h']. The derivative ddtr2\frac{d}{dt}r^2 requires the chain rule: 2rr′2r r'. So, V′(t)=π[2rr′h+r2h′]V'(t) = \pi [2rr'h + r^2h']. We are given V′=0V' = 0, r=2r=2, h=5h=5, r′=0.1r'=0.1. Plugging in: 0=π[2(2)(0.1)(5)+(2)2h′]=π[2+4h′]0 = \pi [2(2)(0.1)(5) + (2)^2 h'] = \pi[2 + 4h']. Dividing by π\pi: 0=2+4h′0 = 2 + 4h'. Solving gives 4h′=−24h' = -2, so h′=−0.5h' = -0.5 in/s.

Question 17

The line tangent to the graph of f(x)f(x) at x=2x=2 is y=−x+5y = -x+5, and the line tangent to the graph of g(x)g(x) at x=2x=2 is y=3x−4y=3x-4. If p(x)=f(x)g(x)p(x) = f(x)g(x), find the slope of the line tangent to the graph of p(x)p(x) at x=2x=2.

  1. −3-3
  2. 33
  3. 77 (correct answer)
  4. 1111
Explanation: The slope of the tangent line to p(x)p(x) at x=2x=2 is p′(2)p'(2). From the tangent line information, we can deduce the values of the functions and their derivatives at x=2x=2. For f(x)f(x), the slope is f′(2)=−1f'(2) = -1 and the point is (2,−2+5)=(2,3)(2, -2+5) = (2,3), so f(2)=3f(2)=3. For g(x)g(x), the slope is g′(2)=3g'(2) = 3 and the point is (2,3(2)−4)=(2,2)(2, 3(2)-4) = (2,2), so g(2)=2g(2)=2. Now, use the product rule for p(x)p(x): p′(x)=f′(x)g(x)+f(x)g′(x)p'(x) = f'(x)g(x) + f(x)g'(x). At x=2x=2, p′(2)=f′(2)g(2)+f(2)g′(2)=(−1)(2)+(3)(3)=−2+9=7p'(2) = f'(2)g(2) + f(2)g'(2) = (-1)(2) + (3)(3) = -2 + 9 = 7.
  • A is incorrect because it is the product of the slopes, f′(2)g′(2)f'(2)g'(2), a common error.
  • B is incorrect as it may result from misassigning the function and derivative values, for example, f′(2)f(2)+g′(2)g(2)=(−1)(3)+(3)(2)=3f'(2)f(2) + g'(2)g(2) = (-1)(3)+(3)(2) = 3.
  • D is incorrect due to a sign error on f′(2)f'(2): (1)(2)+(3)(3)=11(1)(2) + (3)(3) = 11.

Question 18

The derivative of a function h(x)h(x) is given by h′(x)=2xsin⁡(x)+x2cos⁡(x)h'(x) = 2x\sin(x) + x^2\cos(x). Which of the following could be the function h(x)h(x)?

  1. h(x)=x2cos⁡(x)h(x) = x^2\cos(x)
  2. h(x)=−x2cos⁡(x)h(x) = -x^2\cos(x)
  3. h(x)=x2sin⁡(x)h(x) = x^2\sin(x) (correct answer)
  4. h(x)=2xcos⁡(x)h(x) = 2x\cos(x)
Explanation: The expression h′(x)=2xsin⁡(x)+x2cos⁡(x)h'(x) = 2x\sin(x) + x^2\cos(x) has the form of the Product Rule, f′(x)g(x)+f(x)g′(x)f'(x)g(x) + f(x)g'(x). We need to identify functions f(x)f(x) and g(x)g(x) such that their derivatives match the given terms. Let's test the option h(x)=x2sin⁡(x)h(x) = x^2\sin(x). Let f(x)=x2f(x) = x^2 and g(x)=sin⁡(x)g(x) = \sin(x). Then f′(x)=2xf'(x) = 2x and g′(x)=cos⁡(x)g'(x) = \cos(x). According to the Product Rule, the derivative of h(x)h(x) would be: h′(x)=f′(x)g(x)+f(x)g′(x)=(2x)(sin⁡(x))+(x2)(cos⁡(x))h'(x) = f'(x)g(x) + f(x)g'(x) = (2x)(\sin(x)) + (x^2)(\cos(x)). This matches the given h′(x)h'(x). Therefore, h(x)=x2sin⁡(x)h(x) = x^2\sin(x) is a possible function (any function x2sin⁡(x)+Cx^2\sin(x) + C would work, but this is the form given in the options).

Question 19

What is the equation of the tangent line to the graph of h(x)=x2ln⁡(x)h(x) = x^2 \ln(x) at the point where x=ex=e?

  1. y=3ex−2e2y = 3ex - 2e^2 (correct answer)
  2. y=2x−2e+e2y = 2x - 2e + e^2
  3. y=2ex−e2y = 2ex - e^2
  4. y=exy = ex
Explanation: First, find the point of tangency by evaluating h(e)h(e): h(e)=e2ln⁡(e)=e2(1)=e2h(e) = e^2 \ln(e) = e^2(1) = e^2. The point is (e,e2)(e, e^2). Next, find the slope of the tangent line by finding the derivative h′(x)h'(x) using the Product Rule. Let f(x)=x2f(x) = x^2 and g(x)=ln⁡(x)g(x) = \ln(x). h′(x)=f′(x)g(x)+f(x)g′(x)=(2x)(ln⁡(x))+(x2)(1x)=2xln⁡(x)+xh'(x) = f'(x)g(x) + f(x)g'(x) = (2x)(\ln(x)) + (x^2)(\frac{1}{x}) = 2x\ln(x) + x. Now, evaluate the slope at x=ex=e: h′(e)=2eln⁡(e)+e=2e(1)+e=3eh'(e) = 2e\ln(e) + e = 2e(1) + e = 3e. The equation of the tangent line in point-slope form is y−y1=m(x−x1)y - y_1 = m(x - x_1): y−e2=3e(x−e)y - e^2 = 3e(x - e) y−e2=3ex−3e2y - e^2 = 3ex - 3e^2 y=3ex−2e2y = 3ex - 2e^2.

Question 20

For which non-zero value of xx is the tangent line to the graph of f(x)=x2e−xf(x) = x^2 e^{-x} horizontal?

  1. x=−2x = -2
  2. x=−1x = -1
  3. x=1x = 1
  4. x=2x = 2 (correct answer)
Explanation: A horizontal tangent line occurs when the derivative of the function is zero. We must find f′(x)f'(x) and set it to 0. Using the Product Rule on f(x)=x2e−xf(x) = x^2 e^{-x}: f′(x)=(ddxx2)e−x+x2(ddxe−x)f'(x) = (\frac{d}{dx} x^2) e^{-x} + x^2 (\frac{d}{dx} e^{-x}) f′(x)=(2x)e−x+x2(−e−x)f'(x) = (2x)e^{-x} + x^2(-e^{-x}) f′(x)=2xe−x−x2e−xf'(x) = 2xe^{-x} - x^2e^{-x}. Now, set f′(x)=0f'(x) = 0: 2xe−x−x2e−x=02xe^{-x} - x^2e^{-x} = 0 Factor out the common term xe−xxe^{-x}: xe−x(2−x)=0xe^{-x}(2-x) = 0. Since e−xe^{-x} is never zero, the solutions are x=0x=0 or 2−x=02-x=0. This gives x=0x=0 and x=2x=2. The question asks for the non-zero value, which is x=2x=2.