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AP Calculus AB Quiz

AP Calculus AB Quiz: The Product Rule

Practice The Product Rule in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

A profit function is p(x)=(2x−1)sin⁡(3x)p(x)=(2x-1)\sin(3x)p(x)=(2x−1)sin(3x). What is p′(x)p'(x)p′(x)?

Select an answer to continue

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 AB.

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

A profit function is p(x)=(2x−1)sin⁡(3x)p(x)=(2x-1)\sin(3x)p(x)=(2x−1)sin(3x). What is p′(x)p'(x)p′(x)?

  1. 2sin⁡(3x)+(2x−1)cos⁡(3x)2\sin(3x)+(2x-1)\cos(3x)2sin(3x)+(2x−1)cos(3x)
  2. 2sin⁡(3x)+(2x−1)⋅3cos⁡(3x)2\sin(3x)+(2x-1)\cdot 3\cos(3x)2sin(3x)+(2x−1)⋅3cos(3x) (correct answer)
  3. 2sin⁡(3x)2\sin(3x)2sin(3x)
  4. (2x−1)⋅3cos⁡(3x)(2x-1)\cdot 3\cos(3x)(2x−1)⋅3cos(3x)
  5. 2+3cos⁡(3x)2+3\cos(3x)2+3cos(3x)

Explanation: p(x) = (2x - 1) sin(3x) needs the product rule: 2 sin(3x) + (2x - 1) (3 cos(3x)), incorporating the chain rule for sin(3x). Without chain rule, one might get choice A. A shortcut is omitting the chain, leading to errors. Proper method includes it. Strategy: For products with composite trig, combine product and chain rules.

Question 2

A model is k(x)=(x2+1) ln⁡(2x+1)k(x)=(x^2+1)\,\ln(2x+1)k(x)=(x2+1)ln(2x+1). What is k′(x)k'(x)k′(x)?

  1. 2xln⁡(2x+1)+2(x2+1)2x+12x\ln(2x+1)+\dfrac{2(x^2+1)}{2x+1}2xln(2x+1)+2x+12(x2+1)​ (correct answer)
  2. 2x+22x+12x+\dfrac{2}{2x+1}2x+2x+12​
  3. 2xln⁡(2x+1)2x\ln(2x+1)2xln(2x+1)
  4. (x2+1)⋅22x+1(x^2+1)\cdot\dfrac{2}{2x+1}(x2+1)⋅2x+12​
  5. ln⁡(2x+1)⋅22x+1\ln(2x+1)\cdot\dfrac{2}{2x+1}ln(2x+1)⋅2x+12​

Explanation: k(x) = (x² + 1) ln(2x + 1) uses product: 2x ln(2x + 1) + (x² + 1) (2/(2x + 1)). Chain for log. Shortcut: Missing chain, like E. Strategy: Log with linear argument needs chain.

Question 3

A rate is q(t)=(t2+2t+1) sin⁡(t2)q(t)=(t^2+2t+1)\,\sin(t^2)q(t)=(t2+2t+1)sin(t2). What is q′(t)q'(t)q′(t)?

  1. (2t+2)sin⁡(t2)(2t+2)\sin(t^2)(2t+2)sin(t2)
  2. (t2+2t+1)⋅2tcos⁡(t2)(t^2+2t+1)\cdot 2t\cos(t^2)(t2+2t+1)⋅2tcos(t2)
  3. (2t+2)sin⁡(t2)+(t2+2t+1)⋅2tcos⁡(t2)(2t+2)\sin(t^2)+(t^2+2t+1)\cdot 2t\cos(t^2)(2t+2)sin(t2)+(t2+2t+1)⋅2tcos(t2) (correct answer)
  4. (2t+2)cos⁡(t2)(2t+2)\cos(t^2)(2t+2)cos(t2)
  5. sin⁡(t2)+cos⁡(t2)\sin(t^2)+\cos(t^2)sin(t2)+cos(t2)

Explanation: q(t) = (t² + 2t + 1) sin(t²) requires product: (2t + 2) sin(t²) + (t² + 2t + 1) (2t cos(t²)). Chain for sin(t²). Error: Missing chain, like A. Strategy: Composite sin needs chain in product.

Question 4

A cost function is C(x)=(x)(x2+4)C(x)=(\sqrt{x})(x^2+4)C(x)=(x​)(x2+4). What is C′(x)C'(x)C′(x) for x>0x>0x>0?

  1. 12x+2x\dfrac{1}{2\sqrt{x}}+2x2x​1​+2x
  2. 12x(x2+4)\dfrac{1}{2\sqrt{x}}(x^2+4)2x​1​(x2+4)
  3. 12x(x2+4)+x(2x)\dfrac{1}{2\sqrt{x}}(x^2+4)+\sqrt{x}(2x)2x​1​(x2+4)+x​(2x) (correct answer)
  4. x(2x)\sqrt{x}(2x)x​(2x)
  5. (x2+4)+x(x^2+4)+\sqrt{x}(x2+4)+x​

Explanation: C(x) is a product of √x and (x² + 4), so use the product rule: (1/(2√x))(x² + 4) + √x (2x). This accounts for both parts changing with x. A tempting shortcut is to treat √x as constant or miscompute its derivative, leading to just √x (2x) as in D. Others might forget the chain rule in the square root derivative. The correct method combines them properly. Always verify if the function is a product and differentiate each factor accurately for the strategy.

Question 5

Water height is modeled by h(t)=sin⁡(t) (t2+1)h(t)=\sin(t)\,(t^2+1)h(t)=sin(t)(t2+1). What is h′(t)h'(t)h′(t)?

  1. cos⁡(t)+2t\cos(t)+2tcos(t)+2t
  2. cos⁡(t)(t2+1)+2tsin⁡(t)\cos(t)(t^2+1)+2t\sin(t)cos(t)(t2+1)+2tsin(t) (correct answer)
  3. cos⁡(t)(t2+1)\cos(t)(t^2+1)cos(t)(t2+1)
  4. sin⁡(t)(2t)\sin(t)(2t)sin(t)(2t)
  5. cos⁡(t)(t2+1)+2t\cos(t)(t^2+1)+2tcos(t)(t2+1)+2t

Explanation: Since h(t) is the product of sin(t) and (t² + 1), the product rule is required: derivative is cos(t)(t² + 1) + sin(t)(2t). This captures the rate of change from both factors. Without it, one might wrongly differentiate only the trigonometric part, yielding just cos(t)(t² + 1) as in C. Another shortcut error is adding derivatives without multiplying, like choice A. The full application avoids these pitfalls. For transferable strategy, scan for multiplication of two non-constant functions and apply the product rule to ensure complete differentiation.

Question 6

A displacement is s(t)=(t−5)cos⁡(t)s(t)=(t-5)\cos(t)s(t)=(t−5)cos(t). What is s′(t)s'(t)s′(t)?

  1. cos⁡(t)−(t−5)sin⁡(t)\cos(t)-(t-5)\sin(t)cos(t)−(t−5)sin(t) (correct answer)
  2. cos⁡(t)\cos(t)cos(t)
  3. 1−sin⁡(t)1-\sin(t)1−sin(t)
  4. cos⁡(t)−sin⁡(t)\cos(t)-\sin(t)cos(t)−sin(t)
  5. (t−5)(−sin⁡(t))(t-5)(-\sin(t))(t−5)(−sin(t))

Explanation: The displacement s(t) = (t - 5) cos(t) is a product, necessitating the product rule: 1 · cos(t) + (t - 5) (-sin(t)), or cos(t) - (t - 5) sin(t). This reflects contributions from both linear and trigonometric terms. A common error is omitting the derivative of the linear part, resulting in just (t - 5)(-sin(t)) as in E. Some might add without the negative sign. Proper application ensures accuracy. To decide, recognize products involving trig functions and apply the rule consistently.

Question 7

A growth model is G(x)=(x2+1) e2xG(x)=(x^2+1)\,e^{2x}G(x)=(x2+1)e2x. What is G′(x)G'(x)G′(x)?

  1. (2x)e2x(2x)e^{2x}(2x)e2x
  2. (x2+1)e2x+2e2x(x^2+1)e^{2x}+2e^{2x}(x2+1)e2x+2e2x
  3. 2xe2x+(x2+1)⋅2e2x2x e^{2x}+(x^2+1)\cdot 2e^{2x}2xe2x+(x2+1)⋅2e2x (correct answer)
  4. (2x+1)e2x(2x+1)e^{2x}(2x+1)e2x
  5. 2e2x2e^{2x}2e2x

Explanation: G(x) = (x² + 1) e^{2x} requires the product rule due to the polynomial and exponential product: (2x) e^{2x} + (x² + 1) (2 e^{2x}). Note the chain rule for the exponential derivative. A shortcut mistake is forgetting the chain rule, leading to (2x + 1) e^{2x} as in D. Others might omit one term entirely. The full rule combines them. Strategy: Identify exponential products and incorporate chain rule where needed in differentiation.

Question 8

A temperature function is T(t)=ln⁡(t) (t3+2)T(t)=\ln(t)\,(t^3+2)T(t)=ln(t)(t3+2). What is T′(t)T'(t)T′(t) for t>0t>0t>0?

  1. 1t(t3+2)+ln⁡(t)(3t2)\dfrac{1}{t}(t^3+2)+\ln(t)(3t^2)t1​(t3+2)+ln(t)(3t2) (correct answer)
  2. 1t+3t2\dfrac{1}{t}+3t^2t1​+3t2
  3. ln⁡(t)+t3+2\ln(t)+t^3+2ln(t)+t3+2
  4. t3+2t\dfrac{t^3+2}{t}tt3+2​
  5. ln⁡(t)(3t2)\ln(t)(3t^2)ln(t)(3t2)

Explanation: T(t) = ln(t) (t³ + 2) is a product of logarithmic and polynomial functions, so apply the product rule: (1/t)(t³ + 2) + ln(t)(3t²). This ensures both rates are included. A tempting error is ignoring the log derivative, yielding just ln(t)(3t²) as in E. Some might confuse with quotient rule. Correct use avoids this. For strategy, check for log-polynomial products and systematically differentiate each.

Question 9

A function is J(x)=(x2+4x) sec⁡2(x)J(x)=(x^2+4x)\,\sec^2(x)J(x)=(x2+4x)sec2(x). What is J′(x)J'(x)J′(x)?

  1. (2x+4)sec⁡2(x)+2(x2+4x)sec⁡2(x)tan⁡(x)(2x+4)\sec^2(x)+2(x^2+4x)\sec^2(x)\tan(x)(2x+4)sec2(x)+2(x2+4x)sec2(x)tan(x) (correct answer)
  2. (2x+4)sec⁡2(x)(2x+4)\sec^2(x)(2x+4)sec2(x)
  3. (x2+4x)⋅2sec⁡2(x)tan⁡(x)(x^2+4x)\cdot 2\sec^2(x)\tan(x)(x2+4x)⋅2sec2(x)tan(x)
  4. (2x+4)+2sec⁡2(x)tan⁡(x)(2x+4)+2\sec^2(x)\tan(x)(2x+4)+2sec2(x)tan(x)
  5. (2x+4)sec⁡(x)(2x+4)\sec(x)(2x+4)sec(x)

Explanation: J(x) is the product of (x² + 4x) and sec²(x), so the product rule gives (2x + 4)sec²(x) + (x² + 4x)(2 sec²(x) tan(x)). This is required due to the multiplication of a polynomial and a trigonometric function with a power. The derivative of sec²(x) uses the chain rule: 2 sec(x) · (sec(x) tan(x)). An incorrect shortcut might be to forget the chain rule or omit one product term, like in choice C. Apply the rule carefully to both parts. When faced with trigonometric powers multiplied by other functions, confirm it's a product and use the rule with any necessary chain applications.

Question 10

A volume rate is V(x)=(x2−3x) ln⁡(x2+1)V(x)=(x^2-3x)\,\ln(x^2+1)V(x)=(x2−3x)ln(x2+1). What is V′(x)V'(x)V′(x)?

  1. (2x−3)ln⁡(x2+1)+2x(x2−3x)x2+1(2x-3)\ln(x^2+1)+\dfrac{2x(x^2-3x)}{x^2+1}(2x−3)ln(x2+1)+x2+12x(x2−3x)​ (correct answer)
  2. (2x−3)ln⁡(x2+1)(2x-3)\ln(x^2+1)(2x−3)ln(x2+1)
  3. (x2−3x)⋅2xx2+1(x^2-3x)\cdot\dfrac{2x}{x^2+1}(x2−3x)⋅x2+12x​
  4. (2x−3)+2xx2+1(2x-3)+\dfrac{2x}{x^2+1}(2x−3)+x2+12x​
  5. (2x−3)ln⁡(x2)+1(2x-3)\ln(x^2)+1(2x−3)ln(x2)+1

Explanation: V(x) = (x² - 3x) ln(x² + 1) uses product: (2x - 3) ln(x² + 1) + (x² - 3x) (2x/(x² + 1)). Includes chain for log. Shortcut: Omitting second term, like B. Strategy: Log products with composites require chain in rule.

Question 11

A function is g(t)=(t)(cos⁡(2t))g(t)=(\sqrt{t})(\cos(2t))g(t)=(t​)(cos(2t)). What is g′(t)g'(t)g′(t) for t>0t>0t>0?

  1. 12tcos⁡(2t)−2tsin⁡(2t)\dfrac{1}{2\sqrt{t}}\cos(2t)-2\sqrt{t}\sin(2t)2t​1​cos(2t)−2t​sin(2t) (correct answer)
  2. 12tcos⁡(2t)−tsin⁡(2t)\dfrac{1}{2\sqrt{t}}\cos(2t)-\sqrt{t}\sin(2t)2t​1​cos(2t)−t​sin(2t)
  3. 12t−2sin⁡(2t)\dfrac{1}{2\sqrt{t}}-2\sin(2t)2t​1​−2sin(2t)
  4. 12tcos⁡(2t)\dfrac{1}{2\sqrt{t}}\cos(2t)2t​1​cos(2t)
  5. −2tsin⁡(2t)-2\sqrt{t}\sin(2t)−2t​sin(2t)

Explanation: g(t) = √t cos(2t) is product: (1/(2√t)) cos(2t) + √t (-2 sin(2t)), or (1/(2√t)) cos(2t) - 2 √t sin(2t). Chain for cos(2t). But wait, verifying: derivative of cos(2t) is -sin(2t) * 2 = -2 sin(2t), yes. Marked A is correct. Tempting to miss the 2 in chain, like B. Strategy: Double chain in products.

Question 12

A population model is P(t)=(t2+3t)(et)P(t)=(t^2+3t)(e^t)P(t)=(t2+3t)(et). What is P′(t)P'(t)P′(t)?

  1. (2t+3)et(2t+3)e^t(2t+3)et
  2. (t2+3t)et(t^2+3t)e^t(t2+3t)et
  3. (t2+3t+2t+3)et(t^2+3t+2t+3)e^t(t2+3t+2t+3)et
  4. (2t+3)et+(t2+3t)et(2t+3)e^t+(t^2+3t)e^t(2t+3)et+(t2+3t)et (correct answer)
  5. (2t+3)et+t2+3t(2t+3)e^t+t^2+3t(2t+3)et+t2+3t

Explanation: The product rule is essential here because P(t) is the product of two functions: a polynomial (t² + 3t) and an exponential (e^t), both depending on t. To find P'(t), differentiate each part and apply the formula: if f(t) = u(t)v(t), then f'(t) = u'(t)v(t) + u(t)v'(t). Here, u(t) = t² + 3t with u'(t) = 2t + 3, and v(t) = e^t with v'(t) = e^t, yielding (2t + 3)e^t + (t² + 3t)e^t. A tempting incorrect shortcut is to differentiate only one factor, like treating it as just the polynomial times e^t without adding the second term, which might lead to choice A or B. Another error could be combining terms prematurely without proper application. When deciding on differentiation rules, identify if the function is a clear product of two variable-dependent parts, and apply the product rule systematically to avoid omitting terms.

Question 13

A revenue function is R(x)=(x^3-1)\ln(x). What is R′(x)R'(x)R′(x) for x>0x>0x>0?

  1. 3x2ln⁡(x)+x3−1x3x^2\ln(x)+\dfrac{x^3-1}{x}3x2ln(x)+xx3−1​ (correct answer)
  2. 3x2+1x3x^2+\dfrac{1}{x}3x2+x1​
  3. (x3−1)⋅1x(x^3-1)\cdot\dfrac{1}{x}(x3−1)⋅x1​
  4. 3x2ln⁡(x)3x^2\ln(x)3x2ln(x)
  5. (3x2−1)ln⁡(x)(3x^2-1)\ln(x)(3x2−1)ln(x)

Explanation: The product rule applies because R(x) is the product of (x³ - 1) and ln(x), both functions of x. The derivative is u'(x)v(x) + u(x)v'(x), where u(x) = x³ - 1, u'(x) = 3x², v(x) = ln(x), v'(x) = 1/x, resulting in 3x² ln(x) + (x³ - 1)(1/x). This simplifies to the given expression in A, but the rule ensures both terms are included. A common mistake is forgetting the second term, leading to just 3x² ln(x) as in D, or misapplying the derivative of ln(x). Some might incorrectly treat it as a quotient. To decide, check for products of differentiable functions and always compute both contributions to the derivative.

Question 14

A signal is S(t)=(e−t)(sin⁡t)S(t)=(e^{-t})(\sin t)S(t)=(e−t)(sint). What is S′(t)S'(t)S′(t)?

  1. e−tcos⁡te^{-t}\cos te−tcost
  2. −e−tsin⁡t-e^{-t}\sin t−e−tsint
  3. −e−tsin⁡t+e−tcos⁡t-e^{-t}\sin t+e^{-t}\cos t−e−tsint+e−tcost (correct answer)
  4. e−t(sin⁡t+cos⁡t)e^{-t}(\sin t+\cos t)e−t(sint+cost)
  5. −e−t(sin⁡t+cos⁡t)-e^{-t}(\sin t+\cos t)−e−t(sint+cost)

Explanation: S(t) = e−tsin⁡te^{-t} \sin te−tsint is a product: (−e−tsin⁡t)+e−tcos⁡t(-e^{-t} \sin t) + e^{-t} \cos t(−e−tsint)+e−tcost, or −e−tsin⁡t+e−tcos⁡t-e^{-t} \sin t + e^{-t} \cos t−e−tsint+e−tcost. Note chain for exponential. Error: Forgetting chain, like B. Rule with chain is correct. Strategy: Exponential-trig products need both rules.

Question 15

A position is x(t)=(t2) sec⁡(t)x(t)=(t^2)\,\sec(t)x(t)=(t2)sec(t). What is dxdt\dfrac{dx}{dt}dtdx​?

  1. 2tsec⁡(t)2t\sec(t)2tsec(t)
  2. t2sec⁡(t)tan⁡(t)t^2\sec(t)\tan(t)t2sec(t)tan(t)
  3. 2tsec⁡(t)+t2sec⁡(t)tan⁡(t)2t\sec(t)+t^2\sec(t)\tan(t)2tsec(t)+t2sec(t)tan(t) (correct answer)
  4. 2t+t2tan⁡(t)2t+t^2\tan(t)2t+t2tan(t)
  5. sec⁡(t)+tan⁡(t)\sec(t)+\tan(t)sec(t)+tan(t)

Explanation: x(t) = t² sec(t) requires product: 2t sec(t) + t² sec(t) tan(t). Captures both. Error: Forgetting trig derivative, like A. Strategy: Secant products need trig derivatives in rule.

Question 16

A rate is r(t)=(t4+1) cos⁡(t)r(t)=(t^4+1)\,\cos(t)r(t)=(t4+1)cos(t). What is r′(t)r'(t)r′(t)?

  1. 4t3cos⁡(t)−(t4+1)sin⁡(t)4t^3\cos(t)-(t^4+1)\sin(t)4t3cos(t)−(t4+1)sin(t) (correct answer)
  2. 4t3−sin⁡(t)4t^3-\sin(t)4t3−sin(t)
  3. 4t3cos⁡(t)4t^3\cos(t)4t3cos(t)
  4. (t4+1)(−sin⁡(t))(t^4+1)(-\sin(t))(t4+1)(−sin(t))
  5. (4t3+1)cos⁡(t)(4t^3+1)\cos(t)(4t3+1)cos(t)

Explanation: r(t) = (t⁴ + 1) cos(t) is a product: (4t³) cos(t) + (t⁴ + 1) (-sin(t)). This is the full derivative. Tempting to forget the negative, yielding something like E. Others omit polynomial derivative. Rule prevents this. Strategy: Recognize cos products and include signs carefully.

Question 17

A model is H(x)=(x3) cos⁡(x2)H(x)=(x^3)\,\cos(x^2)H(x)=(x3)cos(x2). What is H′(x)H'(x)H′(x)?

  1. 3x2cos⁡(x2)−2x4sin⁡(x2)3x^2\cos(x^2)-2x^4\sin(x^2)3x2cos(x2)−2x4sin(x2) (correct answer)
  2. 3x2cos⁡(x2)−x3sin⁡(x2)3x^2\cos(x^2)-x^3\sin(x^2)3x2cos(x2)−x3sin(x2)
  3. 3x2cos⁡(x2)3x^2\cos(x^2)3x2cos(x2)
  4. −2xsin⁡(x2)-2x\sin(x^2)−2xsin(x2)
  5. x3(−sin⁡(x2))x^3(-\sin(x^2))x3(−sin(x2))

Explanation: The product rule states that the derivative of a product of two functions, f(x) and g(x), is f'(x)g(x) + f(x)g'(x). Here, H(x) is the product of x³ and cos(x²), so the product rule is required to differentiate it correctly. First, differentiate x³ to get 3x², multiply by cos(x²), then add x³ times the derivative of cos(x²), which involves the chain rule to yield -sin(x²) · 2x. A tempting incorrect shortcut might be to treat it as a simple power or forget the chain rule in the second term, leading to errors like choice B. Instead, always apply the product rule fully and compute each derivative accurately. For any product of functions, identify the two parts, compute their derivatives separately, and combine using the product rule formula to ensure correctness.

Question 18

A function is d(x)=(x2−1) arcsin⁡(x)d(x)=(x^2-1)\,\arcsin(x)d(x)=(x2−1)arcsin(x). What is d′(x)d'(x)d′(x)?

  1. 2xarcsin⁡(x)+x2−11−x22x\arcsin(x)+\dfrac{x^2-1}{\sqrt{1-x^2}}2xarcsin(x)+1−x2​x2−1​ (correct answer)
  2. 2x+11−x22x+\dfrac{1}{\sqrt{1-x^2}}2x+1−x2​1​
  3. 2xarcsin⁡(x)2x\arcsin(x)2xarcsin(x)
  4. (x2−1)⋅11−x2(x^2-1)\cdot\dfrac{1}{\sqrt{1-x^2}}(x2−1)⋅1−x2​1​
  5. (2x−1)arcsin⁡(x)(2x-1)\arcsin(x)(2x−1)arcsin(x)

Explanation: Differentiate d(x) = (x² - 1) arcsin(x) with product rule: 2x arcsin(x) + (x² - 1)(1/√(1 - x²)). Required for product of polynomial and inverse trig. Arcsin derivative is 1/√(1 - x²). Incorrect shortcut: omitting product, like choice D. Include all terms. Identify products with inverse functions and apply rule consistently.

Question 19

A rate model uses w(x)=(2x+1) 3xw(x)=(2x+1)\,3^{x}w(x)=(2x+1)3x. What is w′(x)w'(x)w′(x)?

  1. 2⋅3x2\cdot 3^{x}2⋅3x
  2. (2x+1) 3xln⁡3(2x+1)\,3^{x}\ln 3(2x+1)3xln3
  3. 2+3xln⁡32+3^{x}\ln 32+3xln3
  4. 2⋅3x+(2x+1) 3xln⁡32\cdot 3^{x}+(2x+1)\,3^{x}\ln 32⋅3x+(2x+1)3xln3 (correct answer)
  5. (2x+1) 3x(2x+1)\,3^{x}(2x+1)3x

Explanation: The function w(x)=(2x+1)⋅3xw(x) = (2x+1) \cdot 3^xw(x)=(2x+1)⋅3x requires the product rule for a product involving an exponential function. Let u=2x+1u = 2x+1u=2x+1 with u′=2u' = 2u′=2, and v=3xv = 3^xv=3x with v′=3xln⁡3v' = 3^x \ln 3v′=3xln3. Applying the product rule: w′(x)=2⋅3x+(2x+1)⋅3xln⁡3w'(x) = 2 \cdot 3^x + (2x+1) \cdot 3^x \ln 3w′(x)=2⋅3x+(2x+1)⋅3xln3. This gives us w′(x)=2⋅3x+(2x+1)3xln⁡3w'(x) = 2 \cdot 3^x + (2x+1)3^x \ln 3w′(x)=2⋅3x+(2x+1)3xln3. A common mistake with exponential functions like 3x3^x3x is forgetting the chain rule factor of ln⁡3\ln 3ln3. Remember that (ax)′=axln⁡a(a^x)' = a^x \ln a(ax)′=axlna for any positive constant a≠1a \neq 1a=1. When products involve exponential functions, use the product rule just as with any other functions, but be careful with the exponential derivatives.

Question 20

A model uses P(t)=(1+t)ln⁡tP(t)=(1+t)\ln tP(t)=(1+t)lnt for t>0t>0t>0. What is P′(t)P'(t)P′(t)?

  1. ln⁡t+1+tt\ln t+\frac{1+t}{t}lnt+t1+t​ (correct answer)
  2. 1+tt\frac{1+t}{t}t1+t​
  3. 1+1t1+\frac{1}{t}1+t1​
  4. ln⁡t\ln tlnt
  5. (1+t)⋅1t+1(1+t)\cdot\frac{1}{t}+1(1+t)⋅t1​+1

Explanation: This function P(t)=(1+t)ln⁡tP(t) = (1+t)\ln tP(t)=(1+t)lnt is a product of two functions, so we must use the product rule. Let u=1+tu = 1+tu=1+t with u′=1u' = 1u′=1, and v=ln⁡tv = \ln tv=lnt with v′=1tv' = \frac{1}{t}v′=t1​. The product rule gives us: P′(t)=1⋅ln⁡t+(1+t)⋅1tP'(t) = 1 \cdot \ln t + (1+t) \cdot \frac{1}{t}P′(t)=1⋅lnt+(1+t)⋅t1​. Simplifying: P′(t)=ln⁡t+1+ttP'(t) = \ln t + \frac{1+t}{t}P′(t)=lnt+t1+t​. A common error would be to differentiate each part separately and add them as 1+1t1 + \frac{1}{t}1+t1​, which misses the crucial cross-terms from the product rule. The key insight is recognizing that even simple-looking products like (1+t)ln⁡t(1+t)\ln t(1+t)lnt require the full product rule machinery. Always check: are two variable expressions being multiplied? If yes, use the product rule.