AP Calculus BC Quiz: Chain Rule
20 questions · exam conditions
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Chain RuleQuestion 1 of 20

A tank's temperature is T(t)=(3t2−5t+1)4T(t)=\big(3t^2-5t+1\big)^4 (°C). What is T′(t)T'(t)?

4(3t2−5t+1)34(3t^2-5t+1)^3
12t(3t2−5t+1)412t(3t^2-5t+1)^4
4(6t−5)(3t2−5t+1)34(6t-5)(3t^2-5t+1)^3
(6t−5)4(6t-5)^4
(6t−5)(3t2−5t+1)4(6t-5)(3t^2-5t+1)^4
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AP Calculus BC Quiz

AP Calculus BC Quiz: Chain Rule

Practice Chain Rule in AP Calculus BC 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 Chain 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.

All questions

Question 1

A tank's temperature is T(t)=(3t2−5t+1)4T(t)=\big(3t^2-5t+1\big)^4 (°C). What is T′(t)T'(t)?

  1. 4(3t2−5t+1)34(3t^2-5t+1)^3
  2. 12t(3t2−5t+1)412t(3t^2-5t+1)^4
  3. 4(6t−5)(3t2−5t+1)34(6t-5)(3t^2-5t+1)^3 (correct answer)
  4. (6t−5)4(6t-5)^4
  5. (6t−5)(3t2−5t+1)4(6t-5)(3t^2-5t+1)^4
Explanation: This problem requires the chain rule to differentiate the composite function T(t) = (3t² - 5t + 1)⁴. The outer function is u⁴ and the inner function is u = 3t² - 5t + 1. By the chain rule, we differentiate the outer function to get 4u³, then multiply by the derivative of the inner function, which is 6t - 5. This gives us T'(t) = 4(3t² - 5t + 1)³ · (6t - 5) = 4(6t - 5)(3t² - 5t + 1)³. Choice A incorrectly omits the derivative of the inner function, treating it as if the chain rule doesn't apply. To recognize when to use the chain rule, look for a function raised to a power or any composite structure where one function is inside another.

Question 2

A model for pressure is P(x)=(x2+1x)5P(x)=\big(\frac{x^2+1}{x}\big)^5; what is P′(x)P'(x)?

  1. 5(x2+1x)4(1−1x2)5\left(\frac{x^2+1}{x}\right)^4\left(1-\frac{1}{x^2}\right) (correct answer)
  2. 5(x2+1x)4(1+1x2)5\left(\frac{x^2+1}{x}\right)^4\left(1+\frac{1}{x^2}\right)
  3. 5(x2+1x)5(1−1x2)5\left(\frac{x^2+1}{x}\right)^5\left(1-\frac{1}{x^2}\right)
  4. (x2+1x)4(1−1x2)\left(\frac{x^2+1}{x}\right)^4\left(1-\frac{1}{x^2}\right)
  5. 5(x2+1x)45\left(\frac{x^2+1}{x}\right)^4
Explanation: To differentiate P(x)=(x2+1x)5P(x)=(\frac{x^2+1}{x})^5, we use the chain rule. The outer function is u5u^5 with derivative 5u45u^4, and the inner function is u=x2+1x=x+1xu=\frac{x^2+1}{x}=x+\frac{1}{x} with derivative 1−1x21-\frac{1}{x^2}. Applying the chain rule gives P′(x)=5(x2+1x)4(1−1x2)P'(x)=5(\frac{x^2+1}{x})^4(1-\frac{1}{x^2}). Choice B has the wrong sign in the inner derivative. When the inner function is a quotient, simplify it first if possible, then differentiate—this often reveals a cleaner derivative.

Question 3

A tank's temperature is modeled by T(t)=(3t2−5t+1)4T(t)=\big(3t^2-5t+1\big)^4; what is T′(t)T'(t)?

  1. 4(3t2−5t+1)34(3t^2-5t+1)^3
  2. 12t−512t-5
  3. 4(3t2−5t+1)3(6t−5)4(3t^2-5t+1)^3(6t-5) (correct answer)
  4. (3t2−5t+1)4(6t−5)(3t^2-5t+1)^4(6t-5)
  5. 4(6t−5)3(3t2−5t+1)4(6t-5)^3(3t^2-5t+1)
Explanation: This problem requires the chain rule to differentiate T(t)=(3t2−5t+1)4T(t)=(3t^2-5t+1)^4. The outer function is u4u^4 and the inner function is u=3t2−5t+1u=3t^2-5t+1. Using the chain rule, we differentiate the outer function to get 4u34u^3, then multiply by the derivative of the inner function, which is 6t−56t-5. This gives us T′(t)=4(3t2−5t+1)3(6t−5)T'(t)=4(3t^2-5t+1)^3(6t-5). Choice B incorrectly gives only the derivative of the inner function without applying the chain rule. When you see a composite function like (expression)n(expression)^n, always identify the inner expression first, then multiply the power rule result by the inner derivative.

Question 4

If T(t)=(3t2−5t+1)4T(t)=\big(3t^2-5t+1\big)^4 models tank pressure, what is T′(t)T'(t)?

  1. 4(3t2−5t+1)34\big(3t^2-5t+1\big)^3
  2. (12t−20)(3t2−5t+1)4\big(12t-20\big)\big(3t^2-5t+1\big)^4
  3. 4(12t−5)(3t2−5t+1)34\big(12t-5\big)\big(3t^2-5t+1\big)^3
  4. 4(6t−5)(3t2−5t+1)34\big(6t-5\big)\big(3t^2-5t+1\big)^3 (correct answer)
  5. 16(3t2−5t+1)316\big(3t^2-5t+1\big)^3
Explanation: This problem requires the chain rule to differentiate a composite function. The function T(t) = (3t² - 5t + 1)⁴ has an outer function f(u) = u⁴ and an inner function u = 3t² - 5t + 1. By the chain rule, T'(t) = f'(u) · u'(t) = 4u³ · (6t - 5) = 4(3t² - 5t + 1)³ · (6t - 5). A common error would be to forget the inner derivative and choose option A, which only shows 4(3t² - 5t + 1)³. When applying the chain rule, always identify both the outer and inner functions, differentiate each separately, then multiply the results together.

Question 5

A tank's temperature is modeled by T(t)=(3t2−5t+4)7T(t)=\big(3t^2-5t+4\big)^7. What is T′(t)T'(t)?

  1. 7(3t2−5t+4)67\big(3t^2-5t+4\big)^6
  2. (21t−5)(3t2−5t+4)6\big(21t-5\big)\big(3t^2-5t+4\big)^6
  3. 7(6t−5)(3t2−5t+4)77\big(6t-5\big)\big(3t^2-5t+4\big)^7
  4. 7(6t−5)(3t2−5t+4)67\big(6t-5\big)\big(3t^2-5t+4\big)^6 (correct answer)
  5. (6t−5)(3t2−5t+4)7\big(6t-5\big)\big(3t^2-5t+4\big)^7
Explanation: This problem requires the chain rule to differentiate a composite function. The function T(t) = (3t² - 5t + 4)⁷ has an outer function f(u) = u⁷ and an inner function u = 3t² - 5t + 4. By the chain rule, T'(t) = f'(u) · u'(t) = 7u⁶ · (6t - 5) = 7(3t² - 5t + 4)⁶ · (6t - 5). A common error is forgetting to multiply by the derivative of the inner function, which would give answer A: 7(3t² - 5t + 4)⁶. To recognize when to use the chain rule, look for a function nested inside another function—here, a polynomial inside a power function.

Question 6

A sensor's temperature is modeled by T(t)=(3t2−5t+1)4T(t)=\big(3t^2-5t+1\big)^4; what is T′(t)T'(t)?

  1. 4(3t2−5t+1)34\big(3t^2-5t+1\big)^3
  2. (12t−5)(3t2−5t+1)4\big(12t-5\big)\big(3t^2-5t+1\big)^4
  3. 4(3t2−5t+1)3(6t−5)4\big(3t^2-5t+1\big)^3\big(6t-5\big) (correct answer)
  4. (6t−5)4\big(6t-5\big)^4
  5. 12t2−20t+412t^2-20t+4
Explanation: This problem requires the chain rule to differentiate a composite function. The outer function is raising to the fourth power, while the inner function is the quadratic 3t² - 5t + 1. To differentiate, first take the derivative of the outer function, which is 4 times the inner function raised to the third power. Then, multiply by the derivative of the inner function, which is 6t - 5, resulting in 4(3t² - 5t + 1)³(6t - 5). A tempting distractor like choice A forgets to multiply by the inner derivative, leading to an incomplete answer. Always look for compositions where a function is plugged into another, such as a polynomial inside a power, to recognize when to apply the chain rule.

Question 7

The position of a cart is s(t)=(2t−7)5s(t)=\big(2t-7\big)^5 meters. What is s′(t)s'(t)?

  1. 5(2t−7)45(2t-7)^4
  2. 10(2t−7)410(2t-7)^4 (correct answer)
  3. 5(2)(2t−7)55(2)(2t-7)^5
  4. 10(2t−7)510(2t-7)^5
  5. (2t−7)4(2t-7)^4
Explanation: This problem requires the chain rule to differentiate s(t) = (2t - 7)⁵. The outer function is u⁵ where u = 2t - 7 is the inner function. Applying the chain rule, we first differentiate the outer function to get 5u⁴, then multiply by the derivative of the inner function, which is 2. Therefore, s'(t) = 5(2t - 7)⁴ · 2 = 10(2t - 7)⁴. Choice A incorrectly forgets to multiply by the derivative of the inner function (2), which is a common chain rule error. Remember that the chain rule always requires multiplying by the derivative of the inner function, even when that derivative is a constant.

Question 8

A model uses P(x)=(1−x3)−2P(x)=\big(1-x^3\big)^{-2} for pressure. What is P′(x)P'(x)?

  1. −2(1−x3)−3-2(1-x^3)^{-3}
  2. 6x2(1−x3)−36x^2(1-x^3)^{-3} (correct answer)
  3. −6x2(1−x3)−3-6x^2(1-x^3)^{-3}
  4. 6x2(1−x3)−26x^2(1-x^3)^{-2}
  5. −2(1−x3)−2-2(1-x^3)^{-2}
Explanation: This problem requires the chain rule to differentiate P(x) = (1 - x³)⁻². The outer function is u⁻² and the inner function is u = 1 - x³. By the chain rule, we differentiate the outer function to get -2u⁻³, then multiply by the derivative of the inner function, which is -3x². This gives us P'(x) = -2(1 - x³)⁻³ · (-3x²) = 6x²(1 - x³)⁻³. Choice C incorrectly includes a negative sign, likely from mishandling the product of two negative values. When applying the chain rule with negative exponents, carefully track the signs: negative times negative equals positive.

Question 9

A wave height is H(x)=sin⁡ ⁣(2x+3)H(x)=\sin\!\big(\sqrt{2x+3}\big); what is H′(x)H'(x)?

  1. cos⁡ ⁣(2x+3)\cos\!\big(\sqrt{2x+3}\big)
  2. cos⁡ ⁣(2x+3)2x+3\frac{\cos\!\big(\sqrt{2x+3}\big)}{\sqrt{2x+3}} (correct answer)
  3. cos⁡ ⁣(2x+3)22x+3\frac{\cos\!\big(\sqrt{2x+3}\big)}{2\sqrt{2x+3}}
  4. sin⁡ ⁣(2x+3)2x+3\frac{\sin\!\big(\sqrt{2x+3}\big)}{\sqrt{2x+3}}
  5. cos⁡ ⁣(2x+3)2x+3\frac{\cos\!\big(2x+3\big)}{\sqrt{2x+3}}
Explanation: For H(x)=sin⁡(2x+3)H(x)=\sin(\sqrt{2x+3}), we apply the chain rule. The outer function is sin⁡(u)\sin(u) with derivative cos⁡(u)\cos(u), and the inner function is u=2x+3=(2x+3)1/2u=\sqrt{2x+3}=(2x+3)^{1/2} with derivative 12(2x+3)−1/2⋅2=12x+3\frac{1}{2}(2x+3)^{-1/2}\cdot 2=\frac{1}{\sqrt{2x+3}}. The chain rule gives H′(x)=cos⁡(2x+3)⋅12x+3H'(x)=\cos(\sqrt{2x+3})\cdot\frac{1}{\sqrt{2x+3}}. Choice C incorrectly includes an extra factor of 12\frac{1}{2}. When you have nested functions like sin⁡(expression)\sin(\sqrt{expression}), work from outside to inside, carefully tracking each derivative in the chain.

Question 10

A signal's intensity is I(x)=an(5x3−x)I(x)= an\big(5x^3-x\big); what is I′(x)I'(x)?

  1. an(5x3−x)(15x2−1) an(5x^3-x)\big(15x^2-1\big)
  2. (15x2−1)an2(5x3−x)\big(15x^2-1\big) an^2(5x^3-x)
  3. (15x2−1)(1+an2(5x3−x))\big(15x^2-1\big)\big(1+ an^2(5x^3-x)\big) (correct answer)
  4. 1+an2(5x3−x)1+ an^2(5x^3-x)
  5. (15x2−1)(1+an2x)\big(15x^2-1\big)\big(1+ an^2 x\big)
Explanation: To find I′(x)I'(x) for I(x)=tan⁡(5x3−x)I(x)=\tan(5x^3-x), we need the chain rule. The outer function is tan⁡(u)\tan(u) with derivative sec⁡2(u)=1+tan⁡2(u)\sec^2(u)=1+\tan^2(u), and the inner function is u=5x3−xu=5x^3-x with derivative 15x2−115x^2-1. Applying the chain rule gives I′(x)=(15x2−1)(1+tan⁡2(5x3−x))I'(x)=(15x^2-1)(1+\tan^2(5x^3-x)). Choice A incorrectly uses tan⁡\tan instead of sec⁡2\sec^2 as the derivative of tangent. Remember that ddx[tan⁡(u)]=sec⁡2(u)⋅u′=(1+tan⁡2(u))⋅u′\frac{d}{dx}[\tan(u)]=\sec^2(u)\cdot u'=(1+\tan^2(u))\cdot u' when differentiating composite trigonometric functions.

Question 11

If g(x)=tan⁡ ⁣(πx2)g(x)=\tan\!\big(\pi x^2\big) models an angle, what is g′(x)g'(x)?

  1. sec⁡2(πx2)\sec^2(\pi x^2)
  2. 2xsec⁡2(πx2)2x\sec^2(\pi x^2)
  3. 2πxsec⁡2(πx2)2\pi x\sec^2(\pi x^2) (correct answer)
  4. πx2sec⁡2(πx2)\pi x^2\sec^2(\pi x^2)
  5. 2πsec⁡2(πx2)2\pi\sec^2(\pi x^2)
Explanation: This problem requires the chain rule to differentiate a composite trigonometric function. The function g(x) = tan(πx²) has an outer function f(u) = tan(u) and an inner function u = πx². By the chain rule, g'(x) = sec²(u) · u'(x) = sec²(πx²) · 2πx = 2πx sec²(πx²). Option B shows 2x sec²(πx²), forgetting to include the π factor from differentiating πx². When the inner function contains a constant multiplier, that constant must be included in the final derivative through the chain rule multiplication.

Question 12

A cost model is C(x)=9x2+4x+1C(x)=\sqrt{9x^2+4x+1}. What is C′(x)C'(x)?

  1. 18x+49x2+4x+1\dfrac{18x+4}{\sqrt{9x^2+4x+1}}
  2. 18x+429x2+4x+1\dfrac{18x+4}{2\sqrt{9x^2+4x+1}} (correct answer)
  3. 129x2+4x+1\dfrac{1}{2\sqrt{9x^2+4x+1}}
  4. 9x2+4x+129x2+4x+1\dfrac{9x^2+4x+1}{2\sqrt{9x^2+4x+1}}
  5. 18x+42(9x2+4x+1)\dfrac{18x+4}{2\left(9x^2+4x+1\right)}
Explanation: This problem requires the chain rule to differentiate C(x) = √(9x² + 4x + 1) = (9x² + 4x + 1)^(1/2). The chain rule tells us to multiply the derivative of the outer function by the derivative of the inner function. The outer function is u^(1/2) with derivative (1/2)u^(-1/2), and the inner function is u = 9x² + 4x + 1 with derivative 18x + 4. Applying the chain rule: C'(x) = (1/2)(9x² + 4x + 1)^(-1/2) · (18x + 4) = (18x + 4)/(2√(9x² + 4x + 1)). Choice C incorrectly gives 1/(2√(9x² + 4x + 1)), forgetting to multiply by the derivative of the inner function (18x + 4). When differentiating square roots, remember to apply the chain rule: the derivative includes both the power rule for the outer function and the derivative of what's inside.

Question 13

A growth factor is G(x)=ln⁡ ⁣(x2+6x+10)G(x)=\ln\!\left(x^2+6x+10\right). What is G′(x)G'(x)?

  1. 1x2+6x+10\dfrac{1}{x^2+6x+10}
  2. 2x+6x2+6x+10\dfrac{2x+6}{x^2+6x+10} (correct answer)
  3. x2+6x+102x+6\dfrac{x^2+6x+10}{2x+6}
  4. ln⁡ ⁣(x2+6x+10)(2x+6)\ln\!\left(x^2+6x+10\right)\left(2x+6\right)
  5. 2x+6ln⁡ ⁣(x2+6x+10)\dfrac{2x+6}{\ln\!\left(x^2+6x+10\right)}
Explanation: This problem requires the chain rule to differentiate G(x) = ln(x² + 6x + 10). The chain rule states that when differentiating a composite function, we multiply the derivative of the outer function by the derivative of the inner function. The outer function is ln(u) with derivative 1/u, and the inner function is u = x² + 6x + 10 with derivative 2x + 6. Applying the chain rule: G'(x) = (1/(x² + 6x + 10)) · (2x + 6) = (2x + 6)/(x² + 6x + 10). Choice A incorrectly gives 1/(x² + 6x + 10), omitting the derivative of the inner function. Remember that d/dx[ln(f(x))] = f'(x)/f(x), which is a direct application of the chain rule to logarithmic functions.

Question 14

A cost function is K(x)=(2x−1)7K(x)=\big(2x-1\big)^7. What is K′(x)K'(x)?

  1. 7(2x−1)67(2x-1)^6
  2. 14(2x−1)614(2x-1)^6 (correct answer)
  3. 7(2x−1)77(2x-1)^7
  4. 14(2x−1)714(2x-1)^7
  5. 7(2x−1)6\dfrac{7}{(2x-1)^6}
Explanation: This problem requires the chain rule to differentiate a composite function. The outer function is raising to the 7th power, and the inner function is 2x - 1. To differentiate, take the derivative of the outer function, which is 7 times the inner function raised to the 6th power. Then multiply by the derivative of the inner function, which is 2, resulting in 7(2x - 1)^6 * 2 = 14(2x - 1)^6. A tempting distractor is choice A, which forgets to multiply by the derivative of the inner function, which is 2. To recognize when to use the chain rule, look for compositions where one function is substituted into another, such as a linear expression inside a power function.

Question 15

A signal is modeled by S(x)=sin⁡(4x3−x)S(x)=\sin(4x^3-x). What is S′(x)S'(x)?

  1. cos⁡(4x3−x)\cos(4x^3-x)
  2. (12x2−1)sin⁡(4x3−x)(12x^2-1)\sin(4x^3-x)
  3. (12x2−1)cos⁡(4x3−x)(12x^2-1)\cos(4x^3-x) (correct answer)
  4. sin⁡(4x3−x)(12x2−1)\sin(4x^3-x)(12x^2-1)
  5. cos⁡(4x3−x)(12x3−x)\cos(4x^3-x)(12x^3-x)
Explanation: This problem requires the chain rule to differentiate a composite function. The outer function is the sine function, and the inner function is 4x³ - x. To differentiate, take the derivative of the outer function, which is cosine evaluated at the inner function. Then multiply by the derivative of the inner function, which is 12x² - 1, resulting in cos(4x³ - x) * (12x² - 1). A tempting distractor is choice B, which incorrectly uses sine instead of cosine for the outer derivative. To recognize when to use the chain rule, look for compositions where one function is substituted into another, such as a polynomial inside a trigonometric function.

Question 16

A population model is P(t)=(1+t4)−3P(t)=\big(1+t^4\big)^{-3}. What is P′(t)P'(t)?

  1. −3(1+t4)−4-3(1+t^4)^{-4}
  2. −12t3(1+t4)−4-12t^3(1+t^4)^{-4} (correct answer)
  3. −12t3(1+t4)−3-12t^3(1+t^4)^{-3}
  4. 12t3(1+t4)−412t^3(1+t^4)^{-4}
  5. −3t4(1+t4)−4-3t^4(1+t^4)^{-4}
Explanation: This problem requires the chain rule to differentiate a composite function. The outer function is raising to the -3 power, and the inner function is 1 + t⁴. To differentiate, take the derivative of the outer function, which is -3 times the inner function raised to the -4 power. Then multiply by the derivative of the inner function, which is 4t³, resulting in -3(1 + t⁴)^{-4} * 4t³ = -12t³ (1 + t⁴)^{-4}. A tempting distractor is choice C, which uses an exponent of -3 instead of lowering it to -4. To recognize when to use the chain rule, look for compositions where one function is substituted into another, such as a polynomial inside a negative power function.

Question 17

A force is given by F(x)=ln⁡(5x2−4x+1)F(x)=\ln\big(5x^2-4x+1\big). What is F′(x)F'(x)?

  1. 5x2−4x+110x−4\dfrac{5x^2-4x+1}{10x-4}
  2. 15x2−4x+1\dfrac{1}{5x^2-4x+1}
  3. 10x−45x2−4x+1\dfrac{10x-4}{5x^2-4x+1} (correct answer)
  4. 10x−4ln⁡(5x2−4x+1)\dfrac{10x-4}{\ln(5x^2-4x+1)}
  5. ln⁡(10x−4)\ln(10x-4)
Explanation: This problem requires the chain rule to differentiate a composite function. The outer function is the natural logarithm, and the inner function is 5x² - 4x + 1. To differentiate, take the derivative of the outer function, which is 1 over the inner function. Then multiply by the derivative of the inner function, which is 10x - 4, resulting in (10x - 4) / (5x² - 4x + 1). A tempting distractor is choice A, which incorrectly places the derivative in the numerator without the logarithmic structure. To recognize when to use the chain rule, look for compositions where one function is substituted into another, such as a polynomial inside a logarithmic function.

Question 18

A current is I(t)=tan⁡(πt2)I(t)=\tan\big(\pi t^2\big). What is I′(t)I'(t)?

  1. sec⁡2(πt2)\sec^2(\pi t^2)
  2. 2πtsec⁡2(πt2)2\pi t\sec^2(\pi t^2) (correct answer)
  3. πtsec⁡2(πt2)\pi t\sec^2(\pi t^2)
  4. 2πttan⁡(πt2)2\pi t\tan(\pi t^2)
  5. sec⁡(πt2)tan⁡(πt2)\sec(\pi t^2)\tan(\pi t^2)
Explanation: This problem requires the chain rule to differentiate a composite function. The outer function is the tangent function, and the inner function is π t². To differentiate, take the derivative of the outer function, which is sec² evaluated at the inner function. Then multiply by the derivative of the inner function, which is 2π t, resulting in sec²(π t²) * 2π t. A tempting distractor is choice D, which incorrectly uses tangent instead of sec² for the outer derivative. To recognize when to use the chain rule, look for compositions where one function is substituted into another, such as a polynomial inside a trigonometric function like tangent.

Question 19

A model for pressure is P(t)=(cos⁡(3t))5P(t)=\big(\cos(3t)\big)^5; what is P′(t)P'(t)?

  1. 5(cos⁡(3t))4sin⁡(3t)5\big(\cos(3t)\big)^4\sin(3t)
  2. −15(cos⁡(3t))4sin⁡(3t)-15\big(\cos(3t)\big)^4\sin(3t) (correct answer)
  3. −3(cos⁡(3t))5sin⁡(3t)-3\big(\cos(3t)\big)^5\sin(3t)
  4. 5(cos⁡(3t))45\big(\cos(3t)\big)^4
  5. 15(cos⁡(3t))4sin⁡(3t)15\big(\cos(3t)\big)^4\sin(3t)
Explanation: This problem requires the chain rule to differentiate a composite function. The outer function is raising to the fifth power, while the inner function is cos(3t). To differentiate, first take the derivative of the outer function, which is 5 times the inner raised to the fourth power. Then, multiply by the derivative of the inner function, which is -sin(3t) times 3 or -3 sin(3t), resulting in -15 (cos(3t))⁴ sin(3t). A tempting distractor like choice A forgets the negative sign and the factor of 3 from the inner derivative. Always look for compositions where a function is plugged into another, such as a trigonometric function inside a power, to recognize when to apply the chain rule.

Question 20

A medication level is M(t)=sin⁡(2t3−t)M(t)=\sin\big(2t^3-t\big); what is M′(t)M'(t)?

  1. cos⁡(2t3−t)\cos\big(2t^3-t\big)
  2. cos⁡(2t3−t)(6t2−1)\cos\big(2t^3-t\big)\big(6t^2-1\big) (correct answer)
  3. sin⁡(2t3−t)(6t2−1)\sin\big(2t^3-t\big)\big(6t^2-1\big)
  4. cos⁡(2t3)−cos⁡(t)\cos(2t^3)-\cos(t)
  5. (6t2−1)sin⁡(2t3−t)\big(6t^2-1\big)\sin\big(2t^3-t\big)
Explanation: This problem requires the chain rule to differentiate a composite function. The outer function is the sine function, while the inner function is the cubic 2t³ - t. To differentiate, first take the derivative of the outer function, which is cosine of the inner function. Then, multiply by the derivative of the inner function, which is 6t² - 1, yielding cos(2t³ - t)(6t² - 1). A tempting distractor like choice E swaps cosine for sine, which is the derivative of cosine instead of sine. Always look for compositions where a function is plugged into another, such as a polynomial inside a trigonometric function, to recognize when to apply the chain rule.