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

Let q(x)=sin⁡ ⁣(ln⁡(x2+1))q(x)=\sin\!\big(\ln(x^2+1)\big). What is q′(x)q'(x)?

cos⁡(ln⁡(x2+1))\cos(\ln(x^2+1))
2xx2+1cos⁡(ln⁡(x2+1))\dfrac{2x}{x^2+1}\cos(\ln(x^2+1))
1x2+1cos⁡(ln⁡(x2+1))\dfrac{1}{x^2+1}\cos(\ln(x^2+1))
2xx2+1sin⁡(ln⁡(x2+1))\dfrac{2x}{x^2+1}\sin(\ln(x^2+1))
2xcos⁡(ln⁡(x2+1))2x\cos(\ln(x^2+1))
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AP Calculus AB Quiz

AP Calculus AB Quiz: Chain Rule

Practice Chain 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.

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 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

Let q(x)=sin⁡ ⁣(ln⁡(x2+1))q(x)=\sin\!\big(\ln(x^2+1)\big). What is q′(x)q'(x)?

  1. cos⁡(ln⁡(x2+1))\cos(\ln(x^2+1))
  2. 2xx2+1cos⁡(ln⁡(x2+1))\dfrac{2x}{x^2+1}\cos(\ln(x^2+1)) (correct answer)
  3. 1x2+1cos⁡(ln⁡(x2+1))\dfrac{1}{x^2+1}\cos(\ln(x^2+1))
  4. 2xx2+1sin⁡(ln⁡(x2+1))\dfrac{2x}{x^2+1}\sin(\ln(x^2+1))
  5. 2xcos⁡(ln⁡(x2+1))2x\cos(\ln(x^2+1))
Explanation: q(x) = sin(ln(x² + 1)) chains sin(u) with u(x) = ln(x² + 1). Derivative: cos(u) · u' = cos(ln(x² + 1)) · (2x/(x² + 1)). Inner log derivative is key. Omission: Missing the 2x/(x² + 1). Some use sin derivative without chain. Pattern: Trig of logs requires chaining both derivatives.

Question 2

For J(x)=sec⁡(ln⁡x)J(x)=\sec(\ln x), what is J′(x)J'(x)?

  1. sec⁡(ln⁡x)tan⁡(ln⁡x)\sec(\ln x)\tan(\ln x)
  2. sec⁡(ln⁡x)tan⁡(ln⁡x)x\dfrac{\sec(\ln x)\tan(\ln x)}{x} (correct answer)
  3. sec⁡xtan⁡xx\dfrac{\sec x\tan x}{x}
  4. 1xsec⁡(ln⁡x)\dfrac{1}{x}\sec(\ln x)
  5. xsec⁡(ln⁡x)tan⁡(ln⁡x)x\sec(\ln x)\tan(\ln x)
Explanation: J(x) = sec(ln x), outer sec(u), u = ln x. Derivative: sec(u) tan(u) · (1/x). Log inner. Omission: Missing 1/x. Some forget tan. Pattern: Trig of logs chain the 1/x derivative.

Question 3

Let v(x)=excos⁡xv(x)=e^{x\cos x}. What is v′(x)v'(x)?

  1. excos⁡x(cos⁡x−xsin⁡x)e^{x\cos x}(\cos x- x\sin x) (correct answer)
  2. excos⁡x(cos⁡x)e^{x\cos x}(\cos x)
  3. excos⁡x(xcos⁡x)e^{x\cos x}(x\cos x)
  4. exsin⁡x(cos⁡x−xsin⁡x)e^{x\sin x}(\cos x- x\sin x)
  5. excos⁡x(sin⁡x+xcos⁡x)e^{x\cos x}(\sin x+ x\cos x)
Explanation: v(x) = e^{x cos x}, outer e^u, u = x cos x. u' needs product rule: cos x - x sin x. Derivative: e^u · (cos x - x sin x). Composite exponent. Omission: Missing product rule in u'. Some simplify wrong. Pattern: Exponentials with products chain including product derivative.

Question 4

Let A(x)=ln⁡ ⁣((x−2)2+5)A(x)=\ln\!\big((x-2)^2+5\big). What is A′(x)A'(x)?

  1. 1(x−2)2+5\dfrac{1}{(x-2)^2+5}
  2. 2(x−2)(x−2)2+5\dfrac{2(x-2)}{(x-2)^2+5} (correct answer)
  3. 2x(x−2)2+5\dfrac{2x}{(x-2)^2+5}
  4. (x−2)2+52(x−2)\dfrac{(x-2)^2+5}{2(x-2)}
  5. 2(x−2)ln⁡((x−2)2+5)2(x-2)\ln((x-2)^2+5)
Explanation: A(x) = ln⁡((x−2)2+5)\ln((x - 2)^2 + 5), outer ln⁡(u)\ln(u), u=(x−2)2+5u = (x - 2)^2 + 5. Derivative: 1u⋅2(x−2)\frac{1}{u} \cdot 2(x - 2). Inner quadratic. Omission: Missing 2(x−2)2(x - 2). Some treat as ln⁡(x)\ln(x). Pattern: Logs of quadratics chain the quadratic derivative.

Question 5

A function is Z(t)=ln⁡ ⁣(t4+9)Z(t)=\ln\!\big(\sqrt{t^4+9}\big). What is Z′(t)Z'(t)?

  1. 1t4+9\dfrac{1}{\sqrt{t^4+9}}
  2. 2t3t4+9\dfrac{2t^3}{t^4+9} (correct answer)
  3. 4t3t4+9\dfrac{4t^3}{t^4+9}
  4. 4t3t4+9\dfrac{4t^3}{\sqrt{t^4+9}}
  5. 2t3t4+9\dfrac{2t^3}{\sqrt{t^4+9}}
Explanation: Z(t) = ln(√(t4t^4 + 9)) simplifies to (1/2) ln(t4t^4 + 9), chain rule: (1/2) * (1/(t4t^4 + 9)) * 4t^3 = 2t^3/(t4t^4 + 9). Structure: Log of root. Omission: Forgetting the 1/2 factor. Matches choice B. Correct. Pattern: Logs of roots simplify but still need chain.

Question 6

A cost is K(x)=(ln⁡x)4K(x)=(\ln x)^4. What is K′(x)K'(x)?

  1. 4(ln⁡x)34(\ln x)^3
  2. 4(ln⁡x)3x\dfrac{4(\ln x)^3}{x} (correct answer)
  3. (ln⁡x)4x\dfrac{(\ln x)^4}{x}
  4. 4x3ln⁡x4x^3\ln x
  5. 4ln⁡x\dfrac{4}{\ln x}
Explanation: To find K′(x)K'(x) for K(x)=(ln⁡x)4K(x) = (\ln x)^4, apply the chain rule since this is a power function (outer) composed with the natural logarithm (inner). The outer function is u4u^4 with derivative 4u34u^3, where u=ln⁡xu = \ln x, and the inner derivative is 1x\frac{1}{x}. Thus, K′(x)=4(ln⁡x)3⋅1xK'(x) = 4(\ln x)^3 \cdot \frac{1}{x}. A common omission is neglecting the inner derivative 1/x1/x, leading to just 4(ln⁡x)34(\ln x)^3. This matches choice B directly. Independently verifying, the calculation confirms the marked answer. Recognize this pattern in powers of logarithms, always multiplying by the inner function's derivative.

Question 7

A decay model is D(t)=e−2t3+1D(t)=e^{-2t^3+1}. What is D′(t)D'(t)?

  1. e−2t3+1e^{-2t^3+1}
  2. (−6t2)e−2t3+1(-6t^2)e^{-2t^3+1} (correct answer)
  3. (−2t3+1)e−2t3+1(-2t^3+1)e^{-2t^3+1}
  4. (−6t2)e2t3−1(-6t^2)e^{2t^3-1}
  5. −6t2+e−2t3+1-6t^2+e^{-2t^3+1}
Explanation: D(t) = e^{-2t³ + 1} is exponential with inner u(t) = -2t³ + 1. Derivative: e^u · u' = e^{-2t³ + 1} · (-6t²). The cubic requires careful differentiation. Common omission: Forgetting the -6t². Some flip the sign. Pattern recognition: Exponentials with polynomials inside signal chain rule, multiply by polynomial derivative.

Question 8

A current is I(t)=cos⁡(et)I(t)=\cos(e^{t}). What is I′(t)I'(t)?

  1. −sin⁡(et)-\sin(e^{t})
  2. −etsin⁡(et)-e^{t}\sin(e^{t}) (correct answer)
  3. etcos⁡(et)e^{t}\cos(e^{t})
  4. −sin⁡(t)et-\sin(t)e^{t}
  5. −sin⁡(et)cos⁡(et)-\sin(e^{t})\cos(e^{t})
Explanation: I(t) = cos(ete^t), outer cos(u), u = e^t. Derivative: -sin(u) · e^t = -e^t sin(ete^t). Exponential inner. Omission: Forgetting e^t. Some miss negative. Pattern: Trig of exponentials chain the exponential derivative.

Question 9

A revenue function is R(x)=x2+4x3R(x)=\sqrt[3]{x^2+4x}. What is R′(x)R'(x)?

  1. 13(x2+4x)−2/3\dfrac{1}{3}(x^2+4x)^{-2/3}
  2. 2x+43(x2+4x)−2/3\dfrac{2x+4}{3}(x^2+4x)^{-2/3} (correct answer)
  3. (2x+4)(x2+4x)−2/3(2x+4)(x^2+4x)^{-2/3}
  4. 2x+43(x2+4x)2/3\dfrac{2x+4}{3}(x^2+4x)^{2/3}
  5. 23(x+2)(x2+4x)−1/3\dfrac{2}{3}(x+2)(x^2+4x)^{-1/3}
Explanation: R(x) = ∛(x² + 4x) is (u)^{1/3} with u(x) = x² + 4x, so chain rule gives (1/3) u^{-2/3} · u' = (1/3)(x² + 4x)^{-2/3} · (2x + 4). Simplifying, it's (2x + 4)/3 · (x² + 4x)^{-2/3}. The fractional power is key. Common omission: Forgetting the (2x + 4) factor. Some mishandle the exponent. Pattern: Look for roots of polynomials and chain their derivatives.

Question 10

A path is y(x)=ln⁡ ⁣(1−3x)y(x)=\ln\!\big(\sqrt{1-3x}\big). What is y′(x)y'(x)?

  1. 11−3x\dfrac{1}{\sqrt{1-3x}}
  2. −32(1−3x)\dfrac{-3}{2(1-3x)} (correct answer)
  3. −31−3x\dfrac{-3}{\sqrt{1-3x}}
  4. −321−3x\dfrac{-3}{2\sqrt{1-3x}}
  5. 12(1−3x)\dfrac{1}{2(1-3x)}
Explanation: y(x) = ln(√(1 - 3x)) simplifies to (1/2) ln(1 - 3x), so chain rule on log: (1/2) · 1/(1 - 3x) · (-3) = -3/(2(1 - 3x)). Outer log, inner sqrt then linear. Common omission: Forgetting the 1/2 from sqrt or the -3. Some don't simplify first. Pattern: Logs of roots suggest rewriting for easier chaining.

Question 11

For B(t)=cos⁡3(t2)B(t)=\cos^3(t^2), what is B′(t)B'(t)?

  1. 3cos⁡2(t2)3\cos^2(t^2)
  2. −3sin⁡(t2)cos⁡2(t2)-3\sin(t^2)\cos^2(t^2)
  3. −6tsin⁡(t2)cos⁡2(t2)-6t\sin(t^2)\cos^2(t^2) (correct answer)
  4. −6tcos⁡2(t2)-6t\cos^2(t^2)
  5. −3sin⁡(t)cos⁡2(t2)-3\sin(t)\cos^2(t^2)
Explanation: B(t) = cos^3(t2t^2) is [cos(u)]^3 where u = t^2, so chain rule for power (outer) and trig (inner). Derivative: 3 cos^2(u) * (-sin(u)) * 2t = -6t sin(t2t^2) cos^2(t2t^2). Outer-inner: Power on trig of quadratic. Omission: Missing the 2t or the negative sign. Matches choice C. Confirmed correct. Pattern: Powers on trig functions of polynomials need extended chain rule.

Question 12

A volume is V(r)=(1+r2)−3V(r)=\big(1+r^2\big)^{-3}. What is V′(r)V'(r)?

  1. −3(1+r2)−4-3(1+r^2)^{-4}
  2. −6r(1+r2)−4-6r(1+r^2)^{-4} (correct answer)
  3. −6r(1+r2)−3-6r(1+r^2)^{-3}
  4. −3(1+r2)−3-3(1+r^2)^{-3}
  5. 6r(1+r2)−46r(1+r^2)^{-4}
Explanation: For V(r) = (1 + r^2)^{-3}, the chain rule applies to the power (outer) of the binomial (inner). Derivative: -3(1 + r^2)^{-4} * 2r = -6r (1 + r^2)^{-4}. The outer-inner structure is clear with u = 1 + r^2 and outer u^{-3}. A common omission is forgetting the 2r from the inner derivative. This is choice B. Independent solving confirms it. Pattern: Negative exponents on binomials signal chain rule for rates in physics contexts.

Question 13

A function is H(x)=1ln⁡(x2+4)H(x)=\dfrac{1}{\ln(x^2+4)}. What is H′(x)H'(x)?

  1. −1(ln⁡(x2+4))2-\dfrac{1}{(\ln(x^2+4))^2}
  2. −2x(ln⁡(x2+4))2-\dfrac{2x}{(\ln(x^2+4))^2}
  3. −2x(x2+4)(ln⁡(x2+4))2-\dfrac{2x}{(x^2+4)(\ln(x^2+4))^2} (correct answer)
  4. −1(x2+4)(ln⁡(x2+4))-\dfrac{1}{(x^2+4)(\ln(x^2+4))}
  5. 2x(x2+4)(ln⁡(x2+4))\dfrac{2x}{(x^2+4)(\ln(x^2+4))}
Explanation: H(x) = 1ln⁡(x2+4)\frac{1}{\ln(x^2 + 4)} is (ln⁡u)−1(\ln u)^{-1} with u=x2+4u = x^2 + 4. Chain rule: −1(ln⁡u)2⋅1u⋅2x=−2x(x2+4)(ln⁡(x2+4))2-\frac{1}{(\ln u)^2} \cdot \frac{1}{u} \cdot 2x = -\frac{2x}{(x^2 + 4) (\ln(x^2 + 4))^2}. Structure: Reciprocal of log of quadratic. Common omission: Forgetting the 2x. Choice C matches. Verification confirms. Recognize reciprocals of logs for chain rule in rates.

Question 14

Let z(x)=sin⁡(x2+1)z(x)=\sqrt{\sin(x^2+1)}. What is z′(x)z'(x)?

  1. cos⁡(x2+1)2sin⁡(x2+1)\dfrac{\cos(x^2+1)}{2\sqrt{\sin(x^2+1)}}
  2. 2xcos⁡(x2+1)2sin⁡(x2+1)\dfrac{2x\cos(x^2+1)}{2\sqrt{\sin(x^2+1)}} (correct answer)
  3. 2xsin⁡(x2+1)2sin⁡(x2+1)\dfrac{2x\sin(x^2+1)}{2\sqrt{\sin(x^2+1)}}
  4. cos⁡(x2+1)sin⁡(x2+1)\dfrac{\cos(x^2+1)}{\sqrt{\sin(x^2+1)}}
  5. xcos⁡(x2+1)sin⁡(x2+1)\dfrac{x\cos(x^2+1)}{\sin(x^2+1)}
Explanation: The chain rule is crucial for differentiating z(x)=sin⁡(x2+1)z(x) = \sqrt{\sin(x^2 + 1)}, as it involves a composition of functions: the square root as the outer function applied to the sine, which itself is applied to the inner polynomial x2+1x^2 + 1. Identify the outermost function as the square root, with derivative 12u\frac{1}{2\sqrt{u}} where u=sin⁡(v)u = \sin(v) and v=x2+1v = x^2 + 1. Then, multiply by the derivative of u, which is cos⁡(v)\cos(v) times the derivative of v, which is 2x2x. This yields 12sin⁡(x2+1)⋅cos⁡(x2+1)⋅2x\frac{1}{2\sqrt{\sin(x^2 + 1)}} \cdot \cos(x^2 + 1) \cdot 2x, simplifying to xcos⁡(x2+1)sin⁡(x2+1)\frac{x \cos(x^2 + 1)}{\sqrt{\sin(x^2 + 1)}}. A common omission is forgetting the derivative of the innermost function, such as the 2x2x from v'. The marked answer B is correct as it matches this after simplification. For pattern recognition, spot compositions where trigonometric functions enclose polynomials, indicating multiple chain rule applications.

Question 15

For θ(t)=1sin⁡t\theta(t)=\dfrac{1}{\sqrt{\sin t}}, what is θ′(t)\theta'(t)?

  1. −12(sin⁡t)−3/2-\dfrac{1}{2}(\sin t)^{-3/2}
  2. −cos⁡t2(sin⁡t)−3/2-\dfrac{\cos t}{2}(\sin t)^{-3/2} (correct answer)
  3. −sin⁡t2(cos⁡t)−3/2-\dfrac{\sin t}{2}(\cos t)^{-3/2}
  4. cos⁡t2(sin⁡t)−1/2\dfrac{\cos t}{2}(\sin t)^{-1/2}
  5. −cos⁡t(sin⁡t)1/2-\dfrac{\cos t}{(\sin t)^{1/2}}
Explanation: θ(t)=1sin⁡t=(sin⁡t)−1/2\theta(t) = \dfrac{1}{\sqrt{\sin t}} = (\sin t)^{-1/2}. Derivative: −12(sin⁡t)−3/2cos⁡t-\dfrac{1}{2} (\sin t)^{-3/2} \cos t. Outer: Negative power of sin. Common omission: Sign or exponent error. Matches choice B. Verified. Recognize negative roots as powers for chain rule application.

Question 16

A function is Q(x)=cos⁡(ln⁡(5x))Q(x)=\cos(\ln(5x)). What is Q′(x)Q'(x)?

  1. −sin⁡(ln⁡(5x))-\sin(\ln(5x))
  2. −1xsin⁡(ln⁡(5x))-\dfrac{1}{x}\sin(\ln(5x)) (correct answer)
  3. −5xsin⁡(ln⁡(5x))-\dfrac{5}{x}\sin(\ln(5x))
  4. 1xcos⁡(ln⁡(5x))\dfrac{1}{x}\cos(\ln(5x))
  5. −15xsin⁡(ln⁡(5x))-\dfrac{1}{5x}\sin(\ln(5x))
Explanation: Q(x) = cos(ln(5x)) is cos outer of log inner of linear. Derivative: -sin(ln(5x)) * (1/(5x)) * 5 = - (1/x) sin(ln(5x)). Outer-inner: Trig of log. Common omission: Mishandling the 5/5x to 1/x. Matches choice B. Correct per verification. Strategy: Trig of logs for oscillatory decay, apply chain carefully.

Question 17

A chemical's concentration is modeled by C(t)=5t3−2t+9C(t)=\sqrt{5t^3-2t+9}. What is C′(t)C'(t)?

  1. 15t2−225t3−2t+9\dfrac{15t^2-2}{2\sqrt{5t^3-2t+9}} (correct answer)
  2. 125t3−2t+9\dfrac{1}{2\sqrt{5t^3-2t+9}}
  3. 15t225t3−2t+9\dfrac{15t^2}{2\sqrt{5t^3-2t+9}}
  4. 32(5t3−2t+9)1/2\dfrac{3}{2}(5t^3-2t+9)^{1/2}
  5. (15t2−2)5t3−2t+9(15t^2-2)\sqrt{5t^3-2t+9}
Explanation: This problem requires the chain rule to differentiate C(t)=5t3−2t+9=(5t3−2t+9)1/2C(t)=\sqrt{5t^3-2t+9}=(5t^3-2t+9)^{1/2}. The outer function is the square root (or power of 1/2), and the inner function is 5t3−2t+95t^3-2t+9. Applying the chain rule: first take the derivative of the outer function to get 12(5t3−2t+9)−1/2\frac{1}{2}(5t^3-2t+9)^{-1/2}, then multiply by the derivative of the inner function, which is 15t2−215t^2-2. This gives us C′(t)=12(5t3−2t+9)−1/2⋅(15t2−2)=15t2−225t3−2t+9C'(t)=\frac{1}{2}(5t^3-2t+9)^{-1/2}\cdot(15t^2-2)=\frac{15t^2-2}{2\sqrt{5t^3-2t+9}}. A common error is forgetting to differentiate the inner function, which would incorrectly yield just 125t3−2t+9\frac{1}{2\sqrt{5t^3-2t+9}}. When you see a composite function with a radical, always identify what's inside the radical as your inner function and remember to multiply by its derivative.

Question 18

For d(x)=arcsin⁡(2x−3)d(x)=\arcsin(2x-3), what is d′(x)d'(x)?

  1. 11−(2x−3)2\dfrac{1}{\sqrt{1-(2x-3)^2}}
  2. 21−(2x−3)2\dfrac{2}{\sqrt{1-(2x-3)^2}} (correct answer)
  3. 2x−31−(2x−3)2\dfrac{2x-3}{\sqrt{1-(2x-3)^2}}
  4. 11−(2)2\dfrac{1}{\sqrt{1-(2)^2}}
  5. 121−(2x−3)2\dfrac{1}{2\sqrt{1-(2x-3)^2}}
Explanation: d(x) = arcsin(2x - 3) is inverse sine outer of linear. Derivative: 1/√(1 - (2x - 3)^2) * 2. Structure: Inverse trig of linear. Omission: Forgetting the 2. Matches choice B. Confirmed. Pattern: Inverse trig derivatives always include chain for argument.

Question 19

For F(x)=(sin⁡x)5F(x)=(\sin x)^5, what is F′(x)F'(x)?

  1. 5(sin⁡x)45(\sin x)^4
  2. 5(sin⁡x)4cos⁡x5(\sin x)^4\cos x (correct answer)
  3. (cos⁡x)5(\cos x)^5
  4. 5sin⁡(x4)cos⁡x5\sin(x^4)\cos x
  5. 4(sin⁡x)3cos⁡x4(\sin x)^3\cos x
Explanation: F(x) = (sin x)^5 is power with inner sin x. Chain rule: 5 (sin x)^4 · cos x. Power reduces, multiply by cos x. Omission: Forgetting cos x. Some use 4 instead of 5. Pattern: Powers of trig functions need chaining the trig derivative.

Question 20

Let W(x)=(5−sin⁡x)10W(x)=\big(5-\sin x\big)^{10}. What is W′(x)W'(x)?

  1. 10(5−sin⁡x)910(5-\sin x)^9
  2. 10(5−sin⁡x)9cos⁡x10(5-\sin x)^9\cos x
  3. −10(5−sin⁡x)9cos⁡x-10(5-\sin x)^9\cos x (correct answer)
  4. −9(5−sin⁡x)10cos⁡x-9(5-\sin x)^{10}\cos x
  5. −10(5−cos⁡x)9sin⁡x-10(5-\cos x)^9\sin x
Explanation: W(x) = (5 - sin x)^{10} is power outer of 5 - sin inner. Derivative: 10(5 - sin x)^9 * (-cos x). Outer-inner clear. Common omission: Missing negative from sin derivative. Choice C matches. Verified. Strategy: Powers of trig subtractions use chain with signs.