Drill 1 · Math · Chain Rule
AP Calculus AB: Chain Rule (Drill 1) is a Math practice drill covering Chain Rule. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice applying the chain rule to differentiate composite functions involving polynomial, trigonometric, exponential, and logarithmic expressions. Questions target the most common chain rule errors, including forgetting the inner derivative and misidentifying the inside function.
Question 1. Let \( f(x) = (3x^2 + 1)^5 \). Find \( f'(x) \).
Explanation: Choice B is correct. The chain rule gives \( f'(x) = 5(3x^2+1)^4 \cdot (3x^2+1)' = 5(3x^2+1)^4 \cdot 6x = 30x(3x^2+1)^4 \). Choice A is incorrect because it forgets the inner derivative entirely, the most common chain rule error. Choice C is incorrect because it differentiates \( 3x^2 + 1 \) as if it were a constant-coefficient linear term, using 6 instead of \( 6x \), forgetting to apply the power rule to \( 3x^2 \). Choice D is incorrect because it uses \( 3x^2 \) as the inner derivative instead of \( 6x \), making a power-rule error on the inside function.
Question 2. Let \( g(x) = \sin(4x^3) \). Find \( g'(x) \).
Explanation: Choice D is correct. The chain rule gives \( g'(x) = \cos(4x^3) \cdot (4x^3)' = 12x^2 \cos(4x^3) \). Choice A is incorrect because it forgets the inner derivative entirely, the most direct chain rule omission. Choice B is incorrect because it puts the inner derivative inside the cosine argument rather than multiplying outside. Choice C is incorrect because it introduces a spurious negative sign, confusing \( \frac{d}{dx}[\sin u] = \cos u \cdot u' \) with \( \frac{d}{dx}[\cos u] = -\sin u \cdot u' \).
Question 3. Let \( h(x) = e^{\sin x} \). Find \( h'(x) \).
Explanation: Choice C is correct. Since \( \frac{d}{dx}[e^u] = e^u \cdot u' \), with \( u = \sin x \) and \( u' = \cos x \): \( h'(x) = e^{\sin x} \cdot \cos x \). Choice A is incorrect because it forgets the chain rule entirely, treating \( e^{\sin x} \) as if the exponent were simply \( x \). Choice B is incorrect because it substitutes the derivative of \( \sin x \) into the exponent rather than multiplying it outside. Choice D is incorrect because it incorrectly applies the power rule, bringing the exponent down as a coefficient and reducing it by 1, a procedure that applies to \( x^n \), not to \( e^u \).
Question 4. Which of the following is the derivative of \( y = \ln(\cos x) \)?
Explanation: Choice B is correct. Using \( \frac{d}{dx}[\ln u] = \frac{u'}{u} \) with \( u = \cos x \) and \( u' = -\sin x \): \( y' = \dfrac{-\sin x}{\cos x} = -\tan x \). Choice A is incorrect because it forgets the inner derivative, using only \( \frac{1}{\cos x} \) without multiplying by \( (\cos x)' \). Choice C is incorrect because it has the correct structure but drops the negative sign from \( \frac{d}{dx}[\cos x] = -\sin x \). Choice D is incorrect because it confuses differentiation with antidifferentiation, arriving at a logarithmic expression rather than a trigonometric one.
Question 5. Let \( f(x) = \sin^3(2x) \). Which of the following equals \( f'(x) \)?
Explanation: Choice C is correct. Write \( f(x) = [\sin(2x)]^3 \), a three-layer composition. Applying the chain rule twice: \( f'(x) = 3[\sin(2x)]^2 \cdot \frac{d}{dx}[\sin(2x)] = 3\sin^2(2x) \cdot \cos(2x) \cdot 2 = 6\sin^2(2x)\cos(2x) \). Choice A is incorrect because it applies the outer power rule but forgets both inner derivatives, the derivative of sine and the factor of 2 from \( 2x \). Choice B is incorrect because it correctly handles the power rule and the derivative of sine but forgets the factor of 2 from differentiating the innermost function \( 2x \). Choice D is incorrect because it results from misapplying the chain rule as \( \frac{d}{dx}[\sin^3 u] = 3\cos^2 u \cdot u' \), confusing the derivative of \( \sin^3 u \) with \( (\cos u)^3 \) differentiated, and losing the \( \sin^2(2x) \) factor entirely.