AP Calculus AB: Basic Differentiation Rules (Drill 1)

Question 1. What is \( \dfrac{d}{dx}\left[4x^3 - 7x + 2\right] \)?

• A) \( 12x^2 - 7 \) ✓
• B) \( 12x^2 - 7x \)
• C) \( 4x^2 - 7 \)
• D) \( 12x^3 - 7 \)

Explanation: Choice A is correct. Applying the power rule term by term: \( \dfrac{d}{dx}[4x^3] = 12x^2 \), \( \dfrac{d}{dx}[-7x] = -7 \), and \( \dfrac{d}{dx}[2] = 0 \). The result is \( 12x^2 - 7 \). Choice B is incorrect because the student kept the \( x \) in the \( -7x \) term instead of recognizing that the derivative of \( -7x \) is the constant \( -7 \). Choice C is incorrect because the student reduced the exponent without multiplying by it, writing \( 4x^2 \) instead of \( 12x^2 \). Choice D is incorrect because the student multiplied by 3 but forgot to reduce the exponent, writing \( 12x^3 \) instead of \( 12x^2 \).

Question 2. If \( f(x) = 3\sin x - 5\cos x \), then \( f'(x) = \)

• A) \( 3\cos x + 5\sin x \) ✓
• B) \( 3\cos x - 5\sin x \)
• C) \( -3\cos x + 5\sin x \)
• D) \( -3\sin x - 5\cos x \)

Explanation: Choice A is correct. Using standard rules: \( \dfrac{d}{dx}[3\sin x] = 3\cos x \) and \( \dfrac{d}{dx}[-5\cos x] = 5\sin x \) (since \( \dfrac{d}{dx}[\cos x] = -\sin x \), the negatives cancel). So \( f'(x) = 3\cos x + 5\sin x \). Choice B is incorrect because the student applied \( \dfrac{d}{dx}[\cos x] = +\sin x \) (wrong sign), yielding \( -5\sin x \) instead of \( +5\sin x \). Choice C is incorrect because the student negated the derivative of \( \sin x \), treating \( \dfrac{d}{dx}[\sin x] = -\cos x \). Choice D is incorrect because the student failed to apply the derivative rules for either term and instead reproduced the original functions with incorrect signs, as if differentiating negated both.

Question 3. If \( g(x) = \dfrac{6}{x^2} + 4\sqrt{x} \), then \( g'(x) = \)

• A) \( -\dfrac{12}{x^3} + \dfrac{2}{\sqrt{x}} \) ✓
• B) \( \dfrac{12}{x^3} + \dfrac{2}{\sqrt{x}} \) for the function given in the stem
• C) \( -\dfrac{12}{x^3} + 2\sqrt{x} \)
• D) \( -\dfrac{6}{x} + \dfrac{1}{2\sqrt{x}} \)

Explanation: Choice A is correct. Rewrite as \( g(x) = 6x^{-2} + 4x^{1/2} \). Then \( \dfrac{d}{dx}[6x^{-2}] = -12x^{-3} = -\dfrac{12}{x^3} \) and \( \dfrac{d}{dx}[4x^{1/2}] = 2x^{-1/2} = \dfrac{2}{\sqrt{x}} \). Choice B is incorrect because the student forgot the negative sign when differentiating \( 6x^{-2} \); the power rule brings down \( -2 \), making the term negative. Choice C is incorrect because the student correctly differentiated \( 6x^{-2} \) but failed to reduce the exponent on \( 4x^{1/2} \), writing \( 2x^{1/2} = 2\sqrt{x} \) instead of \( 2x^{-1/2} \). Choice D is incorrect because the student reduced the exponent of \( 6x^{-2} \) to get \( x^{-1} \) without multiplying by \( -2 \), and mishandled the coefficient on the square root term.

Question 4. Let \( h(x) = e^x + \ln x \). Which of the following correctly states \( h'(x) \) and evaluates \( h'(1) \)?

• A) \( h'(x) = e^x + \dfrac{1}{x} \), and \( h'(1) = e + 1 \) ✓
• B) \( h'(x) = xe^{x-1} + \dfrac{1}{x} \), and \( h'(1) = 2 \)
• C) \( h'(x) = e^x + \dfrac{1}{x} \), and \( h'(1) = e \)
• D) \( h'(x) = e^x - \dfrac{1}{x^2} \), and \( h'(1) = e - 1 \)

Explanation: Choice A is correct. Standard rules give \( \dfrac{d}{dx}[e^x] = e^x \) and \( \dfrac{d}{dx}[\ln x] = \dfrac{1}{x} \), so \( h'(x) = e^x + \dfrac{1}{x} \) and \( h'(1) = e^1 + \dfrac{1}{1} = e + 1 \). Choice B is incorrect because the student applied the power rule to \( e^x \), writing \( xe^{x-1} \); the power rule does not apply to exponential functions, \( e^x \) is its own derivative. Choice C is incorrect because the student correctly found \( h'(x) \) but evaluated \( h'(1) \) as just \( e \), forgetting to add the \( \dfrac{1}{1} = 1 \) contribution from the \( \ln x \) term. Choice D is incorrect because the student differentiated \( \ln x \) as \( -\dfrac{1}{x^2} \), which is the derivative of \( x^{-1} \), not \( \ln x \).

Question 5. Let \( f(x) = 2\sin x + e^x - x^3 \). A student claims that \( f'(\pi) = e^\pi \) because "the derivative of \( \sin x \) is zero at \( x = \pi \) since \( \sin\pi = 0 \), and the derivative of \( x^3 \) is zero because \( \pi \) is just a constant being plugged in, so that term disappears." Which of the following is the correct value of \( f'(\pi) \)?

• A) \( e^\pi - 3\pi^2 \)
• B) \( e^\pi + 2 - 3\pi^2 \)
• C) \( e^\pi - 2 - 3\pi^2 \) ✓
• D) \( e^\pi \)

Explanation: Choice C is correct. We have \( f'(x) = 2\cos x + e^x - 3x^2 \). At \( x = \pi \): \( \cos\pi = -1 \), so \( 2\cos\pi = -2 \). Therefore \( f'(\pi) = -2 + e^\pi - 3\pi^2 = e^\pi - 2 - 3\pi^2 \). Choice D is incorrect and matches the student's flawed reasoning. The student's first error is confusing \( \sin\pi = 0 \) (the value of \( \sin x \) at \( \pi \)) with \( \dfrac{d}{dx}[2\sin x]\big|_{x=\pi} = 2\cos\pi = -2 \); the derivative rule produces \( \cos x \), not \( \sin x \). The student's second error is treating \( \pi \) as a symbolic constant that eliminates the \( x^3 \) term, when in fact \( \dfrac{d}{dx}[-x^3] = -3x^2 \) must be evaluated at \( x = \pi \), giving \( -3\pi^2 \). Choice A is incorrect because the student correctly differentiated \( -x^3 \) and evaluated at \( \pi \), but forgot the coefficient 2 when computing \( 2\cos\pi \), omitting the \( -2 \). Choice B is incorrect because the student evaluated \( 2\cos\pi \) as \( +2 \) rather than \( -2 \), forgetting that \( \cos\pi = -1 \).