AP Calculus AB: Product Rule, Quotient Rule, and Differentiability (Drill 1)

Question 1. Let \( f(x) = x^3 \sin x \). Find \( f'(x) \).

• A) \( 3x^2 \cos x \)
• B) \( 3x^2 \sin x + x^3 \cos x \) ✓
• C) \( 3x^2 \sin x \cdot x^3 \cos x \)
• D) \( x^3 \cos x \)

Explanation: Choice B is correct. The product rule states \( (uv)' = u'v + uv' \). With \( u = x^3 \) and \( v = \sin x \), we get \( u' = 3x^2 \) and \( v' = \cos x \), so \( f'(x) = 3x^2 \sin x + x^3 \cos x \). Choice A is incorrect because it differentiates only the first factor while replacing \( \sin x \) with \( \cos x \), ignoring the second term entirely. Choice C is incorrect because it multiplies the two derivatives together rather than adding them, a classic product-rule misconception. Choice D is incorrect because it differentiates only \( \sin x \) while treating \( x^3 \) as a constant.

Question 2. Let \( g(x) = \dfrac{e^x}{x^2} \). Find \( g'(x) \).

• A) \( -\dfrac{2e^x}{x^3} \)
• B) \( \dfrac{x^2 e^x + 2x e^x}{x^4} \)
• C) \( \dfrac{x^2 e^x - 2x e^x}{x^4} \) ✓
• D) \( \dfrac{e^x \cdot 2x}{x^4} \)

Explanation: Choice C is correct. The quotient rule gives \( g'(x) = \dfrac{v \cdot u' - u \cdot v'}{v^2} \). With \( u = e^x \), \( v = x^2 \), \( u' = e^x \), \( v' = 2x \): \( g'(x) = \dfrac{x^2 e^x - 2x e^x}{x^4} \), which simplifies to \( \dfrac{e^x(x-2)}{x^3} \). Choice B is incorrect because it adds instead of subtracts in the numerator, reversing the sign, the most common quotient-rule error. Choice A is incorrect because it comes from rewriting \( g(x) = e^x \cdot x^{-2} \) and differentiating only the \( x^{-2} \) factor while treating \( e^x \) as a constant, giving \( e^x \cdot (-2x^{-3}) \). Choice D is incorrect because it omits \( u' \) entirely, placing only \( u \cdot v' \) in the numerator.

Question 3. Which of the following statements is always true?

• A) If \( f \) is continuous at \( x = a \), then \( f \) is differentiable at \( x = a \).
• B) If \( f \) is differentiable at \( x = a \), then \( f \) is continuous at \( x = a \). ✓
• C) If \( f \) is not differentiable at \( x = a \), then \( f \) is not continuous at \( x = a \).
• D) Differentiability and continuity are independent properties with no implication in either direction.

Explanation: Choice B is correct. Differentiability implies continuity: if \( f'(a) \) exists, then \( f \) must be continuous at \( x = a \). Choice A is incorrect because it reverses the implication; the classic counterexample is \( f(x) = |x| \) at \( x = 0 \), which is continuous but not differentiable due to a corner. Choice C is incorrect because a function can fail to be differentiable (corner, cusp, or vertical tangent) while still being continuous. Choice D is incorrect because the one-directional implication, differentiable implies continuous, is a theorem.

Question 4. The function \( f \) is defined as \( f(x) = x^2 + 1 \) for \( x < 2 \) and \( f(x) = 3x - 1 \) for \( x \geq 2 \). Which of the following correctly describes \( f \) at \( x = 2 \)?

• A) \( f \) is neither continuous nor differentiable at \( x = 2 \).
• B) \( f \) is continuous but not differentiable at \( x = 2 \), because the one-sided derivatives are unequal. ✓
• C) \( f \) is both continuous and differentiable at \( x = 2 \).
• D) \( f \) is differentiable but not continuous at \( x = 2 \).

Explanation: Choice B is correct. Both one-sided limits equal 5 and \( f(2) = 5 \), so \( f \) is continuous at \( x = 2 \). The left-hand derivative is \( 2(2) = 4 \) and the right-hand derivative is \( 3 \); since these are unequal, \( f \) has a corner and is not differentiable there. Choice A is incorrect because continuity holds. Choice C is incorrect because the one-sided derivatives do not match. Choice D is incorrect because differentiability requires continuity, a function cannot be differentiable at a point where it is not continuous.

Question 5. Suppose \( h(x) = f(x) \cdot g(x) \), where \( f(2) = 3 \), \( f'(2) = -1 \), \( g(2) = 4 \), and \( g'(2) = 5 \). What is \( h'(2) \)?

• A) \( -5 \)
• B) \( 11 \) ✓
• C) \( 17 \)
• D) \( -4 \)

Explanation: Choice B is correct. By the product rule, \( h'(x) = f'(x)g(x) + f(x)g'(x) \). At \( x = 2 \): \( h'(2) = (-1)(4) + (3)(5) = -4 + 15 = 11 \). Choice A is incorrect because it results from computing \( f'(2) \cdot g'(2) = (-1)(5) = -5 \), multiplying the two derivatives together instead of applying the product rule. Choice C is incorrect because it drops the negative sign on \( f'(2) \), computing \( (1)(4) + (3)(5) = 17 \). Choice D is incorrect because it uses only the first term of the product rule, \( f'(2) \cdot g(2) = -4 \), and omits the second term entirely.