These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assumptions for every question: no calculator, angles in radians, ln means the natural logarithm, and exact answers unless stated. The functions f and g in the table questions are invented for practice and are differentiable everywhere. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
What is d/dx (x⁴ cos x)?
- (A) −4x³ sin x
- (B) 4x³ cos x − x⁴ sin x
- (C) 4x³ cos x + x⁴ sin x
- (D) 4x³ cos x
Answer and explanation
Answer: (B). With f(x) = x⁴ and g(x) = cos x: f′(x)g(x) + f(x)g′(x) = 4x³ cos x + x⁴(−sin x) = 4x³ cos x − x⁴ sin x.
- (A) multiplies the derivatives, 4x³ · (−sin x). The derivative of a product is not the product of the derivatives.
- (C) uses d/dx cos x = +sin x, losing the minus sign.
- (D) treats cos x as a constant and keeps only the first term.
Question 2 (multiple choice · core)
The table gives values of f, g and their derivatives at x = 1.
| x | f(x) | f′(x) | g(x) | g′(x) |
|---|---|---|---|---|
| 1 | 2 | −3 | −1 | 4 |
If P(x) = f(x) · g(x), what is P′(1)?
- (A) −12
- (B) −5
- (C) 11
- (D) −2
Answer and explanation
Answer: (C). P′(1) = f′(1)g(1) + f(1)g′(1) = (−3)(−1) + (2)(4) = 3 + 8 = 11.
- (A) is f′(1) · g′(1) = (−3)(4), the product of the derivatives.
- (B) subtracts the two terms, 3 − 8. That minus sign belongs to the quotient rule, not the product rule.
- (D) is P(1) = f(1) · g(1) = (2)(−1), the value of the product rather than its slope.
Question 3 (multiple choice · core)
If y = x² ln x, what is the value of dy/dx at x = e?
- (A) 2
- (B) e
- (C) 3e
- (D) e²
Answer and explanation
Answer: (C). dy/dx = 2x · ln x + x² · (1/x) = 2x ln x + x. At x = e: 2e(1) + e = 3e.
- (A) multiplies the derivatives: 2x · (1/x) = 2.
- (B) subtracts the terms, 2e − e. The product rule adds.
- (D) is y(e) = e² · ln e = e², the value of the function, not the slope.
Question 4 (multiple choice · core)
For 0 < x < π, at which value of x does the graph of y = eˣ sin x have a horizontal tangent line?
- (A) π/4
- (B) π/2
- (C) 3π/4
- (D) π
Answer and explanation
Answer: (C). dy/dx = eˣ sin x + eˣ cos x = eˣ(sin x + cos x). Since eˣ > 0, dy/dx = 0 only when sin x + cos x = 0, that is sin x = −cos x. In 0 < x < π this is x = 3π/4.
- (A) solves sin x − cos x = 0. That comes from the sign error d/dx sin x = −cos x, which turns the derivative into eˣ(sin x − cos x).
- (B) is where sin x alone has a horizontal tangent. The factor eˣ also changes, so its derivative contributes a term.
- (D) is where y = 0 (and it is outside the open interval). A zero of y is not a zero of the slope.
Question 5 (constructed response · core)
Let m(x) = (2x + 1)eˣ.
(a) Show that m′(x) = (2x + 3)eˣ. (b) Find the exact coordinates of the point where the graph of m has a horizontal tangent line. (c) Find the equation of the tangent line to the graph of m at x = 0.
Worked solution and suggested Marlbridge rubric
(a) Factors: 2x + 1 (derivative 2) and eˣ (derivative eˣ). m′(x) = 2 · eˣ + (2x + 1) · eˣ = eˣ(2 + 2x + 1) = (2x + 3)eˣ.
(b) eˣ is never 0, so m′(x) = 0 only when 2x + 3 = 0, which gives x = −3/2. Then m(−3/2) = (2(−3/2) + 1)e^(−3/2) = (−3 + 1)e^(−3/2) = −2e^(−3/2), about −0.446. The point is (−3/2, −2e^(−3/2)).
(c) m(0) = (0 + 1)e⁰ = 1. m′(0) = (0 + 3)e⁰ = 3. Tangent line: y = 1 + 3x.
| Point | What earns it |
|---|---|
| 1 | Correct use of the product rule, with both terms shown, and factoring to (2x + 3)eˣ |
| 1 | Sets m′(x) = 0, reasons that eˣ ≠ 0, and finds x = −3/2 |
| 1 | Correct y-coordinate −2e^(−3/2) (a decimal alone does not earn the point; exact form is required) |
| 1 | Tangent line y = 1 + 3x (any equivalent form) |
Question 6 (constructed response · core)
The table gives values of f, g and their derivatives at x = 2.
| x | f(x) | f′(x) | g(x) | g′(x) |
|---|---|---|---|---|
| 2 | −1 | 3 | 5 | −4 |
(a) Let P(x) = f(x) · g(x). Find P′(2). (b) Let R(x) = 3x · f(x). Find R′(2). (c) Let S(x) = [f(x)]². Find S′(2) by writing S as a product. (d) A student claims P′(2) = f′(2) · g′(2). Use your answer to (a) to show the claim is false.
Worked solution and suggested Marlbridge rubric
(a) P′(2) = f′(2)g(2) + f(2)g′(2) = (3)(5) + (−1)(−4) = 15 + 4 = 19.
(b) The factors are 3x (derivative 3) and f(x). R′(x) = 3 · f(x) + 3x · f′(x). So R′(2) = 3(−1) + 3(2)(3) = −3 + 18 = 15.
(c) S(x) = f(x) · f(x). By the product rule, S′(x) = f′(x)f(x) + f(x)f′(x) = 2f(x)f′(x). So S′(2) = 2(−1)(3) = −6.
(d) f′(2) · g′(2) = (3)(−4) = −12, but P′(2) = 19. The two values are different, so (f · g)′ ≠ f′ · g′ in general. One counterexample is enough to show a general claim is false.
| Point | What earns it |
|---|---|
| 1 | P′(2) = 19 with the product rule written out |
| 1 | R′(2) = 15, treating 3x as a factor with derivative 3 |
| 1 | S′(2) = −6 via S = f · f |
| 1 | Computes f′(2)g′(2) = −12 and compares it with 19 to reject the claim |
Question 7 (constructed response · stretch)
(a) Use the product rule twice to show that for differentiable u, v and w, (u · v · w)′ = u′ · v · w + u · v′ · w + u · v · w′. (b) Let y = x eˣ cos x. Find dy/dx. (c) Find the equation of the tangent line to y = x eˣ cos x at x = 0.
Worked solution and suggested Marlbridge rubric
(a) Group the product as (u · v) · w. Apply the product rule with first factor u · v and second factor w:
(u · v · w)′ = (u · v)′ · w + (u · v) · w′.
Apply the product rule again to (u · v)′ = u′ · v + u · v′:
(u · v · w)′ = (u′ · v + u · v′) · w + u · v · w′ = u′vw + uv′w + uvw′.
(b) u = x, v = eˣ, w = cos x, with u′ = 1, v′ = eˣ, w′ = −sin x.
dy/dx = eˣ cos x + x eˣ cos x − x eˣ sin x
(c) At x = 0: y = 0 · 1 · 1 = 0. dy/dx = (1)(1) + 0 − 0 = 1. The tangent line is y = x.
| Point | What earns it |
|---|---|
| 1 | Groups two factors and applies the product rule once, correctly |
| 1 | Applies it a second time and reaches the three-term formula |
| 1 | Correct dy/dx in (b), including the minus sign from d/dx cos x |
| 1 | Point (0, 0), slope 1 and the line y = x |
Acceptable alternative for (a): group as u · (v · w) instead. The result is the same.
How did you do?
- Q1 or Q3 wrong: revisit “The rule” and “Why the derivative of a product is not f′ · g′” in the study guide.
- Q2 or Q6 wrong: redo Worked example 2 (tables). Keep the value f(a)g(a) separate from the slope.
- Q4 or Q5 wrong: redo Worked examples 1 and 3. Factor out eˣ and use the fact that it is never 0.
- Q7 wrong: reread “Three factors” under “When to use it, and when not to”.
Then tick off the topic checklist.