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, and exact answers unless stated. All contexts and data are invented. Notation: f′(x) and dy/dx both mean the derivative with respect to x; sin³x means (sin x)³. This set is for both Calculus AB and Calculus BC students.
Questions 3 and 6 use this table of values for two differentiable functions f and g.
| x | f(x) | f′(x) | g(x) | g′(x) |
|---|---|---|---|---|
| 0 | 2 | −3 | 1 | 4 |
| 1 | 3 | 5 | 2 | −2 |
| 2 | 0 | 1 | 0 | 6 |
Question 1 (multiple choice · foundation)
What is d/dx [(5 − 2x)⁶]?
- (A) 6(5 − 2x)⁵
- (B) −12(5 − 2x)⁵
- (C) 12(5 − 2x)⁵
- (D) −12(5 − 2x)⁶
Answer and explanation
Answer: (B). Outer: u⁶, inner: u = 5 − 2x. The derivative of the outer is 6(5 − 2x)⁵, and the derivative of the inside is −2. Multiply: 6(5 − 2x)⁵ · (−2) = −12(5 − 2x)⁵.
- (A) forgets to multiply by the derivative of the inside.
- (C) uses +2 as the derivative of 5 − 2x, losing the minus sign.
- (D) multiplies by 6 and by −2 but forgets to lower the power from 6 to 5.
Question 2 (multiple choice · core)
Let f(x) = ln(x² + 4). What is f′(2)?
- (A) 1/8
- (B) 1/2
- (C) 4
- (D) ln 8
Answer and explanation
Answer: (B). Outer: ln u, inner: u = x² + 4. f′(x) = (1/(x² + 4)) · 2x = 2x/(x² + 4). At x = 2: 4/8 = 1/2.
- (A) is 1/(x² + 4) at x = 2: the derivative of the outer function without the factor 2x.
- (C) is 2x at x = 2: only the derivative of the inside.
- (D) is f(2), the value of the function, not its derivative.
Question 3 (multiple choice · core)
Use the table above. Let h(x) = f(g(x)). What is h′(0)?
- (A) −18
- (B) −12
- (C) 5
- (D) 20
Answer and explanation
Answer: (D). h′(0) = f′(g(0)) · g′(0). Inside first: g(0) = 1, so f′(g(0)) = f′(1) = 5. Then g′(0) = 4. So h′(0) = 5 × 4 = 20.
- (A) reverses the composition: g′(f(0)) · f′(0) = g′(2) · (−3) = 6 × (−3) = −18. That is the derivative of g(f(x)).
- (B) evaluates the outer derivative at 0 instead of at g(0): f′(0) · g′(0) = (−3)(4) = −12.
- (C) is f′(g(0)) alone: it forgets to multiply by g′(0).
Question 4 (multiple choice · core)
In a fictional model, the cost C (in dollars) of heating a tank of water depends on the water temperature T (in °C), with dC/dT = 0.40 dollars per °C at the current temperature. The temperature is rising at 1.5 °C per minute. At what rate is the cost changing with respect to time?
- (A) 0.27 dollars per minute
- (B) 0.60 dollars per minute
- (C) 1.90 dollars per minute
- (D) 3.75 dollars per minute
Answer and explanation
Answer: (B). Chain rule in Leibniz form: dC/dt = (dC/dT) · (dT/dt) = 0.40 × 1.5 = 0.60. Units: (dollars/°C) × (°C/minute) = dollars/minute.
- (A) divides, 0.40/1.5 ≈ 0.27. The units would be dollars·minute/°C², not dollars per minute.
- (C) adds the rates, 0.40 + 1.5. Rates with different units cannot be added.
- (D) divides the other way, 1.5/0.40 = 3.75, giving °C² per dollar-minute.
Question 5 (error analysis · core)
A student is asked to differentiate y = e^(cos 2x). The student writes:
dy/dx = e^(cos 2x) · (−sin 2x)
(a) Identify the layers of y, from outside to inside. (b) Explain the student’s error and write the correct derivative. (c) Find the exact value of dy/dx at x = π/4, and compare it with the value given by the student’s answer.
Worked solution and suggested Marlbridge rubric
(a) Three layers: the exponential eᵘ (outside), then cos v, then 2x (inside).
(b) The student differentiated the outer two layers correctly but stopped one layer too early. The innermost function 2x has derivative 2, and that factor is missing. Correct:
dy/dx = e^(cos 2x) · (−sin 2x) · 2 = −2 sin(2x) e^(cos 2x)
(c) At x = π/4: 2x = π/2, cos(π/2) = 0 and sin(π/2) = 1. So dy/dx = −2 × 1 × e⁰ = −2. The student’s answer gives −1 × e⁰ = −1, which is half the correct value.
| Point | What earns it |
|---|---|
| 1 | Names the missing factor: the derivative of the innermost function 2x, which is 2 |
| 1 | Correct derivative −2 sin(2x) e^(cos 2x) (any equivalent form) |
| 1 | dy/dx = −2 at x = π/4, with cos(π/2) = 0 and sin(π/2) = 1 shown |
Part (a) supports part (b) and is not scored separately.
Question 6 (constructed response · core)
Use the table above.
(a) Let p(x) = [f(x)]³. Find p′(1). (b) Let q(x) = g(x²). Find q′(1). (c) Let r(x) = x · f(g(x)). Find r′(1).
Worked solution and suggested Marlbridge rubric
(a) Outer: u³, inner: f(x). p′(x) = 3[f(x)]² · f′(x). So p′(1) = 3 × 3² × 5 = 135.
(b) Outer: g, inner: x². q′(x) = g′(x²) · 2x. So q′(1) = g′(1) × 2 = (−2)(2) = −4.
(c) r is a product, and its second factor is a composite. Product rule, then chain rule:
r′(x) = 1 · f(g(x)) + x · f′(g(x)) · g′(x)
At x = 1: g(1) = 2, so f(g(1)) = f(2) = 0 and f′(g(1)) = f′(2) = 1. With g′(1) = −2:
r′(1) = 0 + 1 × 1 × (−2) = −2
| Point | What earns it |
|---|---|
| 1 | (a) Uses 3[f(x)]² f′(x) and gets 135 |
| 1 | (b) Includes the factor 2x from the inner function x² and gets −4 |
| 1 | (c) Applies the product rule with the chain rule in the second term |
| 1 | (c) Evaluates f and f′ at g(1) = 2 (not at 1) and gets −2 |
Common error in (b): writing q′(1) = g′(1) = −2, which drops the factor 2x. Common error in (c): using f(1) = 3 and f′(1) = 5, which evaluates at x instead of at g(x).
Question 7 (constructed response · stretch)
Let g(x) = (x² − 4x)³.
(a) Find g′(x) and write it as a product of linear factors and a constant. (b) Find every x-value where the graph of g has a horizontal tangent line, and give the point on the graph for each. (c) Find the equation of the tangent line to the graph of g at x = 1. (d) Explain why expanding (x² − 4x)³ before differentiating is a poor strategy here, even though it gives the same answer.
Worked solution and suggested Marlbridge rubric
(a) Outer: u³, inner: u = x² − 4x, with derivative 2x − 4.
g′(x) = 3(x² − 4x)² · (2x − 4)
Factor: x² − 4x = x(x − 4), so (x² − 4x)² = x²(x − 4)², and 2x − 4 = 2(x − 2). So
g′(x) = 6x²(x − 4)²(x − 2)
(b) g′(x) = 0 when x = 0, x = 4 or x = 2. Points: g(0) = 0, g(4) = 0, g(2) = (4 − 8)³ = (−4)³ = −64. So (0, 0), (2, −64) and (4, 0).
(c) g(1) = (1 − 4)³ = −27. g′(1) = 6 × 1 × 9 × (−1) = −54. Tangent line: y = −27 − 54(x − 1), which is y = 27 − 54x.
(d) Expanding gives x⁶ − 12x⁵ + 48x⁴ − 64x³. Differentiating that is correct, but it takes longer, gives more chances for arithmetic slips, and hides the factored form you need in (b). The chain rule keeps the factor (x² − 4x) visible.
| Point | What earns it |
|---|---|
| 1 | Correct chain rule: 3(x² − 4x)²(2x − 4) |
| 1 | Correct fully factored form 6x²(x − 4)²(x − 2) |
| 1 | All three x-values and the points (0, 0), (2, −64), (4, 0) |
| 1 | Tangent line at x = 1 with point (1, −27) and slope −54 |
Part (d) is for discussion and is not scored. An answer to (b) that finds only x = 2 (setting 2x − 4 = 0) loses the third point: the factor (x² − 4x)² is also 0 at x = 0 and x = 4.
How did you do?
- Q1 or Q2 wrong: reread “The chain rule” in the study guide and say “times the derivative of the inside” aloud.
- Q3 or Q6 wrong: redo Worked example 2 (tables). Find g(a) before you look up f′.
- Q4 wrong: see “Leibniz form and units” and Figure 1.
- Q5 or Q7 wrong: see “More than two layers” and Worked example 3 on combining rules.
Then tick off the topic checklist.