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, standard ranges for the inverse trigonometric functions, and exact answers unless stated. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
What is d/dx arctan(5x)?
- (A) 1/(1 + 25x²)
- (B) 5/(1 + 25x²)
- (C) 5/(1 + 5x²)
- (D) 5/√(1 − 25x²)
Answer and explanation
Answer: (B). With u = 5x and u′ = 5, d/dx arctan u = u′/(1 + u²) = 5/(1 + (5x)²) = 5/(1 + 25x²).
- (A) forgets the chain-rule factor u′ = 5.
- (C) squares only the x, writing 5x² instead of (5x)² = 25x².
- (D) uses the arcsin formula instead of the arctan formula.
Question 2 (multiple choice · core)
Let f(x) = arccos(2x). What is f′(1/4)?
- (A) −4/√3
- (B) 4/√3
- (C) −2/√3
- (D) −8/√15
Answer and explanation
Answer: (A). f′(x) = −2/√(1 − (2x)²) = −2/√(1 − 4x²). At x = 1/4, 1 − 4/16 = 3/4 and √(3/4) = √3/2. So f′(1/4) = −2/(√3/2) = −4/√3, which is −4√3/3.
- (B) drops the minus sign. arccos is decreasing, so its derivative is negative.
- (C) forgets the chain-rule factor 2: −1/(√3/2) = −2/√3.
- (D) keeps the factor 2 on top but puts x instead of 2x under the root: −2/√(1 − x²) at x = 1/4 is −2/√(15/16) = −8/√15.
Question 3 (multiple choice · core)
Sine, restricted to [−π/2, π/2], has inverse arcsin. Using the inverse-function rule, which expression equals the derivative of arcsin x at x = √3/2?
- (A) 1/cos(π/3)
- (B) cos(π/3)
- (C) 1/cos(√3/2)
- (D) 1/sin(π/3)
Answer and explanation
Answer: (A). arcsin(√3/2) = π/3, because sin(π/3) = √3/2. The inverse-function rule gives 1/sin′(π/3) = 1/cos(π/3) = 1/(1/2) = 2. The formula 1/√(1 − 3/4) = 1/(1/2) = 2 agrees.
- (B) forgets to take the reciprocal.
- (C) evaluates cosine at the input √3/2 instead of at the matching angle π/3.
- (D) uses sin as if it were the derivative of sin. The derivative of sin is cos.
Question 4 (multiple choice · core)
Let g(x) = arcsin(x − 2). For which values of x does g′(x) exist?
- (A) 1 ≤ x ≤ 3
- (B) 1 < x < 3
- (C) −1 < x < 1
- (D) all real x
Answer and explanation
Answer: (B). g′(x) = 1/√(1 − (x − 2)²). This needs 1 − (x − 2)² > 0, so −1 < x − 2 < 1, which gives 1 < x < 3.
- (A) is where g is defined. At the endpoints x = 1 and x = 3 the denominator is 0, and the graph has vertical tangents.
- (C) is the interval for arcsin x itself. It ignores the shift by 2.
- (D) confuses arcsin with arctan, which is differentiable everywhere.
Question 5 (constructed response · core)
Let f(x) = arctan(x/3).
(a) Find f′(x), and simplify it to a single fraction with no fractions inside it. (b) Write an equation for the line tangent to the graph of f at x = 3. (c) Explain why f is increasing for all x. (d) Find the greatest value of f′(x), and say where it occurs. What happens to f′(x) as x becomes very large?
Worked solution and suggested Marlbridge rubric
(a) With u = x/3 and u′ = 1/3: f′(x) = (1/3)/(1 + x²/9). Multiply top and bottom by 9: f′(x) = 3/(9 + x²).
(b) f(3) = arctan 1 = π/4. f′(3) = 3/(9 + 9) = 1/6. The tangent line is y = π/4 + (1/6)(x − 3).
(c) The numerator 3 is positive and 9 + x² ≥ 9 > 0, so f′(x) > 0 for every x. A function with a positive derivative everywhere is increasing.
(d) The fraction 3/(9 + x²) is largest when the denominator is smallest, which is at x = 0. The greatest value is f′(0) = 3/9 = 1/3. As x becomes very large (in either direction), 9 + x² grows without bound, so f′(x) gets close to 0: the graph levels off.
| Point | What earns it |
|---|---|
| 1 | Correct chain-rule derivative, in any equivalent form |
| 1 | Simplified to 3/(9 + x²) |
| 1 | f(3) = π/4 and slope 1/6 |
| 1 | Tangent line y = π/4 + (1/6)(x − 3), or equivalent |
| 1 | f′ > 0 for all x because numerator and denominator are both positive |
| 1 | Maximum slope 1/3 at x = 0, and f′ → 0 as |x| grows |
Question 6 (constructed response · core)
(a) Let y = arccos x, for −1 < x < 1, so that cos y = x with 0 < y < π. Use implicit differentiation to show that dy/dx = −1/√(1 − x²). Justify the sign of the square root. (b) Confirm your result using the identity arcsin x + arccos x = π/2. (c) Explain why arccos is not differentiable at x = 1. (d) Find the derivative of arccos x at x = 0, and explain what it means for the graph.
Worked solution and suggested Marlbridge rubric
(a) Differentiate cos y = x with respect to x: −sin y · dy/dx = 1, so dy/dx = −1/sin y. From sin²y + cos²y = 1, sin y = ±√(1 − x²). Since 0 < y < π, sin y > 0, so sin y = +√(1 − x²). Therefore dy/dx = −1/√(1 − x²).
(b) Differentiate the identity: d/dx arcsin x + d/dx arccos x = 0. So d/dx arccos x = −d/dx arcsin x = −1/√(1 − x²). This agrees with (a).
(c) At x = 1 the denominator √(1 − x²) is 0, so the formula is undefined. Geometrically, arccos(1) = 0, and cos has a horizontal tangent at y = 0 (since −sin 0 = 0). The reflection of a horizontal tangent is vertical, so the graph of arccos has a vertical tangent at (1, 0).
(d) At x = 0: −1/√1 = −1. The graph of arccos crosses the y-axis at (0, π/2) with slope −1.
| Point | What earns it |
|---|---|
| 1 | Implicit differentiation giving −sin y · dy/dx = 1 |
| 1 | Replaces sin y by √(1 − x²) with the reason sin y > 0 for 0 < y < π |
| 1 | Correct confirmation from the identity |
| 1 | Explains non-differentiability at x = 1 (zero denominator or vertical tangent) |
| 1 | Derivative −1 at x = 0, with the point (0, π/2) |
Acceptable alternative for (a): the inverse-function rule, 1/cos′(arccos x) = 1/(−sin(arccos x)) = −1/√(1 − x²), with the same sign argument.
Question 7 (constructed response · stretch)
A function h is differentiable for all x. You are told that h(2) = 1/2, h′(2) = 3 and h(5) = 1.2. Let
- g(x) = arcsin(h(x))
- k(x) = arctan(h(x))
(a) Find g′(2), and give your answer in the form a√3. (b) Find k′(2). (c) Write an equation for the line tangent to the graph of g at x = 2. (d) Explain why g(5) is not defined, but k(5) is.
Worked solution and suggested Marlbridge rubric
(a) By the chain rule, g′(x) = h′(x)/√(1 − [h(x)]²). At x = 2: g′(2) = 3/√(1 − 1/4) = 3/(√3/2) = 6/√3 = 2√3.
(b) k′(x) = h′(x)/(1 + [h(x)]²). At x = 2: k′(2) = 3/(1 + 1/4) = 3/(5/4) = 12/5.
(c) g(2) = arcsin(1/2) = π/6. The tangent line is y = π/6 + 2√3(x − 2).
(d) arcsin only accepts inputs from −1 to 1, because sine never goes outside that interval. Since h(5) = 1.2 > 1, arcsin(1.2) does not exist, so g(5) is not defined. arctan accepts every real number, so k(5) = arctan(1.2) is defined.
| Point | What earns it |
|---|---|
| 1 | Chain-rule form for g′ with h′(x) on top and [h(x)]² under the root |
| 1 | g′(2) = 2√3 |
| 1 | k′(2) = 12/5 |
| 1 | Tangent line through (2, π/6) with slope 2√3 |
| 1 | Domain argument: arcsin needs inputs in [−1, 1]; arctan needs no restriction |
A common error in (a) is 1/√(1 − 1/4) = 2/√3, which forgets the factor h′(2) = 3.
How did you do?
- Q1, Q2 or Q7(a)–(b) wrong: revisit “Using the chain rule” and Worked example 1 in the study guide. Square the whole inner function.
- Q3 or Q6 wrong: reread “Deriving the derivative of arcsin x” and Worked example 3. The inverse-function rule needs the matching angle.
- Q4 or Q7(d) wrong: see “Why the ranges are restricted” and “Where the derivatives fail”.
- Q5 wrong: redo Worked examples 1 and 2, including the tangent line.
Then tick off the topic checklist.