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. Notation: f″(x), y″ and d²y/dx² are second derivatives; f⁽ⁿ⁾(x) and dⁿy/dxⁿ are nth derivatives. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
Let f(x) = 4x⁵ − 3x² + 7x. What is f″(1)?
- (A) 21
- (B) 74
- (C) 80
- (D) 86
Answer and explanation
Answer: (B). f′(x) = 20x⁴ − 6x + 7. f″(x) = 80x³ − 6. So f″(1) = 80 − 6 = 74.
- (A) is f′(1) = 20 − 6 + 7. It stops after one derivative.
- (C) loses the −6, as if −6x had derivative 0 instead of −6.
- (D) gets the sign of the −6 wrong.
Question 2 (multiple choice · core)
If y = cos(2x), what is d⁴y/dx⁴?
- (A) cos(2x)
- (B) 8 cos(2x)
- (C) 16 cos(2x)
- (D) −16 cos(2x)
Answer and explanation
Answer: (C). Step by step: y′ = −2 sin(2x), y″ = −4 cos(2x), y‴ = 8 sin(2x), y⁽⁴⁾ = 16 cos(2x). Four steps bring the cosine cycle back to cos, and each step multiplies by 2, so the factor is 2⁴ = 16.
- (A) follows the cycle correctly but drops the chain factor 2 at every step.
- (B) multiplies 2 by 4 instead of raising 2 to the power 4.
- (D) has a sign slip: after four steps the cycle returns to +cos, not −cos.
Question 3 (multiple choice · core)
The derivative of a function f is f′(x) = √(x² + 7). What is f″(3)?
- (A) 1/8
- (B) 3/4
- (C) 3/2
- (D) 4
Answer and explanation
Answer: (B). f″ is the derivative of f′. Chain rule: f″(x) = (1/2)(x² + 7)^(−1/2) × 2x = x/√(x² + 7). At x = 3: 3/√16 = 3/4.
- (A) forgets the inner derivative 2x: 1/(2√16) = 1/8.
- (C) forgets the 1/2 from the power rule: 2x/√(x² + 7) = 6/4.
- (D) is f′(3) = √16. It evaluates the given derivative instead of differentiating it.
Question 4 (multiple choice · core)
Let f(x) = e^(−3x). What is f⁽⁵⁾(0)?
- (A) 243
- (B) −15
- (C) −243
- (D) 1
Answer and explanation
Answer: (C). Each derivative multiplies by −3, so f⁽⁵⁾(x) = (−3)⁵ e^(−3x) = −243 e^(−3x). At x = 0, e⁰ = 1, so f⁽⁵⁾(0) = −243.
- (A) loses the sign. An odd power of a negative number is negative.
- (B) multiplies −3 by 5 instead of raising it to the power 5.
- (D) is f(0). It reads f⁽⁵⁾ as the function itself, or forgets the chain factors.
Question 5 (constructed response · core)
A curve is defined by xy + y² = 10.
(a) Show that (3, 2) lies on the curve, and find dy/dx at that point. (b) Differentiate your differentiated equation from (a) once more with respect to x, to obtain an equation involving y″. Then find d²y/dx² at (3, 2). (c) Explain why you cannot treat x + 2y as a constant when differentiating the second time.
Worked solution and suggested Marlbridge rubric
(a) 3 × 2 + 2² = 6 + 4 = 10, so the point is on the curve. Differentiate implicitly. The term xy needs the product rule, and y² the chain rule:
y + x y′ + 2y y′ = 0, so y′ = −y/(x + 2y).
At (3, 2): y′ = −2/(3 + 4) = −2/7.
(b) Differentiate y + x y′ + 2y y′ = 0 again, term by term:
- d/dx [y] = y′
- d/dx [x y′] = y′ + x y″ (product rule)
- d/dx [2y y′] = 2(y′)² + 2y y″ (product rule)
So 2y′ + 2(y′)² + (x + 2y) y″ = 0, which gives
y″ = −2y′(1 + y′)/(x + 2y).
At (3, 2), with y′ = −2/7: 1 + y′ = 5/7, so −2y′(1 + y′) = −2 × (−2/7) × (5/7) = 20/49. Divide by x + 2y = 7:
d²y/dx² = 20/343.
(c) y is a function of x, so x + 2y changes as x changes. Its derivative is 1 + 2y′, not 0. Ignoring this loses terms and gives a wrong y″.
| Point | What earns it |
|---|---|
| 1 | Verifies the point and finds y′ = −2/7 with correct implicit differentiation |
| 1 | Second differentiation uses the product rule on both x y′ and 2y y′ |
| 1 | Correct expression for y″ (any equivalent form) |
| 1 | d²y/dx² = 20/343 at (3, 2) |
| 1 | Explains in (c) that y depends on x, so its derivative is y′, not 0 |
Acceptable alternative for (b): apply the quotient rule to y′ = −y/(x + 2y), then substitute y′. This gives the same value, 20/343.
Question 6 (constructed response · core)
Let f(x) = x|x|.
(a) Write f(x) without absolute value signs, using one formula for x ≥ 0 and one for x < 0. (b) Find f′(x) for x > 0 and for x < 0. Use the definition of the derivative to show that f′(0) = 0. Hence show that f′(x) = 2|x| for all x. (c) Find f″(x) for x ≠ 0. (d) Does f″(0) exist? Justify your answer.
Worked solution and suggested Marlbridge rubric
(a) f(x) = x² for x ≥ 0, and f(x) = −x² for x < 0.
(b) For x > 0, f′(x) = 2x. For x < 0, f′(x) = −2x. At 0, use the definition:
f′(0) = lim (h → 0) [f(h) − f(0)]/h = lim (h → 0) h|h|/h = lim (h → 0) |h| = 0.
So f′(x) = 2x when x > 0, −2x when x < 0, and 0 at x = 0. This is exactly f′(x) = 2|x|.
(c) f″(x) = 2 for x > 0 and f″(x) = −2 for x < 0.
(d) No. Use the definition on f′:
[f′(h) − f′(0)]/h = 2|h|/h, which equals 2 for h > 0 and −2 for h < 0.
The one-sided limits are 2 and −2. They are different, so the limit does not exist and f″(0) does not exist. Graphically, f′ = 2|x| has a corner at x = 0. So f is differentiable everywhere, but f′ is not differentiable at 0.
| Point | What earns it |
|---|---|
| 1 | Correct piecewise form of f |
| 1 | Correct f′ on each side and f′(0) = 0 from the definition |
| 1 | f″(x) = 2 for x > 0 and −2 for x < 0 |
| 1 | Concludes f″(0) does not exist, with unequal one-sided limits (2 and −2) or the corner in f′ as the reason |
Acceptable alternative for (d): an argument that f′(x) = 2|x| has a corner at 0, provided it names the different slopes on each side.
Question 7 (constructed response · stretch)
Let g(x) = x e^(2x).
(a) Find g′(x), g″(x) and g‴(x). Write each in the form (ax + b)e^(2x). (b) Use your answers to suggest a formula for g⁽ⁿ⁾(x). (c) Show that differentiating your formula for g⁽ⁿ⁾(x) gives your formula for g⁽ⁿ⁺¹⁾(x). (d) Find g⁽¹⁰⁾(0).
Worked solution and suggested Marlbridge rubric
(a) Product rule, with the chain rule on e^(2x):
- g′(x) = e^(2x) + 2x e^(2x) = (2x + 1)e^(2x)
- g″(x) = 2e^(2x) + 2(2x + 1)e^(2x) = (4x + 4)e^(2x)
- g‴(x) = 4e^(2x) + 2(4x + 4)e^(2x) = (8x + 12)e^(2x)
(b) The x-coefficients are 2, 4, 8, which is 2ⁿ. The constants are 1, 4, 12, which is n × 2ⁿ⁻¹. So
g⁽ⁿ⁾(x) = (2ⁿx + n·2ⁿ⁻¹)e^(2x) = 2ⁿ⁻¹(2x + n)e^(2x).
(c) Differentiate 2ⁿ⁻¹(2x + n)e^(2x) with the product rule:
2ⁿ⁻¹[2e^(2x) + 2(2x + n)e^(2x)] = 2ⁿ[1 + 2x + n]e^(2x) = 2ⁿ(2x + (n + 1))e^(2x).
This is the formula with n replaced by n + 1, as required.
(d) g⁽¹⁰⁾(0) = 2⁹ × (0 + 10) × e⁰ = 512 × 10 = 5120.
| Point | What earns it |
|---|---|
| 1 | Correct g′ and g″ in the requested form |
| 1 | Correct g‴ = (8x + 12)e^(2x) |
| 1 | A correct general formula, consistent with n = 1, 2, 3 |
| 1 | Correct differentiation of the formula in (c), showing it matches n + 1 |
| 1 | g⁽¹⁰⁾(0) = 5120 |
Note: part (c) shows that the pattern carries on from each step to the next. A full proof by induction is not required in this course.
How did you do?
- Q1 or Q3 wrong: revisit “The idea: differentiate the derivative” in the study guide.
- Q2 or Q4 wrong: reread “Patterns in higher derivatives” and Figure 1. Track the chain factor at every step.
- Q5 wrong: redo Worked example 2, including the alternative method.
- Q6 wrong: see the x^(4/3) example under “The idea”, then compare it with the corner in Q6.
- Q7 wrong: practise the product rule with e^(kx), then write out three derivatives before guessing a pattern.
Then tick off the topic checklist.