What this is for. Use this diagnostic to find which topics of Unit 3, Differentiation: Composite, Implicit, and Inverse Functions, to revisit. There is at least one question per topic, and two each for Topics 3.1, 3.2, 3.5 and 3.6. These are original Marlbridge practice questions, not past exam questions. They are not calibrated, and your result is not a predicted score.
Rules. No calculator; about 30 minutes. Angles are in radians, and every answer is exact. Answer everything before opening any answer. f⁻¹ means the inverse function, not 1/f. The unit is shared by Calculus AB and Calculus BC, and no question here is BC only: all six topics are examined in both courses.
Question 1 (multiple choice · 3.1)
Let f(x) = √(1 + e^(2x)). What is f′(0)?
- (A) 1/(2√2)
- (B) 1/√2
- (C) √2
- (D) 2
Answer and explanation
Answer: (B). Three layers: square root, then 1 + e^(2x), then 2x. f′(x) = 1/(2√(1 + e^(2x))) · e^(2x) · 2 = e^(2x)/√(1 + e^(2x)). At x = 0: 1/√2.
- (A) drops the factor 2 from the innermost layer 2x.
- (C) drops the 1/2 from the square root. (It also happens to equal f(0).)
- (D) is only the derivative of the inside, 2e^(2x), at x = 0.
If you missed this: “More than two layers” in the Topic 3.1 study guide.
Question 2 (multiple choice · 3.1)
A differentiable function f has f′(2) = 4 and f′(5) = −2. Let h(x) = f(3x − 1). What is h′(2)?
- (A) −6
- (B) −2
- (C) 12
- (D) −10
Answer and explanation
Answer: (A). h′(x) = f′(3x − 1) · 3. At x = 2 the inside is 3(2) − 1 = 5, so h′(2) = f′(5) × 3 = −6.
- (B) forgets to multiply by the derivative of the inside, 3.
- (C) evaluates f′ at x = 2 instead of at the inside value 5.
- (D) multiplies by the inside, 5, instead of by its derivative.
If you missed this: the table example in the Topic 3.1 study guide. Find the inside value first.
Question 3 (multiple choice · 3.2)
The point (2, 1) lies on the curve x³ + xy² = 10. What is dy/dx at (2, 1)?
- (A) −13/4
- (B) −3
- (C) −13/2
- (D) −4/13
Answer and explanation
Answer: (A). Differentiate both sides. The term xy² needs the product rule and then the chain rule on y²:
3x² + y² + 2xy · dy/dx = 0, so dy/dx = −(3x² + y²)/(2xy).
At (2, 1): −(12 + 1)/4 = −13/4.
- (B) loses the y² term from the product rule: −12/4.
- (C) loses the factor x in 2xy · dy/dx, dividing by 2y = 2 instead of 2xy = 4.
- (D) is the reciprocal, from dividing the wrong way.
If you missed this: the table of terms under “The key idea” in the Topic 3.2 study guide.
Question 4 (short answer · 3.2)
Consider the curve x·e^y + y = 2.
(a) Show that (2, 0) is on the curve, and find the tangent line there. (b) Find the exact coordinates of the point on the curve where the tangent line is vertical.
Worked answer
(a) 2 × e⁰ + 0 = 2. ✓ Differentiate, with the product rule on x·e^y and the chain rule on e^y:
e^y + x·e^y · dy/dx + dy/dx = 0, so dy/dx = −e^y/(x·e^y + 1).
At (2, 0): dy/dx = −1/(2 + 1) = −1/3. Tangent line: y = −(1/3)(x − 2).
(b) Vertical means the denominator is 0: x·e^y + 1 = 0, so x·e^y = −1. Substitute into the curve: −1 + y = 2, so y = 3 and x = −e^(−3). The numerator −e³ is not 0. The point is (−e^(−3), 3).
If you missed this: “Horizontal and vertical tangent lines” in the Topic 3.2 study guide. Always combine the condition with the curve’s equation.
Question 5 (multiple choice · 3.3)
A function f is differentiable and one-to-one, with f(1) = 4, f(4) = 1, f′(1) = −5 and f′(4) = −2/3. What is the derivative of f⁻¹ at x = 1?
- (A) −1/5
- (B) −3/2
- (C) 3/2
- (D) −2/3
Answer and explanation
Answer: (B). The input of f that gives 1 is 4, since f(4) = 1. So (f⁻¹)′(1) = 1/f′(4) = 1/(−2/3) = −3/2.
- (A) is 1/f′(1). It uses x = 1 as an input of f, but 1 is an output of f.
- (C) changes the sign. Reflecting in y = x keeps the sign of a slope.
- (D) forgets the reciprocal.
If you missed this: the three-step method in the Topic 3.3 study guide.
Question 6 (multiple choice · 3.4)
What is the slope of the graph of y = arcsin(3x) at the point where y = π/6?
- (A) 2/√3
- (B) 2√3
- (C) −2√3
- (D) 12/5
Answer and explanation
Answer: (B). First find x: arcsin(3x) = π/6 means 3x = sin(π/6) = 1/2, so x = 1/6. Then dy/dx = 3/√(1 − 9x²). At x = 1/6: 3/√(3/4) = 3 × 2/√3 = 2√3.
- (A) forgets the chain-rule factor 3.
- (C) uses the arccos formula, which has a minus sign.
- (D) uses the arctan formula, 3/(1 + 9x²).
If you missed this: “Using the chain rule” in the Topic 3.4 study guide.
Question 7 (multiple choice · 3.5)
What is lim (h → 0) [arctan(1 + h) − π/4]/h?
- (A) 0
- (B) 1/2
- (C) π/4
- (D) It does not exist, because h = 0 gives 0/0.
Answer and explanation
Answer: (B). Since arctan 1 = π/4, this is the definition of the derivative of arctan x at x = 1. So it equals 1/(1 + 1²) = 1/2.
- (A) treats 0/0 as 0.
- (C) is arctan 1, the value of the function, not its derivative.
- (D) 0/0 means “find another procedure”, not “no limit”.
If you missed this: “A limit that is a derivative in disguise” in the Topic 3.5 study guide.
Question 8 (short answer · 3.5)
Let k(x) = x² · arctan(√x) for x > 0.
(a) What is the last operation in k(x)? Name every rule you need, in order. (b) Find k′(1).
Worked answer
(a) The last operation is a product, so start with the product rule. The second factor is a composite (arctan of √x), so it needs the arctan formula with the chain rule. The derivative of √x = x^(1/2) uses the power rule.
(b) k′(x) = 2x · arctan(√x) + x² · [1/(1 + x)] · [1/(2√x)].
At x = 1: 2 × π/4 + 1 × (1/2) × (1/2) = π/2 + 1/4.
If you missed this: “Step 2: find the last operation” in the Topic 3.5 study guide.
Question 9 (multiple choice · 3.6)
Let f(x) = (2x + 1)⁵. What is f‴(0)?
- (A) 60
- (B) 120
- (C) 480
- (D) 80
Answer and explanation
Answer: (C). Each derivative multiplies by the chain factor 2. f′(x) = 10(2x + 1)⁴, f″(x) = 80(2x + 1)³, f‴(x) = 480(2x + 1)². At x = 0: 480.
- (A) is 5 × 4 × 3, with no chain factors at all.
- (B) includes the factor 2 only once.
- (D) is f″(0): it stops one derivative early.
If you missed this: “Patterns in higher derivatives” in the Topic 3.6 study guide.
Question 10 (short answer · 3.6)
Let y = e^(kx), where k is a constant.
(a) Find dy/dx and d²y/dx². (b) Find every value of k for which d²y/dx² − dy/dx − 6y = 0 for all x. (c) For each value of k from (b), find d³y/dx³ at x = 0.
Worked answer
(a) dy/dx = k·e^(kx) and d²y/dx² = k²·e^(kx).
(b) Substitute: (k² − k − 6)e^(kx) = 0. Since e^(kx) is never 0, k² − k − 6 = 0, so (k − 3)(k + 2) = 0: k = 3 or k = −2.
(c) d³y/dx³ = k³·e^(kx), which is k³ at x = 0: 27 for k = 3 and −8 for k = −2.
If you missed this: the notation table and “Exponentials repeat with a factor” in the Topic 3.6 study guide.
Your next step
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 3.1 The chain rule | 1, 2 | Guide 3.1 |
| 3.2 Implicit differentiation | 3, 4 | Guide 3.2 |
| 3.3 Inverse functions | 5 | Guide 3.3 |
| 3.4 Inverse trig functions | 6 | Guide 3.4 |
| 3.5 Selecting procedures | 7, 8 | Guide 3.5 |
| 3.6 Higher-order derivatives | 9, 10 | Guide 3.6 |
How to use your result
- Mark each topic secure, shaky (unsure or a slip) or gap (wrong).
- Fix a gap in Topic 3.1 first. Every other topic in the unit uses the chain rule, so a slip there spreads.
- For a gap, read the guide, then do the topic’s practice set.
- For a shaky topic, look at which wrong option you chose. Each explanation names the slip behind it.
- Then try the Unit 3 mixed review, where each question combines topics.