Use this page after studying the Unit 2 topics or after the Unit 2 diagnostic. The unit is shared by Calculus AB and Calculus BC; no question here is BC only. These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric, not official scoring. No calculator.
Big ideas of the unit
- A derivative is a limit of average rates. The difference quotient (f(a + h) − f(a))/h is an average rate over a small interval; its limit as h → 0 is f′(a), the rate at an instant (Topic 2.1).
- The derivative is also a function. Letting x vary gives f′(x), written dy/dx, f′(x) or y′. Its value at a is the slope of the tangent line at (a, f(a)) (Topic 2.2).
- Without a formula, you estimate. From a table, use the closest points around a; from a graph, the tangent slope; with technology, the numerical derivative (Topic 2.3).
- Differentiable implies continuous, not the other way round. A break rules out a derivative. Corners, cusps and vertical tangents are continuous points with no derivative (Topic 2.4).
- Rules replace the limit. The power rule handles xʳ for any real r once you rewrite roots and fractions as powers (Topic 2.5). Constants, sums and constant multiples then give every polynomial (Topic 2.6).
- Four special derivatives: sin x → cos x, cos x → −sin x, eˣ → eˣ, ln x → 1/x. Seeing a difference quotient as one of these lets you find a limit fast (Topic 2.7).
- Products and quotients have their own rules. (fg)′ = f′g + fg′ and (f/g)′ = (gf′ − fg′)/g². Neither is f′g′ or f′/g′ (Topic 2.8, Topic 2.9).
- The other four trig derivatives follow from the quotient rule. Write tan, cot, sec and csc using sin and cos, then simplify with sin²x + cos²x = 1 (Topic 2.10).
Key relationships and methods
| You see or need | What it means or what to do | Topics |
|---|---|---|
| (f(b) − f(a))/(b − a) | Average rate of change over [a, b] | 2.1 |
| lim (h → 0) (f(a + h) − f(a))/h or lim (x → a) (f(x) − f(a))/(x − a) | f′(a); if you know the derivative of f, read the limit as f′(a) | 2.1, 2.7 |
| Tangent line at x = a | Point (a, f(a)), slope f′(a): y − f(a) = f′(a)(x − a) | 2.2 |
| A table, no formula | Difference quotient on the shortest interval containing a, with units | 2.3 |
| Piecewise function or absolute value at a join | Check continuity first, then both one-sided limits of the difference quotient | 2.4 |
| Roots, or x in a denominator | Rewrite as xʳ, then power rule | 2.5 |
| Sum or constant multiple | Differentiate term by term; a constant term gives 0 | 2.6 |
| sin, cos, eˣ, ln x | Use the four known derivatives | 2.7 |
| f · g | f′g + fg′ | 2.8 |
| f / g | (gf′ − fg′)/g²; but if the bottom is a single power, split or rewrite first | 2.9, 2.5 |
| tan, cot, sec, csc | sec²x, −csc²x, sec x tan x, −csc x cot x | 2.10 |
| “Horizontal tangent” | Solve f′(x) = 0 in the domain | 2.2, 2.6–2.10 |
Question 1 (multiple choice · mixed)
What is lim (h → 0) ((2 + h)e^(2 + h) − 2e²)/h?
- (A) e²
- (B) 2e²
- (C) 3e²
- (D) It does not exist, because substitution gives 0/0.
Answer and explanation
Answer: (C). The limit is the h-form of f′(2) for f(x) = xeˣ, since f(2) = 2e². By the product rule, f′(x) = 1·eˣ + x·eˣ = (x + 1)eˣ, so f′(2) = 3e².
- (A) keeps only the first product-rule term, 1·eˣ.
- (B) is f(2), the value, not the slope (it is also the second product-rule term on its own).
- (D) 0/0 is normal for a derivative written as a limit; it does not mean the limit fails.
Topics: 2.1, 2.7, 2.8.
Question 2 (multiple choice · mixed)
Let k(x) = ∛x · (x − 4), defined for all real x. Which statement is true?
- (A) k has horizontal tangents at both x = 0 and x = 1.
- (B) k has a horizontal tangent at x = 1, and k′(0) does not exist because the graph has a vertical tangent at x = 0.
- (C) k′(0) does not exist because k is not continuous at x = 0.
- (D) k′(x) = (1/3)x^(−2/3), which is never 0, so k has no horizontal tangent.
Answer and explanation
Answer: (B). Expand: k(x) = x^(4/3) − 4x^(1/3). By the power and sum rules, k′(x) = (4/3)x^(1/3) − (4/3)x^(−2/3) = 4(x − 1)/(3x^(2/3)) for x ≠ 0. This is 0 at x = 1. At x = 0, k is continuous (k(0) = 0), but the difference quotient k(h)/h = (h − 4)/h^(2/3) tends to −∞ from both sides: a vertical tangent, so k′(0) does not exist.
- (A) At x = 0 the derivative is undefined, not 0.
- (C) k is a product of continuous functions, so it is continuous at 0.
- (D) multiplies the derivatives of the two factors, (1/3)x^(−2/3) × 1.
Topics: 2.4, 2.5, 2.6, 2.8.
Question 3 (multiple choice · mixed)
Which is an equation of the line tangent to y = cos x/(1 + sin x) at x = 0?
- (A) y = 1 + x
- (B) y = −x
- (C) y = 1
- (D) y = 1 − x
Answer and explanation
Answer: (D). Quotient rule:
dy/dx = [(1 + sin x)(−sin x) − cos x · cos x]/(1 + sin x)² = −(sin x + sin²x + cos²x)/(1 + sin x)² = −(1 + sin x)/(1 + sin x)² = −1/(1 + sin x).
At x = 0: y = 1/1 = 1 and dy/dx = −1. So y − 1 = −1(x − 0), which is y = 1 − x.
- (A) reverses the top of the quotient rule, giving slope +1.
- (B) uses the slope but the point (0, 0) instead of (0, 1).
- (C) divides the derivatives, −sin x / cos x, which is 0 at x = 0.
Topics: 2.2, 2.7, 2.9, 2.10.
Question 4 (constructed response · mixed)
A function f is differentiable for all x. Selected values are shown (invented data). You are also told that f′(3) = 2 exactly.
| x | 0 | 2 | 3 | 5 | 6 |
|---|---|---|---|---|---|
| f(x) | 1 | 5 | 6 | 10 | 16 |
(a) Find the average rate of change of f over [0, 6]. (b) Use the table to estimate f′(4). Show the difference quotient you use. (c) Let g(x) = f(x) · ln x for x > 0. Find g′(3). (d) Let q(x) = f(x)/(x + 1). Find q′(3). (e) Write an equation of the line tangent to the graph of q at x = 3.
Worked solution and suggested Marlbridge rubric
(a) (f(6) − f(0))/(6 − 0) = (16 − 1)/6 = 5/2.
(b) The shortest interval in the table containing 4 is [3, 5]: f′(4) ≈ (f(5) − f(3))/(5 − 3) = (10 − 6)/2 = 2.
(c) Product rule: g′(x) = f′(x) · ln x + f(x) · (1/x). At x = 3: g′(3) = 2 ln 3 + 6/3 = 2 + 2 ln 3.
(d) Quotient rule, with bottom x + 1 and its derivative 1: q′(3) = [(3 + 1) f′(3) − f(3) · 1]/(3 + 1)² = (4 × 2 − 6)/16 = 2/16 = 1/8.
(e) q(3) = 6/4 = 3/2. Tangent line: y − 3/2 = (1/8)(x − 3).
| Point | What earns it |
|---|---|
| 1 | (a): 5/2, from f(6) and f(0) |
| 1 | (b): an interval containing 4, with the quotient shown (2) |
| 1 | (c): product rule set up with f(3) and f′(3) in the right places |
| 1 | (c): d/dx ln x = 1/x used, and g′(3) = 2 + 2 ln 3 |
| 1 | (d): q′(3) = 1/8, with the quotient rule numerator in the correct order |
| 1 | (e): tangent line through (3, 3/2) with slope 1/8 |
Total: 6 points. Topics: 2.1, 2.2, 2.3, 2.7, 2.8, 2.9.
Question 5 (constructed response · mixed)
For constants a and b, a function f is defined by
- f(x) = a√x for 0 < x ≤ 4
- f(x) = x²/8 + b for x > 4
(a) Use the definition of the derivative to show that the derivative of √x at x = 4 is 1/4. (b) Write an equation in a and b that must hold if f is continuous at x = 4. (c) Find a and b so that f is differentiable at x = 4. Justify. (d) With these values, write an equation of the line tangent to the graph of f at x = 4. (e) Now let a = 4 and b = 5. Is f differentiable at x = 4? Explain.
Worked solution and suggested Marlbridge rubric
(a) lim (h → 0) (√(4 + h) − 2)/h. Multiply top and bottom by √(4 + h) + 2: the top becomes (4 + h) − 4 = h, so the quotient is 1/(√(4 + h) + 2) for h ≠ 0. The limit is 1/(2 + 2) = 1/4.
(b) f(4) = 2a, and the right-hand limit is 16/8 + b = 2 + b. Continuity needs 2a = 2 + b.
(c) For x < 4, f′(x) = a/(2√x), so the left-hand slope at 4 is a/4 (this is a times the limit in (a)). For x > 4, f′(x) = x/4, so the right-hand slope is 4/4 = 1. Differentiability needs continuity (from (b)) and equal one-sided slopes: a/4 = 1, so a = 4, and then b = 2a − 2 = 6.
(d) f(4) = 4 × 2 = 8 and f′(4) = 1. Tangent line: y − 8 = 1(x − 4), or y = x + 4.
(e) No. f(4) = 8 but the right-hand limit is 2 + 5 = 7, so f is not continuous at 4, and so not differentiable there, even though the one-sided slopes match.
| Point | What earns it |
|---|---|
| 1 | (a): conjugate algebra and the limit 1/4 |
| 1 | (b): 2a = 2 + b |
| 1 | (c): one-sided slopes a/4 and 1 |
| 1 | (c): a = 4 and b = 6, with both conditions used |
| 1 | (d): tangent line through (4, 8) with slope 1 |
| 1 | (e): “no”, because continuity fails (8 ≠ 7) |
Total: 6 points. Topics: 2.1, 2.2, 2.4, 2.5, 2.6.
Question 6 (constructed response · mixed)
In a fictional model, the power output of a small test generator t hours after start-up is
P(t) = 50t²/eᵗ kilowatts, for 0 ≤ t ≤ 6.
(a) Use the quotient rule to show that P′(t) = 50t(2 − t)/eᵗ. (b) Find P′(1). Give units and interpret the value in context. (c) Find the time in 0 < t < 6 at which P′(t) = 0, and the power output at that time. What does the sign of P′(3) tell you? (d) Write P(t) = 50t² · (1/eᵗ). Use the quotient rule to find the derivative of 1/eᵗ, then the product rule to check (a). (e) Find the average rate of change of P over [0, 2], with units.
Worked solution and suggested Marlbridge rubric
(a) Top 50t², derivative 100t. Bottom eᵗ, derivative eᵗ. P′(t) = [eᵗ · 100t − 50t² · eᵗ]/(eᵗ)² = eᵗ(100t − 50t²)/e^(2t) = 50t(2 − t)/eᵗ.
(b) P′(1) = 50(1)(1)/e = 50/e kilowatts per hour (about 18.4). One hour after start-up, the power output is increasing at 50/e kilowatts per hour.
(c) eᵗ is never 0, so P′(t) = 0 only when t = 0 or t = 2. In 0 < t < 6 that is t = 2, when P(2) = 200/e² kilowatts. P′(3) = 50(3)(−1)/e³ = −150/e³ < 0, so at t = 3 the power output is decreasing.
(d) d/dt (1/eᵗ) = [eᵗ · 0 − 1 · eᵗ]/(eᵗ)² = −1/eᵗ. Product rule: P′(t) = 100t · (1/eᵗ) + 50t² · (−1/eᵗ) = (100t − 50t²)/eᵗ = 50t(2 − t)/eᵗ, which agrees with (a).
(e) (P(2) − P(0))/(2 − 0) = (200/e² − 0)/2 = 100/e² kilowatts per hour (about 13.5).
| Point | What earns it |
|---|---|
| 1 | (a): quotient rule with correct parts and simplification |
| 1 | (b): 50/e with units kilowatts per hour |
| 1 | (b): interpretation: rate at t = 1, increasing |
| 1 | (c): t = 2 and P(2) = 200/e², with the reason eᵗ ≠ 0 |
| 1 | (c): P′(3) < 0 means decreasing |
| 1 | (d): derivative of 1/eᵗ is −1/eᵗ and product rule agrees |
| 1 | (e): 100/e² with units |
Total: 7 points. Topics: 2.1, 2.2, 2.5, 2.7, 2.8, 2.9.
Question 7 (constructed response · mixed)
For a constant k, let g be defined on −π/2 < x < π/2 by
- g(x) = x + cos x for x ≤ 0
- g(x) = 1 + k tan x for x > 0
(a) Show that g is continuous at x = 0 for every value of k. (b) Find lim (h → 0⁻) (g(h) − g(0))/h. Recognise (cos h − 1)/h as a derivative to help. (c) Find lim (h → 0⁺) (g(h) − g(0))/h in terms of k. Recognise tan h/h as a derivative to help. (d) For which value of k is g differentiable at x = 0? Write g′(x) for x < 0 and for x > 0 with that k. (e) With that k, write an equation of the line tangent to the graph of g at x = π/4.
Worked solution and suggested Marlbridge rubric
(a) g(0) = 0 + cos 0 = 1. Left-hand limit: 0 + 1 = 1. Right-hand limit: 1 + k tan 0 = 1. All three agree for any k, so g is continuous at 0.
(b) (h + cos h − 1)/h = 1 + (cos h − 1)/h. The second part is the difference quotient of cos at 0, so it tends to −sin 0 = 0. The limit is 1.
(c) (1 + k tan h − 1)/h = k · tan h/h. Since tan 0 = 0, tan h/h is the difference quotient of tan at 0, so it tends to sec²0 = 1. The limit is k.
(d) g is continuous at 0, so it is differentiable there exactly when the one-sided limits agree: k = 1, with g′(0) = 1. Then g′(x) = 1 − sin x for x < 0 and g′(x) = sec²x for x > 0.
(e) g(π/4) = 1 + tan(π/4) = 2, and g′(π/4) = sec²(π/4) = (√2)² = 2. Tangent line: y − 2 = 2(x − π/4).
| Point | What earns it |
|---|---|
| 1 | (a): g(0) and both one-sided limits equal 1 |
| 1 | (b): limit 1, using (cos h − 1)/h → 0 |
| 1 | (c): limit k, using tan h/h → 1 |
| 1 | (d): k = 1, with continuity and equal one-sided limits as the reason |
| 1 | (d): g′(x) correct on both sides |
| 1 | (e): tangent line through (π/4, 2) with slope 2 |
Total: 6 points. Topics: 2.1, 2.4, 2.7, 2.10.
How did you do?
Add your points from Questions 4–7 (25 in total) and your correct answers to Questions 1–3. The total is only a rough guide, not a predicted exam score. More useful: note which topics your lost points came from (each answer lists them), then work through those topic checklists:
2.1 · 2.2 · 2.3 · 2.4 · 2.5 · 2.6 · 2.7 · 2.8 · 2.9 · 2.10
If many topics need work, start from the “Your next step” table in the Unit 2 diagnostic.