Use this page after you have studied the topics of Unit 3, Differentiation: Composite, Implicit, and Inverse Functions, or after the Unit 3 diagnostic. The unit is shared by Calculus AB and Calculus BC, and every question here is for both courses. The questions below 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 official scoring. No calculator for any question; angles are in radians.
Big ideas of the unit
- Rates multiply through a composition. If y depends on u and u depends on x, then dy/dx = (dy/du)(du/dx). The outer derivative is taken at the inside value, never at x (Topic 3.1).
- Count the layers. Each layer gives one factor. Most slips in this unit are a missing factor from the innermost layer.
- Implicit differentiation is the chain rule applied to y. Treat y as a function of x, so every y term gains a factor dy/dx. Then collect and solve (Topic 3.2).
- Implicit slopes depend on both coordinates. Horizontal tangents need numerator 0; vertical tangents need denominator 0. Either way, solve together with the curve’s equation.
- An inverse swaps rise and run. Differentiating f(g(x)) = x gives g′(b) = 1/f′(a) where f(a) = b. You need the matching input a, not a formula for the inverse (Topic 3.3).
- A horizontal tangent on f means a vertical tangent on f⁻¹, so the inverse has no derivative there.
- Inverse trig derivatives come from the same idea. sin y = x, differentiated implicitly, gives 1/√(1 − x²); the restricted ranges fix the sign of each root (Topic 3.4).
- Choose the procedure from the form. Rewrite first where you can, then let the last operation pick the first rule (Topic 3.5).
- Higher derivatives repeat the process. f″ exists only where f′ is differentiable, and implicit second derivatives must remember that y still depends on x (Topic 3.6).
Key relationships and methods
| You see or need | What to do | Topics |
|---|---|---|
| f(g(x)), or a power, root, exp, log or trig of an expression | Chain rule: f′(g(x)) · g′(x), one factor per layer | 3.1 |
| f(g(a)) from a table | Find g(a) first, then read f′ at that value | 3.1, 3.5 |
| Rates with units | dy/dt = (dy/du)(du/dt); the units cancel in the same way | 3.1 |
| An equation in x and y, not solved for y | Differentiate both sides, attach dy/dx to y terms, collect, solve | 3.2 |
| Horizontal or vertical tangent on an implicit curve | Numerator 0 or denominator 0, then solve with the curve | 3.2 |
| Derivative of f⁻¹ at b | Solve f(a) = b, then 1/f′(a), provided f′(a) ≠ 0 | 3.3 |
| arcsin u, arccos u, arctan u | u′/√(1 − u²), −u′/√(1 − u²), u′/(1 + u²) | 3.4 |
| A limit shaped like [f(a + h) − f(a)]/h | It is f′(a) | 3.5 |
| Single-term denominator, root or log of a product | Rewrite as powers or a sum of logs first | 3.5 |
| f″, f‴, f⁽ⁿ⁾ | Differentiate again; simplify before each step; look for a pattern | 3.6 |
| d²y/dx² for an implicit curve | Differentiate the differentiated equation again; substitute dy/dx | 3.2, 3.6 |
Question 1 (multiple choice · mixed)
Let f(x) = e^(2x) + x. Then f is increasing, and g is its inverse. Let h(x) = g(x²). What is h′(1)?
- (A) 1/3
- (B) 2/3
- (C) 6
- (D) 2/(2e² + 1)
Answer and explanation
Answer: (B). By the chain rule, h′(x) = g′(x²) · 2x, so h′(1) = 2g′(1). For g′(1), find the matching input: f(0) = 1 + 0 = 1, so g(1) = 0. Then f′(x) = 2e^(2x) + 1, so f′(0) = 3 and g′(1) = 1/3. So h′(1) = 2/3.
- (A) is g′(1) alone: it forgets the chain factor 2x.
- (C) is 2f′(0): it forgets the reciprocal.
- (D) is 2/f′(1): it evaluates f′ at the output 1, not the matching input 0.
Topics: 3.1, 3.3.
Question 2 (multiple choice · mixed)
Let f(x) = arctan x. What is f″(1)?
- (A) −1/2
- (B) 1/2
- (C) −1/4
- (D) −1
Answer and explanation
Answer: (A). f′(x) = 1/(1 + x²) = (1 + x²)⁻¹. Chain rule: f″(x) = −(1 + x²)⁻² · 2x = −2x/(1 + x²)². At x = 1: −2/4 = −1/2.
- (B) loses the minus sign from the power −1.
- (C) forgets the inner derivative 2x: −1/(1 + 1)².
- (D) does not square the denominator: −2x/(1 + x²) at x = 1.
Topics: 3.1, 3.4, 3.6.
Question 3 (multiple choice · mixed)
The point (2, 1) lies on the curve ln y + xy = 2. Use the tangent line at (2, 1) to approximate the value of y on the curve near (2, 1) when x = 2.3.
- (A) 0.9
- (B) 0.7
- (C) 1
- (D) 1.1
Answer and explanation
Answer: (A). Check the point: ln 1 + 2 = 2. ✓ Differentiate: (1/y) · dy/dx + y + x · dy/dx = 0, so dy/dx = −y/(1/y + x) = −y²/(1 + xy). At (2, 1): −1/3. The tangent line is y = 1 − (1/3)(x − 2), which gives 1 − 0.1 = 0.9 at x = 2.3.
- (B) uses slope −1, from writing d/dx [ln y] as 1/y without dy/dx.
- (C) uses slope 0, from dropping the product rule on xy.
- (D) loses the minus sign of the slope.
Topics: 3.1, 3.2.
Question 4 (constructed response · mixed)
The functions f and g are differentiable for all x. The function g is increasing, and w is the inverse of g. Selected values are shown.
| x | f(x) | f′(x) | g(x) | g′(x) |
|---|---|---|---|---|
| 0 | 3 | −3 | 1 | 1/2 |
| 1 | 1 | −3/2 | 2 | 3/2 |
| 2 | 0 | −1/2 | 4 | 5/2 |
(a) Let h(x) = f(g(x)). Find h′(0). (b) Write an equation for the line tangent to the graph of w at x = 4. (c) Let m(x) = arctan(f(x)). Find m′(0). (d) Let p(x) = x · g(2x). Find p′(1).
Worked solution and suggested Marlbridge rubric
(a) Chain rule: h′(0) = f′(g(0)) · g′(0). Inside first: g(0) = 1, so f′(g(0)) = f′(1) = −3/2. Then h′(0) = (−3/2)(1/2) = −3/4.
(b) Find 4 in the g(x) column: g(2) = 4, so w(4) = 2. Then w′(4) = 1/g′(2) = 1/(5/2) = 2/5. The tangent line is y = 2 + (2/5)(x − 4).
(c) m′(x) = f′(x)/(1 + [f(x)]²). At x = 0: −3/(1 + 9) = −3/10.
(d) Product rule, with the chain rule on g(2x): p′(x) = g(2x) + x · g′(2x) · 2. At x = 1: g(2) + 2g′(2) = 4 + 5 = 9.
| Point | What earns it |
|---|---|
| 1 | (a) −3/4, with f′ evaluated at g(0) = 1 |
| 1 | (b) the point (4, 2) from g(2) = 4 |
| 1 | (b) slope 1/g′(2) = 2/5 and a correct tangent line |
| 1 | (c) −3/10 from the arctan chain-rule form |
| 1 | (d) 9, with both the product rule and the chain factor 2 |
Total: 5 points. Topics: 3.1, 3.3, 3.4, 3.5.
Common errors: f′(0)g′(0) = −3/2 in (a); 1/g′(4), which is not in the table, in (b); 13/2 in (d), which drops the chain factor 2.
Question 5 (constructed response · mixed)
Consider the curve y³ + 3y = 2x² + 4.
(a) Find dy/dx in terms of x and y. (b) Explain why the curve has no vertical tangent lines. (c) Find every point on the curve where the tangent line is horizontal. (d) Find d²y/dx² at the point you found in (c). (e) Show that (√5, 2) is on the curve, and find the slope of the curve there.
Worked solution and suggested Marlbridge rubric
(a) Differentiate both sides; y³ and 3y need dy/dx:
3y² · dy/dx + 3 · dy/dx = 4x, so dy/dx = 4x/(3y² + 3).
(b) A vertical tangent needs the denominator to be 0. But 3y² + 3 ≥ 3 for every y, so it is never 0.
(c) Horizontal needs 4x = 0, so x = 0. Then y³ + 3y = 4, or y³ + 3y − 4 = 0. This factors as (y − 1)(y² + y + 4) = 0. The quadratic has discriminant 1 − 16 = −15 < 0, so y = 1 is the only solution. The point is (0, 1).
(d) Differentiate (3y² + 3) · dy/dx = 4x again. The left side is a product, and 3y² depends on x:
6y · (dy/dx)² + (3y² + 3) · d²y/dx² = 4.
At (0, 1), dy/dx = 0, so 6 · d²y/dx² = 4 and d²y/dx² = 2/3.
(e) 2³ + 3(2) = 14 and 2(5) + 4 = 14. ✓ Slope: 4√5/(12 + 3) = 4√5/15.
| Point | What earns it |
|---|---|
| 1 | (a) dy/dx attached to both y terms, with dy/dx = 4x/(3y² + 3) |
| 1 | (b) Denominator at least 3, so never 0 |
| 1 | (c) x = 0 substituted into the curve |
| 1 | (c) Only (0, 1), with the quadratic factor shown to have no real roots |
| 1 | (d) Second differentiation with the product rule on (3y² + 3) · dy/dx |
| 1 | (d) d²y/dx² = 2/3 |
| 1 | (e) Point checked and slope 4√5/15 |
Total: 7 points. Topics: 3.1, 3.2, 3.6.
Acceptable alternative for (d): apply the quotient rule to 4x/(3y² + 3), then substitute dy/dx = 0. It gives the same value, 2/3.
Question 6 (constructed response · mixed)
Let f(x) = x³ + arctan x, and let g be the inverse of f.
(a) Explain why g exists. (b) Find g(0) and g′(0). (c) Find the exact value of g′(1 + π/4). (d) Let H(x) = arctan(g(x)). Find H′(1 + π/4). (e) Differentiate f′(g(x)) · g′(x) = 1 to show that g″(x) = −f″(g(x)) · g′(x)/[f′(g(x))]². Then find g″(1 + π/4).
Worked solution and suggested Marlbridge rubric
(a) f′(x) = 3x² + 1/(1 + x²). Since 3x² ≥ 0 and 1/(1 + x²) > 0, f′(x) > 0 for all x. So f is increasing, hence one-to-one, and g exists.
(b) f(0) = 0 + 0 = 0, so g(0) = 0. f′(0) = 1, so g′(0) = 1/1 = 1.
(c) f(1) = 1 + π/4, so g(1 + π/4) = 1. f′(1) = 3 + 1/2 = 7/2. So g′(1 + π/4) = 2/7.
(d) Chain rule with the arctan formula: H′(x) = g′(x)/(1 + [g(x)]²). At x = 1 + π/4: (2/7)/(1 + 1) = 1/7.
(e) The left side is a product, and its first factor is a composite. Product rule, then chain rule:
f″(g(x)) · g′(x) · g′(x) + f′(g(x)) · g″(x) = 0.
Solve: g″(x) = −f″(g(x)) · [g′(x)]²/f′(g(x)). Since g′(x) = 1/f′(g(x)), this equals −f″(g(x)) · g′(x)/[f′(g(x))]², as required.
f″(x) = 6x − 2x/(1 + x²)², so f″(1) = 6 − 2/4 = 11/2. At x = 1 + π/4, where g = 1:
g″(1 + π/4) = −(11/2)(2/7)/(7/2)² = −(11/7) × (4/49) = −44/343.
| Point | What earns it |
|---|---|
| 1 | (a) f′(x) > 0 for all x, so f is one-to-one |
| 1 | (b) g(0) = 0 and g′(0) = 1 |
| 1 | (c) 2/7, using the matching input 1 and f′(1) = 7/2 |
| 1 | (d) 1/7, with g(1 + π/4) = 1 in the denominator |
| 1 | (e) Correct product and chain rule when differentiating f′(g(x)) · g′(x) |
| 1 | (e) f″(1) = 11/2 and g″(1 + π/4) = −44/343 |
Total: 6 points. Topics: 3.1, 3.3, 3.4, 3.6.
Question 7 (constructed response · mixed)
Let f(x) = x · arcsin x + √(1 − x²), for −1 < x < 1.
(a) Name the rule you use first on each term of f. Then show that f′(x) = arcsin x. (b) Find f″(x) and f″(1/2). (c) Find f‴(x), and show that f‴(0) = 0. (d) Write an equation for the line tangent to the graph of y = f′(x) at x = 1/2.
Worked solution and suggested Marlbridge rubric
(a) The first term is a product, so use the product rule. The second is a root of an expression, so use the chain rule.
- d/dx [x · arcsin x] = arcsin x + x/√(1 − x²).
- d/dx [√(1 − x²)] = (−2x)/(2√(1 − x²)) = −x/√(1 − x²).
Add: the two fractions cancel, so f′(x) = arcsin x.
(b) f″(x) = 1/√(1 − x²). At x = 1/2: 1/√(3/4) = 2/√3 = 2√3/3.
(c) Rewrite f″(x) = (1 − x²)^(−1/2). Chain rule: f‴(x) = (−1/2)(1 − x²)^(−3/2) · (−2x) = x/(1 − x²)^(3/2). At x = 0 the numerator is 0, so f‴(0) = 0.
(d) The point is (1/2, f′(1/2)) = (1/2, π/6). The slope of f′ is f″(1/2) = 2√3/3. Tangent line: y = π/6 + (2√3/3)(x − 1/2).
| Point | What earns it |
|---|---|
| 1 | (a) Product rule on x · arcsin x, with the arcsin derivative |
| 1 | (a) Chain rule on the root, and the cancellation to arcsin x |
| 1 | (b) f″(x) = 1/√(1 − x²) and f″(1/2) = 2√3/3 |
| 1 | (c) f‴(x) = x/(1 − x²)^(3/2), with f‴(0) = 0 |
| 1 | (d) Point (1/2, π/6) and slope 2√3/3 in a correct line |
Total: 5 points. Topics: 3.1, 3.4, 3.5, 3.6.
A common error in (d) is to use f(1/2) as the y-coordinate. The graph is of f′, so the point uses f′(1/2) = π/6.
How did you do?
Add up your points from Questions 4–7 (23 in total) and your correct answers to Questions 1–3. The total is only a guide to how secure you are across the unit; it is not a predicted exam score. More useful is to look at which topics your lost points came from (each answer lists them), then tick off those topic checklists:
3.1 · 3.2 · 3.3 · 3.4 · 3.5 · 3.6
If many topics need work, go back to the Unit 3 diagnostic and use its “Your next step” table to choose where to start.