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, and exact answers unless stated. Notation: f′(x), dy/dx and d/dx[…] all mean the derivative with respect to x; fractional exponents are written in brackets, e.g. x^(3/4). This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
What is d/dx[1/x⁵]?
- (A) −5/x⁶
- (B) −5/x⁴
- (C) 1/(5x⁴)
- (D) 5/x⁶
Answer and explanation
Answer: (A). Rewrite 1/x⁵ as x⁻⁵. Bring down −5 and lower the exponent by 1: −5 − 1 = −6. So the derivative is −5x⁻⁶ = −5/x⁶.
- (B) lowers the exponent the wrong way: it treats −5 − 1 as −4.
- (C) differentiates only the denominator, x⁵ → 5x⁴, and keeps it in the denominator. A reciprocal must be rewritten as a negative power first.
- (D) loses the minus sign that comes down with the exponent −5. The function 1/x⁵ is decreasing for x > 0, so its derivative there must be negative.
Question 2 (multiple choice · core)
Let f(x) = ∜(x³), the fourth root of x³, for x > 0. What is f′(16)?
- (A) 3/8
- (B) 3/2
- (C) 96
- (D) 8
Answer and explanation
Answer: (A). Rewrite: f(x) = x^(3/4). New exponent: 3/4 − 1 = −1/4. So f′(x) = (3/4)x^(−1/4). Since 16^(1/4) = 2, 16^(−1/4) = 1/2. So f′(16) = (3/4)(1/2) = 3/8.
- (B) uses exponent +1/4 instead of −1/4: (3/4)·16^(1/4) = (3/4)(2) = 3/2. That is a sign slip in 3/4 − 1.
- (C) adds 1 to the exponent instead of subtracting it: (3/4)·16^(7/4) = (3/4)(128) = 96.
- (D) is f(16) = 16^(3/4) = 8, the value of the function, not its derivative.
Question 3 (multiple choice · core)
Which function has derivative −(1/2)x^(−3/2) for x > 0?
- (A) √x
- (B) 1/√x
- (C) −1/√x
- (D) 1/x^(3/2)
Answer and explanation
Answer: (B). 1/√x = x^(−1/2). The power rule gives (−1/2)x^(−1/2 − 1) = −(1/2)x^(−3/2). This matches.
- (A) √x = x^(1/2) has derivative (1/2)x^(−1/2): wrong sign and wrong exponent.
- (C) −1/√x = −x^(−1/2) has derivative +(1/2)x^(−3/2). The two minus signs cancel.
- (D) x^(−3/2) has derivative −(3/2)x^(−5/2). This option copies the exponent of the target answer instead of working backwards from it.
Question 4 (multiple choice · core)
At which x-values does the graph of y = x³ have a tangent line with slope 12?
- (A) x = 2 only
- (B) x = −2 and x = 2
- (C) x = ∛12 only
- (D) x = 4 only
Answer and explanation
Answer: (B). The slope of the tangent line is dy/dx = 3x². Set 3x² = 12, so x² = 4 and x = ±2. The points are (2, 8) and (−2, −8); the tangent at each has slope 12.
- (A) forgets the negative square root. 3(−2)² = 12 as well.
- (C) sets the function, x³, equal to 12 instead of the derivative.
- (D) uses 3x as the derivative (bringing the 3 down but lowering the exponent by 2 instead of 1), then solves 3x = 12.
Question 5 (constructed response · core)
Let f(x) = 1/x².
(a) Use the definition f′(x) = lim (h → 0) [f(x + h) − f(x)]/h to find f′(x). Show the algebra. (b) Confirm your answer with the power rule. (c) Explain why f′(0) does not exist.
Worked solution and suggested Marlbridge rubric
(a) Combine the fractions on top over the common denominator x²(x + h)²:
1/(x + h)² − 1/x² = [x² − (x + h)²] / [x²(x + h)²] = (−2xh − h²) / [x²(x + h)²].
Divide by h and cancel h (allowed, since h ≠ 0 inside the limit):
[f(x + h) − f(x)]/h = (−2x − h) / [x²(x + h)²].
Now let h → 0: f′(x) = −2x / (x²·x²) = −2x/x⁴ = −2/x³.
(b) f(x) = x⁻². Power rule: f′(x) = −2x⁻³ = −2/x³. The two methods agree.
(c) 0 is not in the domain of f, because 1/0² is undefined. A function cannot be differentiable at a point where it is not even defined (differentiable implies continuous, Topic 2.4). The formula −2/x³ also has a zero denominator at x = 0.
| Point | What earns it |
|---|---|
| 1 | Correct difference quotient with the fractions combined over a common denominator |
| 1 | Expands and cancels h correctly to reach (−2x − h)/[x²(x + h)²] or equivalent |
| 1 | Takes the limit, keeping “lim” until h is replaced, to get −2/x³ |
| 1 | Rewrites as x⁻² and applies the power rule correctly |
| 1 | Explains that 0 is not in the domain of f, so f is not continuous (and so not differentiable) there |
A table of difference quotients alone earns no marks for (a), because the question asks for the definition.
Question 6 (constructed response · core)
Rewrite each function as a single power of x, then find its derivative. Assume x > 0.
(a) g(x) = x³·∛x (b) k(x) = 1/(x²√x) (c) m(x) = (x⁴)³ / x⁵ (d) A student writes: “The derivative of x³·∛x is 3x² · (1/3)x^(−2/3) = x^(4/3).” Show that the student’s answer is wrong at x = 8, and explain the error.
Worked solution and suggested Marlbridge rubric
(a) x³·x^(1/3) = x^(10/3). Derivative: (10/3)x^(7/3).
(b) x²·x^(1/2) = x^(5/2), so k(x) = x^(−5/2). New exponent: −5/2 − 1 = −7/2. Derivative: −(5/2)x^(−7/2).
(c) (x⁴)³ = x¹², and x¹²/x⁵ = x⁷. Derivative: 7x⁶.
(d) At x = 8 the correct derivative is (10/3)·8^(7/3) = (10/3)(128) = 1280/3. The student’s formula gives 8^(4/3) = 16. These differ, so the student is wrong. The error: the student differentiated each factor and multiplied the results. The derivative of a product is not the product of the derivatives. Combine the powers first (as in (a)), or use the product rule from Topic 2.8.
| Point | What earns it |
|---|---|
| 1 | (a) x^(10/3) and derivative (10/3)x^(7/3) |
| 1 | (b) x^(−5/2) and derivative −(5/2)x^(−7/2), with the exponent lowered correctly |
| 1 | (c) x⁷ and derivative 7x⁶ |
| 1 | (d) Shows the two values at x = 8 differ (1280/3 and 16) and names the error: multiplying derivatives of factors |
Accept equivalent forms, e.g. −5/(2x^(7/2)) in (b).
Question 7 (constructed response · stretch)
Let f(x) = x^(2/3), defined for all real x as the square of the cube root of x.
(a) Find f′(x) for x ≠ 0. (b) Find the equation of the tangent line to the graph at x = 8. (c) Find the point on the graph where the tangent line has slope −1/3. (d) Using the definition of the derivative, explain whether f is differentiable at x = 0, and describe the graph there.
Worked solution and suggested Marlbridge rubric
(a) New exponent: 2/3 − 1 = −1/3. So f′(x) = (2/3)x^(−1/3) = 2/(3∛x), for x ≠ 0.
(b) f(8) = (∛8)² = 2² = 4. f′(8) = 2/(3·2) = 1/3. Tangent: y − 4 = (1/3)(x − 8), or y = x/3 + 4/3.
(c) Solve 2/(3∛x) = −1/3. Cross-multiply: 6 = −3∛x, so ∛x = −2 and x = −8. Then f(−8) = (−2)² = 4. The point is (−8, 4). (By symmetry, the graph is the mirror image of the right-hand side, so the slope there is the negative of f′(8).)
(d) f(0) = 0, so the difference quotient at 0 is f(h)/h = h^(2/3)/h = 1/∛h for h ≠ 0. As h → 0⁺ this tends to +∞; as h → 0⁻ it tends to −∞. The limit does not exist, so f is not differentiable at 0. f is continuous at 0, though, so the graph has a cusp: a sharp point at the origin where both sides become vertical, pointing downward.
| Point | What earns it |
|---|---|
| 1 | f′(x) = (2/3)x^(−1/3) |
| 1 | f(8) = 4 and f′(8) = 1/3 |
| 1 | Correct tangent line equation |
| 1 | x = −8 found from f′(x) = −1/3, with point (−8, 4) |
| 1 | Difference quotient 1/∛h with the two one-sided limits (+∞ and −∞) |
| 1 | Concludes not differentiable at 0, describes a cusp, and notes f is continuous there |
Pointing to the zero denominator in f′(x) at x = 0 is acceptable supporting evidence for (d), but the full mark needs the definition, as the question asks.
How did you do?
- Q1 or Q6(b) wrong: revisit “Subtracting 1 from a negative or fractional exponent” in the study guide.
- Q2, Q3 or Q6 wrong: redo Worked example 1 and the “Rewrite before you differentiate” table.
- Q4 or Q7(b)–(c) wrong: redo Worked example 2 (tangent lines) and remember the slope is the derivative, not the function value.
- Q5 or Q7(d) wrong: reread “Building the pattern from the definition” and “When the power rule gives ‘undefined’”.
Then tick off the topic checklist.