Differentiation is the backbone of AP Calculus AB, and by the time students reach the exam, they're expected to move fluidly between the power rule, chain rule, product rule, and quotient rule — often combining several of them within a single problem. Beyond the mechanical rules, the AP exam consistently tests derivatives of exponential, logarithmic, and inverse trigonometric functions, higher-order derivatives (second, third, and even fourth derivatives), and the tangent line approximation, which uses a derivative to estimate a function's value near a known point.
Many of the trickiest AP Calculus AB questions aren't hard because any single rule is difficult — they're hard because they require correctly identifying which rule (or combination of rules) applies, and in what order. A function like esin4x requires the chain rule twice in a row: once for the exponential's inner function, and again for the sine's inner function. A function like tan²(ex) requires the chain rule stacked three layers deep. Quotient rule problems often simplify dramatically at a specific point, and questions involving a table of values for two abstract functions f and g test whether you can apply the chain rule correctly without ever knowing an explicit formula for either function.
Below are 12 AP Calculus AB-style practice problems covering the chain rule, product rule, quotient rule, higher-order derivatives, derivatives from a table of values, and tangent line approximation. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus AB QBank.
If y = ln(sin x) and 0 ≤ x ≤ π, then dy/dx is
Use the chain rule for logarithms: d/dx[ln(u)] = u′/u, where u = sin x and u′ = cos x.
dy/dx = cos x / sin x
This ratio is exactly the definition of cotangent: cos x / sin x = cot x.
The correct answer is B. A common error is mixing up the sign or flipping the ratio to get −tan x or tan x.
If y = sin⁻¹(e²ˣ), then dy/dx is
The derivative of sin⁻¹(u) is u′ / √(1 − u²). Here, u = e²ˣ and u′ = 2e²ˣ (chain rule on the exponent).
Substitute into the formula: dy/dx = 2e²ˣ / √(1 − (e²ˣ)²).
Simplify the squared term: (e²ˣ)² = e⁴ˣ, giving dy/dx = 2e²ˣ / √(1 − e⁴ˣ).
The correct answer is A. Watch out for two common errors: forgetting the chain-rule factor of 2 (choice B), and flipping the sign under the radical (choices C and D).
Given the functions f(x) and g(x) that are both continuous and differentiable, and that have values given on the table below.
| x | f(x) | f′(x) | g(x) | g′(x) |
|---|---|---|---|---|
| 2 | 4 | −2 | 8 | 1 |
| 4 | 10 | 8 | 4 | 3 |
| 8 | 6 | −12 | 2 | 4 |
Given that h(x) = g(g(x)), h′(8) =
Since h(x) = g(g(x)) is a composition, apply the chain rule: h′(x) = g′(g(x)) · g′(x).
Evaluate at x = 8: h′(8) = g′(g(8)) · g′(8). From the table, g(8) = 2 and g′(8) = 4.
So h′(8) = g′(2) · 4. From the table, g′(2) = 1.
h′(8) = 1 × 4 = 4. The correct answer is C. The f-column is a distractor here — only g and g′ values are needed since h is built entirely from g.
Keep the momentum going.
Explore 1,500+ Free AP Calculus AB ProblemsIf g(x) = tan²(eˣ), then g′(x) is
This requires the chain rule twice, since g(x) is a power of a trig function of an exponential function. Think of it as three layers: squaring, tangent, and eˣ.
Outer layer (power rule on tan(eˣ)): 2tan(eˣ) · d/dx[tan(eˣ)].
Middle layer (derivative of tangent): d/dx[tan(eˣ)] = sec²(eˣ) · d/dx[eˣ].
Inner layer (derivative of the exponential): d/dx[eˣ] = eˣ.
Multiplying all three layers together: g′(x) = 2tan(eˣ) · sec²(eˣ) · eˣ = 2eˣtan(eˣ)sec²(eˣ). The correct answer is B — every layer of the chain rule must contribute a factor, so an answer missing the eˣ, the tan, or the sec² factor is incomplete.
Let f(x) be the function with f(2) = 4 and f′(x) = √(x³ + 1). Using the tangent line approximation to the graph of f(x) at x = 2, estimate f(2.2).
The tangent line approximation says f(x) ≈ f(a) + f′(a)(x − a) for x close to a. Here a = 2, so we need f′(2) first.
f′(2) = √(2³ + 1) = √9 = 3
Apply the approximation with x = 2.2: f(2.2) ≈ f(2) + f′(2)(2.2 − 2) = 4 + 3(0.2)
= 4 + 0.6 = 4.6. The correct answer is D.
Which of the following statements must be true?
Statement I: Using the chain rule, d/dx√(u) = u′/(2√u) with u = eˣ + 3 and u′ = eˣ gives exactly eˣ/(2√(eˣ+3)). This statement is true.
Statement II: Using the chain rule, d/dx[ln(cos x)] = (−sin x)/(cos x) = −tan x. The statement claims the result is tan x (missing the negative sign), so this statement is false.
Statement III: Differentiate term by term: d/dx[6x³] = 18x²; d/dx[−π] = 0; d/dx[x⁸⁄³] = (8/3)x⁵⁄³ = (8/3)∛(x⁵); d/dx[−2x⁻³] = 6x⁻⁴ = 6/x⁴. All terms match, so this statement is true.
Since I and III are true and II is false, the correct answer is D.
The value of the derivative of y = (x² − 3)³ / (5x − 9)² at x = 2 is
Let u = (x²−3)³ and v = (5x−9)², so u′ = 3(x²−3)²(2x) = 6x(x²−3)² and v′ = 2(5x−9)(5) = 10(5x−9).
Evaluate the building blocks at x = 2: x²−3 = 1, so u = 1³ = 1 and u′ = 6(2)(1)² = 12. Also 5x−9 = 1, so v = 1² = 1 and v′ = 10(1) = 10.
Apply the quotient rule, y′ = (u′v − uv′)/v²:
y′(2) = (12 · 1 − 1 · 10)/1² = (12 − 10)/1 = 2. The correct answer is C.
7 down, 5 to go — keep practicing.
Explore 1,500+ Free AP Calculus AB ProblemsFind the following derivative:
Rewrite every root as a fractional exponent first, then differentiate term by term with the power rule. x7 differentiates to 7x6.
48√(x7) = 4x7/8 differentiates to 4(7/8)x−1/8 = (7/2)x−1/8 (kept negative, since it's subtracted in the original).
7x is an exponential with a constant base, not the power rule — it differentiates to 7xln7.
1/7√(x4) = x−4/7 differentiates to (−4/7)x−11/7; since this term is subtracted in the original, it contributes +(4/7)x−11/7 to the derivative.
1/(5x) = (1/5)x−1 differentiates to −(1/5)x−2 = −1/(5x2).
Putting it all together: 7x6 − (7/2)x−1/8 + 7xln7 + (4/7)x−11/7 − 1/(5x2). The correct answer is A.
Find Dx[e³ˣ² cos4x].
This is a product of two functions, so use the product rule: (uv)′ = u′v + uv′, with u = e³ˣ² and v = cos4x.
Find each derivative separately. By the chain rule, u′ = 6x · e³ˣ² (the inner derivative of 3x² is 6x). And v′ = −4sin4x.
Combine using the product rule: D_x = (6xe³ˣ²)(cos4x) + (e³ˣ²)(−4sin4x) = 6xe³ˣ²cos4x − 4e³ˣ²sin4x.
Factor out the common e³ˣ² term: e³ˣ²(6x cos4x − 4sin4x). The correct answer is A.
f(x) = esin4x; find the exact value of f″(π/4).
Find the first derivative using the chain rule: f′(x) = esin4x · 4cos4x (the inner derivative of sin4x is 4cos4x).
Differentiate again, applying the product rule to 4cos4x · esin4x:
f″(x) = 4(−4sin4x)esin4x + 4cos4x(esin4x · 4cos4x) = esin4x(16cos²4x − 16sin4x)
Evaluate at x = π/4, where 4x = π: sinπ = 0 and cosπ = −1.
f″(π/4) = e⁰(16(−1)² − 16(0)) = 1 × (16 − 0) = 16. The correct answer is C.
10 down, 2 to go.
Explore 1,500+ Free AP Calculus AB ProblemsProblem 11 refers to the following information. A fourth differentiable function is defined for all real numbers and satisfies each of the following: g(2) = 5, g′(2) = −2, and g″(2) = 3. The function f is given by f(x) = ek(x−1) + g(2x), where k is a constant.
Find f(1), f′(1), and f″(1).
Find f(1) by substituting x = 1 directly: f(1) = ek(1−1) + g(2 · 1) = e⁰ + g(2) = 1 + 5 = 6.
Differentiate f(x) using the chain rule on each term: f′(x) = k · ek(x−1) + 2g′(2x) (the inner derivative of k(x−1) is k, and the inner derivative of 2x is 2).
Evaluate at x = 1: f′(1) = k · e⁰ + 2g′(2) = k(1) + 2(−2) = k − 4.
Differentiate again: f″(x) = k² · ek(x−1) + 4g″(2x) (each additional derivative brings down another factor of k or another factor of 2). At x = 1: f″(1) = k²(1) + 4g″(2) = k² + 4(3) = k² + 12.
The correct answer is A.
Which expression correctly shows the fourth derivative of f?
Look at the pattern from each differentiation in part (a). Every time ek(x−1) is differentiated, the chain rule brings down one more factor of k, since the inner function k(x−1) has derivative k. After n derivatives, that term becomes kⁿek(x−1).
Similarly, every time g(2x) is differentiated, the chain rule brings down one more factor of 2, since the inner function 2x has derivative 2. After n derivatives, that term becomes 2ⁿg(n)(2x).
For the fourth derivative (n = 4): the first term becomes k⁴ek(x−1), and the second term becomes 2⁴gIV(2x) = 16gIV(2x) (since 2⁴ = 16).
Putting both terms together: fIV(x) = k⁴ek(x−1) + 16gIV(2x), exactly as given. The correct answer is D.
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