AP Calc AB Differentiation Definition and Rules — Questions and Answers
Question 1: If f(x) = x² sin x, what is f′(π)?
- π²
- 2π
- −π² (Correct answer)
- 0
Correct answer: −π²
By the product rule, f′(x) = 2x sin x + x² cos x. At x = π this is 2π(0) + π²(−1) = −π².
The product rule says (uv)′ = u′v + uv′. With u = x² and v = sin x, we get f′(x) = 2x sin x + x² cos x. At x = π, sin π = 0 and cos π = −1. The first term vanishes and the second becomes −π². A common error is to forget that cos π is negative, which gives +π².
Question 2: Let f(x) = x³. What is the value of lim_{h→0} [(2 + h)³ − 8] / h?
Correct answer: 12
The limit is the definition of f′(2) for f(x) = x³. Since f′(x) = 3x², f′(2) = 12.
The limit definition of the derivative at a is f′(a) = lim_{h→0} [f(a + h) − f(a)] / h. Here a = 2 and f(2) = 8, so the expression is exactly f′(2). The power rule gives f′(x) = 3x², so f′(2) = 12. You can also expand (2 + h)³ = 8 + 12h + 6h² + h³, subtract 8, divide by h, and let h → 0 to get 12.
Question 3: What is d/dx of √(3x² + 1)?
- 6x / √(3x² + 1)
- 1 / (2√(3x² + 1))
- 3 / (2√(3x² + 1))
- 3x / √(3x² + 1) (Correct answer)
Correct answer: 3x / √(3x² + 1)
By the chain rule, the derivative is (1/2)(3x² + 1)^(−1/2) · 6x = 3x / √(3x² + 1).
Write √(3x² + 1) as (3x² + 1)^(1/2). The outer derivative is (1/2)(3x² + 1)^(−1/2), and the inner derivative is 6x. Multiplying gives 6x / (2√(3x² + 1)), which simplifies to 3x / √(3x² + 1). The common mistakes are leaving out the inner factor 6x or leaving out the 1/2.
Question 4: The point (1, 2) lies on the curve x² + xy + y² = 7. What is dy/dx at that point?
- −4/5 (Correct answer)
- 4/5
- −5/4
- −2/5
Correct answer: −4/5
Differentiating implicitly gives 2x + y + x(dy/dx) + 2y(dy/dx) = 0. At (1, 2) this is 4 + 5(dy/dx) = 0, so dy/dx = −4/5.
Differentiate each term with respect to x: x² gives 2x, xy gives y + x·y′, y² gives 2y·y′, and the constant 7 gives 0. Collect the y′ terms: y′(x + 2y) = −(2x + y). At (1, 2), x + 2y = 5 and 2x + y = 4, so y′ = −4/5. Check that (1, 2) lies on the curve: 1 + 2 + 4 = 7.
Question 5: Let f(x) = x³ + x + 1, which is one-to-one. If g is the inverse function of f, what is g′(3)?
- 4
- 1/3
- 1/4 (Correct answer)
- 3/4
Correct answer: 1/4
Since f(1) = 3, g(3) = 1. Then g′(3) = 1 / f′(1) = 1/(3·1² + 1) = 1/4.
The inverse function derivative formula is g′(b) = 1 / f′(g(b)). To find g(3), solve x³ + x + 1 = 3, which gives x = 1. Next, f′(x) = 3x² + 1, so f′(1) = 4. Therefore g′(3) = 1/4. Using f′(3) = 28 instead is a common mistake, because the derivative must be evaluated at g(3), not at 3.
Question 6: If f(x) = x² / (x + 1), what is f′(1)?
- 3/4 (Correct answer)
- 1/2
- 1/4
- 3/2
Correct answer: 3/4
By the quotient rule, f′(x) = [2x(x + 1) − x²] / (x + 1)² = (x² + 2x) / (x + 1)². At x = 1 this is 3/4.
Let u = x² and v = x + 1, so u′ = 2x and v′ = 1. The quotient rule gives (u′v − uv′)/v² = (2x(x + 1) − x²)/(x + 1)². The numerator simplifies to x² + 2x. At x = 1 the numerator is 3 and the denominator is 4, so f′(1) = 3/4. Reversing the order of the subtraction in the numerator would give the wrong sign.
Question 7: Which of the following functions is continuous at x = 0 but not differentiable at x = 0?
- f(x) = x^(4/3)
- f(x) = x^(2/3) (Correct answer)
- f(x) = x³
- f(x) = sin x
Correct answer: f(x) = x^(2/3)
For f(x) = x^(2/3), f′(x) = (2/3)x^(−1/3), which is unbounded near 0, so there is a cusp at x = 0. The other three functions are differentiable at 0.
A function can be continuous without being differentiable, for example at a corner, a cusp, or a vertical tangent. The function x^(2/3) is defined and continuous at 0, but the difference quotient h^(2/3)/h = h^(−1/3) has no finite limit as h → 0. That gives a cusp. By contrast, x^(4/3) has derivative (4/3)x^(1/3), which is 0 at x = 0, and x³ and sin x are smooth everywhere.
Question 8: What is d/dx [arctan(3x)]?
- 3 / (1 + 3x²)
- 1 / (1 + 9x²)
- 9 / (1 + 9x²)
- 3 / (1 + 9x²) (Correct answer)
Correct answer: 3 / (1 + 9x²)
The derivative of arctan u is u′ / (1 + u²). With u = 3x, this gives 3 / (1 + 9x²).
The derivative of arctan x is 1/(1 + x²). For a composite arctan(u), the chain rule gives u′/(1 + u²). With u = 3x, u′ = 3 and u² = 9x², so the result is 3/(1 + 9x²). Forgetting the chain-rule factor 3 gives 1/(1 + 9x²). Failing to square the 3 inside u² gives 3/(1 + 3x²).
Question 9: What is d/dx [5^x]?
- x · 5^(x−1)
- 5^x
- 5^x ln 5 (Correct answer)
- 5^x / ln 5
Correct answer: 5^x ln 5
For a constant base a > 0, d/dx [a^x] = a^x ln a. With a = 5, the derivative is 5^x ln 5.
Write 5^x as e^(x ln 5). By the chain rule, its derivative is e^(x ln 5) · ln 5, which equals 5^x ln 5. The power rule x·5^(x−1) applies only when the base is the variable and the exponent is a constant. Writing 5^x alone would be correct only for base e, and dividing by ln 5 is the rule for antiderivatives, not derivatives.
Question 10: Suppose f(2) = 1 and f′(2) = 4. If h(x) = [f(x)]³, what is h′(2)?
Correct answer: 12
By the chain rule, h′(x) = 3[f(x)]² · f′(x). At x = 2, h′(2) = 3(1)²(4) = 12.
The outer function is u³, with derivative 3u². The inner function is f(x), with derivative f′(x). So h′(x) = 3[f(x)]² f′(x). Substituting f(2) = 1 and f′(2) = 4 gives 3 · 1 · 4 = 12. Leaving out the f′(2) factor gives 3, and using f(2) = 4 in place of f′(2) mixes up function values and derivative values.
