Calculus Mastery: 100 Multiple-Choice Questions on Differentiation:
Techniques, Applications, and Optimization Problems with Detailed
Solutions
1. What is the derivative of f(x) = 3x^2 - 2x + 5?
a) 6x - 2
b) 3x^2 - 2
c) 6x - 1
d) 3x - 2
Answer: a) 6x - 2
Solution: Use the power rule and constant rule:
f'(x) = 3(2x) - 2(1) + 0 = 6x - 2
2. Find the derivative of g(x) = e^(3x).
a) 3e^(3x)
b) e^(3x)
c) 3xe^(3x)
d) e^x
Answer: a) 3e^(3x)
Solution: Use the chain rule:
g'(x) = e^(3x) * 3 = 3e^(3x)
3. What is the second derivative of h(x) = 4x^3 - 2x^2 + 3x - 1?
a) 12x - 4
b) 24x - 4
c) 12x^2 - 4x + 3
d) 24x - 4x + 3
Answer: b) 24x - 4
Solution: First derivative: h'(x) = 12x^2 - 4x + 3
Second derivative: h''(x) = 24x - 4
4. Find the critical points of f(x) = x^3 - 3x^2 + 2.
a) x = 0 and x = 2
b) x = 1 and x = 2
c) x = 0 and x = 1
d) x = 1 and x = 3
Answer: b) x = 1 and x = 2
Solution: f'(x) = 3x^2 - 6x
Set f'(x) = 0: 3x^2 - 6x = 0
Factor: 3x(x - 2) = 0
Solve: x = 0 or x = 2
5. What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
What is the derivative of sin(2x)?
a) cos(2x)
b) 2cos(2x)
c) -2sin(2x)
d) -sin(2x)
Answer: b) 2cos(2x)
Solution: Use the chain rule:
d/dx[sin(2x)] = cos(2x) * d/dx[2x] = cos(2x) * 2 = 2cos(2x)
6. Find the absolute maximum of f(x) = x^2 - 4x + 3 on the interval [0, 4].
a) 3
b) 4
c) 5
d) 7
Answer: c) 5
Solution: Find critical points: f'(x) = 2x - 4, set to 0: x = 2
Evaluate f(0), f(2), and f(4):
f(0) = 3, f(2) = -1, f(4) = 3
The maximum value is f(0) = f(4) = 3
7. What is the derivative of ln(x^2 + 1)?
a) 2x / (x^2 + 1)
b) 1 / (x^2 + 1)
c) 2 / (x^2 + 1)
d) x / (x^2 + 1)
Answer: a) 2x / (x^2 + 1)
Solution: Use the chain rule:
d/dx[ln(x^2 + 1)] = 1/(x^2 + 1) * d/dx[x^2 + 1] = 1/(x^2 + 1) * 2x = 2x/(x^2 + 1)
8. Find the inflection point of f(x) = x^3 - 3x^2 + 2x.
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: b) x = 1
Solution: f'(x) = 3x^2 - 6x + 2
f''(x) = 6x - 6
Set f''(x) = 0: 6x - 6 = 0, x = 1
9. What is the derivative of tan(x)?
a) sec(x)
b) sec^2(x)
c) -sin(x)
d) cos(x)
Answer: b) sec^2(x)
Solution: Use the quotient rule on sin(x)/cos(x):
d/dx[tan(x)] = [cos(x)*cos(x) + sin(x)*sin(x)] / cos^2(x) = 1 / cos^2(x) = sec^2(x)
10. Find the local minimum of f(x) = x^3 - 3x^2 - 9x + 5.
a) x = -1
b) x = 0
c) x = 1
d) x = 3
Answer: c) x = 1
Solution: f'(x) = 3x^2 - 6x - 9
Set f'(x) = 0: 3x^2 - 6x - 9 = 0
Factor: 3(x^2 - 2x - 3) = 0
Solve: x = -1 or x = 3
Check second derivative: f''(x) = 6x - 6
f''(-1) = -12 (maximum), f''(3) = 12 (minimum)
11. What is the derivative of arcsin(x)?
a) 1 / sqrt(1 - x^2)
b) -1 / sqrt(1 - x^2)
c) 1 / (1 + x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / sqrt(1 - x^2)
Solution: The derivative of arcsin(x) is 1 / sqrt(1 - x^2)
12. Find the critical points of f(x) = x^4 - 4x^3 + 4x^2.
a) x = 0 and x = 2
b) x = 0, x = 1, and x = 2
c) x = 1 and x = 2
d) x = 0 and x = 1
Answer: b) x = 0, x = 1, and x = 2
Solution: f'(x) = 4x^3 - 12x^2 + 8x
Factor: 4x(x^2 - 3x + 2) = 0
Solve: x = 0 or x^2 - 3x + 2 = 0
x = 0, x = 1, or x = 2
13. What is the derivative of cosh(x)?
a) sinh(x)
b) -sinh(x)
c) cosh(x)
d) -cosh(x)
Answer: a) sinh(x)
Solution: The derivative of cosh(x) is sinh(x)
14. Find the absolute minimum of f(x) = x^3 - 3x^2 + 3 on the interval [-1, 3].
a) -4
b) -1
c) 0
d) 3
Answer: b) -1
Solution: Find critical points: f'(x) = 3x^2 - 6x, set to 0: x = 0 or x = 2
Evaluate f(-1), f(0), f(2), and f(3):
f(-1) = -1, f(0) = 3, f(2) = -1, f(3) = 3
The minimum value is f(-1) = f(2) = -1
15. What is the derivative of x^x?
a) x^x * (1 + ln(x))
b) x^x * ln(x)
c) x^(x-1)
d) x * x^(x-1)
Answer: a) x^x * (1 + ln(x))
Solution: Use the formula d/dx[a^x] = a^x * ln(a) and the product rule:
d/dx[x^x] = x^x * ln(x) + x^x * 1/x = x^x * (ln(x) + 1)
16. Find the points of horizontal tangency for f(x) = x^3 - 6x^2 + 9x + 1.
a) x = 1 and x = 2
b) x = 0 and x = 3
c) x = 1 and x = 3
d) x = 2 and x = 3
Answer: c) x = 1 and x = 3
Solution: f'(x) = 3x^2 - 12x + 9
Set f'(x) = 0: 3x^2 - 12x + 9 = 0
Factor: 3(x^2 - 4x + 3) = 0
Solve: x = 1 or x = 3
17. What is the derivative of sec(x)?
a) tan(x) * sec(x)
b) -tan(x) * sec(x)
c) csc(x) * cot(x)
d) -csc(x) * cot(x)
Answer: a) tan(x) * sec(x)
Solution: Use the quotient rule on 1/cos(x):
d/dx[sec(x)] = d/dx[1/cos(x)] = [-1/cos^2(x)] * (-sin(x)) = sin(x)/cos^2(x) = tan(x) * sec(x)
18. Find the maximum rate of change of f(x, y) = 3x^2y - 2xy^2 at the point (1, 2).
a) 5
b) 7
c) 9
d) 11
Answer: c) 9
Solution: Compute partial derivatives:
fx = 6xy - 2y^2, fy = 3x^2 - 4xy
At (1, 2): fx = 6(1)(2) - 2(2^2) = 4, fy = 3(1^2) - 4(1)(2) = -5
Maximum rate of change = sqrt(fx^2 + fy^2) = sqrt(4^2 + (-5)^2) = sqrt(81) = 9
19. What is the derivative of arctan(x)?
a) 1 / (1 + x^2)
b) -1 / (1 + x^2)
c) 1 / sqrt(1 - x^2)
d) 1 / (1 - x^2)
Answer: a) 1 / (1 + x^2)
Solution: The derivative of arctan(x) is 1 / (1 + x^2)
20. Find the interval where f(x) = x^3 - 3x^2 - 24x + 8 is increasing.
a) (-4, 2)
b) (-2, 4)
c) (-∞, -4) ∪ (2, ∞)
d) (-4, -2) ∪ (4, ∞)
Answer: c) (-∞, -4) ∪ (2, ∞)
Solution: f'(x) = 3x^2 - 6x - 24
Factor: 3(x^2 - 2x - 8) = 3(x + 2)(x - 4)
Critical points: x = -2 and x = 4
Test intervals: f is increasing when f'(x) > 0
f'(x) > 0 when x < -4 or x > 2
21. What is the derivative of (sin x)^(cos x)?
a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
b) (sin x)^(cos x) * [cos x * ln(sin x) + sin x * cos x]
c) (cos x)^(sin x) * [sin x * ln(cos x) - cos x * sin x]
d) (cos x)^(sin x) * [sin x * ln(cos x) + cos x * sin x]
Answer: a) (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
Solution: Use the formula d/dx[u^v] = u^v * [v * (du/dx)/u + ln(u) * dv/dx]:
u = sin x, v = cos x
du/dx = cos x, dv/dx = -sin x
d/dx[(sin x)^(cos x)] = (sin x)^(cos x) * [cos x * (cos x / sin x) + ln(sin x) * (-sin x)]
= (sin x)^(cos x) * [cos x * cot x - sin x * ln(sin x)]
= (sin x)^(cos x) * [cos x * ln(sin x) - sin x * cos x]
22. Find the points of vertical tangency for f(x) = (x^2 - 1)^(1/3).
a) x = -1 and x = 1
b) x = 0
c) x = -1, x = 0, and x = 1
d) No points of vertical tangency
Answer: a) x = -1 and x = 1
Solution: f'(x) = (1/3) * (x^2 - 1)^(-2/3) * 2x
Vertical tangents occur when the denominator is zero:
x^2 - 1 = 0
Solve: x = -1 or x = 1
23. What is the derivative of ln(sec x + tan x)?
a) 1
b) sec x
c) tan x
d) csc x
Answer: b) sec x
Solution: Use the chain rule and the identity d/dx[ln(sec x + tan x)] = sec x:
d/dx[ln(sec x + tan x)] = 1/(sec x + tan x) * d/dx[sec x + tan x]
= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)
= (sec x * tan x + sec^2 x) / (sec x + tan x)
= sec x * (tan x + sec x) / (sec x + tan x)
= sec x
24. Find the absolute extrema of f(x) = x^3 - 3x on the interval [-2, 2].
a) Min: -4 at x = -2, Max: 4 at x = 2
b) Min: -4 at x = -1, Max: 4 at x = 1
c) Min: -2 at x = -1, Max: 2 at x = 1
d) Min: -8 at x = -2, Max: 8 at x = 2
Answer: b) Min: -4 at x = -1, Max: 4 at x = 1
Solution: f'(
