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AP Calculus Formulas

Instructions:

Answer 50 questions in 15 minutes.
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Match each statement with the correct term.
Don't refresh. All questions and answers are randomly picked and ordered every time you load a test.

This is a study tool. The 3 wrong answers for each question are randomly chosen from answers to other questions. So, you might find at times the answers obvious, but you will see it re-enforces your understanding as you take the test each time.

1. Volume of a right circular cylinder
-csc² x
pr²h
1
n ln m

2. Average speed
-sin x
e^x
?s/?t
ln m + ln n

3. sin p/3
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
1/2
v3/2
v2/2

4. How to get from precalculus to calculus
2 sin x cos x
Limits
v2/2
sin x / cos x

5. Derivative of a constant
If two functions - f and g - are differentiable - then d/dx[ f(x) / g(x) ] = [g(x)f'(x) - f(x) g'(x)] / [g(x)]²
d/dx[c] = 0
csc²x
?s/?t

6. ln mn
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
pr²h/3
v3/2
n ln m

7. cos p/3
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
0
1/2
u'/u - u > 0

8. d/dx[ln u]

9. ln 1
-1/v(1-x²)
-1/(1+x²)
0
Derivative of Position - s'(t)

10. d/dx[log_a x]
v3s² / 4
1. Given - Want - Sketch 2. Write an equation using variables given/to be determined 3. Differentiate w.r.t. time (using chain rule) 4. Plug in & solve
-sin x
1/((ln a) x)

11. cos²x
-1
2pr
(1 + cos 2x) / 2
1. Differentiate both sides w.r.t. x 2. Move all y' terms to one side & other terms to the other 3. Factor out y' 4. Divide to solve for y'

12. Extreme Value Theorem
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
u'/((ln a) u)
1/(1+x²)
2 sin x cos x

13. Velocity - v(t)

14. sin p/2
1 / sin x
v2/2
1
-1

15. 1 + tan²x
sec²x
Let f be continuous on [a -b] and differentiable on (a -b) and if f(a)=f(b) then there is at least one number c on (a -b) such that f'(c)=0 (If the slope of the secant is 0 - the derivative must = 0 somewhere in the interval).
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
?s/?t

16. sec x
e^x
cos²x - sin²x
0/0
1 / cos x

17. Position function of a falling object (with acceleration in ft/s²)
1
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
1/((ln a) x)
v3s² / 4

18. cot x
Slope of a function at a point/slope of the tangent line to a function at a point
sec x tan x
(1 - cos 2x) / 2
1 / tan x = cos x / sin x

19. sin(2x)
0
if f(x) is continuous and differentiable - slope of tangent line equals slope of secant line at least once in the interval (a - b) - f '(c) = [f(b) - f(a)]/(b - a)
2 sin x cos x
u' (ln a) a^u

20. cos 3p/2
1 / sin x
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
0
f(x) g'(x) + g(x) f'(x)

21. d/dx[sin x]
cos x
S = 4 pi r^2
1
1/((ln a) x)

22. d/dx[log_a u]

23. Limit Definition of a Derivative

24. The Quotient Rule

25. Sum and Difference Rules for Derivatives

26. d/dx[arctan x]
1/(1+x²)
pr²h/3
pr²
u' e^u

27. The Product Rule

28. Continuity & differentiability
Derivative of Position - s'(t)
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x
ln m - ln n

29. 1 + cot²x
csc²x
1 / tan x = cos x / sin x
0
sec²x

30. If f(-x) = f(x)
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
f is an even function
1
f(x) g'(x) + g(x) f'(x)

31. sin p/4
1/2
(ln a) a^x
-1/(1+x²)
v2/2

32. cos p/6
u' e^u
1
v3/2
u'/((ln a) u)

33. cos p/2
0
1 / cos x
sec x tan x
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].

34. Surface Area of a Sphere
1. Differentiate both sides w.r.t. x 2. Move all y' terms to one side & other terms to the other 3. Factor out y' 4. Divide to solve for y'
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
1. f(x) is defined at f(c) 2. The limit as x approaches c of f(x) exists 3. The limit as x approaches c of f(x) = f(c)
S = 4 pi r^2

35. ln (mn)
Derivative of position at a point
ln m + ln n
0
Differentiability implies continuity - but continuity does not necessarily imply differentiability.

36. d/dx[sec x]
0
1. f(x) is defined at f(c) 2. The limit as x approaches c of f(x) exists 3. The limit as x approaches c of f(x) = f(c)
sec x tan x
0

37. d/dx[a^u]

38. cos p
1
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
0
-1

39. d/dx[cos x]
d/dx[f(x) ± g(x)] = f'(x) ± g'(x)
-sin x
f'(g(x))g'(x)
(1 + cos 2x) / 2

40. Guidelines for solving related rates problems
1. Given - Want - Sketch 2. Write an equation using variables given/to be determined 3. Differentiate w.r.t. time (using chain rule) 4. Plug in & solve
if f(x) is continuous and differentiable - slope of tangent line equals slope of secant line at least once in the interval (a - b) - f '(c) = [f(b) - f(a)]/(b - a)
f is an odd function
Limits

41. Area of a circle
V = 4/3 pi r^3
n ln m
?s/?t
pr²

42. The limit as x approaches 0 of (1 - cos x) / x
if f(x) is continuous and differentiable - slope of tangent line equals slope of secant line at least once in the interval (a - b) - f '(c) = [f(b) - f(a)]/(b - a)
1/(|x|v(x²-1))
0
Limits

43. Volume of a Sphere
V = 4/3 pi r^3
2pr
u'/u - u > 0
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²

44. d/dx[e^x]
e^x
1
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
1/(|x|v(x²-1))

45. d/dx[cot x]
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x
-1/(|x|v(x²-1))
1 / cos x
-csc² x

46. Derivative of an inverse (if g(x) is the inverse of f(x))

47. d/dx[arccsc x]
0
1
x values where f'(x) is zero or undefined.
-1/(|x|v(x²-1))

48. The limit as x approaches 0 of sin x / x
1 / tan x = cos x / sin x
sec² x
n ln m
1

49. Circumference of a circle
2pr
-1/(1+x²)
1/(|x|v(x²-1))
1. f(x) is defined at f(c) 2. The limit as x approaches c of f(x) exists 3. The limit as x approaches c of f(x) = f(c)

50. cos 0
1. f(x) is defined at f(c) 2. The limit as x approaches c of f(x) exists 3. The limit as x approaches c of f(x) = f(c)
sec²x
1
-1/v(1-x²)