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AP Calculus Formulas

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Answer 50 questions in 15 minutes.
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Match each statement with the correct term.
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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 cone
Derivative of Position - s'(t)
u' e^u
1 / tan x = cos x / sin x
pr²h/3

2. d/dx[x]
v2/2
1
1. f(x) is continuous on the closed interval (a -b) 2. The limit from the right as x approaches a of f(x) is f(a) 3. The limit from the left as x approaches b of f(x) is f(b)
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0

3. cos(2x)
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x
Slope of a function at a point/slope of the tangent line to a function at a point
cos x
cos²x - sin²x

4. Continuity on a closed interval - [a -b]
1. f(x) is continuous on the closed interval (a -b) 2. The limit from the right as x approaches a of f(x) is f(a) 3. The limit from the left as x approaches b of f(x) is f(b)
f'(g(x))g'(x)
0/0
d/dx[c] = 0

5. d/dx[ln x]
1/x - x>0
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'
1/(1+x²)
pr²

6. If f(-x) = f(x)
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
f is an even function
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x
v3/2

7. Area of an equilateral triangle
2pr
ln m + ln n
1
v3s² / 4

8. csc x
-1/v(1-x²)
csc²x
-csc x cot x
1 / sin x

9. Volume of a Sphere
V = 4/3 pi r^3
1
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
1/((ln a) x)

10. Continuity at a point (x = c)
sin x / cos x
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)
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
u' e^u

11. Surface Area of a Sphere
f is an odd function
1/((ln a) x)
1
S = 4 pi r^2

12. The Product Rule

13. d/dx[cos x]
0
-sin x
?s/?t
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]

14. Area of a circle
pr²
sin x / cos x
d/dx[c] = 0
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'

15. cos²x
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x
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).
1/2
(1 + cos 2x) / 2

16. Alternate Limit Definition of a derivative

17. Position function of a falling object (with acceleration in m/s²)
0
1/(|x|v(x²-1))
s(t) = -4.9t²+ 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)

18. Guidelines for solving related rates problems
1 / cos x
-csc x cot x
u' e^u
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

19. d/dx[sec x]
Derivative of position at a point
sec x tan x
1
2pr

20. Continuity on an open interval - (a -b)
(1 + cos 2x) / 2
ln m + ln n
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(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.

21. cos p/4
v2/2
pr²h
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
1/2

22. Mean Value Theorem

23. Power Rule for Derivatives
u' (ln a) a^u
-1
d/dx[x^n]=nx^(n-1)
f is an even function

24. Instantaneous velocity
-1
1/(1+x²)
Derivative of position at a point
Limits

25. d/dx[ f(x) g(x) ]

26. sin p/4
0
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
Derivative of position at a point
v2/2

27. Sum and Difference Rules for Derivatives

28. Rolle's Theorem

29. cos p
Derivative of Position - s'(t)
-1
1/(|x|v(x²-1))
(1 + cos 2x) / 2

30. tan x
sin x / cos x
-1/(|x|v(x²-1))
1
Derivative of Position - s'(t)

31. d/dx[a^u]

32. d/dx[tan x]
-sin x
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)
sec² x

33. 1 + tan²x
v2/2
sec²x
-csc x cot x
u' e^u

34. sin 3p/2
-1/(|x|v(x²-1))
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
-1
f is an odd function

35. Intermediate Value Theorem
?s/?t
f'(g(x))g'(x)
sec² x
If f(x) is continuous on a closed interval [a -b] and k is any number between f(a) and f(b) - then there is at least one number c in [a -b] such that f(c) = k.

36. d/dx[arctan x]
1/(1+x²)
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
1/(|x|v(x²-1))
v3s² / 4

37. d/dx[a^x]
(ln a) a^x
e^x
0/0
(1 + cos 2x) / 2

38. Limit Definition of a Derivative

39. Derivative of a constant
pr²h
d/dx[c] = 0
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).
1

40. ln e
Limits
ln m - ln n
1
0

41. Chain Rule: d/dx[f(g(x))] =

42. d/dx[log_a 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
1
1/((ln a) x)

43. d/dx[arccot x]
1/2
v2/2
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)
-1/(1+x²)

44. ln (mn)
1
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
ln m + ln n
d/dx[c] = 0

45. d/dx[arccos x]
0/0
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
-1/v(1-x²)
sec x tan x

46. sin p/6
v3s² / 4
1
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
1/2

47. cos p/6
f'(g(x))g'(x)
v3/2
1
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height

48. d/dx[cot x]
-csc² x
S = 4 pi r^2
d/dx[x^n]=nx^(n-1)
1/((ln a) x)

49. d/dx[arccsc x]
1
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'
1/v(1-x²)
-1/(|x|v(x²-1))

50. d/dx[ln u]