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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. How to get from precalculus to calculus
prh/3
Limits
1
cscx

2. Continuity at a point (x = c)
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 + cos 2x) / 2
Slope of a function at a point/slope of the tangent line to a function at a point
prh

3. Indeterminate form
-csc x
1/((ln a) x)
0/0
sec x tan x

4. cosx + sinx
-1/v(1-x)
x values where f'(x) is zero or undefined.
sec x
1

5. sin p/4
v2/2
-sin x
S = 4 pi r^2
2pr

6. sec x
ln m - ln n
sec x tan x
1 / cos x
d/dx[cf(x)] = c f'(x)

7. Area of a circle
pr
d/dx[f(x) g(x)] = f'(x) g'(x)
v3/2
v3/2

8. d/dx[arcsin x]
ln m + ln n
prh
1/v(1-x)
cscx

9. d/dx[arccot x]
-1/(1+x)
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
s(t) = -4.9t+ v0t + s0 - v0 = initial velocity - s0 = initial height

10. Velocity - v(t)
prh/3
Derivative of Position - s'(t)
u'/((ln a) u)
0

11. Volume of a Sphere
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)
ln m - ln n
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
V = 4/3 pi r^3

12. d/dx[arcsec x]
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)
0
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
1/(|x|v(x-1))

13. Position function of a falling object (with acceleration in ft/s)
prh/3
1
f is an even function
s(t) = -16t+ v0t + s0 - v0 = initial velocity - s0 = initial height

14. d/dx[csc x]
-csc x cot x
1/2
0
s(t) = -16t+ v0t + s0 - v0 = initial velocity - s0 = initial height

15. cot x
v3/2
-csc x
1 / tan x = cos x / sin x
e^x

16. Position function of a falling object (with acceleration in m/s)
s(t) = -4.9t+ v0t + s0 - v0 = initial velocity - s0 = initial height
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
v2/2

17. Circumference of a circle
2pr
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).
[g(x)f'(x) - f(x) g'(x)] / [g(x)]
-csc x

18. sin(2x)
1 / tan x = cos x / sin x
s(t) = -16t+ v0t + s0 - v0 = initial velocity - s0 = initial height
-sin x
2 sin x cos x

19. d/dx[tan 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'
sec x
(1 + cos 2x) / 2

20. cos p/6
prh/3
v3/2
d/dx[f(x) g(x)] = f'(x) g'(x)
(1 + cos 2x) / 2

21. d/dx[e^x]
1/v(1-x)
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]
(ln a) a^x
e^x

22. Extreme Value Theorem
sec x tan x
u' e^u
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [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)

23. cos(2x)
cosx - sinx
ln m + ln n
v2/2
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)

24. The Quotient Rule
V = 4/3 pi r^3
0
f'(g(x))g'(x)
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)]

25. Chain Rule: d/dx[f(g(x))] =
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]
f'(g(x))g'(x)
1/x - x>0
pr

26. Sum and Difference Rules for Derivatives
d/dx[f(x) g(x)] = f'(x) g'(x)
1 / cos x
u'/((ln a) u)
1/2

27. tan x
1
sin x / cos x
1
e^x

28. d/dx[cos x]
2pr
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
-sin x
-1/v(1-x)

29. Derivative of an inverse (if g(x) is the inverse of f(x))
1 / tan x = cos x / sin x
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
1
cosx - sinx

30. d/dx[a^x]
1/v(1-x)
f is an odd function
(1 - cos 2x) / 2
(ln a) a^x

31. Limit Definition of a Derivative
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x
v2/2
1 / sin x
Limits

32. sin p/2
1/2
1
secx
f(x) g'(x) + g(x) f'(x)

33. 1 + tanx
1/2
secx
d/dx[f(x) g(x)] = f'(x) g'(x)
u'/u - u > 0

34. sin p
u'/((ln a) u)
0
S = 4 pi r^2
v3/2

35. Area of an equilateral triangle
v3s / 4
1
1/(1+x)
s(t) = -4.9t+ v0t + s0 - v0 = initial velocity - s0 = initial height

36. Volume of a cone
prh/3
(ln a) a^x
1
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]

37. d/dx[arctan x]
u'/u - u > 0
1/(1+x)
Slope of a function at a point/slope of the tangent line to a function at a point
cscx

38. sin 3p/2
cscx
f(x) g'(x) + g(x) f'(x)
-1
1

39. ln mn
1
-1
0
n ln m

40. d/dx[cot x]
Limits
-csc x
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
v2/2

41. The Product Rule
If two functions - f and g - are differentiable - then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(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).
s(t) = -4.9t+ v0t + s0 - v0 = initial velocity - s0 = initial height
2pr

42. sin 0
(1 + cos 2x) / 2
0
-1
Derivative of position at a point

43. d/dx[log_a x]
f is an odd function
f(x) g'(x) + g(x) f'(x)
-1/(|x|v(x-1))
1/((ln a) x)

44. cos p/3
sec x
0
(1 + cos 2x) / 2
1/2

45. cos p
1/x - x>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).
2 sin x cos x
-1

46. d/dx[e^u]
u' e^u
u' (ln a) a^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
cscx

47. If f(-x) = -f(x)
s(t) = -4.9t+ v0t + s0 - v0 = initial velocity - s0 = initial height
f is an odd function
pr
-1/v(1-x)

48. d/dx[arccsc 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
-1/(|x|v(x-1))
0

49. Instantaneous velocity
1
Derivative of position at a point
sin x / cos x
pr

50. d/dx[log_a u]
sec x tan x
u'/((ln a) u)
[g(x)f'(x) - f(x) g'(x)] / [g(x)]
s(t) = -16t+ v0t + s0 - v0 = initial velocity - s0 = initial height

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