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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.
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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. cos(2x)
v3/2
cos²x - sin²x
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)
sec²x

2. d/dx[log_a u]

3. Power Rule for Derivatives
d/dx[x^n]=nx^(n-1)
0
2 sin x cos x
1 / cos x

4. ln 1
csc²x
pr²
0
u' (ln a) a^u

5. ln (m/n)
Derivative of Position - s'(t)
-sin x
ln m - ln n
f is an even function

6. d/dx[tan x]
Derivative of position at a point
1
sin x / cos x
sec² x

7. Area of an equilateral triangle
d/dx[f(x) ± g(x)] = f'(x) ± g'(x)
v3s² / 4
pr²h/3
2pr

8. d/dx[arctan x]
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
csc²x
1
1/(1+x²)

9. Guidelines for solving related rates problems
ln m - ln n
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
Slope of a function at a point/slope of the tangent line to a function at a point
Derivative of Position - s'(t)

10. Volume of a right circular cylinder
pr²h
?s/?t
-csc x cot x
-1/(1+x²)

11. Extreme Value Theorem
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
d/dx[c] = 0
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
sec² x

12. cos²x
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
2 sin x cos x
(1 + cos 2x) / 2
S = 4 pi r^2

13. The Quotient Rule

14. sin 3p/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
d/dx[cf(x)] = c f'(x)

15. Alternate Limit Definition of a derivative

16. sin p/2
v2/2
1
?s/?t
-sin x

17. sin 0
0
pr²h/3
1
v3/2

18. d/dx[cot x]
d/dx[cf(x)] = c f'(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.
f is an even function
-csc² x

19. Area of a circle
v3s² / 4
pr²
f is an even function
1

20. d/dx[e^x]
pr²h/3
d/dx[x^n]=nx^(n-1)
e^x
2pr

21. Sum and Difference Rules for Derivatives

22. cos p/2
d/dx[f(x) ± g(x)] = f'(x) ± g'(x)
0
u'/((ln a) u)
f is an odd function

23. Continuity on an open interval - (a -b)
-csc x cot x
1/2
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
f'(g(x))g'(x)

24. Intermediate Value Theorem
-csc x cot 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.
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)
Derivative of position at a point

25. sec x
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
1 / cos x
V = 4/3 pi r^3
cos x

26. sin p
0
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
-1/v(1-x²)
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

27. 1 + cot²x
d/dx[f(x) ± g(x)] = f'(x) ± g'(x)
v2/2
csc²x
e^x

28. csc x
n ln m
1 / sin x
Derivative of position at a point
v3s² / 4

29. Average speed
-1
?s/?t
1/(1+x²)
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).

30. d/dx[arccsc x]
d/dx[f(x) ± g(x)] = f'(x) ± g'(x)
sin x / cos x
-1/(|x|v(x²-1))
d/dx[x^n]=nx^(n-1)

31. d/dx[arccos x]
-1/v(1-x²)
ln m + ln n
1/v(1-x²)
v2/2

32. tan x
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
0/0
u' (ln a) a^u
sin x / cos x

33. Position function of a falling object (with acceleration in m/s²)
f(x) g'(x) + g(x) f'(x)
(ln a) a^x
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
0/0

34. cos p/3
v3s² / 4
u'/u - u > 0
1/2
(1 - cos 2x) / 2

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

36. Volume of a Sphere
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
1/x - x>0
f'(g(x))g'(x)
V = 4/3 pi r^3

37. d/dx[cos x]
-sin x
0
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 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)

38. d/dx[a^x]
1
d/dx[cf(x)] = c f'(x)
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x
(ln a) a^x

39. Velocity - v(t)

40. Continuity & differentiability
pr²h/3
d/dx[c] = 0
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]
Differentiability implies continuity - but continuity does not necessarily imply differentiability.

41. 1 + tan²x
1 / cos x
sec²x
1/2
f'(g(x))g'(x)

42. ln e
-1
d/dx[cf(x)] = c f'(x)
1
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.

43. Continuity on a closed interval - [a -b]
1/((ln a) x)
(ln a) a^x
1/x - x>0
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)

44. Derivative of a constant
d/dx[c] = 0
1/x - x>0
1
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height

45. cos 0
0/0
0
u' (ln a) a^u
1

46. d/dx[x]
e^x
1
0/0
-sin x

47. If f(-x) = f(x)
v2/2
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
f is an even function
1 / tan x = cos x / sin x

48. d/dx[a^u]

49. d/dx[e^u]

50. cot x
1 / tan x = cos x / sin x
-sin 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) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height