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

2. d/dx[a^u]

3. d/dx[a^x]
v3s² / 4
(ln a) a^x
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height

4. cot x
1 / tan x = cos x / sin x
0
x values where f'(x) is zero or undefined.
Derivative of position at a point

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

6. d/dx[ln u]

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

8. ln (m/n)
1/v(1-x²)
ln m - ln n
If two functions - f and g - are differentiable - then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)
1 / tan x = cos x / sin x

9. d/dx[log_a 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.
1/((ln a) x)
csc²x

10. sin p/2
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
f(x) g'(x) + g(x) f'(x)
1

11. cos p/3
0
-csc² x
1/2
(1 - cos 2x) / 2

12. d/dx[cot x]
d/dx[cf(x)] = c f'(x)
V = 4/3 pi r^3
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).
-csc² x

13. ln (mn)
ln m + ln n
f(x) g'(x) + g(x) f'(x)
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
1

14. d/dx[tan x]
sec² x
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
v3/2
v2/2

15. Continuity at a point (x = c)
d/dx[x^n]=nx^(n-1)
pr²h
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)

16. d/dx[arcsin x]
1/x - x>0
e^x
1/v(1-x²)
-1/v(1-x²)

17. Instantaneous velocity
Derivative of position at a point
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)
1
0

18. d/dx[log_a u]

19. The Product Rule

20. sin²x
If two functions - f and g - are differentiable - then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(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)
Derivative of Position - s'(t)
(1 - cos 2x) / 2

21. Critical number

22. sin p
0
u' (ln a) a^u
2pr
v3/2

23. Constant Multiple Rule for Derivatives

24. d/dx[cos x]
pr²h/3
-sin x
1
S = 4 pi r^2

25. d/dx[sec x]
sec x tan x
-1
n ln m
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.

26. cos²x + sin²x
1 / tan x = cos x / sin x
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
cos²x - sin²x
1

27. d/dx[x]
1
0
1 / tan x = cos x / sin x
(1 - cos 2x) / 2

28. sin p/4
(1 + cos 2x) / 2
x values where f'(x) is zero or undefined.
-1
v2/2

29. sin(2x)
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
2 sin x cos x
Limits

30. d/dx[e^u]

31. Area of a circle
ln m - ln n
sec²x
pr²
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height

32. Sum and Difference Rules for Derivatives

33. Continuity on a closed interval - [a -b]
sin x / cos 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)
Derivative of position at a point
0

34. Circumference of a circle
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
1 / sin x
-1/v(1-x²)
2pr

35. sec x
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]
(1 + cos 2x) / 2
1 / cos x
sec² x

36. Guidelines for solving related rates problems
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
csc²x
(ln a) a^x
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

37. Area of an equilateral triangle
sec x tan x
-1
v3s² / 4
d/dx[cf(x)] = c f'(x)

38. d/dx[e^x]
1/(1+x²)
e^x
1/(|x|v(x²-1))
-sin x

39. d/dx[arcsec x]
1/(|x|v(x²-1))
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.
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x

40. The limit as x approaches 0 of sin x / x
u'/u - u > 0
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
1
v3/2

41. csc x
1 / sin x
sec² 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)]²
2 sin x cos x

42. cos 3p/2
x values where f'(x) is zero or undefined.
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
0
V = 4/3 pi r^3

43. Derivative of a constant
d/dx[c] = 0
Limits
d/dx[cf(x)] = c f'(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)

44. sin p/3
sec x tan x
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'
v3/2
1

45. Volume of a Sphere
V = 4/3 pi r^3
f(x) g'(x) + g(x) f'(x)
u'/u - u > 0
Differentiability implies continuity - but continuity does not necessarily imply differentiability.

46. sin 0
pr²
0
sec² x
v3s² / 4

47. The limit as x approaches 0 of (1 - cos x) / x
1/(1+x²)
f(x) g'(x) + g(x) f'(x)
1 / cos x
0

48. How to get from precalculus to calculus
v2/2
sec x tan x
0
Limits

49. cos 0
-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.
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
1

50. The Quotient Rule