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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. Continuity & differentiability
e^x
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
pr²h/3
2 sin x cos x

2. Derivative
V = 4/3 pi r^3
Slope of a function at a point/slope of the tangent line to a function at a point
-1/(1+x²)
u' (ln a) a^u

3. sin²x
ln m + ln n
pr²h
-1
(1 - cos 2x) / 2

4. Derivative of a constant
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x
-1
cos²x - sin²x
d/dx[c] = 0

5. d/dx[sec x]
e^x
sec x tan 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
1 / sin x

6. cot x
-1/v(1-x²)
1 / tan x = cos x / sin x
cos²x - sin²x
1/2

7. 1 + cot²x
pr²
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
-1/(|x|v(x²-1))
csc²x

8. cos²x
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
0
(1 + cos 2x) / 2

9. d/dx[a^u]

10. sin p/3
v3/2
e^x
ln m + ln n
(1 - cos 2x) / 2

11. d/dx[log_a x]
d/dx[x^n]=nx^(n-1)
1/((ln a) x)
u'/((ln a) u)
1

12. ln 1
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 two functions - f and g - are differentiable - then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)
0
(1 + cos 2x) / 2

13. cos(2x)
u' e^u
csc²x
v2/2
cos²x - sin²x

14. cos 3p/2
0
-1/(1+x²)
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
pr²h

15. ln (mn)
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
d/dx[x^n]=nx^(n-1)
ln m + ln n
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. Extreme Value Theorem
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].
u'/u - u > 0
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0

17. The Product Rule

18. ln e
-1/(|x|v(x²-1))
csc²x
-1
1

19. If f(-x) = f(x)
-1/(1+x²)
sec²x
f is an even function
sec² x

20. cos p/6
v3/2
0
0
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height

21. sin p/2
-1
-1
1
u' (ln a) a^u

22. ln mn
1 / sin x
n ln m
sec²x
e^x

23. d/dx[cot x]
sec x tan x
v2/2
-csc² x
f is an odd function

24. Volume of a cone
u' (ln a) a^u
pr²h/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).
(1 + cos 2x) / 2

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

26. d/dx[ln x]
1/x - x>0
(1 - cos 2x) / 2
1/v(1-x²)
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].

27. sin 0
Derivative of position at a point
1
0
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].

28. sec x
1 / cos x
Slope of a function at a point/slope of the tangent line to a function at a point
1
-csc² x

29. Guidelines for implicit differentiation

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

31. Area of a circle
0
2 sin x cos x
-1/(1+x²)
pr²

32. d/dx[log_a u]

33. Continuity on a closed interval - [a -b]
(1 - cos 2x) / 2
x values where f'(x) is zero or undefined.
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)

34. If f(-x) = -f(x)
f is an odd function
(1 - cos 2x) / 2
1/(|x|v(x²-1))
S = 4 pi r^2

35. sin p/6
1/2
1
1
S = 4 pi r^2

36. Intermediate Value Theorem
1/2
ln m - ln n
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.
-sin x

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

38. ln (m/n)
cos x
1
ln m - ln n
u' e^u

39. Rolle's Theorem

40. Volume of a right circular cylinder
0
pr²h
pr²
If two functions - f and g - are differentiable - then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)

41. Velocity - v(t)

42. d/dx[tan x]
sec² 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).
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
0

43. Sum and Difference Rules for Derivatives

44. Position function of a falling object (with acceleration in ft/s²)
(1 - cos 2x) / 2
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).
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height

45. d/dx[sin x]
1/((ln a) x)
cos x
n ln m
v3/2

46. d/dx[csc x]
v2/2
(1 - cos 2x) / 2
1/2
-csc x cot x

47. d/dx[arcsec x]
1/(|x|v(x²-1))
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[x^n]=nx^(n-1)
-sin x

48. d/dx[ln u]

49. d/dx[arctan x]
1 / tan x = cos x / sin x
1/(1+x²)
Slope of a function at a point/slope of the tangent line to a function at a point
f is an even function

50. Guidelines for solving related rates problems
1/(|x|v(x²-1))
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
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
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²