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Test your basic knowledge 
AP Calculus Formulas
Start Test
Study First
Subjects
:
math
,
ap
,
calculus
Instructions:
Answer 50 questions in 15 minutes.
If you are not ready to take this test, you can
study here
.
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 reenforces your understanding as you take the test each time.
1. Guidelines for implicit differentiation
2. sin p
0
1/v(1x²)
v3s² / 4
Limits
3. cos p/6
v3/2
s(t) = 4.9t²+ v0t + s0  v0 = initial velocity  s0 = initial height
u'/u  u > 0
x values where f'(x) is zero or undefined.
4. d/dx[arctan x]
u' e^u
ln m  ln n
1/(1+x²)
[g(x)f'(x)  f(x) g'(x)] / [g(x)]²
5. Intermediate Value Theorem
f'(g(x))g'(x)
1 / sin x
1 / cos 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.
6. Position function of a falling object (with acceleration in m/s²)
Derivative of position at a point
csc² x
s(t) = 4.9t²+ v0t + s0  v0 = initial velocity  s0 = initial height
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).
7. Extreme Value Theorem
If f is continuous on the closed interval [a b] then it must have both a minimum and maximum on [a b].
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/(xv(x²1))
1 / cos x
8. d/dx[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
1
Differentiability implies continuity  but continuity does not necessarily imply differentiability.
1/v(1x²)
9. Guidelines for solving related rates problems
2 sin x cos x
f'(x) = lim as x ? c of [ f(x)  f(c) ] / [ x  c]
g'(x) = 1/f'(g(x))  f'(g(x)) cannot = 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
10. The limit as x approaches 0 of sin x / x
1
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/v(1x²)
11. 1 + tan²x
sec²x
1/v(1x²)
1
sin x / cos x
12. d/dx[ln u]
13. Volume of a Sphere
V = 4/3 pi r^3
1/v(1x²)
sec x tan x
1
14. sin p/4
d/dx[c] = 0
v2/2
V = 4/3 pi r^3
1/(1+x²)
15. d/dx[a^x]
x values where f'(x) is zero or undefined.
v3s² / 4
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
(ln a) a^x
16. d/dx[cos x]
sin x
1
csc² x
(1 + cos 2x) / 2
17. ln 1
u'/((ln a) u)
0
u'/u  u > 0
1 / tan x = cos x / sin x
18. sin p/3
v3/2
If two functions  f and g  are differentiable  then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)
ln m + ln n
1
19. d/dx[arcsin x]
1/v(1x²)
f is an even function
sin x / cos x
Differentiability implies continuity  but continuity does not necessarily imply differentiability.
20. Constant Multiple Rule for Derivatives
21. d/dx[a^u]
22. ln (mn)
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
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)
ln m + ln n
23. Critical number
24. sin(2x)
1
(1 + cos 2x) / 2
2 sin x cos x
u'/((ln a) u)
25. sec x
1 / cos x
d/dx[c] = 0
f is an odd function
0
26. d/dx[arccsc x]
1/2
1/(xv(x²1))
1
d/dx[cf(x)] = c f'(x)
27. d/dx[tan x]
2pr
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)
sec² x
28. d/dx[csc x]
1/v(1x²)
sec x tan x
d/dx[x^n]=nx^(n1)
csc x cot x
29. cos p/4
1/(xv(x²1))
v2/2
2pr
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)
30. The limit as x approaches 0 of (1  cos x) / x
v3/2
cos²x  sin²x
0
1/((ln a) x)
31. d/dx[arcsec x]
If two functions  f and g  are differentiable  then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)
V = 4/3 pi r^3
0
1/(xv(x²1))
32. d/dx[e^u]
33. Surface Area of a Sphere
pr²
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 = 4 pi r^2
v3s² / 4
34. Circumference of a circle
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)
2pr
s(t) = 16t²+ v0t + s0  v0 = initial velocity  s0 = initial height
1
35. Position function of a falling object (with acceleration in ft/s²)
If f is continuous on the closed interval [a b] then it must have both a minimum and maximum on [a b].
f is an even function
s(t) = 16t²+ v0t + s0  v0 = initial velocity  s0 = initial height
sec²x
36. Limit Definition of a Derivative
37. Rolle's Theorem
38. cos(2x)
cos²x  sin²x
Derivative of position at a point
sec²x
f(x) g'(x) + g(x) f'(x)
39. sin p/2
pr²h/3
csc x cot x
0
1
40. Chain Rule: d/dx[f(g(x))] =
41. The Product Rule
42. Volume of a right circular cylinder
v3/2
csc x cot x
pr²h
Differentiability implies continuity  but continuity does not necessarily imply differentiability.
43. cos p
1
u'/u  u > 0
1/(1+x²)
v3s² / 4
44. Continuity at a point (x = c)
1
If two functions  f and g  are differentiable  then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)
csc² 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)
45. Continuity on an open interval  (a b)
If two functions  f and g  are differentiable  then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)
pr²h
f(x) is continuous if for every point on the interval (a b) the conditions for continuity at a point are satisfied.
cos²x  sin²x
46. ln (m/n)
S = 4 pi r^2
1/2
ln m  ln n
g'(x) = 1/f'(g(x))  f'(g(x)) cannot = 0
47. cos p/2
n ln m
s(t) = 16t²+ v0t + s0  v0 = initial velocity  s0 = initial height
ln m  ln n
0
48. ln mn
1
n ln m
1/v(1x²)
0
49. d/dx[log_a x]
s(t) = 4.9t²+ v0t + s0  v0 = initial velocity  s0 = initial height
1/((ln a) x)
1
1/(xv(x²1))
50. cos 3p/2
cos x
0
Limits
v2/2