SUBJECTS
|
BROWSE
|
CAREER CENTER
|
POPULAR
|
JOIN
|
LOGIN
Business Skills
|
Soft Skills
|
Basic Literacy
|
Certifications
About
|
Help
|
Privacy
|
Terms
|
Email
Search
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 re-enforces your understanding as you take the test each time.
1. Volume of a cone
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
pr²h/3
?s/?t
If two functions - f and g - are differentiable - then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)
2. d/dx[arccot x]
-1
-csc² x
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
-1/(1+x²)
3. cos 0
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
e^x
1
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
4. Circumference of a circle
1
v3/2
2pr
-1/v(1-x²)
5. d/dx[ f(x) / g(x) ]
6. Surface Area of a Sphere
n ln m
u'/((ln a) u)
S = 4 pi r^2
v3/2
7. Critical number
8. Sum and Difference Rules for Derivatives
9. d/dx[e^x]
Derivative of Position - s'(t)
x values where f'(x) is zero or undefined.
e^x
S = 4 pi r^2
10. The Product Rule
11. Derivative of a constant
1 / cos x
d/dx[c] = 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)
d/dx[cf(x)] = c f'(x)
12. d/dx[sin x]
0
sin x / cos x
sec²x
cos x
13. ln (mn)
d/dx[cf(x)] = c f'(x)
ln m + ln n
1 / tan x = cos x / sin x
V = 4/3 pi r^3
14. d/dx[arccos x]
0
-1/v(1-x²)
1
If two functions - f and g - are differentiable - then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)
15. d/dx[sec x]
sec x tan x
-1/v(1-x²)
1 / tan x = cos x / sin x
1
16. d/dx[a^x]
-1/(|x|v(x²-1))
(ln a) a^x
1 / tan x = cos x / sin x
-1
17. Position function of a falling object (with acceleration in ft/s²)
-csc x cot x
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
u'/((ln a) u)
1
18. d/dx[log_a x]
1
1
-csc x cot x
1/((ln a) x)
19. Continuity on a closed interval - [a -b]
Slope of a function at a point/slope of the tangent line to a function at a point
v2/2
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 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)
20. sin(2x)
1/((ln a) x)
f is an odd function
1
2 sin x cos x
21. Volume of a right circular cylinder
pr²h
Limits
v3/2
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]
22. Guidelines for solving related rates problems
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
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
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).
23. ln mn
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
d/dx[x^n]=nx^(n-1)
n ln m
24. cos p/2
1/(|x|v(x²-1))
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
0
25. ln e
1
cos²x - sin²x
-1
-1/(|x|v(x²-1))
26. Area of an equilateral triangle
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
-csc x cot x
v3s² / 4
27. cos(2x)
1
cos²x - sin²x
Derivative of position at a point
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]
28. Indeterminate form
0/0
f'(g(x))g'(x)
1/(1+x²)
1
29. d/dx[arcsec x]
1/(|x|v(x²-1))
f is an even function
-1
-sin x
30. sin p/2
1
ln m - ln n
2pr
If two functions - f and g - are differentiable - then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)
31. d/dx[arctan x]
v3s² / 4
0
1/(1+x²)
sec²x
32. 1 + tan²x
(1 - cos 2x) / 2
1/((ln a) x)
1 / sin x
sec²x
33. d/dx[arccsc x]
e^x
0/0
1/x - x>0
-1/(|x|v(x²-1))
34. Alternate Limit Definition of a derivative
35. d/dx[tan x]
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
sec x tan x
u' e^u
sec² x
36. How to get from precalculus to calculus
Limits
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
2 sin x cos x
0
37. Instantaneous velocity
If two functions - f and g - are differentiable - then d/dx[ f(x) g(x) ] = f(x) g'(x) + g(x) f'(x)
f is an odd function
-1
Derivative of position at a point
38. sin 0
-csc² x
cos x
0
1
39. d/dx[ln 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/x - x>0
-1/v(1-x²)
0
40. Mean Value Theorem
41. Area of a circle
pr²
v2/2
(1 - cos 2x) / 2
1/2
42. d/dx[cos 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
-sin x
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
u'/((ln a) u)
43. ln 1
0
d/dx[cf(x)] = c f'(x)
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]
f is an even function
44. sin p/3
pr²
v2/2
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
v3/2
45. sin p
(ln a) a^x
sin x / cos x
0
f(x) g'(x) + g(x) f'(x)
46. Derivative
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
Slope of a function at a point/slope of the tangent line to a function at a point
?s/?t
1
47. sin p/6
pr²h/3
1/x - x>0
1 / tan x = cos x / sin x
1/2
48. d/dx[cot x]
-csc² x
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
-1/(|x|v(x²-1))
-csc x cot x
49. d/dx[log_a u]
50. sin p/4
v2/2
d/dx[c] = 0
Derivative of position at a point
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]