AP Calculus Formulas

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. Alternate Limit Definition of a derivative

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2. d/dx[arccot x]
d/dx[f(x) ± g(x)] = f'(x) ± g'(x)
-1/(1+x²)
f(x) g'(x) + g(x) f'(x)
pr²


3. cos²x
0
(1 + cos 2x) / 2
?s/?t
-1


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

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

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6. d/dx[arctan x]
u' (ln a) a^u
1
1
1/(1+x²)


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

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8. d/dx[log_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
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
1
1/((ln a) x)


9. Position function of a falling object (with acceleration in m/s²)
1
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
0


10. cos p/6
v3/2
(1 - cos 2x) / 2
d/dx[c] = 0
1


11. ln 1
sec² x
0
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
-csc x cot x


12. d/dx[arccos x]
Limits
pr²
-1/v(1-x²)
2 sin x cos x


13. Average speed
Derivative of position at a point
?s/?t
-1/(1+x²)
u'/u - u > 0


14. The Quotient Rule

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15. sin 0
ln m + ln n
0/0
d/dx[x^n]=nx^(n-1)
0


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

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17. Position function of a falling object (with acceleration in ft/s²)
(1 - cos 2x) / 2
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


18. Volume of a right circular cylinder
1 / tan x = cos x / sin x
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)
pr²h


19. ln mn
1
sec² x
1
n ln m


20. cos 0
1
ln m - ln n
1/((ln a) x)
(ln a) a^x


21. Volume of a cone
pr²h/3
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
d/dx[f(x) ± g(x)] = f'(x) ± g'(x)
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x


22. Constant Multiple Rule for Derivatives

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23. sin p/4
v3/2
v2/2
cos x
(ln a) a^x


24. Continuity on a closed interval - [a -b]
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)
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)
pr²h
u' (ln a) a^u


25. cos p/4
v2/2
-1
cos x
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)


26. d/dx[ln u]

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27. cot x
0
1 / tan x = cos x / sin x
1
d/dx[cf(x)] = c f'(x)


28. Indeterminate form
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
0/0
1
1


29. Continuity at a point (x = c)
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
0
1/v(1-x²)
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. If f(-x) = -f(x)
e^x
1 / sin x
cos x
f is an odd function


31. If f(-x) = f(x)
1
(1 - cos 2x) / 2
1 / cos x
f is an even function


32. The limit as x approaches 0 of (1 - cos x) / x
Derivative of position at a point
0
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
1/x - x>0


33. 1 + tan²x
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).
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
d/dx[c] = 0
sec²x


34. d/dx[a^x]
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]
1
(ln a) a^x
(1 - cos 2x) / 2


35. d/dx[x]
1/2
d/dx[c] = 0
1/(1+x²)
1


36. sin²x
(1 - cos 2x) / 2
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
2pr
u' e^u


37. sin p/3
v3s² / 4
v3/2
?s/?t
-1


38. Area of a circle
d/dx[cf(x)] = c f'(x)
pr²
cos²x - sin²x
sec²x


39. cos²x + sin²x
sec x tan x
(ln a) a^x
pr²h
1


40. ln (m/n)
u'/((ln a) u)
v3/2
ln m - ln n
1 / tan x = cos x / sin x


41. sec x
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)
1 / cos x
1/2
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).


42. Critical number

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43. d/dx[sin x]
2pr
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)
cos x
f'(x) = lim as x ? c of [ f(x) - f(c) ] / [ x - c]


44. sin p/2
1
n ln m
u'/u - u > 0
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height


45. sin(2x)
-1/(|x|v(x²-1))
2 sin x cos x
1/2
0


46. Velocity - v(t)

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47. Circumference of a circle
(ln a) a^x
d/dx[f(x) ± g(x)] = f'(x) ± g'(x)
1/2
2pr


48. Continuity & differentiability
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
sec x tan x
u'/((ln a) u)


49. Intermediate Value Theorem
v2/2
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
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.
v3/2


50. sin 3p/2
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
-1
pr²h
2pr