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. Continuity on an open interval - (a -b)
Derivative of position at a point
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).
pr²h/3
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.


2. d/dx[log_a u]

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3. sin p/4
1/2
1/v(1-x²)
v2/2
cos x


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

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5. d/dx[ln u]

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6. cos p
-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.
1/2
1/v(1-x²)


7. sin(2x)
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
x values where f'(x) is zero or undefined.
2 sin x cos x


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

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9. cos 0
f(x) g'(x) + g(x) f'(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
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).


10. 1 + tan²x
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 / cos x
sec²x
1/v(1-x²)


11. sec x
(ln a) a^x
1 / cos x
v2/2
1 / sin x


12. How to get from precalculus to calculus
Slope of a function at a point/slope of the tangent line to a function at a point
(ln a) a^x
-csc x cot x
Limits


13. csc x
1 / tan x = cos x / sin x
V = 4/3 pi r^3
[g(x)f'(x) - f(x) g'(x)] / [g(x)]²
1 / sin x


14. The Quotient Rule

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15. The limit as x approaches 0 of sin x / x
(1 - cos 2x) / 2
n ln m
1
1 / tan x = cos x / sin x


16. d/dx[arcsec x]
1
-1/v(1-x²)
f'(g(x))g'(x)
1/(|x|v(x²-1))


17. d/dx[arccsc x]
1/2
u'/u - u > 0
-1/(|x|v(x²-1))
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height


18. Constant Multiple Rule for Derivatives

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19. sin 0
0
1
-1/(1+x²)
-csc² x


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

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21. Velocity - v(t)

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22. Area of an equilateral triangle
2pr
0
-1/(|x|v(x²-1))
v3s² / 4


23. Guidelines for solving related rates problems
v3/2
1/2
-1/v(1-x²)
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


24. d/dx[cos x]
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
-sin x
1


25. Position function of a falling object (with acceleration in ft/s²)
pr²h
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
0/0
sec²x


26. Volume of a right circular cylinder
pr²h
V = 4/3 pi r^3
0
1/(|x|v(x²-1))


27. d/dx[sec x]
u'/u - u > 0
pr²h
0
sec x tan x


28. cot x
sec x tan x
u' e^u
S = 4 pi r^2
1 / tan x = cos x / sin x


29. cos²x
1/(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)
(1 + cos 2x) / 2
?s/?t


30. d/dx[a^x]
u'/((ln a) u)
(ln a) a^x
0
1


31. tan x
sin x / cos x
1/(1+x²)
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
cos x


32. Rolle's Theorem

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33. cos p/6
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
0
(1 + cos 2x) / 2
v3/2


34. sin p/6
1
f is an even function
-csc² x
1/2


35. d/dx[log_a x]
sec² x
1/((ln a) x)
cos²x - sin²x
-1


36. Continuity at a point (x = c)
-1/(1+x²)
d/dx[f(x) ± g(x)] = f'(x) ± g'(x)
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?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)


37. The limit as x approaches 0 of (1 - cos x) / 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'
f(x) g'(x) + g(x) f'(x)
d/dx[x^n]=nx^(n-1)
0


38. cos p/3
1/2
0
sec x tan x
v2/2


39. sin 3p/2
-1
pr²
f(x) g'(x) + g(x) f'(x)
-sin x


40. Surface Area of a Sphere
f(x) g'(x) + g(x) f'(x)
2pr
v3s² / 4
S = 4 pi r^2


41. Sum and Difference Rules for Derivatives

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42. d/dx[e^u]

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43. d/dx[arctan x]
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?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)
1/(1+x²)
e^x


44. Area of a circle
-1/(|x|v(x²-1))
sec²x
1/v(1-x²)
pr²


45. Derivative
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.
2 sin x cos x
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


46. d/dx[arccos x]
-1/v(1-x²)
sin x / cos x
u' (ln a) a^u
d/dx[cf(x)] = c f'(x)


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

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48. Mean Value Theorem

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49. Volume of a Sphere
cos²x - sin²x
d/dx[f(x) ± g(x)] = f'(x) ± g'(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


50. d/dx[csc x]
2pr
u' e^u
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
-csc x cot x