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

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2. Critical number

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3. d/dx[a^u]

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4. cos 3p/2
1 / sin x
u'/u - u > 0
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.
0


5. cos p/2
2 sin x cos x
0
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


6. Average speed
?s/?t
1 / tan x = cos x / sin 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'
0


7. sin p/3
v3/2
n ln m
d/dx[cf(x)] = c f'(x)
sec²x


8. Position function of a falling object (with acceleration in ft/s²)
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
s(t) = -16t²+ 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).
-1/(|x|v(x²-1))


9. Surface Area of a Sphere
S = 4 pi r^2
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))
cos²x - sin²x


10. cos²x
Derivative of Position - s'(t)
(1 + cos 2x) / 2
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)
-1/(|x|v(x²-1))


11. d/dx[arcsec x]
1/(|x|v(x²-1))
u'/u - u > 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
?s/?t


12. Continuity & differentiability
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
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.
Derivative of position at a point
Differentiability implies continuity - but continuity does not necessarily imply differentiability.


13. Continuity on an open interval - (a -b)
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
0
1/x - x>0
v3s² / 4


14. sec x
1 / cos x
0
(1 - cos 2x) / 2
V = 4/3 pi r^3


15. d/dx[log_a u]

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16. d/dx[cot x]
If f is continuous on the closed interval [a -b] then it must have both a minimum and maximum on [a -b].
-1/(|x|v(x²-1))
-csc² x
ln m - ln n


17. Constant Multiple Rule for Derivatives

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18. d/dx[cos x]
1/v(1-x²)
-sin x
-1/(1+x²)
-csc x cot x


19. The Quotient Rule

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20. sin 3p/2
1
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
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'
-1


21. How to get from precalculus to calculus
sec x tan x
Limits
v2/2
S = 4 pi r^2


22. cos²x + sin²x
Derivative of Position - s'(t)
d/dx[cf(x)] = c f'(x)
1
u'/u - u > 0


23. ln 1
x values where f'(x) is zero or undefined.
1/2
v2/2
0


24. Continuity on a closed interval - [a -b]
1/(|x|v(x²-1))
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x
0
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)


25. d/dx[arccsc x]
v3/2
1/2
-1/(|x|v(x²-1))
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height


26. d/dx[csc x]
cos²x - sin²x
pr²h/3
ln m + ln n
-csc x cot x


27. cos p
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/x - x>0
d/dx[cf(x)] = c f'(x)


28. Velocity - v(t)

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29. cos p/6
v3s² / 4
d/dx[c] = 0
v3/2
1


30. Rolle's Theorem

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31. d/dx[e^x]
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
1
e^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).


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

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33. cos p/4
v2/2
1/x - x>0
Differentiability implies continuity - but continuity does not necessarily imply differentiability.
s(t) = -16t²+ v0t + s0 - v0 = initial velocity - s0 = initial height


34. d/dx[log_a x]
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?x
1/((ln a) x)
2pr
1/x - x>0


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

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36. sin 0
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
sec² x
0


37. Volume of a Sphere
1/2
0
Slope of a function at a point/slope of the tangent line to a function at a point
V = 4/3 pi r^3


38. Position function of a falling object (with acceleration in m/s²)
s(t) = -4.9t²+ v0t + s0 - v0 = initial velocity - s0 = initial height
csc²x
d/dx[f(x) ± g(x)] = f'(x) ± g'(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.


39. Alternate Limit Definition of a derivative

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40. If f(-x) = f(x)
x values where f'(x) is zero or undefined.
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.
sec² x
f is an even function


41. Guidelines for solving related rates problems
1/v(1-x²)
0
f(x) is continuous if for every point on the interval (a -b) the conditions for continuity at a point are satisfied.
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


42. tan x
pr²
sin x / cos 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'
-1


43. Continuity at a point (x = c)
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/((ln a) x)
d/dx[x^n]=nx^(n-1)
f(x) g'(x) + g(x) f'(x)


44. Mean Value Theorem

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45. Intermediate Value Theorem
f'(x) = lim as ?x ? 0 of [ f(x + ?x) - f(x) ] / ?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.
f is an even function
1


46. ln (mn)
1
g'(x) = 1/f'(g(x)) - f'(g(x)) cannot = 0
f is an even function
ln m + ln n


47. 1 + tan²x
sec²x
1/(1+x²)
0/0
-1


48. d/dx[sin x]
u'/((ln a) u)
n ln m
cos x
1/2


49. sin(2x)
-csc x cot x
f is an odd function
1/(|x|v(x²-1))
2 sin x cos x


50. Indeterminate form
(ln a) a^x
1/x - x>0
0/0
1