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30 Cards in this Set
- Front
- Back
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tan x=
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sin x
----- cos x |
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cot x=
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cos x
----- sin x |
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sec x=
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1
----- cos x |
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csc x=
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1
----- sin x |
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sin^2 x + cos^2 x=
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1
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tan^2 x + 1 =
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sec^2 x
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cot^2 x + 1 =
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csc^2 x
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sin (x + y)=
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sinxcosy+cosxsiny
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cos (x+y)=
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cosx cosy - sinx siny
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sin (x-y)=
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sinxcosy-cosxsiny
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cos (x-y)=
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cosxcosy+sinx siny
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sin 2x=
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2 sinxcosx
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cos2x
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cos^2x -sin^2x
2cos^2x-1 1-2sin^2 x |
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d/dx sinx
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cos x
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d/dx cos x
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-sin x
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d/dx tanx
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sec^2 x
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d/dx cotx
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-csc^2 x
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d/dx sec x
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sec x tan x
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d/dx csc x
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-csc x ctn x
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d/dx arcsin x
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1
-------------- Sqrt (1-x^2) |
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d/dx arccos x
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-1
-------------- Sqrt (1-x^2) |
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d/dx arctan x
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1
-------------- 1 + x^2 |
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d/dx arccot x
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-1
-------------- 1 + x^2 |
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d/dx arcsec x
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1
----------------- x Sqrt (x^2 -1) |
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d/dx arccsc x
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-1
----------------- x Sqrt (x^2 -1) |
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mean value theorem
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suppose y=f(x) is continuous over the closed interval [a,b] and differentiable in its interior points (a,b) If f(a)=f(b) then there is at least one point c in which f'(c)= (f(b)-f(a))/b-a
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the derivative of a function f at a point Xo is
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f(x + h) - f(x)
f'(x)= lim ----------------- x->0 h |
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intermediate value theorm
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a function y=f(x) that is continuous on a closed interval [a,b] takes on every value between f(a) and f(b). In other words if y_0 is any value between f(a) and f(b), then y_0=f(c) for some c in [a,b]
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extreme value theorem
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if f is continuous on a closed interval [a,b] the f attains both an absolute maximum value M and an absolute minimum value m in [a,b] that is, there are numbers x1 and x2 in [a,b] with f(x1)=m, f(x2)=M, and m is less than or equal to f(x) which is less than or equal to M for every other x in [a,b]
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Mean value theorem
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suppose y=f(x)is continuous on a closed interval [a,b] and differentiable on the interval's interior (a,b) then there is at least one point c at which
f(b) - f(a) ----------- = f'(c) b-a |
