<p> Stand and Deliver – AP Calculus AB sin 2x  2sin xcos x 1 cos 2x= cos2 x - sin 2 x cos 2x= 2cos2 x - 1 2 cos 2x= 1 - 2sin2 x 1 cos2x sin 2 x  3 2 1 cos2x cos2 x  4 2 sin 2 x  cos2 x 1 1 tan 2 x  sec2 x 5 1 cot 2 x  csc2 x</p><p>Point-Slope Form : y  y1  m(x  x1 ) 6 Distance Formula: 2 2 d  (x2  x1 )  (y2  y1 ) 7 Existence of a Limit: 8 lim f (x)  L AND xc lim f (x)  L xc sin(x) lim 1 9 x0 x 1 cos(x) lim  0 10 x0 x Definition of Continuity: 11 f is continuous a,b if it is continuous on a,b and lim f (x)  f (a) xa and lim f (x)  f (b) xb Angle sin cos tan 12 0 0 1 0  1 3 3 6 2 2 3  2 2 1 4 2 2  3 1 3 3 2 2  1 0 2  Definition of a Derivative: 13 f (x  x)  f (x) f (x)  lim x0 x d Power Rule: (x n )  nx n1 dx 14 Average Velocity = 15 final position- initial position changein time Product Rule 16 d (f ( x ) g ( x ))= f ( x ) g '( x ) + g ( x ) f '( x ) dx Quotient Rule 17 d骣 f( x ) g ( x ) f '( x )- f ( x ) g '( x ) 琪 = dx桫 g( x ) ( g ( x ))2 “lo d hi – hi d lo” d sin x  cos x dx 18 d cos x  sin x dx 19 d tan x  sec2 x dx 20 d cot x  csc2 x dx 21 d sec x  sec x tan x dx 22 d csc x  csc xcot x 23 dx f (x) = position 24 f (x) = velocity f (x) = acceleration Chain Rule “Skin and Guts” 25 d  f (g(x))  f (g(x)) * g(x) dx “d of skin w/ guts intact times d of guts” Volume of a Cube 26 V  e3 Volume of a Cone 27 r 2 V  3</p><p>Volume of a Sphere 28 4 V  r 3 3 Critical Values: 29 set f( x )= 0 and f’(x) = undefined Mean Value Theorem for Derivatives: 30 If f (x) is continuous from [a,b] and differential from (a,b) then there exists at least one c where f (b)  f (a) f (c)  . b  a If the sign of f (x) changes from + to – 31 then a maximum occurs at (x, f(x)) If the sign of f (x) changes from – to + 32 then a minimum occurs at (x, f(x)) Concavity: 33 If f ’’(x) > 0 (positive): concave up If f ’’(x) < 0 (negative): concave down A point of inflection occurs when 34 1) f (x) is 0 AND 2) the sign of f (x) changes at that point Second Derivative Test 35 1. If f ‘’ (x) > 0, there is a minimum at (c, f(c)) 2. If f ‘’ (x) < 0, there is a maximum at (c, f(c)) 3. If f ’’ (x) = 0, test fails, use f ‘ (x) x n1 x n dx   c ; n 1 36  n  1 sin x dx  cos x  c 37 cos x dx  sin x  c 38 2  sec x dx  tan x  c 39 2  csc x dx  cot x  c 40  sec x tan x dx  sec x  c 41  csc x cot x dx   csc x  c 42 Fundamental Theorem of Calculus – Part I 43 b f (x)dx  F(b)  F(a) a</p><p>Mean Value Theorem for Integrals 44 b zf (x)dx  f (c)(b  a) a Average Value Theorem for Integrals: 45 1 a f (c)  f (x)dx b  a b Fundamental Theorem of Calculus – Part II 46 x dx f (t)dt  f (x) a Trapezoidal Rule: 47 b  a  f (x )  2 f (x )  ...  2 f (x )  f (x ) 2n 0 1 n1 n d 1 d u' ln x  lnu  48 dx x dx u 1 dx  ln x  c 49  x  tan x dx  ln cos x  c 50  cot x dx  ln sin x  c 51  sec x dx  ln sec x  tan x  c 52  csc x dx   ln csc x  cot x  c 53 d e x  e x dx 54 e x dx  e x  c 55 d a x  a x lna 56 dx d 1 log x  57 dx b xlnb a x a x dx   c  ln a 58</p><p> d-1 u ' sin u = 59 dx 1- u 2</p><p> d-1 - u ' cos u = dx 1- u2 60 d u ' tan-1 u = dx1+ u2 61 d- u ' cot-1 u = 62 dx1+ u2</p><p> d-1 u ' sec u = 63 dx u u2 -1</p><p> d-1 - u ' csc u = 64 dx u u2 -1 du u  arcsin  c 65  2 2 a  u a du 1 u  arctan  c 66  a 2  u 2 a a du 1 u  arc sec  c 67  2 2 u u  a a a Volume - Disk Method: 68 a V  (OR 2  IR 2 )dx b Volume - By Known Cross-Section: 69 a V  Area(x) dx b b 2 Arclength = 1 ( f (x)) dx 70 a Law of Growth/Decay: y  Ce kt</p>