Formulario de Cálculo Diferencial e Integral                                                                                                                                                                                                                                                                                                                    Jesús Rubí M.

Formulario de                                                                ( a + b ) ⋅ ( a 2 − ab + b2 ) = a3 + b3                                                        θ         sen        cos            tg           ctg            sec        csc
                                                                                                                                                                                                                                                             Gráfica 4. Las funciones trigonométricas inversas
                                                                                                                                                                                                                                                             arcctg x , arcsec x , arccsc x :                     sen α + sen β
                                                                                                                                                                                                                                                                                                                                               1              1
                                                                                                                                                                                                                                                                                                                                        = 2sen (α + β ) ⋅ cos (α − β )
                                                                             ( a + b ) ⋅ ( a3 − a 2 b + ab2 − b3 ) = a 4 − b4                                             0            0       ∞      ∞          0                                                                                                                             2              2
Cálculo Diferencial
                                                                                                                                                                                       1           1
                                                                                                                                                                                                                                                                                                                                               1              1
                                                                                                                                                                          30          12        3 2 3 2                                                                                                           sen α − sen β         = 2 sen (α − β ) ⋅ cos (α + β )
                                                                                                                                                                                                                                                                     4
                                                                                                                                                                                      3 2 1 3
                                                                             ( a + b ) ⋅ ( a 4 − a3b + a 2 b2 − ab3 + b4 ) = a5 + b5
e Integral    VER.4.3                                                                                                                                                     45     1 2 1 2   1   1    2  2                                                             3
                                                                                                                                                                                                                                                                                                                                               2              2
                                                                             ( a + b ) ⋅ ( a5 − a 4 b + a3b2 − a 2 b3 + ab4 − b5 ) = a 6 − b6
                                                                                                                                                                                                                                                                                                                                               1              1
                                                                                                                                                                          60      3 2 12    3 1 3  2 2 3                                                                                                          cos α + cos β         = 2 cos (α + β ) ⋅ cos (α − β )
Jesús Rubí Miranda (jesusrubim@yahoo.com)                                                                                                                                                                                                                            2                                                                         2              2
                                                                                                                                                                          90      1    0   ∞   0   ∞  1
http://mx.geocities.com/estadisticapapers/                                               ⎛ n                     ⎞                                                                                                                                                                                                cos α − cos β
                                                                                                                                                                                                                                                                                                                                                 1              1
                                                                                                                                                                                                                                                                                                                                        = −2sen (α + β ) ⋅ sen (α − β )
                                                                             ( a + b ) ⋅ ⎜ ∑ ( −1) a n− k b k −1 ⎟ = a n + b n ∀ n ∈
                                                                                                  k +1
                                                                                                                                                             impar                             ⎡ π π⎤                                                                1

http://mx.geocities.com/dicalculus/                                                      ⎝ k =1                  ⎠                                                    y = ∠ sen x          y ∈ ⎢− , ⎥                                                                                                                                            2              2
                                                                                                                                                                                               ⎣ 2 2⎦
                                                                                                                                                                                                                                                                                                                                        sen (α ± β )
                                                                                                                                                                                                                                                                     0
                               VALOR ABSOLUTO                                                ⎛   n
                                                                                                                           ⎞
                                                                             ( a + b ) ⋅ ⎜ ∑ ( −1)                                                                                         y ∈ [ 0, π ]
                                                                                                                k +1
                                                                                                            a n − k b k −1 ⎟ = a n − b n ∀ n ∈               par      y = ∠ cos x                                                                                                                                 tg α ± tg β =
     ⎧a si a ≥ 0                                                                             ⎝ k =1                        ⎠
                                                                                                                                                                                                                                                                 -1
                                                                                                                                                                                                                                                                                                                                        cos α ⋅ cos β
 a =⎨
                                                                                                                                                                                                                                                                                                  arc ctg x

                                                                                                                                                                                                          π π                                                                                     arc sec x

     ⎩− a si a < 0                                                                                                                                                    y = ∠ tg x           y∈ −                                                                                                   arc csc x
                                                                                                        SUMAS Y PRODUCTOS                                                                                  ,                                                                                                                     1
                                                                                                                                                                                                                                                                                                                                   ⎡sen (α − β ) + sen (α + β ) ⎤
                                                                                                                                                                                                                                                                 -2
                                                                                                                                                                                                          2 2                                                      -5                     0                   5
                                                                                                                                                                                                                                                                                                                  sen α ⋅ cos β =
 a = −a                                                                                                               n
                                                                                                                                                                                                                                                                                                                                 2⎣                             ⎦
                                                                             a1 + a2 +               + an = ∑ ak                                                                         1                                                                           IDENTIDADES TRIGONOMÉTRICAS
a ≤ a y −a≤ a                                                                                                                                                         y = ∠ ctg x = ∠ tg                   y ∈ 0, π                                                                                                              1
                                                                                                                  k =1
                                                                                                                                                                                         x                                                                                                                        sen α ⋅ sen β = ⎡cos (α − β ) − cos (α + β ) ⎤
                                                                                                                                                                                                                                                             sen θ + cos2 θ = 1                                                  2⎣                            ⎦
                                                                                                                                                                                                                                                                 2
                                                                              n
 a ≥0y a =0 ⇔ a=0
                                                                             ∑ c = nc                                                                                 y = ∠ sec x = ∠ cos
                                                                                                                                                                                          1
                                                                                                                                                                                             y ∈ [ 0, π ]                                                    1 + ctg 2 θ = csc2 θ                                                1
                                                                             k =1                                                                                                                                                                                                                                 cos α ⋅ cos β = ⎡cos (α − β ) + cos (α + β ) ⎤
                                                                                                                                                                                                                                                                                                                                 2⎣                            ⎦
                                n                     n                                                                                                                                   x
 ab = a b ó                    ∏a             = ∏ ak                          n                   n
                                                                                                                                                                                                   ⎡ π π⎤                                                    tg 2 θ + 1 = sec2 θ
                                                                             ∑ ca           = c ∑ ak                                                                                      1
                                      k
                               k =1               k =1                              k                                                                                 y = ∠ csc x = ∠ sen     y ∈ ⎢− , ⎥                                                                                                                        tg α + tg β
                                          n                   n              k =1                k =1                                                                                     x        ⎣ 2 2⎦                                                    sen ( −θ ) = − sen θ                                 tg α ⋅ tg β =
 a+b ≤ a + b ó                        ∑a              ≤ ∑ ak                  n                           n                    n                                                                                                                                                                                               ctg α + ctg β
                                      k =1
                                                  k
                                                          k =1               ∑(a
                                                                             k =1
                                                                                    k       + bk ) = ∑ ak + ∑ bk
                                                                                                         k =1                 k =1
                                                                                                                                                                     Gráfica 1. Las funciones trigonométricas: sen x ,                                       cos ( −θ ) = cosθ                                               FUNCIONES HIPERBÓLICAS
                                                                                                                                                                     cos x , tg x :
                                      EXPONENTES                              n                                                                                                                                                                              tg ( −θ ) = − tg θ                                            e x − e− x
a p ⋅ a q = a p+q                                                            ∑(a            − ak −1 ) = an − a0                                                                                                                                                                                                   senh x =
                                                                                                                                                                                                                                                             sen (θ + 2π ) = sen θ
                                                                                    k                                                                                      2

                                                                             k =1
                                                                                                                                                                                                                                                                                                                                2
 ap                                                                                                                                                                      1.5
                                                                                                                                                                                                                                                                                                                           e x + e− x
    = a p−q                                                                   n
                                                                                                                          n                                                                                                                                  cos (θ + 2π ) = cosθ                                 cosh x =
 aq                                                                          ∑ ⎡ a + ( k − 1) d ⎤ = 2 ⎡ 2a + ( n − 1) d ⎤
                                                                               ⎣                ⎦     ⎣                 ⎦
                                                                                                                                                                           1
                                                                                                                                                                                                                                                                                                                                2
                                                                             k =1                                                                                                                                                                            tg (θ + 2π ) = tg θ
( a p ) = a pq                                                                                                                                                                                                                                                                                                            senh x e x − e − x
          q                                                                                                                                                              0.5

                                                                                                        n                                                                                                                                                                                                         tgh x =         =
                                                                                                          (a + l )
                                                                                                                =                                                          0                                                                                 sen (θ + π ) = − sen θ                                       cosh x e x + e − x
(a ⋅ b)           = ap ⋅bp
              p
                                                                                                        2                                                               -0.5
                                                                                                                                                                                                                                                             cos (θ + π ) = − cosθ                                            1       e x + e− x
                                                                              n
                                                                                                 1 − r n a − rl                                                                                                                                                                                                   ctgh x =        =
⎛a⎞  ap
          p
                                                                             ∑ ar   k −1
                                                                                              =a        =                                                                 -1
                                                                                                                                                                                                                                                             tg (θ + π ) = tg θ                                            tgh x e x − e − x
⎜ ⎟ = p                                                                      k =1                1− r      1− r
⎝b⎠
                                                                                                                                                                                                                                           sen x
     b                                                                                                                                                                  -1.5
                                                                                                                                                                                                                                                                                                                               1            2
                                                                                                                                                                                                                                                             sen (θ + nπ ) = ( −1) sen θ                          sech x =          =
                                                                                                                                                                                                                                           cos x                                      n

                                                                                      ( n + n)
                                                                              n
                                                                                    1 2
                                                                             ∑k =
                                                                                                                                                                                                                                           tg x

a   p/q
          = a
                  q       p                                                                                                                                               -2
                                                                                                                                                                            -8   -6        -4        -2     0        2        4        6           8                                                                       cosh x e x + e − x
                                                                                    2
                                                                             k =1
                                                                                                                                                                                                                                                             cos (θ + nπ ) = ( −1) cos θ
                                                                                                                                                                                                                                                                                      n
                                                                                                                                                                                                                                                                                                                               1            2
                   LOGARITMOS                                                                                                                                                                                                                                                                                     csch x =          =
                                                                             ∑ k 2 = 6 ( 2n3 + 3n2 + n )
                                                                              n
                                                                                     1                                                                               Gráfica 2. Las funciones trigonométricas csc x ,
log a N = x ⇒ a x = N                                                                                                                                                                                                                                        tg (θ + nπ ) = tg θ                                           senh x e x − e − x
                                                                             k =1                                                                                    sec x , ctg x :
log a MN = log a M + log a N                                                                                                                                                                                                                                                                                      senh :     →
                                                                                                                                                                                                                                                             sen ( nπ ) = 0
                                                                             ∑ k 3 = 4 ( n 4 + 2n3 + n 2 )
                                                                              n
                                                                                     1
      M
                                                                                                                                                                         2.5
                                                                                                                                                                                                                                                                                                                  cosh :     → [1, ∞
          = log a M − log a N                                                                                                                                                                                                                                cos ( nπ ) = ( −1)
                                                                                                                                                                                                                                                                                  n
log a                                                                        k =1                                                                                          2

      N                                                                                                                                                                                                                                                                                                           tgh :    → −1,1
                                                                             ∑ k 4 = 30 ( 6n5 + 15n4 + 10n3 − n )
                                                                              n
                                                                                      1
                                                                                                                                                                                                                                                             tg ( nπ ) = 0
                                                                                                                                                                         1.5

log a N r = r log a N                                                                                                                                                      1
                                                                                                                                                                                                                                                                                                                  ctgh :     − {0} → −∞ , −1 ∪ 1, ∞
                                                                             k =1
                                                                                                                                                                                                                                                                  ⎛ 2n + 1 ⎞
                                                                                                        + ( 2n − 1) = n 2                                                                                                                                                 π ⎟ = ( −1)                                        → 0,1]
                                                                                                                                                                         0.5                                                                                                          n
                          log b N ln N                                       1+ 3 + 5 +                                                                                                                                                                      sen ⎜
log a N =                         =                                                                                                                                                                                                                                                                               sech :
                                                                                                                                                                                                                                                                  ⎝ 2        ⎠
                                                                                                                                                                           0
                           log b a ln a                                                 n
                                                                                                                                                                                                                                                                                                                             − {0} →          − {0}
                                                                             n! = ∏ k
                                                                                                                                                                        -0.5
                                                                                                                                                                                                                                                                                                                  csch :
log10 N = log N y log e N = ln N                                                                                                                                                                                                                                  ⎛ 2n + 1 ⎞
                                                                                    k =1
                                                                                                                                                                          -1
                                                                                                                                                                                                                                                             cos ⎜        π⎟=0
                ALGUNOS PRODUCTOS                                                                                                                                       -1.5                                                                                      ⎝ 2        ⎠                                    Gráfica 5. Las funciones hiperbólicas senh x ,
                                                                             ⎛n⎞         n!                                                                                                                                                csc x

a ⋅ ( c + d ) = ac + ad                                                      ⎜ ⎟=                 , k≤n                                                                   -2                                                               sec x
                                                                                                                                                                                                                                                                ⎛ 2n + 1 ⎞                                        cosh x , tgh x :
                                                                             ⎝ k ⎠ ( n − k )!k !
                                                                                                                                                                                                                                           ctg x
                                                                                                                                                                                                                                                             tg ⎜        π⎟=∞
                                                                                                                                                                                                                                                                ⎝ 2        ⎠
                                                                                                                                                                        -2.5


( a + b) ⋅ ( a − b) = a − b
                                                                                                                                                                            -8   -6        -4        -2     0        2        4        6           8
                                      2           2
                                                                                                                                                                                                                                                                                                                             5
                                                                                           n
                                                                                              ⎛n⎞
                                                                             ( x + y ) = ∑ ⎜ ⎟ xn−k y k                                                                                                                                                                      π⎞
                                                                                      n
                                                                                                                                                                     Gráfica 3. Las funciones trigonométricas inversas                                                   ⎛
( a + b ) ⋅ ( a + b ) = ( a + b ) = a 2 + 2ab + b2
                                                                                                                                                                                                                                                                                                                             4

                                                                                                                                                                                                                                                             sen θ = cos ⎜θ − ⎟
                                 2
                                                                                         k =0 ⎝ k ⎠                                                                  arcsen x , arccos x , arctg x :                                                                     ⎝   2⎠                                              3



( a − b ) ⋅ ( a − b ) = ( a − b ) = a 2 − 2ab + b 2
                                 2                                                                                                                                                                                                                                                                                           2


                                                                             ( x1 + x2 +                + xk )
                                                                                                                  n
                                                                                                                          =∑
                                                                                                                                   n!
                                                                                                                                           x1n1 ⋅ x2 2                                                                                                                  ⎛   π⎞
                                                                                                                                                                                                                                                             cosθ = sen ⎜θ + ⎟
                                                                                                                                                   n     nk
                                                                                                                                                         x                 4


( x + b ) ⋅ ( x + d ) = x 2 + ( b + d ) x + bd
                                                                                                                                                                                                                                                                                                                             1
                                                                                                                                                         k
                                                                                                                             n1 !n2 ! nk !
                                                                                                                                                                           3
                                                                                                                                                                                                                                                                        ⎝   2⎠                                               0


( ax + b ) ⋅ ( cx + d ) = acx 2 + ( ad + bc ) x + bd                                         CONSTANTES                                                                                                                                                      sen (α ± β ) = sen α cos β ± cos α sen β                        -1




( a + b ) ⋅ ( c + d ) = ac + ad + bc + bd                                    π = 3.14159265359…                                                                            2

                                                                                                                                                                                                                                                             cos (α ± β ) = cos α cos β ∓ sen α sen β
                                                                                                                                                                                                                                                                                                                             -2

                                                                                                                                                                                                                                                                                                                                                                se nh x

                                                                             e = 2.71828182846…
                                                                                                                                                                                                                                                                                                                             -3                                 co sh x


( a + b ) = a3 + 3a 2b + 3ab2 + b3                                                                                                                                                                                                                                          tg α ± tg β
         3                                                                                                                                                                 1                                                                                                                                                                                    tgh x


                                                                                                                                                                                                                                                             tg (α ± β ) =
                                                                                                                                                                                                                                                                                                                             -4
                                                                                                                                                                                                                                                                                                                               -5                       0                 5
                                                                                           TRIGONOMETRÍA
                                                                                                                                                                                                                                                                           1 ∓ tg α tg β
( a − b ) = a3 − 3a 2b + 3ab2 − b3
         3                                                                                                                                                                 0
                                                                                      CO                   1                                                                                                                                                                                                               FUNCIONES HIPERBÓLICAS INV
                                                                             sen θ =             cscθ =                                                                                                                                                      sen 2θ = 2sen θ cosθ
( a + b + c ) = a 2 + b2 + c 2 + 2ab + 2ac + 2bc
               2                                                                     HIP
                                                                                      CA
                                                                                                         sen θ
                                                                                                           1
                                                                                                                                                                          -1
                                                                                                                                                                                                                                   arc sen x
                                                                                                                                                                                                                                   arc cos x
                                                                                                                                                                                                                                   arc tg x                  cos 2θ = cos 2 θ − sen 2 θ                                             (
                                                                                                                                                                                                                                                                                                                  senh −1 x = ln x + x 2 + 1 , ∀x ∈ )
                                                                             cosθ =              secθ =
( a − b ) ⋅ ( a + ab + b ) = a − b                                                                                                                                                                                                                                                                                                  (               )
                                                                                                                                                                          -2

                                                                                                                                                                                                                                                                        2 tg θ
                           2                  2           3       3
                                                                                                         cosθ
                                                                                                                                                                            -3        -2        -1          0            1         2               3
                                                                                     HIP                                                                                                                                                                     tg 2θ =                                              cosh −1 x = ln x ± x 2 − 1 , x ≥ 1
                                                                                    sen θ CO                                                                                                                                                                          1 − tg 2 θ
( a − b ) ⋅ ( a3 + a 2 b + ab2 + b3 ) = a 4 − b4                             tg θ =      =       ctg θ =
                                                                                                          1
                                                                                                                                                                                                                                                                                                                             1 ⎛1+ x ⎞
                                                                                    cosθ CA              tg θ                                                                                                                                                           1                                         tgh −1 x = ln ⎜     ⎟, x < 1
( a − b ) ⋅ ( a 4 + a3b + a 2 b2 + ab3 + b 4 ) = a5 − b5                                                                                                                                                                                                     sen 2 θ = (1 − cos 2θ )                                         2 ⎝ 1− x ⎠
                                                                                                                                                                                                                                                                        2
                      ⎛                       ⎞                              π radianes=180                                                                                                                                                                                                                                   1 ⎛ x +1 ⎞
                                                                                                                                                                                                                                                                                                                  ctgh −1 x = ln ⎜
                           n
                                                                                                                                                                                                                                                                        1
( a − b ) ⋅ ⎜ ∑ a n − k b k −1 ⎟ = a n − b n                          ∀n ∈                                                                                                                                                                                   cos 2 θ = (1 + cos 2θ )                                                    ⎟, x > 1
                                                                                                                                                                                                                                                                                                                              2 ⎝ x −1 ⎠
                      ⎝ k =1                  ⎠                                                                                                                                                                                                                         2
                                                                                                                                                                                                                                                                      1 − cos 2θ                                                 ⎛ 1 ± 1 − x2 ⎞
                                                                                                                       HIP                                                                                                                                   tg 2 θ =                                             sech −1 x = ln ⎜             ⎟, 0 < x ≤ 1
                                                                                                                                                CO                                                                                                                    1 + cos 2θ                                                 ⎜     x       ⎟
                                                                                                                                                                                                                                                                                                                                 ⎝             ⎠
                                                                                                              θ                                                                                                                                                                                                                  ⎛1      x2 + 1 ⎞
                                                                                                                                                                                                                                                                                                                  csch −1 x = ln ⎜ +             ⎟, x ≠ 0
                                                                                                                           CA                                                                                                                                                                                                    ⎜x       x ⎟
                                                                                                                                                                                                                                                                                                                                 ⎝               ⎠
Formulario de Cálculo Diferencial e Integral                                                                                                                                                                                                                                                                            Jesús Rubí M.

                                                                                                                                                                                                                                         ∫ tgh udu = ln cosh u
            IDENTIDADES DE FUNCS HIP                         d            dv      du                              DERIVADA DE FUNCS HIPERBÓLICAS                                 INTEGRALES DE FUNCS LOG & EXP
                                                                ( uv ) = u + v
cosh 2 x − senh 2 x = 1                                                                                        d                   du
                                                                                                                                                                          ∫ e du = e
                                                                                                                                                                            u                u
                                                             dx           dx      dx                              senh u = cosh u
1 − tgh 2 x = sech 2 x                                       d               dw       dv       du              dx                  dx                                                                                                    ∫ ctgh udu = ln senh u
                                                                ( uvw) = uv + uw + vw                                                                                                        a u ⎧a > 0
                                                                                                                                                                          ∫ a du = ln a ⎨a ≠ 1                                           ∫ sech udu = ∠ tg ( senh u )
                                                                                                               d                   du                                        u
                                                             dx              dx       dx       dx                 cosh u = senh u
ctgh x − 1 = csch x
       2
                                                                                                                                                                                        ⎩
                                                             d ⎛ u ⎞ v ( du dx ) − u ( dv dx )                 dx                  dx
senh ( − x ) = − senh x                                                                                                                                                                                                                  ∫ csch udu = − ctgh ( cosh u )
                                                                                                                                                                                                                                                                             −1
                                                                ⎜ ⎟=                                           d                   du                                                            au ⎛      1 ⎞
                                                             dx ⎝ v ⎠              v2                             tgh u = sech 2 u                                        ∫ ua du = ln a ⋅ ⎜ u − ln a ⎟
                                                                                                                                                                                 u

cosh ( − x ) = cosh x                                                                                          dx                  dx                                                      ⎝          ⎠                                                         1
                                                                                                                                                                                                                                                       = ln tgh u
                                                             d n
                                                                ( u ) = nu n−1 du
tgh ( − x ) = − tgh x                                                                                          d                      du
                                                                                                                                                                          ∫ ue du = e ( u − 1)                                                                  2
                                                                                                                                                                                 u               u
                                                             dx                dx                                 ctgh u = − csch 2 u
                                                                                                               dx                     dx                                                                                                                INTEGRALES DE FRAC
senh ( x ± y ) = senh x cosh y ± cosh x senh y               dF dF du                                                                                                     ∫ ln udu =u ln u − u = u ( ln u − 1)
                                                                =   ⋅ (Regla de la Cadena)                     d
                                                                                                                  sech u = − sech u tgh u
                                                                                                                                           du                                                                                                 du       1       u
cosh ( x ± y ) = cosh x cosh y ± senh x senh y               dx du dx                                          dx                          dx                                            1                    u                          ∫ u 2 + a 2 = a ∠ tg a
                                                             du   1                                                                                                       ∫ log              ( u ln u − u ) = ( ln u − 1)
                                                                                                                                                                                         udu =
               tgh x ± tgh y                                    =
                                                                                                                                                                                     a
                                                                                                               d                            du                                          ln a                 ln a                                      1        u
tgh ( x ± y ) =                                              dx dx du                                             csch u = − csch u ctgh u                                                                                                         = − ∠ ctg
              1 ± tgh x tgh y                                                                                  dx                           dx                                             u 2

                                                                                                                                                                          ∫ u log a udu = 4 ⋅ ( 2log a u − 1)
                                                                                                                                                                                                                                                       a        a
                                                             dF dF du                                                  DERIVADA DE FUNCS HIP INV                                                                                                        1 u−a
senh 2 x = 2senh x cosh x
                                                                                                                                                                                                                                         ∫ u 2 − a 2 2a u + a ( u > a )
                                                               =                                                                                                                                                                              du
                                                                                                                                                                                                                                                     =     ln         2   2
                                                             dx dx du                                          d                 1       du                                             u2
cosh 2 x = cosh x + senh x                                                                                        senh −1 u =          ⋅
                                                                                                                                                                          ∫ u ln udu = 4 ( 2ln u − 1)
                       2       2

                                                             dy dy dt f 2′ ( t )       ⎪ x = f1 ( t )
                                                                                       ⎧                                       1 + u 2 dx                                                                                                               1 a+u
                                                                                                                                                                                                                                         ∫ a 2 − u 2 = 2a ln a − u ( u < a )
                                                                                                               dx                                                                                                                             du                      2   2
tgh 2 x =
              2 tgh x                                          =      =          donde ⎨
            1 + tgh 2 x                                      dx dx dt   f1′( t )       ⎪ y = f2 ( t )
                                                                                       ⎩                       d                  ±1     du       ⎧+ si cosh -1u > 0
                                                                                                                                                  ⎪                                INTEGRALES DE FUNCS TRIGO
                                                                                                                  cosh −1 u =           ⋅ , u >1 ⎨
           1                                                        DERIVADA DE FUNCS LOG & EXP                dx                u 2 − 1 dx       ⎪− si cosh u < 0
                                                                                                                                                  ⎩
                                                                                                                                                             -1
                                                                                                                                                                          ∫ sen udu = − cos u                                                                 INTEGRALES CON
senh 2 x =   ( cosh 2 x − 1)
           2                                                 d
                                                                ( ln u ) =
                                                                            du dx 1 du
                                                                                    = ⋅
                                                                                                               d
                                                                                                                  tgh −1 u =
                                                                                                                                1 du
                                                                                                                                     ⋅ , u <1                             ∫ cos udu = sen u                                              ∫
                                                                                                                                                                                                                                                   du
                                                                                                                                                                                                                                                            = ∠ sen
                                                                                                                                                                                                                                                                             u
           1                                                 dx               u        u dx                    dx            1 − u 2 dx                                                                                                       a2 − u2                        a
cosh x = ( cosh 2 x + 1)
      2
                                                             d               log e du                          d                 1 du                                     ∫ sec udu = tg u
                                                                                                                                                                                     2

           2                                                    ( log u ) =        ⋅                              ctgh −1 u =         ⋅ , u >1                                                                                                               = −∠ cos
                                                                                                                                                                                                                                                                                  u
          cosh 2 x − 1                                                                                                        1 − u 2 dx                                  ∫ csc udu = − ctg u
                                                                                                                                                                                     2
                                                             dx                u dx                            dx                                                                                                                                                                 a
tgh 2 x =
                                                                                                                                             ⎧−
                                                                                                                                                                                                                                                                  (                              )
                                                                                                                                                      −1
          cosh 2 x + 1                                       d                log e du
                                                                ( log a u ) = a ⋅ a > 0, a ≠ 1                                    ∓1      du ⎪ si sech u > 0, u ∈ 0,1                                                                              du
                                                                                                                                                                          ∫ sec u tg udu = sec u
                                                                                                               d
            senh 2 x                                         dx                 u       dx                     dx
                                                                                                                  sech −1 u =            ⋅ ⎨          −1
                                                                                                                              u 1 − u 2 dx ⎪ + si sech u < 0, u ∈ 0,1
                                                                                                                                                                                                                                         ∫    u 2 ± a2
                                                                                                                                                                                                                                                            = ln u + u 2 ± a 2
tgh x =                                                                                                                                      ⎩
           cosh 2 x + 1
                                                             d u
                                                                ( e ) = eu ⋅ du                                d                      1      du                           ∫ csc u ctg udu = − csc u                                                du    1         u
                                                             dx               dx                                       −1
                                                                                                                  csch u = −                ⋅ , u≠0                                                                                      ∫u            = ln
e x = cosh x + senh x                                                                                          dx                u 1 + u 2 dx                             ∫ tg udu = − ln cos u = ln sec u                                         a2 ± u2
                                                                                                                                                                                                                                                         a a + a2 ± u 2
e − x = cosh x − senh x
                                                             d u
                                                                ( a ) = au ln a ⋅ du                                                                                                                                                                du   1       a
                                                             dx                    dx                           INTEGRALES DEFINIDAS, PROPIEDADES                         ∫ ctg udu = ln sen u                                           ∫ u u 2 − a 2 = a ∠ cos u
                            OTRAS                            d v
                                                                ( u ) = vu v−1 du + ln u ⋅ u v ⋅ dv            ∫ { f ( x ) ± g ( x )} dx = ∫ f ( x ) dx ± ∫ g ( x ) dx
                                                                                                                b                                           b         b

ax + bx + c = 0
   2
                                                             dx                 dx               dx             a                                          a          a   ∫ sec udu = ln sec u + tg u                                                    1
                                                                                                                                                                                                                                                       = ∠ sec
                                                                                                                                                                                                                                                                 u
                                                                                                               ∫ cf ( x ) dx = c ⋅ ∫ f ( x ) dx c ∈                       ∫ csc udu = ln csc u − ctg u
                                                                                                                b                                    b
                                                                    DERIVADA DE FUNCIONES TRIGO                                                                                                                                                          a       a
               −b ± b 2 − 4ac
       ⇒ x=                                                  d                     du
                                                                                                                a                                    a
                                                                                                                                                                                                                                                          u 2        a2  u
                                                                ( sen u ) = cos u                                                                                                                                                        ∫ a − u du = 2 a − u + 2 ∠ sen a
                                                                                                               ∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx
                                                                                                                b                        c                      b                                                                            2     2               2
                      2a                                                                                                                                                              u 1
                                                                                                                                                                          ∫ sen          udu =
                                                                                                                                                                                        − sen 2u
                                                                                                                                                                                     2
                                                             dx                    dx
       b 2 − 4ac = discriminante                                                                                a                    a                          c
                                                                                                                                                                                      2 4
                                                                                                                                                                                                                                                                                                              (                    )
                                                                                                                                                                                                                                                                       2
                                                             d                        du
                                                                ( cos u ) = − sen u                            ∫ f ( x ) dx = −∫ f ( x ) dx
                                                                                                                b                                a
                                                                                                                                                                                                                                                          u 2        a
exp (α ± iβ ) = eα ( cos β ± i sen β ) si α , β ∈                                                                                                                                                                                        ∫ u ± a du = 2 u ± a ± 2 ln u + u ± a
                                                                                                                                                                                                                                             2     2               2       2   2
                                                                                                                                                                                      u 1
                                                                                                                                                                          ∫ cos udu = 2 + 4 sen 2u
                                                                                                                                                                               2
                                                             dx                       dx                        a                            b


                                                                                                               ∫ f ( x ) dx = 0
                                                                                                                a
                           LÍMITES                           d                    du
                                                                ( tg u ) = sec2 u                                                                                                                                                                        MÁS INTEGRALES
                                                                                                                                                                          ∫ tg udu = tg u − u
                                                                                                                a                                                                2
                                                                                                                                                                                                                                                         e au ( a sen bu − b cos bu )
               1
lim (1 + x ) x = e = 2.71828...                              dx                   dx
                                                                                                               m ⋅ ( b − a ) ≤ ∫ f ( x ) dx ≤ M ⋅ ( b − a )
                                                                                                                                     b

                                                                                                                                                                                                                                         ∫e        sen bu du =
                                                                                                                                                                                                                                              au

                                                                                                                                                                          ∫ ctg udu = − ( ctg u + u )
x →0                                                         d                         du
                                                                ( ctg u ) = − csc2 u                                                                                                                                                                                a 2 + b2
                                                                                                                                                                                     2
                                                                                                                                     a
                   x
     ⎛ 1⎞
lim ⎜1 + ⎟ = e
                                                             dx                        dx                       ⇔ m ≤ f ( x ) ≤ M ∀x ∈ [ a, b ] , m, M ∈                                                                                                 e au ( a cos bu + b sen bu )
     ⎝ x⎠                                                                                                                                                                 ∫ u sen udu = sen u − u cos u                                  ∫ e cos bu du =
x →∞                                                                                                                                                                                                                                        au
                                                             d                          du
                                                                ( sec u ) = sec u tg u                         ∫ f ( x ) dx ≤ ∫ g ( x ) dx
                                                                                                                b                        b
                                                                                                                                                                                                                                                                    a 2 + b2
      sen x                                                  dx                         dx
            =1                                                                                                                                                            ∫ u cos udu = cos u + u sen u
                                                                                                                a                    a
lim                                                                                                                                                                                                                                                     ALGUNAS SERIES
x →0                                                         d                              du                  ⇔ f ( x ) ≤ g ( x ) ∀x ∈ [ a, b ]
        x
                                                                ( csc u ) = − csc u ctg u                                                                                            INTEGRALES DE FUNCS TRIGO INV                                                                                            f '' ( x0 )( x − x0 )
                                                                                                                                                                                                                                                                                                                                       2
     1 − cos x                                               dx                             dx                                                                                                                                           f ( x ) = f ( x0 ) + f ' ( x0 )( x − x0 ) +
               =0
                                                                                                               ∫ f ( x ) dx ≤ ∫ f ( x ) dx si a < b
                                                                                                                    b                        b
lim
                                                                                                                                                                          ∫ ∠ sen udu = u∠ sen u + 1 − u                                                                                                                 2!
                                                                                                                                                                                                         2
x →0                                                         d                       du
          x
                                                                ( vers u ) = sen u                              a                        a

                                                                                                                                                                                                                                                                  f(
                                                                                                                                                                                                                                                                       n)
                                                                                                                                                                                                                                                                             ( x0 )( x − x0 )
                                                                                                                                                                                                                                                                                                     n
      ex −1                                                  dx                      dx                                                                  INTEGRALES
             =1                                                                                                                                                           ∫ ∠ cos udu = u∠ cos u − 1 − u                                                +     +
                                                                                                                                                                                                                    2
lim                                                                                                                                                                                                                                                                                                       : Taylor
                                                                                                               ∫ adx =ax
x →0    x                                                             DERIV DE FUNCS TRIGO INVER                                                                                                                                                                                      n!
      x −1                                                                                                                                                                ∫ ∠ tg udu = u∠ tg u − ln 1 + u                                                                                  f '' ( 0 ) x
                                                                                                                                                                                                                   2                                                                                      2
                                                             d                     1        du
lim
 x →1 ln x
            =1                                                  ( ∠ sen u ) =             ⋅
                                                                                                               ∫ af ( x ) dx = a ∫ f ( x ) dx                                                                                            f ( x ) = f ( 0) + f '( 0) x +
                                                             dx                  1 − u 2 dx                                                                                                                                                                                                      2!
                                                                                                                                                                          ∫ ∠ ctg udu = u∠ ctg u + ln 1 + u
                                                                                                                                                                                                                        2


                                                                                                               ∫ ( u ± v ± w ± ) dx = ∫ udx ± ∫ vdx ± ∫ wdx ±                                                                                                                ( 0) x
                       DERIVADAS                                                                                                                                                                                                                                      ( n)             n
                                                             d                       1      du                                                                                                                                                                    f
                                                                                                                                                                          ∫ ∠ sec udu = u∠ sec u − ln ( u + u                        )
               df       f ( x + ∆x ) − f ( x )          ∆y
                                                                ( ∠ cos u ) = −           ⋅                                                                                                                                 2
                                                                                                                                                                                                                                −1                      +     +                             : Maclaurin
Dx f ( x ) =                                                 dx                    1 − u 2 dx
                                                                                                               ∫ udv = uv − ∫ vdu ( Integración por partes )
                  = lim                        = lim                                                                                                                                                                                                                         n!
               dx ∆x→0           ∆x              ∆x → 0 ∆x
                                                             d                  1 du                                                                                                             = u∠ sec u − ∠ cosh u                               x 2 x3     xn
                                                                ( ∠ tg u ) =         ⋅                                      u n +1                                                                                                       ex = 1 + x + + + + +
                                                                                                                                                                          ∫ ∠ csc udu = u∠ csc u + ln ( u +                          )
d
   (c) = 0                                                   dx              1 + u 2 dx                        ∫ u du = n + 1
                                                                                                                        n
                                                                                                                                             n ≠ −1
                                                                                                                                                                                                                         u 2 −1
                                                                                                                                                                                                                                                     2! 3!      n!
dx                                                           d                     1 du                                                                                                                                                              x3 x 5 x 7            x 2 n −1
                                                                ( ∠ ctg u ) = − 2 ⋅                                                                                                                                                      sen x = x − + − + + ( −1)
                                                                                                                                                                                                                                                                   n −1
d
   ( cx ) = c                                                dx                 1 + u dx
                                                                                                                du
                                                                                                               ∫ u = ln u                                                                      = u∠ csc u + ∠ cosh u                                 3! 5! 7!           ( 2n − 1)!
dx
                                                             d                        1      du ⎧ + si u > 1                                                                                 INTEGRALES DE FUNCS HIP
                                                                                                                                                                                                                                                                                                                       x 2n−2
                                                                ( ∠ sec u ) = ± 2 ⋅ ⎨                                                                                                                                                    cos x = 1 −
                                                                                                                                                                                                                                                            x2 x4 x6
                                                                                                                                                                                                                                                              + − +                            + ( −1)
                                                                                                                                                                                                                                                                                                              n −1
d
dx
   ( cx n ) = ncxn−1                                         dx                  u u − 1 dx ⎩ − si u < −1                                                                 ∫ senh udu = cosh u                                                               2! 4! 6!                                                 ( 2n − 2 )!
                                                             d                     1       du ⎧− si u > 1                                                                 ∫ cosh udu = senh u
                                                                ( ∠ csc u ) = ∓                                                                                                                                                                        x 2 x3 x 4               n
d                    du dv dw                                                                                                                                                                                                                                           n −1 x
   (u ± v ± w ± ) = ± ± ±                                                                 ⋅ ⎨                                                                                                                                            ln (1 + x ) = x −+ − + + ( −1)
                                                             dx                 u u 2 − 1 dx ⎩+ si u < −1                                                                                                                                              2 3 4                  n
                                                                                                                                                                          ∫ sech udu = tgh u
dx                   dx dx dx                                                                                                                                                            2
                                                                                                                                                                                                                                                                           2 n −1
d              du                                            d                     1       du                                                                                                                                                        x3 x5 x7
   ( cu ) = c                                                   ( ∠ vers u ) =           ⋅                                                                                                                                               ∠ tg x = x − + − + + ( −1)
                                                                                                                                                                                                                                                                    n −1 x

                                                                                                                                                                          ∫ csch udu = − ctgh u
                                                                                                                                                                                         2
dx             dx                                            dx                  2u − u 2 dx                                                                                                                                                         3 5 7              2n − 1

                                                                                                                                                                          ∫ sech u tgh udu = − sech u
                                                                                                                                                                          ∫ csch u ctgh udu = − csch u
Formulario de Cálculo Diferencial e Integral                         Jesús Rubí M.
               ÁLGEBRA LINEAL
Def. El determinante de una matriz
                       ⎡a    a ⎤
                  A = ⎢ 11 12 ⎥
                       ⎣ a21 a22 ⎦
está dado por
                      a11 a12
         det A =              = a11a22 − a12 a21 .
                      a21 a22
Def. El determinante de una matriz
                          ⎡ a11 a12     a13 ⎤
                          ⎢                 ⎥
                      A = ⎢ a21 a22     a23 ⎥
                          ⎢
                          ⎣ a31 a32         ⎥
                                        a33 ⎦
está dado por
        a11 a12        a13    a11 ⋅ a22 ⋅ a33 + a12 ⋅ a23 ⋅ a31
det A = a21 a22        a23 = + a13 ⋅ a21 ⋅ a32 − a11 ⋅ a23 ⋅ a32 .
        a31     a32     a33   −a12 ⋅ a21 ⋅ a33 − a13 ⋅ a22 ⋅ a31

formulas calculo integral y diferencial

  • 1.
    Formulario de CálculoDiferencial e Integral Jesús Rubí M. Formulario de ( a + b ) ⋅ ( a 2 − ab + b2 ) = a3 + b3 θ sen cos tg ctg sec csc Gráfica 4. Las funciones trigonométricas inversas arcctg x , arcsec x , arccsc x : sen α + sen β 1 1 = 2sen (α + β ) ⋅ cos (α − β ) ( a + b ) ⋅ ( a3 − a 2 b + ab2 − b3 ) = a 4 − b4 0 0 ∞ ∞ 0 2 2 Cálculo Diferencial 1 1 1 1 30 12 3 2 3 2 sen α − sen β = 2 sen (α − β ) ⋅ cos (α + β ) 4 3 2 1 3 ( a + b ) ⋅ ( a 4 − a3b + a 2 b2 − ab3 + b4 ) = a5 + b5 e Integral VER.4.3 45 1 2 1 2 1 1 2 2 3 2 2 ( a + b ) ⋅ ( a5 − a 4 b + a3b2 − a 2 b3 + ab4 − b5 ) = a 6 − b6 1 1 60 3 2 12 3 1 3 2 2 3 cos α + cos β = 2 cos (α + β ) ⋅ cos (α − β ) Jesús Rubí Miranda (jesusrubim@yahoo.com) 2 2 2 90 1 0 ∞ 0 ∞ 1 http://mx.geocities.com/estadisticapapers/ ⎛ n ⎞ cos α − cos β 1 1 = −2sen (α + β ) ⋅ sen (α − β ) ( a + b ) ⋅ ⎜ ∑ ( −1) a n− k b k −1 ⎟ = a n + b n ∀ n ∈ k +1 impar ⎡ π π⎤ 1 http://mx.geocities.com/dicalculus/ ⎝ k =1 ⎠ y = ∠ sen x y ∈ ⎢− , ⎥ 2 2 ⎣ 2 2⎦ sen (α ± β ) 0 VALOR ABSOLUTO ⎛ n ⎞ ( a + b ) ⋅ ⎜ ∑ ( −1) y ∈ [ 0, π ] k +1 a n − k b k −1 ⎟ = a n − b n ∀ n ∈ par y = ∠ cos x tg α ± tg β = ⎧a si a ≥ 0 ⎝ k =1 ⎠ -1 cos α ⋅ cos β a =⎨ arc ctg x π π arc sec x ⎩− a si a < 0 y = ∠ tg x y∈ − arc csc x SUMAS Y PRODUCTOS , 1 ⎡sen (α − β ) + sen (α + β ) ⎤ -2 2 2 -5 0 5 sen α ⋅ cos β = a = −a n 2⎣ ⎦ a1 + a2 + + an = ∑ ak 1 IDENTIDADES TRIGONOMÉTRICAS a ≤ a y −a≤ a y = ∠ ctg x = ∠ tg y ∈ 0, π 1 k =1 x sen α ⋅ sen β = ⎡cos (α − β ) − cos (α + β ) ⎤ sen θ + cos2 θ = 1 2⎣ ⎦ 2 n a ≥0y a =0 ⇔ a=0 ∑ c = nc y = ∠ sec x = ∠ cos 1 y ∈ [ 0, π ] 1 + ctg 2 θ = csc2 θ 1 k =1 cos α ⋅ cos β = ⎡cos (α − β ) + cos (α + β ) ⎤ 2⎣ ⎦ n n x ab = a b ó ∏a = ∏ ak n n ⎡ π π⎤ tg 2 θ + 1 = sec2 θ ∑ ca = c ∑ ak 1 k k =1 k =1 k y = ∠ csc x = ∠ sen y ∈ ⎢− , ⎥ tg α + tg β n n k =1 k =1 x ⎣ 2 2⎦ sen ( −θ ) = − sen θ tg α ⋅ tg β = a+b ≤ a + b ó ∑a ≤ ∑ ak n n n ctg α + ctg β k =1 k k =1 ∑(a k =1 k + bk ) = ∑ ak + ∑ bk k =1 k =1 Gráfica 1. Las funciones trigonométricas: sen x , cos ( −θ ) = cosθ FUNCIONES HIPERBÓLICAS cos x , tg x : EXPONENTES n tg ( −θ ) = − tg θ e x − e− x a p ⋅ a q = a p+q ∑(a − ak −1 ) = an − a0 senh x = sen (θ + 2π ) = sen θ k 2 k =1 2 ap 1.5 e x + e− x = a p−q n n cos (θ + 2π ) = cosθ cosh x = aq ∑ ⎡ a + ( k − 1) d ⎤ = 2 ⎡ 2a + ( n − 1) d ⎤ ⎣ ⎦ ⎣ ⎦ 1 2 k =1 tg (θ + 2π ) = tg θ ( a p ) = a pq senh x e x − e − x q 0.5 n tgh x = = (a + l ) = 0 sen (θ + π ) = − sen θ cosh x e x + e − x (a ⋅ b) = ap ⋅bp p 2 -0.5 cos (θ + π ) = − cosθ 1 e x + e− x n 1 − r n a − rl ctgh x = = ⎛a⎞ ap p ∑ ar k −1 =a = -1 tg (θ + π ) = tg θ tgh x e x − e − x ⎜ ⎟ = p k =1 1− r 1− r ⎝b⎠ sen x b -1.5 1 2 sen (θ + nπ ) = ( −1) sen θ sech x = = cos x n ( n + n) n 1 2 ∑k = tg x a p/q = a q p -2 -8 -6 -4 -2 0 2 4 6 8 cosh x e x + e − x 2 k =1 cos (θ + nπ ) = ( −1) cos θ n 1 2 LOGARITMOS csch x = = ∑ k 2 = 6 ( 2n3 + 3n2 + n ) n 1 Gráfica 2. Las funciones trigonométricas csc x , log a N = x ⇒ a x = N tg (θ + nπ ) = tg θ senh x e x − e − x k =1 sec x , ctg x : log a MN = log a M + log a N senh : → sen ( nπ ) = 0 ∑ k 3 = 4 ( n 4 + 2n3 + n 2 ) n 1 M 2.5 cosh : → [1, ∞ = log a M − log a N cos ( nπ ) = ( −1) n log a k =1 2 N tgh : → −1,1 ∑ k 4 = 30 ( 6n5 + 15n4 + 10n3 − n ) n 1 tg ( nπ ) = 0 1.5 log a N r = r log a N 1 ctgh : − {0} → −∞ , −1 ∪ 1, ∞ k =1 ⎛ 2n + 1 ⎞ + ( 2n − 1) = n 2 π ⎟ = ( −1) → 0,1] 0.5 n log b N ln N 1+ 3 + 5 + sen ⎜ log a N = = sech : ⎝ 2 ⎠ 0 log b a ln a n − {0} → − {0} n! = ∏ k -0.5 csch : log10 N = log N y log e N = ln N ⎛ 2n + 1 ⎞ k =1 -1 cos ⎜ π⎟=0 ALGUNOS PRODUCTOS -1.5 ⎝ 2 ⎠ Gráfica 5. Las funciones hiperbólicas senh x , ⎛n⎞ n! csc x a ⋅ ( c + d ) = ac + ad ⎜ ⎟= , k≤n -2 sec x ⎛ 2n + 1 ⎞ cosh x , tgh x : ⎝ k ⎠ ( n − k )!k ! ctg x tg ⎜ π⎟=∞ ⎝ 2 ⎠ -2.5 ( a + b) ⋅ ( a − b) = a − b -8 -6 -4 -2 0 2 4 6 8 2 2 5 n ⎛n⎞ ( x + y ) = ∑ ⎜ ⎟ xn−k y k π⎞ n Gráfica 3. Las funciones trigonométricas inversas ⎛ ( a + b ) ⋅ ( a + b ) = ( a + b ) = a 2 + 2ab + b2 4 sen θ = cos ⎜θ − ⎟ 2 k =0 ⎝ k ⎠ arcsen x , arccos x , arctg x : ⎝ 2⎠ 3 ( a − b ) ⋅ ( a − b ) = ( a − b ) = a 2 − 2ab + b 2 2 2 ( x1 + x2 + + xk ) n =∑ n! x1n1 ⋅ x2 2 ⎛ π⎞ cosθ = sen ⎜θ + ⎟ n nk x 4 ( x + b ) ⋅ ( x + d ) = x 2 + ( b + d ) x + bd 1 k n1 !n2 ! nk ! 3 ⎝ 2⎠ 0 ( ax + b ) ⋅ ( cx + d ) = acx 2 + ( ad + bc ) x + bd CONSTANTES sen (α ± β ) = sen α cos β ± cos α sen β -1 ( a + b ) ⋅ ( c + d ) = ac + ad + bc + bd π = 3.14159265359… 2 cos (α ± β ) = cos α cos β ∓ sen α sen β -2 se nh x e = 2.71828182846… -3 co sh x ( a + b ) = a3 + 3a 2b + 3ab2 + b3 tg α ± tg β 3 1 tgh x tg (α ± β ) = -4 -5 0 5 TRIGONOMETRÍA 1 ∓ tg α tg β ( a − b ) = a3 − 3a 2b + 3ab2 − b3 3 0 CO 1 FUNCIONES HIPERBÓLICAS INV sen θ = cscθ = sen 2θ = 2sen θ cosθ ( a + b + c ) = a 2 + b2 + c 2 + 2ab + 2ac + 2bc 2 HIP CA sen θ 1 -1 arc sen x arc cos x arc tg x cos 2θ = cos 2 θ − sen 2 θ ( senh −1 x = ln x + x 2 + 1 , ∀x ∈ ) cosθ = secθ = ( a − b ) ⋅ ( a + ab + b ) = a − b ( ) -2 2 tg θ 2 2 3 3 cosθ -3 -2 -1 0 1 2 3 HIP tg 2θ = cosh −1 x = ln x ± x 2 − 1 , x ≥ 1 sen θ CO 1 − tg 2 θ ( a − b ) ⋅ ( a3 + a 2 b + ab2 + b3 ) = a 4 − b4 tg θ = = ctg θ = 1 1 ⎛1+ x ⎞ cosθ CA tg θ 1 tgh −1 x = ln ⎜ ⎟, x < 1 ( a − b ) ⋅ ( a 4 + a3b + a 2 b2 + ab3 + b 4 ) = a5 − b5 sen 2 θ = (1 − cos 2θ ) 2 ⎝ 1− x ⎠ 2 ⎛ ⎞ π radianes=180 1 ⎛ x +1 ⎞ ctgh −1 x = ln ⎜ n 1 ( a − b ) ⋅ ⎜ ∑ a n − k b k −1 ⎟ = a n − b n ∀n ∈ cos 2 θ = (1 + cos 2θ ) ⎟, x > 1 2 ⎝ x −1 ⎠ ⎝ k =1 ⎠ 2 1 − cos 2θ ⎛ 1 ± 1 − x2 ⎞ HIP tg 2 θ = sech −1 x = ln ⎜ ⎟, 0 < x ≤ 1 CO 1 + cos 2θ ⎜ x ⎟ ⎝ ⎠ θ ⎛1 x2 + 1 ⎞ csch −1 x = ln ⎜ + ⎟, x ≠ 0 CA ⎜x x ⎟ ⎝ ⎠
  • 2.
    Formulario de CálculoDiferencial e Integral Jesús Rubí M. ∫ tgh udu = ln cosh u IDENTIDADES DE FUNCS HIP d dv du DERIVADA DE FUNCS HIPERBÓLICAS INTEGRALES DE FUNCS LOG & EXP ( uv ) = u + v cosh 2 x − senh 2 x = 1 d du ∫ e du = e u u dx dx dx senh u = cosh u 1 − tgh 2 x = sech 2 x d dw dv du dx dx ∫ ctgh udu = ln senh u ( uvw) = uv + uw + vw a u ⎧a > 0 ∫ a du = ln a ⎨a ≠ 1 ∫ sech udu = ∠ tg ( senh u ) d du u dx dx dx dx cosh u = senh u ctgh x − 1 = csch x 2 ⎩ d ⎛ u ⎞ v ( du dx ) − u ( dv dx ) dx dx senh ( − x ) = − senh x ∫ csch udu = − ctgh ( cosh u ) −1 ⎜ ⎟= d du au ⎛ 1 ⎞ dx ⎝ v ⎠ v2 tgh u = sech 2 u ∫ ua du = ln a ⋅ ⎜ u − ln a ⎟ u cosh ( − x ) = cosh x dx dx ⎝ ⎠ 1 = ln tgh u d n ( u ) = nu n−1 du tgh ( − x ) = − tgh x d du ∫ ue du = e ( u − 1) 2 u u dx dx ctgh u = − csch 2 u dx dx INTEGRALES DE FRAC senh ( x ± y ) = senh x cosh y ± cosh x senh y dF dF du ∫ ln udu =u ln u − u = u ( ln u − 1) = ⋅ (Regla de la Cadena) d sech u = − sech u tgh u du du 1 u cosh ( x ± y ) = cosh x cosh y ± senh x senh y dx du dx dx dx 1 u ∫ u 2 + a 2 = a ∠ tg a du 1 ∫ log ( u ln u − u ) = ( ln u − 1) udu = tgh x ± tgh y = a d du ln a ln a 1 u tgh ( x ± y ) = dx dx du csch u = − csch u ctgh u = − ∠ ctg 1 ± tgh x tgh y dx dx u 2 ∫ u log a udu = 4 ⋅ ( 2log a u − 1) a a dF dF du DERIVADA DE FUNCS HIP INV 1 u−a senh 2 x = 2senh x cosh x ∫ u 2 − a 2 2a u + a ( u > a ) = du = ln 2 2 dx dx du d 1 du u2 cosh 2 x = cosh x + senh x senh −1 u = ⋅ ∫ u ln udu = 4 ( 2ln u − 1) 2 2 dy dy dt f 2′ ( t ) ⎪ x = f1 ( t ) ⎧ 1 + u 2 dx 1 a+u ∫ a 2 − u 2 = 2a ln a − u ( u < a ) dx du 2 2 tgh 2 x = 2 tgh x = = donde ⎨ 1 + tgh 2 x dx dx dt f1′( t ) ⎪ y = f2 ( t ) ⎩ d ±1 du ⎧+ si cosh -1u > 0 ⎪ INTEGRALES DE FUNCS TRIGO cosh −1 u = ⋅ , u >1 ⎨ 1 DERIVADA DE FUNCS LOG & EXP dx u 2 − 1 dx ⎪− si cosh u < 0 ⎩ -1 ∫ sen udu = − cos u INTEGRALES CON senh 2 x = ( cosh 2 x − 1) 2 d ( ln u ) = du dx 1 du = ⋅ d tgh −1 u = 1 du ⋅ , u <1 ∫ cos udu = sen u ∫ du = ∠ sen u 1 dx u u dx dx 1 − u 2 dx a2 − u2 a cosh x = ( cosh 2 x + 1) 2 d log e du d 1 du ∫ sec udu = tg u 2 2 ( log u ) = ⋅ ctgh −1 u = ⋅ , u >1 = −∠ cos u cosh 2 x − 1 1 − u 2 dx ∫ csc udu = − ctg u 2 dx u dx dx a tgh 2 x = ⎧− ( ) −1 cosh 2 x + 1 d log e du ( log a u ) = a ⋅ a > 0, a ≠ 1 ∓1 du ⎪ si sech u > 0, u ∈ 0,1 du ∫ sec u tg udu = sec u d senh 2 x dx u dx dx sech −1 u = ⋅ ⎨ −1 u 1 − u 2 dx ⎪ + si sech u < 0, u ∈ 0,1 ∫ u 2 ± a2 = ln u + u 2 ± a 2 tgh x = ⎩ cosh 2 x + 1 d u ( e ) = eu ⋅ du d 1 du ∫ csc u ctg udu = − csc u du 1 u dx dx −1 csch u = − ⋅ , u≠0 ∫u = ln e x = cosh x + senh x dx u 1 + u 2 dx ∫ tg udu = − ln cos u = ln sec u a2 ± u2 a a + a2 ± u 2 e − x = cosh x − senh x d u ( a ) = au ln a ⋅ du du 1 a dx dx INTEGRALES DEFINIDAS, PROPIEDADES ∫ ctg udu = ln sen u ∫ u u 2 − a 2 = a ∠ cos u OTRAS d v ( u ) = vu v−1 du + ln u ⋅ u v ⋅ dv ∫ { f ( x ) ± g ( x )} dx = ∫ f ( x ) dx ± ∫ g ( x ) dx b b b ax + bx + c = 0 2 dx dx dx a a a ∫ sec udu = ln sec u + tg u 1 = ∠ sec u ∫ cf ( x ) dx = c ⋅ ∫ f ( x ) dx c ∈ ∫ csc udu = ln csc u − ctg u b b DERIVADA DE FUNCIONES TRIGO a a −b ± b 2 − 4ac ⇒ x= d du a a u 2 a2 u ( sen u ) = cos u ∫ a − u du = 2 a − u + 2 ∠ sen a ∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx b c b 2 2 2 2a u 1 ∫ sen udu = − sen 2u 2 dx dx b 2 − 4ac = discriminante a a c 2 4 ( ) 2 d du ( cos u ) = − sen u ∫ f ( x ) dx = −∫ f ( x ) dx b a u 2 a exp (α ± iβ ) = eα ( cos β ± i sen β ) si α , β ∈ ∫ u ± a du = 2 u ± a ± 2 ln u + u ± a 2 2 2 2 2 u 1 ∫ cos udu = 2 + 4 sen 2u 2 dx dx a b ∫ f ( x ) dx = 0 a LÍMITES d du ( tg u ) = sec2 u MÁS INTEGRALES ∫ tg udu = tg u − u a 2 e au ( a sen bu − b cos bu ) 1 lim (1 + x ) x = e = 2.71828... dx dx m ⋅ ( b − a ) ≤ ∫ f ( x ) dx ≤ M ⋅ ( b − a ) b ∫e sen bu du = au ∫ ctg udu = − ( ctg u + u ) x →0 d du ( ctg u ) = − csc2 u a 2 + b2 2 a x ⎛ 1⎞ lim ⎜1 + ⎟ = e dx dx ⇔ m ≤ f ( x ) ≤ M ∀x ∈ [ a, b ] , m, M ∈ e au ( a cos bu + b sen bu ) ⎝ x⎠ ∫ u sen udu = sen u − u cos u ∫ e cos bu du = x →∞ au d du ( sec u ) = sec u tg u ∫ f ( x ) dx ≤ ∫ g ( x ) dx b b a 2 + b2 sen x dx dx =1 ∫ u cos udu = cos u + u sen u a a lim ALGUNAS SERIES x →0 d du ⇔ f ( x ) ≤ g ( x ) ∀x ∈ [ a, b ] x ( csc u ) = − csc u ctg u INTEGRALES DE FUNCS TRIGO INV f '' ( x0 )( x − x0 ) 2 1 − cos x dx dx f ( x ) = f ( x0 ) + f ' ( x0 )( x − x0 ) + =0 ∫ f ( x ) dx ≤ ∫ f ( x ) dx si a < b b b lim ∫ ∠ sen udu = u∠ sen u + 1 − u 2! 2 x →0 d du x ( vers u ) = sen u a a f( n) ( x0 )( x − x0 ) n ex −1 dx dx INTEGRALES =1 ∫ ∠ cos udu = u∠ cos u − 1 − u + + 2 lim : Taylor ∫ adx =ax x →0 x DERIV DE FUNCS TRIGO INVER n! x −1 ∫ ∠ tg udu = u∠ tg u − ln 1 + u f '' ( 0 ) x 2 2 d 1 du lim x →1 ln x =1 ( ∠ sen u ) = ⋅ ∫ af ( x ) dx = a ∫ f ( x ) dx f ( x ) = f ( 0) + f '( 0) x + dx 1 − u 2 dx 2! ∫ ∠ ctg udu = u∠ ctg u + ln 1 + u 2 ∫ ( u ± v ± w ± ) dx = ∫ udx ± ∫ vdx ± ∫ wdx ± ( 0) x DERIVADAS ( n) n d 1 du f ∫ ∠ sec udu = u∠ sec u − ln ( u + u ) df f ( x + ∆x ) − f ( x ) ∆y ( ∠ cos u ) = − ⋅ 2 −1 + + : Maclaurin Dx f ( x ) = dx 1 − u 2 dx ∫ udv = uv − ∫ vdu ( Integración por partes ) = lim = lim n! dx ∆x→0 ∆x ∆x → 0 ∆x d 1 du = u∠ sec u − ∠ cosh u x 2 x3 xn ( ∠ tg u ) = ⋅ u n +1 ex = 1 + x + + + + + ∫ ∠ csc udu = u∠ csc u + ln ( u + ) d (c) = 0 dx 1 + u 2 dx ∫ u du = n + 1 n n ≠ −1 u 2 −1 2! 3! n! dx d 1 du x3 x 5 x 7 x 2 n −1 ( ∠ ctg u ) = − 2 ⋅ sen x = x − + − + + ( −1) n −1 d ( cx ) = c dx 1 + u dx du ∫ u = ln u = u∠ csc u + ∠ cosh u 3! 5! 7! ( 2n − 1)! dx d 1 du ⎧ + si u > 1 INTEGRALES DE FUNCS HIP x 2n−2 ( ∠ sec u ) = ± 2 ⋅ ⎨ cos x = 1 − x2 x4 x6 + − + + ( −1) n −1 d dx ( cx n ) = ncxn−1 dx u u − 1 dx ⎩ − si u < −1 ∫ senh udu = cosh u 2! 4! 6! ( 2n − 2 )! d 1 du ⎧− si u > 1 ∫ cosh udu = senh u ( ∠ csc u ) = ∓ x 2 x3 x 4 n d du dv dw n −1 x (u ± v ± w ± ) = ± ± ± ⋅ ⎨ ln (1 + x ) = x −+ − + + ( −1) dx u u 2 − 1 dx ⎩+ si u < −1 2 3 4 n ∫ sech udu = tgh u dx dx dx dx 2 2 n −1 d du d 1 du x3 x5 x7 ( cu ) = c ( ∠ vers u ) = ⋅ ∠ tg x = x − + − + + ( −1) n −1 x ∫ csch udu = − ctgh u 2 dx dx dx 2u − u 2 dx 3 5 7 2n − 1 ∫ sech u tgh udu = − sech u ∫ csch u ctgh udu = − csch u
  • 3.
    Formulario de CálculoDiferencial e Integral Jesús Rubí M. ÁLGEBRA LINEAL Def. El determinante de una matriz ⎡a a ⎤ A = ⎢ 11 12 ⎥ ⎣ a21 a22 ⎦ está dado por a11 a12 det A = = a11a22 − a12 a21 . a21 a22 Def. El determinante de una matriz ⎡ a11 a12 a13 ⎤ ⎢ ⎥ A = ⎢ a21 a22 a23 ⎥ ⎢ ⎣ a31 a32 ⎥ a33 ⎦ está dado por a11 a12 a13 a11 ⋅ a22 ⋅ a33 + a12 ⋅ a23 ⋅ a31 det A = a21 a22 a23 = + a13 ⋅ a21 ⋅ a32 − a11 ⋅ a23 ⋅ a32 . a31 a32 a33 −a12 ⋅ a21 ⋅ a33 − a13 ⋅ a22 ⋅ a31