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C3       DIFFERENTIATION                                                                                                           Answers - Worksheet I

          1       −1
1    a    2
              y    2



     b y = x2
     c 2x
          dx           1       −1              1               1
     d            =    2
                           y    2
                                     =                 =
          dy                                  2 y              2x
              1                1                               dy            1
                       =                  = 2x ∴                    =
          ( dy )
            d
              x            ( 21x )                             dx        ( dy )
                                                                           d
                                                                             x


          dy                                                                     dy                                                         dx                         −1        1           1
2    a            = 2e2x − 1                                             b            = 3x2                                             c        =    1
                                                                                                                                                      2
                                                                                                                                                        (ln       y) 2 ×             =
          dx                                                                     dx                                                         dy                                   y        2 y ln y
                                                                                                       1                                              2
         x=        1
                   2
                       (ln y + 1)                                                x = ( y − 2) 3                                             y = ex
          dx           1                      1                                  dx                        −2            1                  dy                    2
                  =              =            2 x−1
                                                                                      = 1 ( y − 2)
                                                                                        3
                                                                                                            3
                                                                                                                    =                            = 2 xe x = 2 y ln y
          dy           2y                2e                                      dy                                     3x 2                dx
          dy        dx                                     1                     dy       dx                     1                          dy       dx                                     1
                  ×        = 2e2x − 1×                     2 x−1
                                                                    =1                ×         = 3x2 ×                  =1                      ×         = 2 y ln y ×                              =1
          dx        dy                                2e                         dx       dy                    3x 2                        dx       dy                                  2 y ln y

          dx                                                                     dx                                                         dx
3    a            = 2y                                                   b            = 3(y − 1)2 × 1                                   c        = sec2 y
          dy                                                                     dy                                                         dy
                  dy           1                                                      dy               1                                         dy
         ∴               =                                                       ∴             =                                            ∴             = cos2 y
                  dx           2y                                                     dx           3( y − 1)2                                    dx

          dx             1                                                       dx                                                         dx        1 × e y − ( y − 2) × e y                3− y
     d            =                      ×3                              e            = 2 sin y cos y = sin 2y f                                 =                                        =
          dy           3y + 2                                                    dy                                                         dy                 (e y ) 2                        ey
                  dy           3y + 2                                                 dy                                                         dy            ey
         ∴               =                                                       ∴             = cosec 2y                                   ∴             =
                  dx             3                                                    dx                                                         dx           3− y

          dx
4    a            = 3y2 − 8y                                                                       5            a ey = ax + b
          dy
                                                                                                                               1
     b y = 3 ∴ x = −9                                                                                                   x=         (ey − b)
                                                                                                                               a
          dx                                                                                                            dx         1
                  =3                                                                                            b              =       ey
          dy                                                                                                            dy         a
                                 dy                                                                                     d                            dy                     dx
         ∴ grad =                         =       1
                                                  3
                                                                                                                c            [ln (ax + b)] =                  =1÷
                                 dx                                                                                     dx                           dx                     dy
                                                                                                                                                      a                 a
         ∴ y−3=                      1
                                     3
                                         (x + 9)                                                                                                 =            =
                                                                                                                                                      ey              ax + b
                             1
                  [y=        3
                                 x+6]

6    a ln y = ln 3x = x ln 3
                         ln y
         ∴ x=
                         ln 3
          dx            1     1                     1
     b            =         ×             =
          dy           ln 3 y                     y ln 3
          dy                     dx
     c            =1÷                     = y ln 3
          dx                     dy
           = 3x ln 3
     d grad = 9 ln 3
       ∴ y − 9 = (9 ln 3)(x − 2)
          [ y = 9x ln 3 + 9 − 18 ln 3 ]

                                                                                   Solomon Press

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C3 differentiation i answers

  • 1. C3 DIFFERENTIATION Answers - Worksheet I 1 −1 1 a 2 y 2 b y = x2 c 2x dx 1 −1 1 1 d = 2 y 2 = = dy 2 y 2x 1 1 dy 1 = = 2x ∴ = ( dy ) d x ( 21x ) dx ( dy ) d x dy dy dx −1 1 1 2 a = 2e2x − 1 b = 3x2 c = 1 2 (ln y) 2 × = dx dx dy y 2 y ln y 1 2 x= 1 2 (ln y + 1) x = ( y − 2) 3 y = ex dx 1 1 dx −2 1 dy 2 = = 2 x−1 = 1 ( y − 2) 3 3 = = 2 xe x = 2 y ln y dy 2y 2e dy 3x 2 dx dy dx 1 dy dx 1 dy dx 1 × = 2e2x − 1× 2 x−1 =1 × = 3x2 × =1 × = 2 y ln y × =1 dx dy 2e dx dy 3x 2 dx dy 2 y ln y dx dx dx 3 a = 2y b = 3(y − 1)2 × 1 c = sec2 y dy dy dy dy 1 dy 1 dy ∴ = ∴ = ∴ = cos2 y dx 2y dx 3( y − 1)2 dx dx 1 dx dx 1 × e y − ( y − 2) × e y 3− y d = ×3 e = 2 sin y cos y = sin 2y f = = dy 3y + 2 dy dy (e y ) 2 ey dy 3y + 2 dy dy ey ∴ = ∴ = cosec 2y ∴ = dx 3 dx dx 3− y dx 4 a = 3y2 − 8y 5 a ey = ax + b dy 1 b y = 3 ∴ x = −9 x= (ey − b) a dx dx 1 =3 b = ey dy dy a dy d dy dx ∴ grad = = 1 3 c [ln (ax + b)] = =1÷ dx dx dx dy a a ∴ y−3= 1 3 (x + 9) = = ey ax + b 1 [y= 3 x+6] 6 a ln y = ln 3x = x ln 3 ln y ∴ x= ln 3 dx 1 1 1 b = × = dy ln 3 y y ln 3 dy dx c =1÷ = y ln 3 dx dy = 3x ln 3 d grad = 9 ln 3 ∴ y − 9 = (9 ln 3)(x − 2) [ y = 9x ln 3 + 9 − 18 ln 3 ]  Solomon Press