Page 15 - E-Modul
P. 15

Contoh Soal:

                        Dengan menggunakan konsep turunan, tentukan turunan pertama dari:
                        1.    (  ) = 10

                            Jawaban:

                            Karena    (  ) = 10  merupakan  gungsi  konstan  (tetap)  maka    (   + ∆  ) =
                            10 (          )
                                           (   + ∆  ) −   (  )     10 − 10          0
                             ′
                               (  ) = lim                   = lim           = lim      = lim 0 = 0
                                    ∆  →0        ∆            ∆  →0   ∆       ∆  →0 ∆    ∆  →0

                        2.    (  ) = 3   + 5

                            Jawaban:
                              (  ) = 3   + 5            (   + ∆  ) = 3(   + ∆  ) + 5 = 3   + 3∆   + 5

                                           (   + ∆  ) −   (  )     3(   + ∆  ) + 5 − (3   + 5)
                             ′
                               (  ) = lim                   = lim
                                    ∆  →0        ∆            ∆  →0           ∆  
                                   3   + 3∆   + 5 − 3   − 5       3∆  
                            = lim                          = lim       = lim 3 = 3
                              ∆  →0          ∆               ∆  →0 ∆     ∆  →0

                        3.    (  ) = 5   + 3
                                      2
                            Jawaban:
                                                                               2
                                                   2
                                                               2
                              (   + ∆  ) = 5(   + ∆  ) + 3 = 5(   + 2  . ∆   + ∆   ) + 3
                                                   2
                                 2
                            = 5   + 10  . ∆   + 5∆   + 3
                                           (   + ∆  ) −   (  )
                             ′
                               (  ) = lim
                                    ∆  →0        ∆  
                                                        2
                                                                    2
                                      2
                                   5   + 10  . ∆   + 5∆   + 3 − (5   + 3)
                            = lim
                              ∆  →0                  ∆  
                                      2
                                                        2
                                                                    2
                                   5   + 10  . ∆   + 5∆   + 3 − (5   + 3)
                            = lim
                              ∆  →0                  ∆  
                                   10  . ∆   + 5∆   2      10  . ∆        5∆   2
                            = lim                  = lim           + lim
                              ∆  →0      ∆            ∆  →0  ∆       ∆  →0 ∆  
                            = lim 10   + lim 10   = 10   + 0 = 10  
                              ∆  →0       ∆  →0










                                                          9
   10   11   12   13   14   15   16   17   18   19   20