Page 24 - TUTORIAL_MAT_IPS_K11-2
P. 24

TURUNAN FUNGSI ALJABAR


                              −  
            19.    (  ) =
                             −  
                               −  
                  →   (1) =      = 1 →    −    = 1 −     → 2   − 1 =    … (  )
                             1−  
                      ′
                  →    (  ) =      = 2  →    = 2
                             1
                  Dari (i) →    = 2.2 − 1 = 3
                             2.2−3
                  →   (2) =       = −1
                              1−2

            20.    (  ) = (  o      )(  )
                  ∴ ingat aturan rantai
                                           ′
                                                  ′
                            ′
                    ′(  ) =    (  (  (  ))) .      (  ).    (  )
                                         ′
                                                                              ′
                                                                    ′
                    ′
                            ′
                                                 ′
                                                          ′
                     (0) =    (  (  (0))).      (0).    (0) =    (  (0)).    (0). 2 =    (0). 2.2 = 2.2.2 = 8

            21.  ℎ(  ) =   (  (  ))
                 ∴ ingat aturan rantai
                                                                         ′
                                                      ′
                                      ′
                   ′
                                             ′
                                                                ′
                            ′
                  ℎ (  ) =    (  (  )).    (  ), ℎ (4) =    (  (4)).    (4) =    (4). 2 = 4.2 = 8

                                     3
                                                     ′
                                                 2
                    ′
            22.     (  ) =   0.(2  (  )−1) −1.3(2  (  )−1) .(2   (  )−0)  ,
                                       (2  (  )−1) 3
                                      2
                                                         2
                    ′
                      (0) =  0−3(2  (0)−1) .2  ′(0)  =   −3(2.1−1) .2.2  = −12
                               (2  (0)−1) 3        (2.1−1) 3

                          ′
                                                    ′
                                  ′
            23.  ∴ (  .   ) (2) =    (2)  (2) +   (2)   (2) = 11  → 3  (2) + 4  (2) = 11 … (  )
                                             ′
                            2 ′
                       2
                                                           ′
                  ∴ (   +    ) (2) = 2  (2)   (2) + 2  (2)   (2) = 2  (2). 3 + 2  (2). 4 = 20
                  → 3  (2) + 4  (2) = 10 … (    )
                  Eliminasi (i) dan (ii) →   (2) = 2 ,   (2) = 1
                     ′        ′
                                (2)  (2)−  (2)  ′(2)  3.1−2.4
                  ( ) (2) =                   =         = −5
                                  (  (2)) 2        1 2

                                    2
            24.  ∴   (  ).   (  ) =    − 3  
                      ′
                                        ′
                  →    (  )  (  ) +   (  )   (  ) = 2   − 3
                      ′
                                        ′
                  →    (1)  (1) +   (1)   (1) = 2.1 − 3 = −1
                             ′
                                      ′
                  ∴   (1) =    (1) =    (1),   (1) = 2
                                                      ′
                                                           2
                                ′
                                                                  ′
                                     ′
                        ′
                  → 2   (1) +    (1)   (1) = −1  → {   (1)} + 2   (1) + 1 = 0
                                           ′
                                                             ′
                       ′
                                2
                  → {   (1) + 1} = 0 →    (1) + 1 = 0 →    (1) = −1

                              2
            25.    (  ) =   (   + 2)
                  ∴ ingat aturan rantai
                            ′
                               2
                    ′(  ) =    (   + 2). 2  
                                  2
                               ′
                      ′
                  →    (1) =    (1 + 2). 2.1
                           ′
                  → 2 =    (3). 2
                      ′
                  →    (3) = 1

                                                                                      ‘LEARNING IS FUN’  23
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