Page 23 - TUTORIAL MATEMATIKA IPS KELAS XI-2
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TURUNAN FUNGSI ALJABAR

                   ′
            12.     (  ) =
                                             3 2 4
                                                                              3
                                                          2
                                 4
                                                                 4
                         2
                  5(((2   + 3  ) − (5   + 3) ) ) . 2((2   + 3  ) − (5   + 3) ).
                                                        2
                                3
                        2
                  (4(2   + 3  ) . (4   + 3) − 3(5   + 3) . 5)

                                        1                             1
                                     1  2             1           1 − 2      1       1
                                                                                   −
                                              ′
            13.    (  ) = (2   + (2  )2)  →    (  ) =  (2   + (2  )2)  . (2 + (2  ) 2. 2)
                                                      2                      2
                                          1
                                       1 − 2            1
                          1
                                                       −
                   ′
                     (  ) = (2   + (2  )2)  . (2 + (2  ) 2)
                          2

                                                2
                                   2 7
            14.    (  ) = (2   − 7   ) . (4   + 3)3
                  ∴   (  ) =   (  ).   (  )
                                      2 7
                                              ′
                                                               2 6
                  ∴   (  ) = (2   − 7   ) →    (  ) = 7(2   − 7   ) . (2 − 14  )
                                     2              2          1            8           1
                                                              −
                                                                                      −
                                            ′
                  ∴   (  ) = (4   + 3)3  →    (  ) =  (4   + 3) 3. (4 + 0) =  (4   + 3) 3
                                                    3                       3
                                                              2                8           1
                                                                                         −
                   ′
                                      2 6
                                                                           2 7
                     (  ) =  7(2   − 7   ) . (2 − 14  ). (4   + 3)3 + (2   − 7   ) . (4   + 3) 3
                                                                               3
                                                  1                          8
                                                 −
                                                                                        2
                   ′
                                    2 6
                     (  ) = (2   − 7   ) . (4   + 3) 8. (7(2 − 14  )(4   + 3) +  (2   − 7   ))
                                                                             3

                                                                        7
                               8
                                                        7
                                          ′
            15.    (  ) = 256   − 1  →    (  ) = 8.256   − 0 =  2048  

                          (  −5) 3
            16.    (  ) =
                            2   2
                                          ′
                                   3
                                                          2
                  ∴   (  ) = (   − 5)  →    (  ) = 5(   − 5)
                                     ′
                               2
                  ∴   (  ) = 2    →    (  ) = 4  
                                           3
                                 2
                                                         2
                                                                              2
                                    2
                   ′
                     (  ) =   5(  −5) .2   −(  −5) .4     =   2  (  −5) (5  −2(  −5))  =  2  (  −5) (3  +10)
                                     2 2
                                  (2   )                  4   4              4   2

                                    1
                        (3  −1) 4   2
            17.     = (        + 5) →
                         (2−  ) 3
                                        1         3      3       4      2
                                      −
                                4
                        1 (3  −1)
                   ′
                     =  (        + 5)   2 . ( 4(3  −1) .3.(2−  ) −(3  −1) .3.(2−  ) .−1   + 0)
                                                             3 2
                        2  (2−  ) 3                    ((2−  ) )
                                                             3
                                                                   2
                                     2
                               3
                    ′
                      =   3(3  −1) (2−  ) (4(2−  )+(3  −1))  =  3(3  −1) (2−  ) (7−  )
                                   (2−  ) 6                  (2−  ) 6

                           2
                       (3−  ) .(3  +1) 3
            18.     =          1    →
                             2
                          (4   )3
                                                            1                    − 2
                                                                          3 1
                                                      2
                                                                               2
                                                                   2
                                       3
                                              2
                        (2(3−  ).−1.(3  +1) + (3−  ) .3(3  +1) ).(4   ) −(3−  ) .(3  +1) . (4   )  3
                                                           2 3
                   ′
                     =                                                     3
                                                       1  2
                                                     2
                                                 ((4   )3)
                                                          1                   − 2
                                                            1
                                     3
                                                    2
                                                                  2
                                             2
                                                                            2
                                                                        3
                                                        2 3
                        (−2(3−  )(3  +1) + 3(3−  ) (3  +1) )(4   ) − (3−  ) (3  +1) (4   )  3
                   ′
                     =                                      3
                                                     1  2
                                                    2
                                                ((4   )3)
                             2
                            −
                                                               1
                                             2
                                         2
                           2
                        (4   )  3 (3−  )(3  +1) ((4   (3(3−  )−2(3  +1)))− (3−  )(3  +1))
                                                               3
                   ′
                     =                             2
                                                  2
                                               (4   )3
                                                                                      ‘LEARNING IS FUN’  22
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