Page 18 - (1-1)eng
P. 18

(1 - 1) The First Derivative of a Function

                                                      n
                 8) Power of a function    y = [ ƒ ( x ) ]    dy   n    x   n 1   ' x  y   1    dy     '    x
                                                                              
                                                                  
                                                            dx                             x  dx     x    2
                                                                                                   
                                                                                                        
                                                                                               dy     ' x
                                                                                    y        
                                                                                           x
                                                                                               dx   2   
                                                                                                         x
                             Example (9)
                                   dy
                              Find     if
                                   dx
                                                       10
                                           3
                              (a)   y  = ( 6 x  + 3 x + 1 )           (b)  y  =  x  2  3x  1
                                        x  2   x  2                           9
                              (c)   y  =                              (d)  y  =
                                             x                                  x     1  4
                                                                                    x
                                                                                 2
                              Solution:
                                              3
                                                        10
                              (a)  y    = ( 6 x  + 3 x + 1 )
                                   dy
                                                             9
                                                                     3
                                                 3
                                        = 10 ( 6 x  + 3 x  + 1 )    ( 6 x  + 3 x + 1 )'
                                   dx
                                                                        9
                                                  2
                                                             3
                                      = 10 ( 18 x  + 3 ) ( 6 x  + 3 x + 1 )
                                                 2
                                                                       9
                                                            3
                                      = 30 ( 6 x  + 1 ) ( 6 x  + 3 x + 1 )
                              (b)  y   =  x  2  3x  1
                                   dy        2x  3
                                       =
                                   dx    2 x  2  3x  1
                                          x  2   x  2 
                              (c)  y   =
                                                x
                                          x  2  4
                                       =
                                            x
                                                      1
                                            2
                                       = ( x  – 4 ) x  2
                                           3       1
                                       =  x  2   4 x  2
                                   dy     3   1       3
                                        =   x  2   2 x  2
                                   dx     2
                                              9
                              (d)  y   =
                                            2
                                          x      1  4
                                               x
                                                        – 4
                                              2
                                       = 9 ( x  + x + 1 )
                                   dy
                                                         – 5
                                                              2
                                                2
                                       = – 36 ( x  + x + 1 )  ( x  + x + 1 )
                                   dx
                                                           – 5
                                                 2
                                       = – 36 ( x  + x + 1 )  ( 2 x + 1 )
                                          36  2 x  1 
                                       =
                                           x  2   x  1  5
                 Calculus                                    17                                      Unit (1)
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