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Mathematics Term 2  STPM  Chapter 2 Differentiation


                                               =   lim    2xx + (x) 2 4
                                                x → 0    x
                                               =   lim  (2x + x)
                                                x → 0
                                               = 2x
                                                        1    _  1
                                       d   1          x + x   x
                                  (b)     1 2   =   lim         4
                                       dx x     x → 0    x
                                               =   lim    x – (x + x) 4
                                                x → 0 (x + x)x (x)
         2
                                               =   lim    –1  4
                                                x → 0 x(x + x)
                                               = –  1
                                                 x 2



                 Example 2

                                  dy
                     2
              If y = 3x  + x + 1, find    from the first principles.
                                  dx
                                                                        2
                                                     2
                                  dy          3(x + x)  + (x + x) + 1 – (3x  + x + 1)
              Solution:                =   lim                                 4
                                  dx    x → 0                x
                                                             2
                                                                              2
                                               2
                                      =  lim    3x  + 6xx + 3(x) + x + x + 1 – 3x  – x – 1 4
                                        x → 0                  x
                                                         2
                                      =  lim    6x x + 3(x)  + x 4
                                        x → 0       x
                                      =  lim  [6x + 3 x + 1]
                                        x → 0
                                      = 6x + 1




                 Exercise 2.1


             Find the derivatives of the following functions from the first principles.
                      3
                                                 4
               1.  y = x                  2.  y = x                   3.  y = 5x 2
                                                 2
                                                                              2
               4.  y =   1                5.  y = x  + 5x             6.  y = x  – x + 3
                     x 2
                                                                               2
                       3
               7.  y = 4x                 8.  y =   1                 9.  y = 2x  + 3
                                                 x 3
              10.  y = 2x  – 3x + 1
                       2



              18





       02 STPM Math(T) T2.indd   18                                                                 02/11/2018   12:43 PM
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