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Basic Mathematics – I




                    Notes                                    dv
                                   differentiating v w.r.t. x, we get   .
                                                             dx
                                     dy  dy du dv
                                           .   .   which is chain rule.
                                     dx  du dv dx



                                     Notes  General Rules

                                          d      n        n  1 d
                                                                  x
                                              x
                                      1.     f  ( )  n f x      f ( )
                                         dx                  dx
                                          d   f  ( )  f ( ) d
                                                     x
                                               x
                                      2.    e      e      f ( ) x
                                         dx            dx
                                          d             1   d
                                                 ( )
                                      3.    log f x           f x   etc.
                                               e
                                         dx            f  ( ) x  dx
                                        Example: Differentiate the following functions w.r.t. x:
                                   1.    x 2  1

                                   Solution: Let  y  x 2  1
                                            dy     1     d   2
                                                            x   1
                                            dx  2 x  2  1 dx

                                                 1
                                                      2x
                                              2 x  2  1
                                                 x
                                               x 2  1

                                   2.    ax  2  bx c  5

                                   Solution: Let  y  ax 2  bx c  5

                                            dy              5 1 d
                                                5 ax  2  bx c     ax 2  bx c
                                            dx                 dx
                                              5 ax 2  bx x  4  2ax b


                                   3.   e 2x  3
                                   Solution: Let  y  e 2x  3

                                                  dy   2x  3 d
                                                      e       2x  3
                                                  dx       dx
                                                   e 2x  3  (2)
                                                   2e 2x  3





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