Page 164 - Foundations Of Differential Calculus
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8. On the Higher Differentiation of Differential Formulas  147
                           2
        Different values for d x will be given if the differentials of other functions
                                                                       2
        of x are held constant. However, it is clear that the formulas in which d x
                                                                       2
        occurs take on quite different values depending on whether in place of d x
                          2               2
        we write zero or −dx /x or − (n − 1) dx /x or some other expression of this
                                             2 2    2              2
        kind. For instance, if the given formula is x d x/dx , then, because d x and
          2
        dx are both infinitely small and homogeneous, the expression should have
                                                                        2
        a finite value. If dx is made constant, the expression becomes zero; if d.x
                                      3                               4
        is constant, it becomes −x;if d.x is constant, it becomes −2x;if d.x is
        constant, it becomes −3x, and so forth. Hence, it can have no determined
        value unless the differential of something is assumed to be constant.
        253. This ambiguity of signification is present, for a similar reason, if the
        third differential is present in some formula. Let us consider the formula
                                       3 3
                                      x d x
                                         2
                                     dx d x ,
        which also has a finite value. If the differential dx is constant, then the
                                                                     2
        formula takes the form 0/0, whose value we will soon see. Let d.x be
                       2       2
        constant. Then d x = −dx /x and after another differentiation we obtain
                                       2
                           3      2dx d x  dx 3   3dx 3
                          d x = −        +     =      ,
                                     x      x 2    x 2
              2        2
        since d x = −dx /x. Hence, for this reason, the given formula
                                       3 3
                                      x d x
                                         2
                                      dx d x
                    2
                                   n
        becomes −3x . However, if d.x is constant, then
                                2    − (n − 1) dx 2
                               d x =             ,
                                          x
        so that
                                                      2
                             2
           3     2(n − 1) dx d x  (n − 1) dx 3  2(n − 1) dx 3  (n − 1) dx 3
          d x = −              +      2     =       2      +      2    ,
                       x             x             x             x
                (2n − 1) (n − 1) dx 3
             =                   .
                       x 2

        Hence for this reason we have
                                  3
                                 d x  =  (2n − 1) dx
                                  2
                                 d x       x
        and
                                3 3
                               x d x             2
                                  2
                              dx d x  = − (2n − 1) x .
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