Page 173 - Foundations Of Differential Calculus
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156    8. On the Higher Differentiation of Differential Formulas
        and so forth. Then, since dy = pdx,wehave

                                                  2
                          2       2    2         p     2
                         d y = qdx + pd x =  q −     dx ,
                                                 y
                                  2      2
        and again substituting −pdx /y for d x we have
                                               3
                             3         4pq   3p     3
                            d y =  r −     +      dx
                                        y    y 2
        and
                                       2      2       4
                      4        7pr   4q    25p q   15p     4
                     d y =  s −    −     +       −       dx ,
                                y     y     y 2     y 3
        and so forth. When these values are substituted for the higher differentials
        of x and y, a given expression is transformed into a form of the kind which
        no longer contains higher differentials. This is accomplished by considering
        some differential to be constant.

        269. Frequently in the application of calculus to curves it may happen
                                        2
                                   2

        that the first differential  dx + dy is assumed to be constant. For this
        reason we now show the way in which for this case the second and higher
        differentials should be eliminated. At the same time, by using the same ar-
        gument, the way will be opened to show the process if any other differential
        is assumed to be constant. Now we let
             dy = p dx,   dp = q dx,   dq = r dx,   dr = sdx,    ....
                                                            2
                                     2
                               2

        Then the differential  dx + dy takes the form dx  1+ p . Since this is
        constant, we have
                                               2
                             2        2   pq dx
                            d x 1+ p +          2  =0,
                                           1+ p
        so that
                                   2     pq dx 2
                                  d x = −     2  ,
                                         1+ p
                                       2
        and we already have the value of d x. Furthermore, we have
                       3    2  3     2 2  3          2
            3      pr dx   q dx    2p q dx    2pq dx d x
           d x = −     2  −    2  +       2  −      2
                                        2
                   1+ p    1+ p    (1 + p )     1+ p
                                                                 2
                                                           2
                            2
                                     2 2

                   pr dx 3  q dx 3  4p q dx 3   pr dx 3    3p − 1 q dx 3
               = −       −       +        2  = −      +           2   .
                                                                2
                                        2
                   1+ p 2  1+ p 2  (1 + p )     1+ p 2     (1 + p )
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