Page 171 - Foundations Of Differential Calculus
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154    8. On the Higher Differentiation of Differential Formulas
        and
                              − (n − 1) (2n − 1) (3n − 1)
                           s =                        .
                                        4 4n−1
                                      n x
                                       n
        Hence, if we let the differential of x be constant, then
                                        2
                         2      (n − 1) dx
                        d x = −           ,
                                    x
                         3    (n − 1) (2n − 1) dx 3
                        d x =                  ,
                                      x 2
                         4      (n − 1) (2n − 1) (3n − 1) dx 4
                        d x = −                          ,
                                           x 3
        and so forth.
        266. If an expression contains two variables x and y, and if the differential
                                           2
        of one, x, is held constant, then since d x = 0, there will be no second or
                                  2   3
        higher differentials besides d y, d y, etc. However, these can be treated
        in the same way as before. They can be removed by letting dy = pdx,
        dp = qdx, dq = rdx, dr = sdx, and so forth. Then we have
                  2       2     3        3     4       4
                 d y = qdx ,   d y = rdx ,    d y = sdx ,    ...,
        and so forth. By means of these substitutions we obtain an expression that
        contains only the differential dx besides the finite quantities x, y, p, q, r,
        s, etc. For example, if the given expression is
                                 4        3      4
                              ydx + xdy d y + xd y  ,
                                            2
                                         2
                                    2
                                   (x + y ) d y
        in which we assume that dx is constant, then we let dy = pdx, dp = qdx,
        dq = rdx, and dr = sdx. When these values are substituted the given
        expression is transformed into
                                               2
                                (y + xpr + xs) dx
                                     2
                                          2
                                    (x + y ) q  ,
        which contains no second or higher differential.
        267. In a similar way the second and higher differentials are removed if
        dy is assumed to be constant. However, if any other first differential dt is
        taken to be constant, then the higher differentials of x are removed from
        the calculation with the method first mentioned before. That is, we let
              dx = p dt,  dp = q dt,   dq = r dt,   dr = sdt,   ...,

        so that
                  2       2      3      3      4       4
                 d x = qdt ,    d x = rdt ,   d x = sdt ,    ....
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