Page 86 - Foundations Of Differential Calculus

4. On the Nature of Differentials of Each Order 69
2
of dx, that is, d x into account. Since the second differential arises from
the difference when pdx is subtracted from its successor, which we obtain
2
by substituting x+dx for x and dx+d x for dx, we suppose that the value
of this successor of p has the form p + qdx, and the successor of pdx has
the form

2 2 2 2
(p + qdx) dx + d x = pdx + pd x + qdx + qdx d x.
When we subtract pdx from this we have the second differential
2 2 2 2 2 2
d y = pd x + qdx + qdx d x = pd x + qdx
2
2
since qdx d x vanishes when compared to pd x.
130. Although it is simplest and most convenient to have the increments
of x equal to each other, nevertheless it is frequently the case that y is
not directly a function of x, but a function of some other quantity that
is a function of x. Furthermore, frequently it is specified that the first
differentials of this other quantity should be equal, but their relation to x
may not be clear. In the previous case the second and following differentials
of x depend on a relationship that x has with that quantity, and we suppose
that the change is by equal increments. In this other case the second and
following differentials of x are considered to be unknowns, and we use the
2
3
4
symbols d x, d x, d x,... .
131. The methods by which these differentiations in the different cases are
to be treated we shall discuss at length later. Now we will proceed under
the assumption that x increases uniformly, so that the first differentials
I II
dx, dx , d ,... are equal to each other, so that the second and higher
differentials are equal to zero. We can state this condition by saying that
the differential of x, that is dx, is assumed to be constant. Let y be any
function of x; since the function is defined by x and constants, its first,
second, third, fourth, and so forth, differentials can be expressed in terms
of x and dx. For example, if in y we substitute x + dx for x and subtract
the original value of y, there remains the first differential dy. If in this
I 2 I
differential we substitute x + dx for x, we obtain dy and d y = dy − dy.
2 2 I
In a similar way, by substituting x + dx for x in d y we obtain d y and
2
3
2 I
d y − d y = d y, and so forth. In all of the calculations dx is always seen
as a constant whose differential vanishes.
132. From the definition of y, a function of x, we determine the value of
the function p, which when multiplied by dx gives the first differential dy.
We can determine p either by the method of finite differences, or by a much
more expeditious method that we will discuss later. Given dy = pdx, the
2
differential of pdx gives the second differential d y. Hence, if dp = qdx,
2 2
since dx is constant, we have d y = qdx , as we have already shown. Taking
another step, since the differential of the second differential gives the third