In like manner for the following examples.

To find the fluent of a + cx × x3Âà

To find the fluent of (a + cx)2x2x.

To find the fluent of (a + cx2)} × dx31⁄2.

38. When there are several Terms, involving riable Quantities, having the Fluxion of eac other Quantity or Quantities:

Take the fluent of each term, as if t
variable quantity in it, namely, that whe
tained in it, supposing all the others to
term; then, if the fluents of all the term
very same quantity in all of them, that..
fluent of the whole. Which is the re
for finding fluxions: Thus, if the give
then the fluent of xy is xy, supposing
fluent of ry is also ry, supposing r co
the required fluent of the given fluxio
In like manner,

The fluent of xyz +ajz + xy
The fluent of 2xyż + x2ż is x
The fluent of ¿¬3¿y2+2.x * y

1. and y for j, 22; then · Dry: xy = F,

Ss of Fluxions and

de most usual forms Jon of problems, site to them; by proposed fluxion dent of it will

Ferms.

ogarithms, in the above forms, are the hyperwhich are found by multiplying the common Ny 2.302585092994. And the arcs, whose sine, &c, are mentioned, have the radius 1, and are e common tables of sines, tangents and secants. numbers m, n, &c, are to be some real quantities, as fail when m = 0, or n= 0, &c.

the foregoing Table of Forms of Fluxions and Fluents. using the foregoing table, it is to be observed, that lumn serves only to show the number of the form; ond column are the several forms of fluxions, which derent kinds or classes; and in the third or last ce the corresponding fluents.

ethod of using the table, is this. Having any given, to find its fluent: First, Compare the given With the several forms of fluxions in the second coChe table, till one of the forms be found that agrees done by comparing the terms of the given parts of the tabular fluxion, namely,

e one, with that of the other; and

the

e exponents of the variable quantities of each, both within and without the vinculum; all which, being found to agree or correspond, will give the particular values of the general quantities in the tabular form: then substitute these particular values in the general or tabular form of the fluent, and the result will be the particular fluent of the given fluxion; after it is multiplied by any co-efficient the proposed fluxion may have.

EXAMPLES.

EXAM. 1. To find the fluent of the fluxion 3.x3. This is found to agree with the first form. And, by comparing the fluxions, it appears that x=x, and n - 1 = {}, or n= ; which being substituted in the tabular fluent, or 1r", gives, after multiplying by 3 the co-efficient, 3 × fr3, or 3, for the fluent sought.

3

8

EXAM. 2. To find the fluent of 5xxx, or 5.x2x(c3 — x3)ž•

This fluxion, it appears, belongs to the 2d tabular form: for ac, and-"= r3, and n = 3 under the vinculum, 1 = 1, or m =, and the exponent n-1 of x3-1 without the vinculum, by using 3 for n, is n

also m

- 12, which agrees with r2 in the given fluxion: so that all the parts of the form are found to correspond. Then, substituting these values into the general fluent, – ma (a — x1)",

EXAM. 3. To find the fluent of

This is found to agree with the 8th form; where ±x"= + x3 in the denominator, or n= 3; and the numerator then becomes r2, which agrees with the numerator in the given fluxion; also a = 1. Hence then, by substituting in the general or tabular fluent, log. of a + x", it becomes log. 1 + x3.

EXAM. 4. To find the fluent of ar^%,

EXAM. 5. To find the fluent of 2 (10 + x^)3xx.

ax

EXAM. 6. To find the fluent of

(c2 + ¿ ̈3)
3.

EXAM. 7. To find the fluent of