41. Many fluents may be found by the Direct Method thus : Take the fluxion again of the given fluxion, or the second fluxion of the fluent sought; into which substitute for x,

for j, &c; that is, make x, *, *, as also y, ÿ, ÿ, &c, to be in continual proportion, or so that *:*::*:*, and y: ÿ ÿÿ, &c; then divide the square of the given fluxional expression by the second fluxion, just found, and the quotient will be the fluent required in many cases.

:

Or the same rule may be otherwise delivered thus In the given fluxion F, write x for x, y for y, &c, and call the result G, taking also the fluxion of this quantity, G; then make GFG: F; so shall the fourth proportional F be the fluent sought in many cases.

It may be proved if this be the true fluent, by taking the fluxion of it again, which, if it agree with the proposed fluxion, will show that the fluent is right; otherwise, it is

wrong.

EXAMPLES.

EXAM. 1. Let it be required to find the fluent of nx1 ̄x. Here Fnxx. Write x for x, then n11r or nx" = G; the fluxion of this is G = n2; therefore G: F:: G: F, becomes n111: nxa ̄ ̄1x :: nx" : "F, the fluent sought.

EXAM. 2. To find the fluent of xy + xỷ.

for, y for y,

Here F = xy+xy; then, writing x for x, and

=

it is xyxy or 2xy G; hence G 2xy + 2xy; then

G: F :: G: F, becomes 2xy + 2xỷ : xy + xỷ :: 2xy: xy =F, the fluent sought.

42. To find Fluents by means of a Table of Forms of Fluxions and Fluents.

In the following Table are contained the most usual forms of fluxions that occur in the practical solution of problems, with their corresponding fluents set opposite to them; by means of which, namely, by comparing any proposed fluxion with the corresponding form in the table, the fluent of it will be found.

Forms.

Note. The logarithms, in the above forms, are the hyperbolic ones, which are found by multiplying the common logarithms by 2·302585092994. And the arcs, whose sine,

or tangent, &c, are mentioned, have the radius 1, and are those in the common tables of sines, tangents, and secants. Also, the numbers m, n, &c, are to be some real quantities, as the forms fail when m = 0, or n= 0, &c.

The Use of the foregoing Table of Forms of Fluxions and Fluents.

43. In using the foregoing table, it is to be observed, that the first column serves only to show the number of the form; in the second column are the several forms of fluxions, which are of different kinds or classes; and in the third or last column, are the corresponding fluents.

The method of using the table, is this. Having any fluxion given, to find its fluent: First, Compare the given fluxion with the several forms of fluxions in the second column of the table, till one of the forms be found that agrees with it; which is done by comparing the terms of the given fluxion with the like parts of the tabular fluxion, namely, the radical quantity of the one, with that of the

other;

other; and the 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.r3x. This is found to agree with the first form. And, by comparing the fluxions, it appears that x=x, and n- i = }, or n= ; which being substituted in the tabular fluent, or 1", gives, after multiplying by 3 the co-efficient, 3 × fr3, or r3, for the fluent sought.

EXAM. 2. To find the fluent of 5æ**«/ ¿3 — ‚μ3‚ or 5x2¿ (c3—‚x3)31®

This fluxion, it appears, belongs to the 2d tabular form: for a = and -=- x3, and n = 3 under the vinculum, also m− 1 = 1⁄2, or m = 2, and the exponent

without the vinculum, by using 3 for n, is n − 1 = 2, which agrees with x2 in the given fluxion: so that all the parts of the form are found to correspond. Then, substituting these values into the general fluent, - (a −x3)",

I

nם

This is found to agree with the 8th form; where ± x11 = + 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. of 1 + x3.

EXAM. 4. To find the fluent of ax*x.

EXAM. 5. To find the fluent of 2 (10 + x2)3⁄43×x.

EXAM.

Exam. 6. To find the fluent of (+2)

EXAM. 7. To find the fluent of.

ax

(c2

(aα − x)4•

EXAM. 8.

1+ 3x

2x4
3x, 2j

EXAM. 10. To find the fluent of (+2)2.

EXAM. 11. To find the fluent of (+3)

EXAM. 12. To find the fluent of

EXAM. 13. To find the fluent of

EXAM. 14. To find the fluent of

EXAM. 15. To find the fluent of

EXAM. 16. To find the fluent of

EXAM. 17. To find the fluent of

EXAM. 18. To find the fluent of.

EXAM. 19. To find the fluent of

EXAM. 20. To find the fluent of

EXAM. 21. To find the fluent of

EXAM. 22. To find the fluent of

EXAM. 23. To find the fluent of

EXAM. 24. To find the fluent of

EXAM. 25. To find the fluent of

EXAM. 26. To find the fluent of

3x 3
or
ax α
ax

3y

2x 3x

2x x2

2x

I

3xx

X 3x3 1 3x22
3xx
1 X4

2xx
1+x4

ax2 x
2+xs
3xx
√1+x4

ax

Y

EXAM. 27.