FLUXIONS. 
thod, belong to such and such kinds of 
flowing quantities : thus, for example, the 
fluent of 2 a; x is known to be .r 2 ; because, 
by the direct method, the fluxion of ,r 2 is 
found to be 2 x x : but the fluent of y x is 
unknown, since no expression has been 
discovered that produces y x for its fluxion. 
Be this as it will, the following rules are 
those' used by the best mathematicians, for 
finding the fluents of given fluxions. 
1. To find the fluent of any simple 
fluxion, you need only write the letters 
without the dots over them : thus, the 
fluent of x is x, and that of a x by, is 
ax- by. 
2. To assign the fluent of any power of 
a variable quantity, multiplied by the 
fluxion of the root ; first divide by the 
fluxion of the root, add unity to the ex- 
ponent of the power, and divide by the 
exponent so increased : for dividing the 
fluxion nx" — 1 x by x, it becomes tux’ 1 — ' ; 
and adding 1 to the exponent (» — l)we 
have n x" ; which, divided by n, gives x n ,< the 
true fluent of nx" — 'x. Hence, by the 
same rule, the fluent of 3,r 2 x will be = x 3 ; 
that of Sx' x = % ; that of y \y = §y 2 ; that 
from, the said variable part, as occasion re- 
quires, will give the fluent truly corrected. 
To make this plainer by an exarnple or two, 
let y = a. -j- a-] 3 x x. Here we first find 
a 4- .r 1 4 , , . ... u-t-al* 
y — — ^ — ; but when y = 0, then — l — 
becomes = ; since x, by hypothesis, i* 
then = 0 : therefore 3 alwavs exceeds 
4 
y by - ; and so the fluent, properly cor- 
rected, will be y = a — = a 3 x -f- 
3 a 2 x 
+ a x' 4 - - 
2 ^ ' 4 
m X tt 4* 1 
Again, let y — 
e we first have y 
; and making y — 0, the lat- 
a, m -\-x’'‘} n x x m — 1 x : here we first have y 
ter part of the equation becomes 
a»l’ , + I 
a' 
m xn 4-1 
whence the equation or flu- 
mxn-\-l 
ent, properly corrected, is y = 
(fjn _J__ r^rh) n + 1 qTh n + n 
of ay\y =: i and that of y” y — 
~ + 1 
V n ny" ■ „ ax 
- . ; that eft , or a x x~ n = 
m m-^-n x n 
\ X _ n i that of u-f zl 5 x a = i and 
that of a m -| -z m \ n x z m ~ ‘z = ( ( H -2 ' iL i. 
m x n l 
In assigning the fluents of given fluxions, 
it ought to be considered, whether the 
flowing quantity, found as above, requires 
the addition or subtraction of some con- 
stant quantity, to render it complete : 
thus, for instance, the fluent of nx n ~'x 
may be either represented by x n or by x 
4; « ; for a being a constant quantity, the 
fluxion of .r’’4 a, as well as of x", is n x”~ l x. 
Hence it appears, that the variable part 
of a fluent only can be assigned by the 
common method, the constant part being 
only assignable from the particular nature 
of the problem. Now to do this, the best 
way is to consider how much the variable 
part of the fluent, first found, differs from 
the truth, when the quantity which the whole 
fluent ought to express, is equal to nothing ; 
then that difference, added to, or subtracted 
in x »4-i 
Hitherto x and y are both supposed equal 
to nothing, at the same time ; which will 
not always be the case : thus, for instance, - 
though the sine and tangent of an arch are 
both equal to nothing, when the arch itself is 
so ; yet the secant is then equal to the ra- 
dius. It will therefore be proper to add 
some examples, wherein the value of y is 
equal to nothing, when that of x is equal to 
any given quantity a. Thus, let the equa- 
tion y — x 2 x, be proposed ; whereof the 
fluent first found is y = - ; but when y = 
d 3 
0, then - = - , by the hypothesis; 
therefore the fluent, corrected, is y — 
x 3 — a' . . 
— £ . Again, suppose y = x n x; 
. X n 4" 1 
then will y = ; which, corrected, 
a" + 1 — + i 
becomes y = . And lastly, 
if j = C 3 -j- b a 2 |l X x X ; then, first, 
V — — ~^ r ? : therefore the fluent cor- 
6 b ' ' 
rected is y — ~ ci ^ — c 3 b a 2 )} 
3b - 
3. To find the fluents of .such fluxionary 
