1. To find the fluent any simple fluxion. you need only write the letters without the dots over them : thus, the fluent of is x, and that of a X. b y, is a x b y.
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 n xn-1.1: by X, it becomes n x17-1; and adding 1 to the exponent (n 1) we have n x a ; which divided by n, gives xn, tke true fluent of n xn-1 X. Hence, by the same rule, the fluent of S will be x3; tt of 2 x5 s ; that of y = ; that,vf a y y= 3 a y5; and 8' nyn that of v-hn —; that OT m+ a xi-n or ha a': x-rt • t at of 3 + 1—n ' 4 a _4 ; and that of amd-zmi X z 4 a at+ • m X n ± In assigning the fluents of given flux ions, it ought to be considered whether the flowing quantity, found as above, re quires the addition or subtraction of some constant quantity, to render it complete : thus, for instance, the fluent of n may be either represented by xn or by x ± a ; for a being a constant quantity, the fluxion of xn ± a, as well as of xa, is 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 from, the said variable part, as occasion requires, will give the fluent truly corrected. To make this plainer by an example or two, let y a 3 X .17. Here we first find y = (774n. X but when y = 0, then a 4 4 a4 becomes - since s, by hypothesis, is then -- 0 : always exceeds y by T and so the fluent, pro perly corrected, will be y = 4 x4 a1 x 3 a= x' 0 x3 4- . Again, 2 let y = here we first art have y = mX n making y = 0, the latter part of the equation becomes /0(0 mX ; whence the equa tion or fluent, properly corrected, is y = amem tn 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 radius. It will there fore be proper to add some examples, wherein the value of y is equal to nothing, when that of x is equal to any given quan tity a. Thus, let the equation g = be proposed; whereof the fluent first found is y = but when y = 0, then x3 a3 by the hypothesis ; therefore 3' the fluent, corrected, is y Again, suppose y = xn. ; then will y xn+i --.; which, corrected, becomes y n+1 n+1 And lastly, if 'y = c3 -Frio X x then, first, y = c3--11: there 3 fore the fluent, corrected, is y c3 bx=4 — c3 b 3. To find the fluents of such fluxiotiary expressions as involve two or more vari able quantities, substitute, instead of such fluxion, its respective flowing quantity ; and, adding all the terms together, di. vide the sum by the number of terms, and the quotient will be the fluent. Thus, the fluent of y X and the fluent oU y z ya" xyz+xyz+xyz 3xyz --x y 2.
3 = 3 But it seldom happens that these kinds of fluxions, which involve two variable quantities in one term, and yet admit of known and perfect fluents, are to be met with in practice.
Having thus shown the manner of find ing such fluents as can be truly exhibited in algebraic terms, it remains now to say something with regard to those other forms of expressions involving one varia ble quantity only ; which yet are so af fected by compound divisors and radical quantities, that their fluents cannot be accurately determined by any method whatsoever. The only method with re gard to these, of which there are innume rable kinds, is to find their fluents by ap proximation, which, by the method of in finite series, may be done to any degree of exactness. See SEItlES: Thus, if it were proposed to find the fluent of - a it becomes necessary to a-x throw the fluxion into an infinite series, by dividing a S by a-x : thus, a a=a x x3 1 x S: +' Now the fluent of each term of this ries may be found by the foregoing x3 x3 , x4 rules to be x -1- +, &c.