xnx where n is any odd number, viz. always by means of the fluent of each preceding term in the series. 77. In a similar manner may the process be for the fluents of the series of fluxions, x .&c... √(a‡x)' √ (a‡x)' √(@± x)` xnx using the fluent of each preceding term in the series, as a part of the next term, and knowing that the fluent of the x first term x) (ax), of the same sign as x. : is given, by the 2d form of fluents, 2√ = Ex. 1. To find the fluent of xx √(x+a) ,having given that of =2✔✅✅(x+α)=▲ suppose. Here it is evident we must assume y=x√(x+a), for then its flux. ÿ= √(x+u) √(x+a) √(x+a) √(x+a) √(x+a)=-3a; and the required fluent is ży Jaṣ=Z x √(x+α)−}α√(x+a)=(x −2a)×}√(x+a). Ex. 2. To find the fluent of Here y must be assumed=x2√(x+a); for then taking the flu. and reducing, there is found ав= x2√(x+α)—fa(x — 2a) × 3 √ (x+a)=(9x2 - 4ax+8a2) X√(x+a). In the same manner the student will find the fluents of being given = c, he will find the fluent of 78. In a similar way we might proceed to find the fluents of other classes of fluxions by means of other fluents in the table of forms: as, for instance, such as xx/(dx − x2), x3 x√ (dx — x2), x3 x √ (dx-x2), &c. depending on the fluent of x(dr-x2), the fluent of which, by the 16th tabular form, is the circular semisegment to diameter d and versed sine x, or the half or trilineal segment contained by an arc with its right sine, and versed sine, the diameter being d. Ex. 1. Putting then said semiseg. or fluent of x √ (dx − x2) =▲, to find the fluent of xx√(dx-x2). Here assuming y= (dx-x), and taking the fluxions, they are y(x-Lxx) (dx-x2); hence ri✓ (dx-x2)=1dx ✓ (dx—x2)—}y= da-y; therefore the required fluent fxx (dx-x2), is jda — fy=jds — (dx-x)=B suppose. y Ex. 2. To find the fluent of x2xdx-x2), having that of xx(dx-x2) given = B. Here assuming y=x (dx-x2), then taking the fluxions, and reducing, there results (dxx-4x3x) ✓ (dx-x2); hence xx (dx-x) = { dx x √ (dx - x2) - fýd-y, the flu. theref. of xx (dx x2) is & dB-4y=&dB-‡x(dx — x2)§. Ex. 3. In the same manner the series may be continued to any extent; so that in general, the flu. of x2-1 ✓ (dx - x2) being given =c, then the next, or the flu. of xx√(dx—x3) -dc- „x”—1 (dx − x2)}. (dx−x2)§. n+2 will be 2n+1 1 79. To find the fluent of such expressions as a case not included in the table of forms. Put the proposed radical ✓ (≈a ± 2ax)=z, or x2± 2ax =23; then, completing the square, x2 +2ax+a2 =z2+a2, and the root is x + a = √(z2 + a2). The fluxion of this is x= theref. x (22 +a2); √(x2±2ax) √(22+a2); the flu ent of which, by the 12th form, is the hyp. log. of +✔ (2+2)=hyp. log. of x+a+(x2+2ax); the fluent required. Ex. 2. To find now the fluent of xx given, by the above example, the fluent of having /(x2+2ax) suppose. Assume✓ (x2 + 2ax) = y; then its fluxion is xx+ax √(x2+2ax) xx =y; theref = √(x2+2ax, =√(x2+2αx) - as, y-a; the fluent of which is y the fluent sought. Ex. 3. xx Ex. 3. Thus also, this fluent of the flu. of the next in the series, or √(x2+2ax) being given will be found, by assuming x (x2 + 2ax) y; and so on for any other of the same form. As, if the fluent of xnx √(x2+2ax) = be given = c; then, by assuming "-1 (x2 + 2ax) y, the fluent 2n-1 -x2-1(x2+2ax) ac. n n found; and thus the series may be continued exactly as in the 3d ex. only taking 2ax for 2ax. x 80. Again having given the fluent of 1 ✓(2ax-x2)' which is × circular arc to radius a and versed sine x, the fluents xnx √(2ux − x2)' √(2αx − x3)' ✓ (2ax-x2)' assigned by the same method of continuation. Thus, xx may be Ex. 1. For the fluent of √(2ax-x), assume√(2ax-x2) = y; the required fluent will be found = —✓√(2ax − x2)+s or arc to radius a and vers x. √(2ux − x2) − x √(2ax — x2)=gas— √ (2ax-x2), A where a denotes the arc mentioned in the last example. Ex. 3. And in general the fluent of ✔ (2ax − x3), where c is the fluent of the next preceding terin in the series. 81. Thus also, the fluent of (x-a) being given, = (x-a), by the 2d form, the fluents of xx(x-0), x2 x (x-a), &c. ...x^x (x-a), may be found. And in general, if the fluent of x-x √(x −α) c be given; then by assuming assuming x^(x—a)2=y, the fluent of xx✔✅/(x—a) is found= 82. Also, given the fluent of (x-a)m which is 1 m+1 (x-a)mt by the 2d form, the fluents of the series (x a)"xx, (x-a)mx2x &c. (x-a)m can be found. And in gec; then by ... neral, the fluent of (x-a)x-1 being given assuming (-a)"'"y, the fluent of (x-a)mxn is found _xn(x—a)+1+nac = m+n+i Also, by the same way of continuation, the fluents of xxv (ax) and of xx (a ±x)" may be found. 83. When the fluxional expression contains a trinomial quantity, as (b + cx + x2), this may be reduced to a binomial, by substituting another letter for the unknown one x, connected with half the co- efficient of the middle term with its sign. Thus, put z x+c; then z2=x2+cx+{c2; theref. - }c2=x2 + cx, and 22+ b-c2=x2+cx+b the given trinomial which is=22+a+a2, by putting a2bc2. 22 Ex. 1. To find the fluent of 3x (5+4x+x)' Here z=x+2; then 22 = x2+4x+4, and 2a + 1 = 5+ 4x + x2, also ; theref. the proposed fluxion re duces to 3x ; the fluent of which, by the 12th form in this vol. is 3 hyp. log. of z + (1+2)=3 hyp. log. x+2 + √ (5 + 4x + x2). Ex. 2. To find the fluent of x/(b+cx+dx3)=x √↓d × √ b C flux. reduces to %✅/dx√(z2+ putting a for b C2 d 4d2 ; and the fluent will be found by a simi lar process to that employed in ex. 1 art. 75. 1 Ex. 3. In like manner, for the flu. of x-1x √ (b+cxn+ dx"), assuming xr+; C = 2d + + -=22, and ✅✔✅ (dx2n+cxn+b)=√/dX √(x2n + b c2 ニュー d 4d2 1. ; hence the given fluxion becomes✔ X √ (za), and its fluent as in the last example. Ex. 4. Also, for the fluent of xn-1x C ; assume x+ I z, then the fluxion may be reduced to the form- -X and the fluent found as before. So far on this subject may suffice on the present occasion. But the student who may wish to see more on this branch, may profitably consult Mr. Dealtry's very methodical and ingenious treatise on Fluxions, lately published, from which several of the foregoing cases and examples have been taken or imitated. OF MAXIMA AND MINIMA; OR, THE GREATEST AND LEAST MAGNITUDE OF VARIABLE OR FLOWING QUANTITIES. 84. MAXIMUM, denotes the greatest state or quantity attainable in any given case, or the greatest value of a variable quantity; by which it stands opposed to Minimum, which is the least possible quantity in any case. Thus the expression or sum a2+hx, evidently increases as x, or the term bx, increases; therefore the given expression will be the greatest, or a maximum, when x is the greatest, or infinite; and the same expression will be a minimum, or the least, when z is the least, or nothing. Again in the algebraic expression a2- bx, where a and b denote constant or invariable quantities, and x a flowing or variable one. Now, it is evident that the value of this remainder or difference, aa—bx、 will increase, as the term bx, or as x, decreases; therefore the former will be the greatest, when the latter is the smallest ; that is a2 - bx is a maximum, when x. is the least, or nothing at all; and the difference is the least, when x is the greatest. 85. Some variable quantities increase continually; and so have no maximum, but what is infinite. Others again decrease continually; and so have no minimum, but what is of no magnitude, or nothing. But, on the other hand, some variable quantities increase only to a certain finite magnitude, called their Maximum, or greatest state, and after that they decrease again. While others decrease to a certain finite magnitude, called |