fluent of will be found from that of and √(a2-12) √(a2-12) comes out • 1x √(a2 x2)+ax cir. arc to radius a and sine x. Ex. 4. In like manner, the fluent of (+) will be a2) Here it is manifest that y found from that of √(x2 + a2)° must be assumed = x3√(x2 + a2), in order that one part of its fluxion, viz, x × flux. of √(x2 + a2) may agree with the proposed fluxion. Thus, by taking the fluxion, and reducing as before, the fluent of will be found = 1.x13 √(x2 + a2) — 3a2 × ƒ Ex. 5. Thus also the fluent of √(x2 + α2) √(x2 is x(x2-a2) √(a2 - x2) is — x3 √ (a2 In like manner the student may find the fluents of √(x2±a2)? √(x2 + a2)3 number, each from the fluent of that which immediately precedes it in the series, by substituting for y as before. 28. In like manner we may proceed for the series of similar expressions where the index of the power of x in the numerator is some odd number. Ex. 1. To find the fluent of y = x2√(x2 + a2), and taking the fluxion, one part of it will be similar to the fluxion proposed. Thus, y = 2xx x3x √(x2 + a2) + √(x2 + a2); hence at once the given fluxion x3% √(x2 + a2) =; 2xx(x2+a2); theref. the required fluent is yf. 2xx√(x2 √(x2 - ƒ . 2xx √(x2 + a2) = x2 √√/ (x2 + a2) — 3(x2 + a2)3 ̧ by the 2d form of fluents. Ex. 2. In like manner the fluent of ¿¿12 u̸/ (-212 ——— α2) — }(·x2 — a2)3. · √(x2 + a2)? Here it is manifest we must assume y = This in fluxions and reduced gives j = √(x2 + a2) √(x2 + a2). 5x5x + = }j 5 √(x2 + a3) is y — a2 × ƒ √(x2 + a3) = {x2 √/ (.x2 + a2) −‡a3×ƒthe fluent of the latter part being as in ex. 1, above. In like manner the student may find the fluents of where n is any odd number, viz, always by means of the Bluent of each preceding term in the series. 29. In a similar manner may the process bé for the fluents of the series of fluxions, √(ax) √(a + 1)' √(a + x)' √(att)" Fusing the fluent of each preceding term in the series, as a part of the next term, and knowing that the fluent of the is given, by the 2d form of fluents, = first term Ex. 1. To find the fluent of of √(x+4) xx √(x+a)' having given that = 2√√(x + a) = A suppose. Here it is evident we must assume y=x(x+a), for then its flux. j = +x√(x+a)=· + + √(x+a) + √(x+α) √(2+ a) √(x+a) =j-zaa; and the required fluent is 3y Za▲ = 3x √(x + a) - ja√(x+a)=(x − 2a) × 3√/ (x+a). In like manner the student will find the fluents of = B. Here y must be assumed = x2 √(x+α); of for then taking the flu. and reducing, there is found (x+a) 户 =zy - ав =2x2√(x + a) √(s+a) = 3 y - ‡AB = }x2√(x + a) – ‡a(x − 2a) × } √(x + a) = (9.212 -4ax +8a2) × 73 √(x + a). In the same manner the student will find the fluents of And in general, the fluent of √(x+a) 2 = being given = c, he will find the fluent of √(x+a) 2n+1 2n 2n+1 -ac. 30. 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 in volume 2; as, for instance, such as xx(dx-x2), x2*✅ (dx − x2), x3x√√(dx − x2), &c, depending on the fluent of x/(dx-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 (dx—x2) =A, to find the fluent of xx√(dx - x2). Here assuming y = (dx − x2), and taking the fluxions, they are, j = 3⁄4 (dx — 2xx)√√(dx-x); hence x*(dx = x2) = ‡ d÷√ (dx − x2) — } j = då – }; therefore the required fluent, fxx√ (dx − x2), is i̟da—‡y =÷da—+(dx-x2)=в suppose. Ex. 2. To find the fluent of x2x (dx − x2), having that of xx✓✅ (dx—x2) given B. Here assuming y=x(dx − x2), then taking the fluxions, and reducing, there results = ({dxx - 4x2x) √ (dx − x2); hence x2+(dx − x2) = ždxx √ (dx—x2)—‡j={dB-4, the flu. theref. of x2/(dx-x2) is {dB — ‡ y = {dB — ‡x(dx — x2)3. -I Ex. 3. In the same manner the series may be continued to any extent; so that in general, the flu. of -1√ (dx − x2) being given = c, then the next, or the flu. of x" will be - idc-x-(dx— x2)}. 31. To find the fluent of such expressions as a case not included in the table of forms in vol. 2. - (dx − x2) Put the proposed radical (± 2ax)=z, or x2+2ax z2; then, completing the square, x2± 2ax+a2 = x2+a2, and the root is x ± a = √(x2 + a2). The fluxion of this is of which, by the 12th form, is the hyp. log. of ≈ +√(x2+a2) = hyp. log. of r±a + √(x2± 2ax), the fluent required. Er. 2. To find now the fluent of xx √(x2+2ar), having given, by the above example, the fluent of suppose. Assume √(x2 + 2ax) xx √(x2+2ax) = A √(x2+2ar) = y; then its fluxion is √(x2+2a1) the fluent of which is yan= √(x2+2ax) — a▲, the fluent sought. Ex. 3. Thus also, this fluent of xx being given, √(x2+2ax) 12x √(x2+2ax) and so on for any other will be found, the flu. of the next in the series, or xn-- - 1x √(x2+2ax) be given = c; then, by assuming x”-1、'(x2 + 2ax) = y, the fluent Ex. 4. In like manner, the fluent of given, as in the first example, that of √(x2-2ax) √(x2-ax) may be found; and thus the series may be continued exactly as in by pa. 321 vol. 2, is —–—× circular arc to radius a and versed sine x, the fluents of may be assigned by the same method of continuation. Thus, Ex. 1. For the fluent of y; the required fluent will be found (2ax − x2)+▲ or arc to radius a and vers. x. where a denotes the arc mentioned in the last example. Ex. 3. And in general the fluent of —x"1 √(2αx − x2), where c is the fluent of n the next preceding term in the series. 33. Thus = 33. Thus also, the fluent of (x − a) being given, (x-a), by the 2d form, the fluents of xx(x-a), x2x√(x-a), &c...x" x√√(x − a), may be found. And in general, if the fluent of "√(x-a) = c be given; then -I by assuming "(x-a)=y, the fluent of x"√(x- a) is found = 2na 2n+3 C. 1 m + 1 34. Also, given the fluent of (r− a)"x, which is (ra)+1 by the 2d form, the fluents of the series (x − a)TMxx, (x− a)TMx2x, &c. . . (x —a)”x”x can be found. And in general, the fluent of (ra)" being given = c; then (x a)"r"-1 by assuming (x —− a)"+1x" = y, the fluent of (x- a)TMx”x is x^(x− a)" + 1 + nac found = m+n+1 Also, by the same way of continuation, the fluents of x'x✅ (a ≈ x) and of x*x(a + x)" may be found. 35. 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 , connected with half the coefficient of the middle term with its sign. Thus, put z=x+c: then z2 = x2+cx+¡c2; theref. z2-4c2 = x2 + cx, and z2 + b-c2 = x2 + cx + b the given trinomial; which is = x2 + a2, by putting a2 = b- c2. Ex. 1. To find the fluent of 3x Here z = + 2; then z2 = x2 + 4x + 4, and z2 + 1 = 5 + 4x + x2, also ; theref. the proposed fluxion ré 3z √(1+x3) duces to (+); the fluent of which, by the 12th form in the 2d vol. is 3 hyp. log. of x +√(1 + z) = 3 hyp. log. x+2+√(5+ 4x + x2). Ex. 2. To find the fluent of (b + cx + dx2) = √/dx √ √ 2/2 + 1/2 x + x2). Here assuming +2=z; then = 2, and the proposed flux. reduces to ż√✓d× √(x2 + - - _—_—_ )=ż √ d× √(z2+a2), ს putting a2 for d A and the fluent will be found by a similar process to that employed in ex. 1 art. 27. - I Ex. 3. In like manner, for the flu. of "~1x√(b + cx” + dx2”), assuming ♫” + —=z, nx”—! ¿=¿, and a”—1÷==ż; -I hence |