of different variables. v=u+ph+qh2 + &c. Let u=Fx, U= F(x+h), therefore v+k+qk2 + &c. v=fy,v=f(y+k) S and assuming that x and y may be considered as dependent quantities, and taking the ratios in their limit, we have du do dy d.uv + u dx dx dy dx - therefore 16. Required to differentiate the quotient of two functions of different variables. 17. The demonstration of these rules may easily be extended to any number of functions containing any number of variables. Thus let u= F(x, y) and let it be required to show that Sd.uv udv+vdu. v=f% CHAPTER IV. On the different orders of Fluxions; the developement of Functions into series; and successive differentiations. 1. If a variable increases uniformly, its increment being a constant quantity, admits not of a fluxion (1.9); but if it increases with an accelerated or a diminished velocity, its increment is not constant, and it must have a fluxion which is to be calculated upon the same principles and by the same rules as the fluxion of the variable itself. If the increment of the increment be not constant there will be a third fluxion; and thus there arise different orders of fluxions, which should be represented by appropriate notation. The principal variable is in general supposed to flow uniformly, and the fluxions of the function are calculated on that supposition. Thus take 2, the function of x; then if x increase by the numbers 1, 3, 5, 7, · its increment being constant, it cannot have a second fluxion; but a2 increases according to the numbers 1, 9, 25, 49, ...· ·, and its successive increments are 8, 16, 24,, and consequently r2 has a second fluxion which is positive (1.9). Since on the same supposition the increment of the increment of x is a constant quantity, it does not admit of a third fluxion. 2. The fluxional coefficients of a function of one variable. Let u= fr, then it has been shown (1.33) Now, p containing the variable a, it may be differentiated; let dpqdx where q is a function of x; hence, we have dp = qdx, or if dx be supposed a constant quantity, d.du dividing by da, we have =q. By differentiating suc cessively on the same supposition, and dividing by dx, we shall have d.d.du =r, d.d.d.du =s, &c. &c. where r, s, &c. are functions of x. The operation may be continued till x ceases to enter into the function. According to the notation explained in Ch. 1. 3., d.du, or the second fluxion of u, is denoted by du; the third, fourth, nth fluxions of u are denoted by d3u, d*u.... dru. These must be distinguished from du2, du3, du.... du”, which represent different powers of du. We must also distinguish between da" and d.r", the latter of which = na"-1 dx; so also d2 y2 and d2. y2 are different, the first being the square of the second fluxion, and the other the second fluxion of the square. ... :: In Newton's notation, which till within these few years has been adopted by all the English mathematicians, the 2d, 3d, 4th fluxions are denoted by u, u, u, or by u2, u3., u*..... u".; and the different powers of the first fluxion by u•2, u·3, u1.... u•*. ... du d2u The quantities p, q, r, &c. or their equals dx' dx2' are called the 1st, 2d, 3d.... nth fluxional coefficients of the function u = =fx. If u is a function of more than one variable, its different fluxional coefficients will depend upon the suppositions that are made with respect to the variables. or the fluxional coefficients of u are equal to nax-1, n.(n − 1) ax12, n.(n−1) (n− 2)ax2¬3, &c. If n is a positive integer, this example admits of only n dru fluxions, for the nth coefficient is dzn.(n-1) (n-2).... 3.2.1 x a, which is a constant quantity. Er. 2. Required the nth fluxion of pq, p and q being functions of the same variable x. Let upq.. du = pdq + qdp .. dp + 9 d x dx dx du dq P dr deu d3q P dx + = .. differentiating and dividing by dæ, 2dp dq qd2 p + dx dx dx2 d2u = pd2q + 2dp dq+qd2.p. or Similarly it may be shown that d3u = pd3q + 3d3q dp + 3dp dq+qd3p; and by the method of induction it may be demonstrated that d"u = pd"q+ nd"—1q dp + n. n-1 2 dn-2 q dip + &c. Ex. 3. du=ydx-xdy ..d2u=dxdy+yd2x-dxdy - xd3y =yd2x-xd2y=-xdy if dr be supposed constant; or yder if dy be constant. Here we may not suppose dy constant, because d'y enters into the function; and making dr constant, we have 3. The partial fluxional coefficients of a function of the sum of two variables are equal to one another. &c. Let u= f(x+y), then shall = &c. du du du du d3u d3u dx dy dx dy dx3- dy For in the expansion of u= f(x+y+h), it can make no difference in the result whether we suppose h to be the increment of x or of y, hence the coefficient of the second term du du is the same on either supposition, or H |