And again, let us suppose the functions, and their first derived, to vanish when x=xo, y = yo, and the second derived not all to vanish, then It will be observed that I have assumed the da's and the dy's to be the same in both functions: and the result must necessarily involve the ratio of these to each other. If however dx = m dy, where m is a constant multiplier, the result will be independent of the infinitesimal variations. The following is a case in which the preceding process of evaluation is frequently required. Let F(x, y) = c be an implicit function of two variables; whence after differentiation we have and suppose to, yo to be values of x and y which simultaneously satisfy the given equation, and are also such that dr dr and dy dy 0 = 0; then becomes indeterminate, and is to be evaluated according to the preceding method. Let us take the total differentials of the numerator and the denominator; then (44) becomes (dr) dy and let us suppose all these second derived-functions not to vanish nor to become infinite when x = x0 and y = yo; whence the ratio of dy to dx may be determined. then we (46) Again, if all the second derived-functions vanish when x = x0 and y = yo, then taking the total differentials of the numerator and the denominator of (45), we have and let us suppose all these third derived-functions not to vanish nor to become infinite when xx。 and y = (47) becomes yo; then dy 0 whence may be found by the solution of a cubic equation. dx It will be observed that in the total differentials of the numerator and denominator in (42) x and y are equicrescent. dy x=0 and y = 0; it is required to determine the values of Ex. 2. If x+ay3-2axy2-3 ax2y = 0, dy = 4x3-6axy-2 ay2 0 = dx 3ax+4axy-3ay2 0 it is required to determine the values of dy dx dx if x = y = 0; if x = y = 0, (12x2 -6 ay) dx − (6ax +4ay) dy (6 ax+4ay) dx+(4ax-6ay) dy = 0 dy has to dx Other examples will occur subsequently in which be evaluated in a similar way, when x = y = ∞∞. 140.] Again, let it be required to find in ascending powers of h and k the value of F(x+h, y+k), F(x, y) and all its partial derived-functions up to those of the nth order inclusive being finite and continuous for all values of x and y between the values x and x+h, y and y+k; h and k being finite increments of x and y. Let us make the same substitutions in terms of t as in Art. 137. Then by Maclaurin's Theorem, Art. 58, (13), we have and substituting for ƒ(t), ƒ(0), ƒ′(0), ƒ"(0), ..... ƒ”(t) the values given in Art. 137, except that we will replace x and yo by the general values x and y, we have which is the required development. The following are examples in which the expansion is applied. 141.] Ex. 1. Given that F(x, y) = x2 (a+y)3; it is required to find (x+h)2 (a+y+k)3. and omitting to write down the subsequent partial derived functions which vanish, we have (; d5 d4F dr2 dy3) = 12 (a+y), (dx dy3 = 12x, (d) = 12; whence, substituting in equation (51), (x + h)2 (a + y + k)3 = x2 (a + y)3 + 2x (a + y)3 h + 3x2 (a + y)2 k 3 +(a + y)3 h2 + 6x (a+y)2 hk + 3x2 (a + y) k2 +3(a+y)2 h3 k+6x (a+y) hk2 + x2 k3 +3(a+y) h2 k2 + 2x hk3 + h2 k3. Ex. 2. The equation to a conic is 2 F(x, y) = ▲ x2 +вxy+cy2+ Ex+GY ÷ K = 0; that (dr) = 0, (dr) = 0, are the it is required to prove that equa tions to two diameters which are conjugate to diameters parallel are two equations of the first degree in x and y, and therefore represent straight lines. Let the origin of coordinates be moved to a point (h, k) by replacing x and y by x+h and y+k respectively. And let it be observed that the given equation is only of two dimensions in terms of x and y; and that therefore all the partial derived-functions of it after the second vanish. In F(x, y) let x and y be replaced by x+h and y+k; and let us expand in powers of x and y; and we have Now if h and k are such that (1) = 0, (52) has no term in dh volving the first power of r, and therefore is not changed when - is substituted for + x. And dr dh =2Ah+вk+E, which is an expression of one dimension in terms of h and k, and therefore represents a straight line, of which h and k are the current coordinates: this line therefore bisects all chords of the conic which are parallel to the axis of x, and is therefore a diameter conjugate to the diameter which is parallel to the axis dr of x. Similarly, if (dx) = 0, (52) has no term involving the first dk dr power of y; and therefore (d) = 0, which is equal to dk 2ck+Bh+G=0, and represents a straight line, is the equation to a diameter of the conic which bisects all chords parallel to the axis of y; and is therefore conjugate to the diameter which |