hence D3u = − sin (y1 + y1) (dy ̧2+2 dy, dy2+ dy ̧2) + cos (y, + y2). (d3y, + đ3⁄4y2). But hence 2 D'u = − sin (x + x2). (dx2 + 4x dx2 + 4x2 dx2) + 2 cos (x + x2) dx2, D'u dx2 sin (x + x2). (1 + 4x + 4x2) + 2 cos (x + x2). We might of course have obtained this result by first giving y, and y, their values in the expression for u and then differentiating : thus We may remark that, when u is expressed entirely in x, as in the latter method of differentiating u, the expressions du dx d'u dx2, u = sin (y,+ y2), and are both zero, being in fact the first and second partial differential coefficients of u with respect to x, a letter not appearing in sin (y, + y2). Successive Differentiation of an implicit Function of a single Variable. 50. Suppose that v = (x, y) = 0, φ y being thus an implicit function of x. Dv variable, we have, now taking the place of v, dx differential coefficients of v, and therefore, from (2), we may in terms of the first and second order of these coeffi cients. The partial differential coefficients of v may be obtained dy d'y in terms of x and y: hence dx' dx2, may be found in terms of these two letters. We might proceed in the same way, by successive differentiation, to determine the third, fourth, fifth, &c. differential coefficients of y. The principles of this article the student will have no difficulty in extending to the successive differentiation of the system of equations considered in Art. (23). Successive Total Differentials. 51. Let u = ƒ (y,, y1), y, and y, being independent variables: then, by Art. (24), But, du being a function of y1, y2, and a constant dy,, we have, by virtue of (1), du occupying the place of u, Proceeding in the same way, we easily see that D3u = d3 ̧u + 3 d3⁄4Â ̧Ã ̧ ̧u + 3 d ̧ ̧ď3⁄4 ̧u + đ3⁄4‚μ‚ and so on, the law of the symbols of differentiation corresponding to the development of the binomial theorem: thus 29 COR. If u = ƒ (y1, Y2, Y ̧‚.......Ym), it may be proved in a similar way, viz. by induction, that Du = (d11 + dy2+ dyz +. ....+dy)" u. ..... m We may however establish this proposition by the following reasoning. By Art. (24), which shews that the symbol of operation D is equal to the sum of the operative symbols d, dy, dyz,. d hence D3u =D (dy2+ dy2 + dyz +....+ d) u = ; ym (d11 + dy2+ d +.....dÿm) (dy1 + dy2+ dyž +...+ dym) u .(3). 3 .... ... But the symbols d11, dv, d dy are subject inter se to all the laws of combination which belong to symbols of quantity: thus for instance hence the product of the operative polynomials in (3) is equivalent to (dy, + d + dy +....+ dym)2: proceeding successively to the higher orders to total differentials, we get the general formula D'u = (dv1 + dv2 + dy2 +....+dy)" U. u. Successive Differentiation of an explicit Function of three Variables one of which is a Function of the other two. 52. Suppose that u = f (x, y, z), z being a function of x and y, two independent variables. Then, by Art. (25), Du = du du dz + dx dx dz dx Du du du dz Differentiating (1) with regard to x, we have being a function of x, y, z, we have, by virtue of (1), putting in place of u, similarly Also, since z is a function of x and y alone, and since the dz and y, only so far as the variation of is affected by the varia dx tion of x when y remains constant, it is plain that d2z where is the second partial differential coefficient of z with dx2 + + dz du d2z + dydz dx dxdz dy dz dx dy dz dxdy' hence D'u d'u dydz dx |