In order to discover this, it is manifest that we must substitute x + h for x in each of the two functions u, and 2, and find the co-efficient of the simple power of h in the developement of the compound function F (u, z). When a becomes x + h let u become u + k, and let z become z + 1. = 4(x) + Ah + B'h2 + . . But u = (x) .. k = A'h + B'h2 + ... (2) (3) It now remains for us to substitute u + k, z + l, for u and z in F (u, z), but it is manifest that if we make these two substitutions in succession, we shall obtain the same result as if we make them both at once, since u and z are considered altogether independent of each other in these substitutions. Let us then, in the first instance, suppose that u becomes u+k, and that z remains constant in the equation y = F (u, z). ...... (4) y' or F (u+k, z) = F (u, z) + A‚ k + B ̧ k2 + Let us now suppose that z becomes z + 1, and that u remains constant, then F (u, z) becomes F (u, z + 1) = F (u, z) + A, 1 + B, 1a + (5) Since A, k involves z, it now becomes a function of z + 1, and being expanded as a function of (A, k + !) A, k becomes = A, k + terms in kl, kľ3, k2l ... B1 k2... =B, k2 + terms in k2l, Substitute then these values of F (u, z), A, k, B, k2, in (4), and we have y' or F (u+k, z + 1) : = F(u, z) + A1 k + A2 l + terms in kl, k2, l2, Substitute for k and 7 their values from (2) and (3) .. y' = F (u, z) + A1 (A'h + B'h2 + · · ·) + A2 (A′′h + B′′ h2 + . . .) + ·· Arranging according to powers of h = F (u, z) + (A, A' + A2 A ́) h + Ph2 + Qh3 + Hence by definition 2 In like manner it might be proved that if y = f (t, u, z) where t = F (x), u = 4 (x), z = ↓ (x) dy dt dy du dy dz + + and so for any number of functions. Hence we deduce the following general conclusion: The first differential co-efficient of a function composed of different particular functions, will be sum of the first differential co-efficients of each of these functions considered separately and independent of each other, according to the rule established in the last article. This principle, combined with the preceding one, will enable us to determine the first differential co-efficients of all functions of one variable, however complicated in form. Ex. 1. Let y = (ax3 + bx2 + cx + d )TM (ex+ + nx3 + rx^)". dy du dy dz dr du dx úz dx = mum-1 z1 (3ax2 + 2bx + c) + nz1--1 um (4ex3 + 5nx1 + 7rx) Substituting for u and z their values = m (ax3 + bx2 + cx + d)m−1 (ex1 +nx2 +rx ̄)" (3ax2+2bx+c) +n(ex1+nx2+rx2)"−1(ax3+bx2+cx+d)TM(4ex3+5nx++7rx®) By dint of practice, however, the student will be able to obtain the first differential co-efficients of all functions without actually performing the process of substitution. On finding the differential co-efficients of equations. We have hitherto supposed the function x to be given under the form y = f (x) but it frequently happens that y is given only by an equation between x and y of the form F(x, y) = ? The resolution indeed of this equation for y would give us y under the form y = f(x) but this solution is seldom possible, and wholly unnecessary for our present purpose. If the equation, then be of the form F(x, y) = 0 and if we suppose y = f (x) to be the value of y, which would be obtained from the solution of the equation, it is manifest that if we substitute this value for y in the proposed equation, we shall arrive at an identical equation u or F {x, ƒ (x)} = 0 whatever may be the value of x, and hence, if we substitute x + h for x, the equation will still be identically =0, whatever may be the value of h. whatever be the value of h, hence necessarily each individual term must be u = 0, Ah = 0, Bh2 = 0 . . . = 0, and But since .. ... du that x was variable and y constant, and then the differential co-efficient taken the supposition that y was variable and ≈ constant, hence dx dy ON FINDING THE SUCCESSIVE DIFFERENTIAL CO-EFFICIENTS OF FUNCTIONS OF ONE VARIABLE. The first differential co-efficient of any function is itself a new function of the variable, and consequently its differential co-efficient may be found according to principles already explained. This differential co-efficient of a differential co-efficient is called the second differential co-efficient of the original function, and if the first differential co-efficient be expressed by the symbol dx the second differential co-efficient is represented by day dy In like manner this second differential co-efficient is itself a new function of the variable and its differential co-efficient may be found, this is called the third differential co-efficient of the original function, and is represented by the symbol d3y Proceeding in the same manner, the differential co-efficient of is called dx3 the fourth differential co-efficient of the original function, and is written d'y da and so on to any extent. Thus if Ex. 1. y = ax + bx + cx2 + dx2 + ex2 + gx + m The first differential co-efficient is dy dx = 6ax+5bx2 + 4cx3 + 3dx2 + 2x + g (1) In order to find the second differential co-efficient of y, we must take the first differential co-efficient of this new function (1), which will be 5.6. ax + 4. 5 bx3 + 3 . 4 cx2 + 2. 3 dx + 1.2.e = 5. 6. ax1 + 4. 5 bx3 + 3.4 cx2 + 2.3. dx + 1. 2. e ... (2) Taking the first differential co-efficient of this new function (2), we shall have d3y ძებ = 4. 5. 6. (x3 + 3. 4. 5 bx2 + 2. 3. 4. cx + 1. 2. 3 d In like manner d'y dx4 d3y dx5 doy dx6 = 3. 4. 5. 6. ax2 + . 2. 3. 4 . 5 bx + 1 . 2.3.4.c |