Let us now develop according to d Alembert s method: a) f(u)98 or y = 3u ; b) f(x) or u = x + ax . y = 3u , (1) f(u) = 3u . (1a) f(u + h) = 3(u + h) , f(u + h) - f(u) = 3(u + h) - 3u = 3u + 6uh + 3h - 3u = 6uh + 3h (2) (f(u + h) - f(u))/h = 6u + 3h . f (u) = 6u, already given complete in (2), is freed of its factor h by means of division. (f(u + 0) - f(u))/0 = 6u , (y1 - y)/(u1 - u) , alias 0/0 = dy/dx = 6u . dy/du = 6(x + ax ) . Since y in a) is differentiated with respect to u, thus (u1 - u) = h or h = (u1 - u) , And so: dy/du = 6(x + ax ) . [We now develop b) in the same manner, so that] f(x + h) = (x + h) + a(x + h) , f(x + h) - f(x) = (x + h) + a(x + h) - x - ax = x + 3x h + 3xh + h + ax + 2axh + ah { - x || - ax = (3x + 2ax)h + (3x + a)h + h , (f(x + h) - f(x))/h = 3x + 2ax + (3x + a)h + h . If we now set h = 0, on the second side: 0/0 or du/dx = 3x + 2ax . The derived function is already contained complete however, in f(x + h) = (x + h) + a(x + h) , x + 3x h + 3xh + h + ax + 2axh + ah . x + ax + (3x + 2ax)h + (3x + a)h + h . The rest of the procedures serve only to liberate the f (x) thus given from the very beginning from its own coefficient h and from all other terms. The equation (f(x + h) - f(x))/h = etc. provides two things: first, it makes it possible to obtain the numerator on the first side as the difference of f(x), presently = f(x); on the second side, however, it provides the algebraic opportunity to extract the original function given in x, x + ax , from the product of (x + h) + a(x + h) etc. ___________ So we continue. We have obtained for a): dy/du = 6(x + ax ) , du/dx = 3x + 2ax . We multiply dy/du by du/dx, so that dy/du du/dx = dy/dx , dy/dx = 6(x + ax )(3x + 2ax) y = f(u); dy/du = df(u)/du, u = f(x); du/dx = df(x)/dx , dy/du du/dx or dy/dx = df(u)/du df(x)/dx If we now substitute h = u1 - u into equation a) and h = x1 - x into equation b), things are so arranged that: y or f(u) = 3u , f(u + (u1 - u)) = 3(u + (u1 - u)) , = 3u + 6u(u1 - u) + 3(u1 - u) , f(u + (u1 - u)) - f(u) = 3u + 6u(u1 - u) + 3(u1 - u)(u1 - u) - 3u , f(u + (u1 - u)) - f(u) = 6u(u1 - u) + 3(u1 - u) , (f(u + (u1 - u)) - f(u))/(u1 - u) = 6u + 3(u1 - u) . dy/du = 6u + 0 = 6u . ____________ This shows that when f(u) from the very beginning becomes f(u + (u1 - u)), then its increment appears as the positive second term of a defined binomial on the second side, and this second term, which is multiplied by (u1 - u) or h by the binomial theorem, immediately becomes the coefficient to be found. If the second term is polynomial, as it is in x + ax , which becomes (x + h) + a(x + h) , (x + (x1 - x)) + a(x + (x1 - x)) , This result shows: 1) that when in d Alembert s development x1 - x = h is put in reverse h = x1 - x, thereby absolutely nothing is changed in the method itself, rather the method simply brings out more clearly how to obtain the binomial by means of f(x + h) or f(x + (x1 - x)) for the algebraic expression on the other side in place of the original function, in place of 3u for example in the given case. The second term which one finds in that manner attached to h or (x1 - x) is the complete first derived function. The problem now consists of freeing it of h or x1 - x, which is easily done. There the derived function is complete; it is therefore not found by setting x1 - x = 0, but rather freed of its factor (x1 - x) and accessories. Just as it is found by simple multiplication (the binomial development) as the second term [with] x1 - x, so it is finally freed of the latter by means of division of both sides by x1 - x. The crucial procedure (Mittelprozedur), however, consists of the development of the equation f(x + h) - f(x) or f(x + (x1 - x)) - f(x) = [...]. Therefore what happens, for example, in (x + h) + a(x + h) - x - ax , The first differentiation on the second side is nothing but the simple subtraction of the original function from its increased expression, which thus gives us the increment by which it has increased and whose first term, multiplied by h, is already the complete derived function. The other terms can only contain h etc. or (x1 - x) etc. as coefficients; they are reduced by one power with the first division of both sides by x1 - x, while the first term emerges without any h. 2) The difference from the method of f(x1) - f(x) = etc. lies in the fact that, when we have for example f(x) or u = x + ax , f(x1) or u1 = x1 + ax1 , f(x1) - f(x) or u1 - u = x1 + ax1 - (x + ax ) . Here by no means is it a matter of extracting the original function again, since x1 + ax1 does not contain x and ax in any form. On the contrary, this first difference equation provides us with an opportunity for development (Entwicklungsmoment), namely the transformation of both of the two original terms into differences of [power of] x1 and x. Namely, = (x1 - x ) + a(x1 - x ) . It is now clear that when we again resolve both of these two terms into factors of x1 - x, we obtain functions in x1 and x as coefficients of x1 - x, namely: f(x1) - f(x) or u1 - u = (x1 - x) (x +x1x + x ) + a(x1 - x) (x1 + x) . (f(x1) - f(x))/(x1 - x) or (u1 - u)/(x1 - x) = (x1 + x1x + x ) + a(x1 + x) . Thus can finally obtain the first derived function in x only when we set x1 = x, so that x1 - x = 0, and then x1 = x , x1x = x , x(x1 + x1x + x ) = 3x and x1 + x = x + x = 2x ; a(2x) = 2xa . df(x)/dx = du/dx = 0/0 . Thus the derived function is here only obtained by setting x1 = x, so that x1 - x = 0. x1 = x provides the final positive result in the real function of x. But x1 = x also leads to x1 - x = 0 and therefore at the same time, beside this positive result, to the symbolic 0/0 or dy/dx on the other side. We could have said from the very beginning: we have to obtain a derivative in x, and x in the end. This can only be transformed into the derivative in x when x1 is set = x; but setting x1 = x is the same as setting x1 - x = 0, which nullification is positively expressed by the formula x1 = x which is necessary for the transformation of the derivative to a function of x, while its negative form, x1 - x = 0, must provide us with the symbol. 3) Even if this treatment of x, where an increment (x1 - x = x, for example, or h) is not independently introduced next to it, was already well-known, something which is very probable and of which I shall convince myself by consulting J[ohn] Landen at the [British] Museum, still its essential difference cannot have been grasped. What distinguishes this method from Lagrange, however, is that it really differentiates, so that the differential expression also originates on the symbolic side, while with him the derivation does not represent the differentiation algebraically, but instead derives the functions algebraically directly from the binomial and simply accepts their differential form by symmetry , since it is known from differential calculus that the first derived function = dy/dx, the second = d y/dx ;, etc.
Marx's Mathematical Manuscripts 1881