1) Let f(x) or y = uz be a function to be differentiated; u and z are both functions dependent on the independent variable x. They are independent variables with respect to the function y, which depends on them, and thus on x. y1 = u1z1, y1 - y = u1z1 - uz = z1(u1 - u) + u(z1 - z), (y1 - y)/(x1 - x) or y/ x = z1 (u1 - u)/(x1 - x) + u (z1 - z)/(x1 - x) = (z1 u)/( x) + (u z)/( x) .* Now on the right-hand side let x1 = x, so that x1 - x = 0, likewise u1 - u = 0, z1 - z = 0; so that the factor z1 in z1 (u1 - u)/(x1 - x) also goes to z; finally on the left-hand side y1 - y = 0. Therefore: A) dy/dx = z du/dx + u dz/dx . Which equation, when all its terms are multiplied by the common denominator dx, becomes B) dy or d(uz) = z du + u dz. 14 2) Consider for the time being the first equation A): dy/dx = z du/dx + u dz/dx . In equations with only one variable dependent on x, the final result has always been dy/dx = f (x) , and f (x), the first derived function*2 of f(x), has been free15 of all symbolic expressions, for example, mxm-1 when xm is the original function of the independent variable x. As a direct result of the process of differentiation which f(x) had to pass through in order to be transformed into f (x), its shadow image (Doppelg nger) 0/0 or dy/dx appeared as the symbolic equivalent on the left-hand side opposite f (x), the real differential coefficient.16 Alternately 0/0 or dy/dx found its real equivalent in f (x). In equation A) by contrast, f (x), the first derivative of uz, itself includes symbolic differential coefficients, which are therefore present on both sides while on neither is there a real value. Since, however, uz has been handled in the same manner as the earlier functions of x with only one independent variable, this contrast is obviously a result of the peculiar character of the beginning function itself, namely uz. A more complete treatment of this is found under 3). For the moment, it remains to be seen whether there are any twists in the derivation of equation A). On the right-hand side (u1 - u)/(x1 - x) or u/ x and (z1 - z)/(x1 - x) or z/ x If we place the primitive problematic form of du/dx, dz/dx on the right-hand side it becomes: z 0/0 + u 0/0. If we then multiply z and u by the numerators of the 0/0 accompanying them, we obtain: 0/0 + 0/0; and since the variables z and u themselves become = 0,17 as are their derivatives as well, so that [we obtain] finally: 0/0 = 0 and not z du/dx + u dz/dx . This procedure, however, is mathematically false. Let us take, for example (u1 - u)/(x1 - x) = u/ x ; Therefore what arises opposite the variables u and z is not 0 but (0/0), whose numerator in this form remains inseparable from its denominator. Consequently as a multiplier 0/0 then could nullify its coefficients only when and so far as 0/0 = 0. Even in the usual algebra it would be false, in the case where a product P m/n takes the form P 0/0 , to conclude immediately that it must be = 0 , although it may be set always = 0 here, since we can begin18 the nullification arbitrarily with numerator or denominator. For example, P (x - a )/(x - a). Let [because x = a] x be set = a , so that x - a = 0; we then obtain: P 0/0 = 0/0, and the last [term] may be set = 0, since 0/0 can just as readily be 0 as any other number. By contrast, let us reduce x - a to its factors, so that we obtain P.((x - a)(x + a))/(x - a) = P(x + a), and since x = a,19 = 2Pa . In our case, however, where the origin of 0/0, 0/0 is known to be the differential expression of z/ x, u/ x respectively, the two deserve, as above, the uniform (die Uniform) dz/dx , du/dx . 3) In the equations, such as y = xm, y = ax etc., which have been treated previously, an original function of x stands opposite a y dependent on it. In y = uz, both sides contain dependent [variables] . While here y depends directly on u and z, so in turn u and z [depend] as well on x. This specific character of the original function uz necessarily stamps on its derivatives as well. That u is a function of x, and z another function of x is represented by: u = f(x), u1 - u = f(x1) - f(x) , z = (x); z1 - z = (x1) - (x) . But neither the beginning equation for f(x) nor for (x) leads to an original function of x, that is, a definite value*3 in x. Consequently u and z figure as mere names, as symbols of functions of x; therefore as well only the general forms of this ratio of dependence (Abh ngigkeitsverh ltnis) : (u1 - u)/(x1 - x) = (f(x1) - f(x))/(x1 - x) , (z1 - z)/(x1 - x) = ( (x1) - (x))/(x1 - x) du/dx = df(x)/dx , dz/dx = d (x)/dx , and the symbolic differential coefficients du/dx , dz/dx become as such incorporated into the derivatives . In equations with only one dependent variable, dy/dx has no other content at all than du/dx , dz/dx have here. It is also merely the symbolic differential expression of (y1 - y)/(x1 - x) = (f(x1) - f(x))/(x1 - x).20 Although the nature of du/dx , dz/dx - that is, of symbolic coefficients in general - is in no way altered when they appear within the derivative itself, and so on the right-hand side of the differential equation as well, nonetheless their role and the character of the equation are thereby altered. Let us represent the original function of uz, in combination, by f(x), and their first derivative by f (x), dy/dx = z du/dx + u dz/dx dy/dx = f (x) . We have obtained this very general form for equations with only one dependent variable. In both cases the beginning forms of dy/dx arose from the process of taking the derivative (Ableitungsprozesse), which transforms f(x) into f (x). So soon, therefore, as f(x) becomes f (x), dy/dx stands opposite the latter as its own symbolic expression, as its shadow image (Dopelg nger) or symbolic equivalent. In both cases, therefore, dy/dx plays the same role. It is otherwise with du/dx , dz/dx. Together with the other elements of f (x), into which they are incorporated, in dy/dx they meet with their symbolic expression or their symbolic equivalent, but they themselves do not stand opposite the f (x), (x) whose symbolic shadow images they would be in turn. They are brought into the world unilaterally, shadow figures lacking the body which cast them, symbolic differential coefficients without the real differential coefficients, that is, without the corresponding equivalent derivative . Thus the symbolic differential coefficient becomes the autonomous starting point whose real equivalent is first to be found. The initiative is thus shifted from the right-hand pole, the algebraic, to the left-hand one, the symbolic. Thereby, however, the differential calculus also appears as a specific type of calculation which already operates independently on its own ground (Boden). For its starting points du/dx , dz/dx belong only to it and are mathematical quantities characteristic of it. And this inversion of the method arose as a result of the algebraic differentiation of uz. The algebraic method therefore inverts itself into its exact opposite, the differential method.*4 Now, what are the corresponding derivatives of the symbolic differential coefficients du/dx , dz/dx ? The beginning equation y = uz provides no data for the resolution of this question. This last [question] may still be answered if one substitutes arbitrary original functions of x for u and z. For example, u = x4 ; z = x + ax . Thereby, however, the symbolic differential coefficients du/dx , dz/dx are suddenly transformed into operational symbols (Operationssymbole), into symbols of the process which must be carried out with x4 and x + ax in order to find their derivatives . Originally having arisen as the symbolic expression of the derivative and thus already finished, the symbolic differential coefficient now plays the rolw of the symbol of the operation of differentiation which is yet to be completed. At the same time the equation dy/dx = z du/dx + u dz/dx , I remark further that*5 from the early part of the 18th century right down to the present day, the general task of the differential calculus has usually been formulated as follows: to find the real equivalent of the symbolic differential coefficient. 4) A) dy/dx = z du/dx + u dz/dx . This is obviously not the simplest expression of equation A), since all its terms have the denominator dx in common. Let this be struck out, and then: B) d(uz) or dy = z du + u dz . Any trace in B) of its origin in A) has disappeared. It is therefore equally as valid when u and z depend on x as when they depend only reciprocally on one another, without any relation to x at all.21 From the beginning it has been a symbolic equation and from the beginning could have served as a symbolic operational equation. In the present case it means, that when For our purpose, namely the further investigation of the differential of y in general, form B) nonetheless will not do. We therefore set: u = x4, z = x + ax . du = 4x dx, dz = (3x + 2ax)dx , A) dy/dx = (x + ax )(4x dx)/dx + x4(3x +2ax)dx/dx ; and then dy/dx = (x + ax )4x + x4(3x + 2ax) ; dy = {(x + ax )4x + x4(3x + 2ax)}dx . dy = f (x)dx . We have already obtained the same result from an arbitrary equation with only one variable. For example: y = xm, dy/dx = mxm-1 = f (x) , dy = f (x)dx . In general we have if y = f(x), whether this function of x is now an original function in x or contains a dependent variable, then always dy = df(x) and df(x) = f (x)dx, and so: B) dy = f (x)dx is the most generally valid form of the differential of y. This would be demonstrable immediately also if given f(x) were f(x, z), that is a function of two mutually independent variables. For our purposes, however, this would be superfluous. 1) The differential dy = f (x)dx dy/dx = f (x) In dy/dx = 0/0 the numerator and denominator are inseparably bound; in dy = f (x)dx they are apparently separated, so that one is forced to the conclusion that it is only a disguised expression for 0 = f (x) 0 or 0 = 0 , A French mathematician of the first third of the 19th century, who is clear in a completely different manner than the well-known [to you] elegant Frenchman,22 has drawn a connection between the differential method and Lagrange s algebraic method: - Boucharlat says: If for example dy/dx = 3x , then dy/dx alias 0/0, or rather its value 3x , is the differential coefficient of the function y. Since dy/dx is thus the symbol which represents the value 3x , dx must always stay (stehn)*6 under dy, but in order to facilitate algebraic operation we treat dy/dx as an ordinary fraction and dy/dx = 3x as an ordinary equation. By removing the denominator from the equation one obtains the result dy = 3x dx, Thus in order to facilitate algebraic operation , one introduces a demonstrably false formula which one baptises the differential . In fact the situation is now so nasty. In 0*7/0 the numerator is inseparable from the denominator, but why? Because both only express a ratio if they are not separated, something like (dans l esp ce) the ratio24 reduced to its absolute minimum: (y1 - y)/(x1 - x) = (f(x1) - f(x))/(x1 - x) , As soon, however, as x1 - x = 0 achieves in dx a form which is manifested without modification as the vanished difference in the independent variable x, so that dy as well is a vanished difference in the function of x or in the dependent [variable] y, then the separation of the denominator from the numerator becomes a completely permissible operation. Wherever dx stands now, such a change of position leaves the ratio of dy to dx undisturbed, dy = f (x)dx thus appears to us to be an alternative form of dy/dx = f (x) 2) The differential dy = f (x)dx arose from A) by means of a direct algebraic derivation (see I,4), while the algebraic derivation of equation A) had already shown that the differential symbol, somewhat like (dans l esp ce) the symbolic differential coefficient which originally emerged as a purely symbolic expression of the algebraically performed process of differentiation, necessarily inverts into an independent starting point, into a symbol of an operation yet to be performed, into an operational symbol, and thus the symbolic equations which have emerged along the algebraic route also invert into symbolic operational equations (Operationsgleichungen). We are thus doubly correct in treating the differential y = f (x)dx as a symbolic operational equation. So we now know a priori, that if y = f(x) [then] dy = df(x) , dy/dx = f (x) . As well, however, from the first moment that the differential functions as the starting point of the calculus, the inversion of the algebraic method of differentiation is complete, and the differential calculus itself therefore appears, a unique, specific method of calculating with variable quantities. In order to make this more graphic I will combine at once all the algebraic methods which I have used, while setting simply f(x) in place of a fixed algebraic expression in x, and the preliminary derivative (see the first manuscript*8) will be designated as f (x) to distinguish from the definitive derivative , f (x). Then, if f(x) = y, f(x1) = y1 , f(x1) - f(x) = y1 - y or y , f (x) (x1 - x) = y1 - y or y. The preliminary derivative must contain expressions in x1 and x exactly like the factor (x1 - x) with the single exception when f(x) is an original function to the first power: f (x) = (y1 - y)/(x1 - x) or y/ x . We now substitute into f (x) x1 = x so that x1 - x = 0 , f (x) = 0/0 or dy/dx and finally f (x)dx = dy or dy = f (x)dx The differential of y is therefore the conclusion of an algebraic development; it becomes the starting point for differential calculus operating on its own ground. dy, the differential26 of y - considered in isolation, that is, without its [real] equivalent - here immediately plays the same role as y in the algebraic method; and the differential of x, dx, the same role as x does there. If we had, in y/ x = f (x) I) y = f (x) x . On the other hand, beginning with the differential calculus as a separate, complete type of calculating - and this point of departure has been itself derived algebraically - we start immediately with the differential expression of I), namely: II) dy = f (x)dx . 3) Since the symbolic differential equation (Gleichung des Differentials) arises simply by the algebraic handling of the most elementary functions with only one independent variable, it appears that the inversion of the method (Umschlag in der Methode) could have been developed in a much more simple manner than happened with the example y = uz . The most elementary functions are those of the first degree; they are: a) y = x, which leads to the differential coefficient dy/dx = 1, so that the differential is dy = dx. b) y = x ab; it leads to the differential coefficient dy/dx = 1, so that again the differential is dy = dx. c) y = ax; it leads to the differential coefficient dy/dx = a, so that the differential is dy = adx. Let us take the simplest case of all (under a)). Then: y = x, y1 = x1 ; y = uz . y1 - y or y = x1 - x or x . I) (y1 - y)/(x1 - x) or y/ x = 1. Thus also y = x. In y/ x x1 is now set = x, or x1 - x = 0, and thus: II) 0/0 or dy/dx = 1; so that dy = dx. Right at the start, as soon, as we obtain I) y/ x = 1, we are forced to operate further on the left-hand side, since on the right-hand side is the constant, 1. And therein the inversion of the method, which throws the initiative from the right-hand side to the left-hand side, once and for all from the group up proves to be in fact the first word of the algebraic method itself. Let s look at the matter more closely. The real result was: I) y/ x = 1. II) 0/0 or dy/dx = 1. Since both I) and II) lead to the same result we may choose between them. The setting of x1 - x = 0 appears in any case to be a superfluous and therefore an arbitrary operation. Further: we operate from here on in II) on the left-hand side, since on the right-hand side ain t no way , so that we obtain: 0/0 or d y/dx = 0 . The final conclusion would be that 0/0 = 0, so that the method is erroneous with which 0/0 was obtained. At the first use*9 it leads to nothing new, and at the second to exactly nothing.27 Finally: we know from algebra that if the second sides of two equations are identical, so also must the first sides be. It therefore follows that: dy/dx = y/ x . For its part the differential dy = dx has no meaning, or more correctly only as much meaning as we have discovered for both differentials in the analysis of dy/dx. Were we to accept the interpretation just given,28 we could then perform miraculous operations with the differential, such as for example showing the role of adx in the determination of the subtangent of the parabola, which by no means requires that the nature of dx and dy really be understood. 4) Before I proceed to section III, which sketches the historical path of development of the differential calculus on an extremely condensed scale, here is one more example of the algebraic method applied previously. In order graphically to distinguish it I will place the given function on the left-hand side, which will always be the side of the initiative, since we always write from left to right, so that the general equation is: xm + Pxm-1 + etc. + Tx + U = 0 , 0 = xm + Pxm-1 + etc. + Tx + U . If the function y and the independent variable x are divided into two equations, of which the first expresses y as a function of the variable u, while on the other hand the second expresses u as a function of x, then both symbolic differential coefficients in combination are to be found.29 Assuming: 1) 3u = y , 3u1 = y1 , then 2) x + ax = u ; x1 + ax1 = u1 . We deal with equation 1) for the present: 3u1 - 3u = y1 - y , 3(u1 - u ) = y1 - y , 3(u1 - u) (u1 + u) = y1 - y , 3(u1 + u) = (y1 - y)/(u1 - u) or y/ u . On the left-hand side u1 is now set = u, so that u1 - u = 0, then: 3(u + u) = dy/du , 3(2u) = dy/du , 6u = dy/du . We now substitute for u its value x + ax , so that: 3) 6(x + ax ) = dy/du . Now applying ourselves to equation 2): x1 + ax1 - x - ax = u1 - u , (x1 - x ) + a(x1 - x ) = u1 - u , (x1 - x) (x1 + x1x + x ) + a(x1 - x) (x1 + x) = u1 - u , (x1 + x1x + x ) + a(x1 + x) = (u1 - u)/(x1 - x) or u/ x . We set x1 = x on the left-hand side, so that x1 - x = 0 . (x + xx + x ) + a(x + x) = du/dx . 4) 3x + 2ax = du/dx . We now multiply equations 3) and 4) together, so that: 5) 6(x + ax ) (3x + 2ax) = dy/du du/dx = dy/dx .30 Thus, by algebraic means the operational formula: dy/dx = dy/du du/dx , The above example shows that it is not witchcraft to transform a development demonstrated from given functions into a completely general form. Assume: so that therefore From the difference under 1) comes: (y1 - y)/(u1 - u) = (f(u1) - f(u))/(u1 - u) ; dy/du = df(u)/du , dy/du = f (u)du/du ; 3) dy/du = f (u) . From the difference under 2) follows: (u1 - u)/(x1 - x) = ( (x1) - (x))/(x1 - x) , du/dx = d (x)/dx , du/dx = (x)dx/dx , 4) du/dx = (x) . We multiply equation 3) by 4), so that: 5) dy/du du/dx or dy/dx31 = f (u) (x) Q.E.D. N. III. The conclusion of this second instalment will follow, as soon as I consult John Landen at the Museum.32
Marx's Mathematical Manuscripts 1881