As soon as we reach the differentiation of f(u,z)[=uz], where the variables u and z are both functions of x, we obtain - in contrast to the earlier cases which had only one dependent variable, namely, y - differential expressions on both sides, as follows: in the first instance dy/dx = z du/dx + u dz/dx ; in the second, reduced form dy = z du + u dz , At the same time, however, as soon as we have achieved this result and we therefore already operate on the ground (Boden) of differential calculus, we can reverse [the process]; if, for example, we have xm = f(x) = y dy = mxm-1dx dy/dx = mxm-1. Thus here we begin with the symbol; it no longer figures as the result of a derivation from the function [of] x; rather instead as already a symbolic expression35 which indicates which operations to perform upon f(x) in order to obtain the real value of dy/dx, i.e. f (x). In the first case 0/0 or dy/dx is obtained as the symbolic equivalent of f (x); and this is necessarily first, in order to reveal the origin of dy/dx; in the second case f (x) is obtained as the real value of the symbol dy/dx. But then, where the symbols dy/dx, d y/dx become the operational formulae (Operationsformeln) of differential calculus,36 they may as such formulae also appear on the right-hand side of the equation, as was already the case in the simplest example dy = f (x)dx. If such an equation in its final form does not immediately give us, as in this case, dy/dx = f (x), etc., then this is proof that it is an equation which simply expresses symbolically which operations are to be performed in application to defined (bestimmtem) functions. And this is the case - and the simplest possible case - immediately in d(uz), where u and z are both variables while both are also functions of the same third variable, i.e. of x.37 Given to be differentiated f(x) or y = uz, where u and z are both variables dependent on x. Then y1 = u1z1 y1 - y = u1z1 - uz . (y1 - y)/(x1 - x) = u1z1/(x1 - x) - uz/(x1 - x) y/ x = (u1z1 - uz)/(x1 - x) . u1z1 - uz = z1(u1 - u) + u(z1 - z) , z1u1 - z1u + uz1 - uz = z1u1 - uz . Therefore: (u1z1 - uz)/(x1 - x) = z1 (u1 - u)/(x1 - x) + u (z1 - z)/(x1 - x) . If now on both sides x1 - x becomes = 0, or x1 = x, then we would have u1 - u = 0, so that u1 = u, and z1 - z = 0, so that z1 = z; we therefore obtain dy/dx = z du/dx + u dz/dx d(uz) or dy = z du + u dz . At this point one may note in this differentiation of uz - in distinction to our earlier cases, where we had only one dependent variable tat here we immediately find differential symbols on both sides of the equation, namely: in the first instance dy/dx = z du/dx + u dz/dx ; in the second d(uz) or dy = z du + u dz It has been shown how, in functions with only one independent variable, from one function of x, for example f(x) = xm, a second function of x, f (x), or, in the given case mxm-1 may be derived by means of actual differentiation and subsequent cancellation alone, and at the same time how from this process the symbolic equivalent 0/0 = dy/dx for the derived function originates on the left-hand side of the equation. Further: the substitution 0/0 = dy/dx here was not only permissible but mathematically necessary. Since 0/0 in its own primitive form may have any magnitude at all, for 0/0 = X always gives 0 = 0. Here, however, 0/0 appears as the symbolic equivalent of a completely defined real value, as above, for example, mxm-1, and is itself only the result of the operations whereby this value was derived from xm; as such a result it is firmly fixed (festgehalten) in the form dy/dx. Here, therefore, where dy/dx ( = 0/0 ) is established in its origin, f (x) is by no means found by using the symbol dy/dx; rather instead of the already derived function of x. Once we have obtained this result, however, we can proceed in reverse. Given an f(x), e.g. xm, to differentiate, we then first look for the value of dy and find dy = mxm-1dx, so that dy/dx = mxm-1. Here the symbolic expression appears (figuriert) as the point of departure. [We] are thus (so) already operating on the ground of differential calculus; that is, dy/dx etc. already perform as formulae which indicate which known differential operations to apply to the function of x. In the first case dy/dx ( = 0/0 ) was obtained as the symbolic equivalent of f (x), in the second f (x) was sought and obtained as the real value of the symbols dy/dx , d y/dx , etc. These symbols having already served as operational formulae (Operationsformeln) of differential calculus, they may then also appear on the right-hand side of the equation, as already happened in the simplest case, dy = f (x)dx. If such an equation in its final form is not immediately reducible, as in the case mentioned, to dy/dx = f (x), that is to a real value, then that is proof that it is an equation which merely expresses symbolically which operations to use as soon as defined functions are treated in place of their undefined [symbols]. The simplest case where this comes in is d(uz), where u and z are both variables, but both at the same time are functions of the same 3rd variable, e.g. of x. If we have here obtained by means of the process of differentiation (Differenzierungsprozessi) (see the beginning of this in Book I, repeated on p.10 of this book*) dy/dx = x du/dx + u dz/dx , From the equation dy/dx = z du/dx + u dz/dx d(uz) or dy = z du + u dz . The simplest possible case would be, for example, u = ax, z = bx . d(uz) or dy = bx adx + ax bdx . We divide both sides by dx, so that: dy/dx = abx + bax = 2abx d y/dx = ab + ba = 2ab . If we take, however, the product from the veryh beginning, y or uz = ax bx = abx , uz or y = abx , dy/dx = 2abx , d y/dx = 2ab . As soon as we obtain a formula such as, for example, [w =] z du/dx , it is clear that the equation, what we might call *2 a general operational equation, [is] a symbolic expression of the differential operation to be performed. If for example we take [the] expression y dx/dy, where y is the ordinate and x the abscissa, then this is the general symbolic expression for the subtangent of an arbitrary curve (exactly as d(uz) = z du + u dz is the same for differentiation of the product of two variables which themselves depend on a third). So long, however, as we leave the expression as it is leads to nothing further, although we have the meaningful representation for dx, that it is the differential of the abscissa, and for dy, that it is the differential of the ordinate. In order to obtain any positive result we must first take the equation of a definite curve, which gives us a definite value for y in x and therefore for dx as well, such as, for example, y = ax, the equation of the usual parabola; and then by means of differentiation we obtain 2ydy = adx; hence dx = 2ydy/a . If we substitute this definite value for dx into the general formula for the subtangent, y dx/dy, we then obtain (y 2ydy/a)/dy = y 2ydy/ady = 2y /a , = 2ax/a = 2x , The difficulty becomes evident if we then substitute the original form 0/0 for dy/dx etc. dy/dx = z du/dx + u dz/dx 0/0 = z 0/0 + u 0/0 , 1) Even in the first exposition with one independent variable, we first obtain 0/0 or dy/dx = f (x) ; so that dy = f (x)dx . dy/dx = 0/0, dy = 0 and dx = 0, so that 0 = 0 . Although we again substitute for dy/dx its indefinite expression 0/0 we nonetheless commit here a positive mistake, for 0/0 is only found here as the symbolic equivalent of the real value f (x), and as such is fixed in the expression dy/dx ,and thus in dy = f (x)dx as well. 2) (u1 - u)/(x1 - x) becomes du/dx or 0/0 , because the variable x becomes = x1, for (u1 - u)/(x1 - x) ; we know however in general that 0/0 can have any value, and that in a specific case it has the specific value (Spezialwert) which appears as soon as a defined function of x enters for u; we are thus not only correct in substituting du/dx for 0/0 , but rather we must do it, since du/dx as well as dz/dx appear here only as symbols for the differential operations to be performed. So long as we stop with the result dy/dx = z du/dx + u dz/dx , dy = z dy + u dz , 3) In the usual algebra 0/0 can appear as the form for expressions which have a real value, even though 0/0 can be a symbol for any quantity. For example, given (x - a )/(x - a), we set x = a so that x - a = 0 and x = a , and therefore x - a = 0. We thus obtain (x - a )/(x - a) = 0/0 ; If we resolve x - a into its factors, then it = (x + a) (x - a) ; so that (x - a )/(x - a) = (x + a) (x - a)/(x - a) = x + a ; If we had the term P(x - a) in an ordinary algebraic equation, then if x = a, so that x - a = 0, then necessarily P(x - a) = P 0 = 0; just as under the same assumptions P(x - a ) = 0. The decomposition of x - a into its factors (x + a) (x - a) would change none of this, for P(x + a) (x - a) = P(x + a) 0 = 0 . By no means, however, does it therefore follow that if the term P (0/0) had been developed by setting x = a, its value must necessarily be = 0. 0/0 may have any value because 0/0 = X always leads to: 0 = X 0 = 0; but just because 0/0 may have any value it need not necessarily have the value 0, and if we are acquainted with its origin we are also able to discover a real value hidden behind it. So for example P (x - a )/(x - a) , if x = a, x - a = 0 and so as well x = a , x - a = 0; thus P.(x -a )/(x - a) = P 0/0 . Although we have obtained this result in a mathematically completely correct manner, it would nonetheless be mathematically false, however, to conclude without further ado that P 0/0 = 0, because such an assumption would imply that 0/0 may necessarily have no value other than 0, so that P 0/0 = P 0 . It would be more relevant to investigate whether any other result arises from resolving x - a into its factors (x + a) (x - a); in fact, this transforms the expression to P (x + a) (x - a)/(x - a) = P (x + a) 1, Thus it is in the equation which we obtain: d(uz)/dx or dy/dx = zdu/dx + udz/dx . The equation therefore only has the value of a general equation which indicates by means of symbols which operations to perform as soon as u and z are given respectively, as dependent variables, two defined functions of x. Only when [we] have defined functions of [x] for u and z may du/dx ( = 0/0 ) and dz/dx ( = 0/0 ) and therefore dy/dx ( = 0/0 ) as well become 0, so that the value 0/0 = 0 cannot be presumed but on the contrary must have arisen from the defined functional equation itself. Let, for example, u = x + ax ; then (0/0) = du/dx = 3x + 2ax , (0/0)1 = d u/dx = 6x + 2a , (0/0)2 = d u/dx = 6 , (0/0)3 = d4u/dx4 = 0 , The long and the short of the story is that here by means of differentiation itself we obtain the differential coefficients in their symbolic form as a result, as the value of [dy/dx in] the differential equation, namely in the equation d(uz)/dx or dy/dx z du/dx + u dz/dx . We now know, however, that u = a defined function of x, say f(x). Therefore (u1 - u)/(x1 - x) , in its differential symbol du/dx, is equal to f (x), the first derived function of f(x). Just so z = (x), say, and so similarly dz/dx = (x), ditto - of (x). The original function itself, however, provides us neither with u nor with z in any defined function of x, such as, for example u = xm, z = sqrt{x} It provides us u and z only as general expressions for any 2 arbitrary functions of x whose product is to be differentiated. The equation states that, if a product, represented by uz, of any two functions of x is to be differentiated, one is first to find the real value corresponding to the symbolic differential coefficient du/dx, that is the first derived function say of f(x), and to multiply this value by (x) = z; then similarly to find the real value of dz/dx and multiply [it] by f(x) = u; and finally to add the two products thus obtained. The operations of differential calculus are here already assumed to be well-known. The equation is thus only a symbolic indication of the operations to be performed, and at the same time the symbolic differential coefficients, du/dx, dz/dx here stand for symbols of differential operations still to be completed in any concrete case, which they themselves were originally derived as symbolic formulae for already completed differential operations. As soon as they have taken on (angenommen) this character, they may themselves become the contents of differential equation, as, for example, in Taylor s Theorem: y1 = y + (dy/dx) h + etc. But then these are also only general, symbolic operational equations. In this case of the differentiation of uz, the interest lies in the fact that it is the simplest case in which - in distinction to the development of those cases where the independent variable x has only one dependent variable y - differential symbols due to the application of the original method itself are placed as well on the right-hand side of the equation (its developed expression), so that at the same time they enter as operational symbols and as such became the contents of the equation itself. This role, in which they indicate operations to be performed and therefore serve as the point of departure, is their characteristic role in a differential calculus already operating (sich bewegenden) on its own ground, but it is certain (sicher) that no mathematician has taken account of this inversion, this reversal of roles, still less has it been necessary to demonstrate it using a totally elementary differential equation. It has only been mentioned as a matter of fact that, while the discoverers of the differential calculus and the major part of their followers make the differential symbol the point of departure for calculus, Lagrange in reverse makes the algebraic derivation40 of actual (wirklichen) functions of the independent variable the point of departure, and the differential symbols into merely symbolic expression of already derived functions. If we once more return to d(uz), we obtained next as the result (Produkt) of setting x1 - x = 0, as the result of the differential operation itself: dy/dx = z du/dx + u dz/dx . Since there is a common denominator here, we thus obtain as a reduced expression dy = z du + u dz . dy/dx =f (x) dy = f (x)dx dy = zdu + udz u = f(x), z = (x) du = f (x)dx dz = (x)dx Therefore: dy = (x)f (x)dx + f(x) (x)dx dy/dx = (x)f (x) + f(x) (x) . In the first case therefore first the differential coefficient dy/dx = f (x) dy = f (x)dx . In the second case first the differential dy and then the differential coefficient dy/dx. In the first case, where the differential symbols themselves first originate from the operations performed with f(x), first the derived function, the true (wirkliche) differential coefficient, must be found, to which dy/dx stands opposite (gegen bertrete) as its symbolic expression; and only after it has been found can the differential (das Differential) dy = f (x)dx be derived. It is turned round (umgenkehrt) in dy = zdu + udz . Since du, dz appear here as operational symbols and clearly indicate operations which we already know, from differential calculus, how to carry out, therefore we must first, in order to find the real value of dy/dx, in every concrete case substitute for u its value in x, and for z ditto - its value in x - in order to find dy = (x)f (x)dx + f(x) (x)dx; dy/dx (x)f (x) + f(x) (x) . What is true for du/dx , dz/dx , dy/dx ,d y/dx etc. is true for all complicated formulae where differential symbols themselves appear within general symbolic operational equations.
Marx's Mathematical Manuscripts 1881