We start with the algebraic derivation of f (x), in order to establish in this way at the same time its symbolic differential expressions 0/0 or dy/dx, and thus also discover its meaning. We must then turn int round, starting with symbolic differential coefficients du/dx , dz/dx as given forms in order to find their respective corresponding real equivalents f (x), (x). And indeed, these different ways of treating the differential calculus, setting out from opposite poles - and two different historical schools - here do not arise from changes in our subjective methods,, but from the nature of the function uz to be dealt with. We deal with it, as with functions of x with a singe dependent variable, by starting with the right-hand pole and operating algebraically with it. I do not believe any mathematician has proved or rather even noticed this necessary reversal from the first method of algebraic derivation (historically the second) whether for so elementary a function as uz or any other. They were too absorbed with the material of the calculus for this. Indeed, we find that in the equation 0/0 or dy/dx = z du/dx + u dz/dx The symbolic differential coefficients thus themselves become already the object or content of the differential operation, instead of as before featuring as its purely symbolic result (als symbolisches Resultat derselben). With these two points, first, that the symbolic differential coefficients as well as the variables become substantial elements of the derivation, become objects of differential operations (Differentialoperationen), second, that the question has changed about, from finding the symbolic expression for the real differential coefficient f (x), to finding the real differential coefficient for its symbolic expression - with both these points the third is given, that instead of appearing as the symbolic result of the previous operation of differentiation on the real function of x, the symbolic differential expressions now conversely (umgekehrt) play the role of symbols which indicate operations of differentiation yet to be performed on the real function of x; that they thus become operational symbols. In our case, where dy/dx = z du/dx + u dz/dx , y = xm , u = sqrt{x}, z = x + 2ax . c) The equation found is not only a symbolic operational equation (Operationsgleichung), but also simply a preparatory symbolic operational equation. Since in [I)] dy/dx = z du/dx + u dz/dx , II) dy or d(uz) = zdu + udz . Straight away this equation says that when a product of two arbitrary variables (and this is generalisable in further applications to the product of an arbitrary number of variables) is to be differentiated, each of two factors is to be multiplied by the differential of the other factor and the two products so obtained are to be added. The first operational equation dy/dx = z du/dx + u dz/dx And here it may be remarked that the process of the original algebraic derivation is again turned its opposite. We first obtained there y = y1 - y dy/dx = f (x) , dy = f (x)dx . The above equation,* however, gives us the differentials dy, du, dz as points of departure (Ausgangspunkte). Thus, were in fact arbitrary defined functions of x to be substituted for u and z, designated only as u = f(x) and z = (x) , The result of this differentiation has the general form: df(x) = f (x)dx d (x) = (x)dx . So that dy = (x)f (x)dx + f(x) (x)dx . dy/dx = (x)f (x) + f(x) (x) . Here, where the differential already plays the role of a ready-made operational symbol, we therefore derive the differential coefficient from it; while on the contrary in the original algebraic development the differential was derived from the equation for the differential coefficient. Let us take the differential itself, as we have developed it in its simplest form, namely, from the function of the first degree: y = ax, dy/dx = a ; dy = adx . The equation of this differential appears to be much more meaningful than that of the differential coefficient, 0/0 or dy/dx = a , Since dy = 0 and dx = 0, dy = adx is identical to 0 = 0. Yet, we are completely correct to use dy and dx for the vanished - but fixed, by means of these symbols, in their disappearance - differences, y1 - y and x1 - x. As long as we stay with the expression dy = adx dy = f (x)dx , dy/dx = f (x) , Thus for example in y = ax d(y ) = d(ax) , 2ydy = adx . The last equation of differentials provides us with two equations of differential coefficients, namely: dy/dx = a/2y and dx/dx = 2y/a . But 2ydy = adx also provides us immediately with the value 2ydy/a for dx, which for instance substitutes into the general formula for the subtangent y dx/dy and finally helps to establish 2x, double the abscissa, as the value of the subtangent of the usual parabola. We now want to take an example in which these symbolic expressions first serve the calculus as ready-made (fertige) operational formulae, so that the real value of the symbolic coefficient is also found and then the reversed elementary algebraic exposition may be followed. 1) The dependent function y and the independent variable x are not united in a single equation, but in such a manner that y appears in a first equation as a direct function of the variable u, and then u in a second equation as a direct function of the variable x. The task: to find the real value of the symbolic differential coefficient, dy/dx . Let a) y = f(u), b) u = (x) . Next, 1) y = f(u) gives: dy/du = df(u)/du = f (u)du/du = f (u) . 2) du/dx = d (x)/dx = (x)dx/dx = (x) . So that dy/du du/dx = f (u) (x) . dy/du du/dx = dy/dx . dy/dx = f (u) (x) . Example. If a) y = 3u , b) u = x + ax , then by the formula dy/du = d(3u )/du = 6u ( = f (u)) ; dy/dx = 6(x + ax ) ( =f (u)) . du/dx = 3x + 2ax (= (x)) . dy/du du/dx or dy/dx = 6(x + ax ) (3x + 2ax) ( =f (u) (x)) . 2) We now take the equations contained in the last example as the starting equations (Ausgangsgleichungen), in order to develop them this time in the first, algebraic, method. a) y = 3u , b) u = x + ax . Since y = 3u , [then] y1 = 3u1 , and y1 - y = 3(u1 - u ) = 3(u1 - u) (u1 + u) . (y1 - y)/(u1 - u) = 3(u1 +u) . If now u1 - u becomes = 0, then u1 = u, and 3(u1 + u) is thus transformed to 3(u + u) = 6u. We substitute for u its value in equation b), so that dy/du = 6(x + ax ) . u = x + ax , [then] u1 = x1 + ax1 ; u1 - u = (x1 + ax1 ) - (x + ax ) = (x1 - x ) + a(x1 - x ) , u1 - u = (x1 - x) (x1 + x1x + x ) + a(x1 - x) (x1 + x) ; (u1 - u)/(x1 - x) = (x1 + x1x + x ) + a(x1 + x) . If now x1 - x becomes = 0 then x1 = x, so that x1 + x1x + x = 3x a(x1 + x) = 2ax . du/dx = 3x + 2ax . If we now multiply both equations together, we then obtain on the right-hand side 6(x + ax ) (3x + 2ax) , dy/du du/dx = dy/dx , In order to bring out the difference in the derivations more clearly, we shall place the defined functions of the variables on the left-hand side and the functions dependent on them on the right-hand side, since one is accustomed, following the general equations in which only 0 stands on the right hand, to thinking that the initiative is on the left-hand side. Thus: a) 3u = y ; b) x + ax = u . 3u = y , 3u1 = y1 , 3(u1 - u ) = y1 - y 3(u1 - u) (u1 + u) = y1 - y , 3(u1 + u) = (y1 - y)/(u1 - u) . If now u1 becomes = u, so that u1 - u = 0, [we] then obtain 3(u + u) or 6u = dy/du . If we substitute in 6u its value from equation b), then 6(x + ax ) = dy/du . Furthermore, if x + ax = u , x1 + ax1 = u1 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) ; 3x + 2ax = du/dx . If we multiply the 2 derived functions together, then 6(x + ax ) (3x + 2ax) = dy/dx , dy/du du/dx = dy/dx = 6(x + ax ) (3x + ax) . It is self-evident that due to its details and the frequently difficult division of the first difference, f(x1) - f(x) , into terms which each contain the factor x1 - x, the latter method is not comparable to the historically older one as a means of calculation. On the other hand one begins this last method with dy, dx and dy/dx as given operational formulae, while one sees them arise in the first one, and in a purely algebraic manner as well. And I maintain nothing more. And there in the [historically] first method, how has the point of departure of the differential symbols as operational formulae been obtained? Either through covertly or through overtly metaphysical assumptions, which themselves lead once more to metaphysical, unmathematical consequences, and so it is at that point that the violent suppression is made certain, the derivation is made to start its way, and indeed quantities made to proceed from themselves. And now, in order to give an historical example of beginning from the two opposing poles, I will compare the solution of the case of d(uz) developed above by Newton and Leibnitz on the one hand, to that by Lagrange on the other hand. 1) Newton. We are first told that when the variable quantities increase x., y. etc. designate the velocities of their fluxions, alias of the increase, respectively, of x, y etc. Since furthermore the numerical sizes of all possible quantities may be represented by means of straight lines, the momenta or infinite small quanta which are produced are equal to the product of the velocities x., y. etc. with the infinitely small time intervals during which they occur, thus = u. , x. and y. .42
Marx's Mathematical Manuscripts 1881