Marx's Mathematical Manuscripts 1881
Written: August, 1881;
Source: Marx's Mathematical Manuscripts, New Park Publications, 1983;
First published: in Russian translation, in Pod znamenem marksizma, 1933.
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), φ’(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
dy/dx again springs in just the same way from the derivative occurring on the right, with uz just as with functions of x with a single dependent variable; but on the other hand the differential symbols du/dx , dz/dx are again incorporated in f’(x) or the first derivative of uz, and therefore form elements of the equivalent of dy/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 ,
we would no longer be able to operate unless we knew not only that z and u are both functions of x but also that, just as with
y = xm ,
real values in x are given for u and z, such as, for example,
u = sqrt{x}, z = x³ + 2ax² .
In that manner, then, du/dx , dz/dx in fact stand as indicators of operations whose performance (Ausführungsweise) is assumed to be well-known for any arbitrary function of x substituted in place of u and z.
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 ,
the denominator dx is found in all terms on both sides, its reduced expression in thus:
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
thus becomes, if the products of two arbitrary variables is to be differentiated, a superfluous preparatory equation which, after it has served its purpose, namely that of a general symbolic operational formula, leads directly to the goal.
And here it may be remarked that the process of the original algebraic derivation is again turned its opposite. We first obtained there
Δy = y1 - y
as the corresponding symbol for f(x1) - f(x), both usual algebraic expressions (since f(x) and f(x1) have been given as defined algebraic functions of x). Then (f(x1) - f(x))/(x1 - x) was replaced by Δy/Δx, whereupon f’(x) - the first derived function of f(x) - became dy/dx , and we at last obtained, from the final equation of the differential coefficient,
dy/dx = f’(x) ,
the differential
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 = φ(x) ,
and this d sign merely indicates differentiation to be performed.
The result of this differentiation has the general form:
df(x) = f’(x)dx
and
dφ(x) = φ’(x)dx .
So that
dy = φ(x)f’(x)dx + f(x)φ’(x)dx .
Finally,
dy/dx = φ(x)f’(x) + f(x)φ’(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 ;
of which the differential is
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 ,
from which it is derived.
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
or, in general,
dy = f’(x)dx ,
it is nothing other than a restatement of the fact that
dy/dx = f’(x) ,
which in the above case, = a, from which we may continue to transform it further. But this ability to be transformed already makes it an operational symbol (Operationssymbole). At once, we see that if we have found dy = f’(x)dx as a result of the process of differentiation, we have only to divide both sides by dx to find dy/dx = f’(x), namely, the differential coefficient.
Thus for example in y² = ax
d(y²) = 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 = φ(x) .
Next, 1) y = f(u) gives:
dy/du = df(u)/du = f’(u)du/du = f’(u) .
2) du/dx = dφ(x)/dx = φ’(x)dx/dx = φ’(x) .
So that
dy/du ⋅ du/dx = f’(u)⋅φ’(x) .
But
dy/du ⋅ du/dx = dy/dx .
So that
dy/dx = f’(u)φ’(x) .
Example. If a) y = 3u², b) u = x³ + ax² , then by the formula
dy/du = d(3u²)/du = 6u ( = f’(u)) ;
but the equation b) says u = x³ + ax². If we substitute this value for u in 6u, then
dy/dx = 6(x³ + ax²) ( =f’(u)) .
Furthermore:
du/dx = 3x² + 2ax (=φ’(x)) .
So that
dy/du ⋅ du/dx or dy/dx = 6(x³ + ax²) (3x² + 2ax) ( =f’(u)⋅φ’(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 = 3u² , b) u = x³ + ax² .
Since y = 3u², [then] y1 = 3u1², and
y1 - y = 3(u1² - u²) = 3(u1 - u) (u1 + u) .
Therefore
(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(x³ + ax²) .
Further; since
u = x³ + ax², [then] u1 = x1³ + ax1² ;
so that
u1 - u = (x1³ + ax1²) - (x³ + ax²) = (x1³ - x³) + a(x1² - x²) ,
u1 - u = (x1 - x) (x1² + x1x + x²) + a(x1 - x) (x1 + x) ;
thus
(u1 - u)/(x1 - x) = (x1² + x1x + x²) + a(x1 + x) .
If now x1 - x becomes = 0 then x1 = x, so that
x1² + x1x + x² = 3x²
and
a(x1 + x) = 2ax .
Thus:
du/dx = 3x² + 2ax .
If we now multiply both equations together, we then obtain on the right-hand side
6(x³ + ax²) (3x² + 2ax) ,
which corresponds to the left-hand side
dy/du ⋅ du/dx = dy/dx ,
just as previously.
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) 3u² = y ; b) x³ + ax² = u .
Since
3u² = y , 3u1² = y1 ,
so that
3(u1² - u²) = y1 - y
or
3(u1 - u) (u1 + u) = y1 - y ,
so that
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(x³ + ax²) = dy/du .
Furthermore, if
x³ + ax² = u ,
then
x1³ + ax1² = u1
and
x1³ + ax1² - x³ - ax² = u1 - u ;
so that
(x1³ - x³) + a(x1² - x²) = u1 - u .
We further separate into factors:
(x1 - x) (x1² + x1x + x²) + a(x1 - x) (x1 + x) = u1 - u .
Therefore
(x1² + x1x + x²) + a(x1 + x) = (u1 - u)/(x1 - x) ;
now if x1 = x, so that x1 - x = 0, then
3x² + 2ax = du/dx .
If we multiply the 2 derived functions together, then
6(x³ + ax²) (3x² + 2ax) = dy/dx ,
and if we put it in the usual order,
dy/du ⋅ du/dx = dy/dx = 6(x³ + ax²) (3x² + 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 τ during which they occur, thus = u.τ, x.τ and y.τ.42
41
The manuscripts of the second and third drafts are in very rough form: they contain many deletions and insertions. The first four pages of the second draft are not preserved, so we begin with the first complete paragraph. These two drafts, less some abridgments, were first published in Russian in 1933 (Under the Banner of Marxism [Pod znamenem marksizma], No.1, and Marxism and Science [Marksizm i estestvoznanie], pp.26-34). See ‘Preliminary Drafts and Variants of the Manuscript, “On the Differential”,’ point a, p.477 [Yanovskaya, 1968].
*
Equation II) - Trans.
42
This entire paragraph (beginning with the words ‘when the variable quantities increase ...’) is Marx’s German translation of a passage in Hind’s book (see T. Hind, 2nd edition, Cambridge, 1831, p.108). The second draft breaks off at this point. The vacant space (more than half a page) which Marx left after this paragraph is apparent evidence that, not finding the desired quotation in Hind, Marx put aside the contemplated research, obviously intending to return to it later.