A) Supplement on the differentiation of uz.47 1) For me the essential thing in the last manuscript on the development of d(uz) was the proof, referring to the equation A) dy/dx = z du/dx + u dz/dx , The form of equation A) lends itself all the more readily to this purpose since it allows a comparison between the du/dx, dz/dx, produced within the derivative f (x), and the dy/dx, which is the symbolic differential coefficient of f (x) and therefore comprises its symbolic equivalent, standing opposite on the left-hand side. Confronting the character of du/dx, dz/dx, as operational formulae, I have been content with the hint that for any symbolic differential coefficient an arbitrary derivative may be found as its real value if one substitutes some f(x), 3x for example, for u and some (x), x + ax for example, for z. I however could also have indicated the geometric applicability of each operational formula, since for example, the general formula for the subtangent of a curve = y dx/dy, which has a form generally identical to z du/dx, u dz/dx, for they are all products of a variable and a symbolic differential coefficient. Finally, it could have been noted that y = uz [is] the simplest elementary function (y here = y1, and uz is the simplest form of the second power) with which our theme could have been developed. A) Differentiation of u/z.48 3) Since d(u/z) is the inverse of d(uz), where one has multiplication the other division, one may use the algebraically obtained operational formula d(uz) = z du + u dz a) y = u/z , b) u = yz . yz = (u/z) z = u . We have thus simply formally concealed u in a product of two factors. Nonetheless, the task is thereby in fact already solved, since the problem has been transformed from the differentiation of a fraction to the differentiation of a product, for which we have the magic formula in our pocket. According to this formula: c) du = z dy + y dz . We see immediately that the first term of the second side, namely z dy, must remain sitting in peace at its post until the crack of doom (genau vor Torschluss), since the task consists precisely in finding the differential of y( = u/z ), and thus its expression in differentials of u and z. For this reason, on the other hand, y dz is to be removed to the left-hand side. Therefore: d) du = y dz = z dy . We now substitute the value of y, namely u/z, into y dz, so that du - (u/z) dz = z dy ; (z du - u dz)/z = z dy . (z du - u dz)/z = dy = d(u/z) . ____________
Marx's Mathematical Manuscripts 1881