“\frac {dy}{dx} = \frac {\omega^2x}{g}…The first derivative, the result of the differentiation of y with respect to x, was written by Leibniz in the form
\frac {dy}{dx}…Leibniz's notation …is both extremely useful and dangerous. Today, as the concepts of limit and derivative are sufficiently clarified, the use of the notation… need not be dangerous. Yet, the situation was different in the 150 years between the discovery of calculus by Newton and Leibniz and the time of Cauchy. The derivative \frac {dy}{dx} was considered as the ratio of two "infinitely small quanitites", of the infinitesimals dy and dx. …it greatly facilitated the systematization of the rules of the calculus and gave intuitive meaning to its formulas. Yet this consideration was also obscure… it brought mathematics into disrepute… some of the best minds… such as… Berkeley, complained that calculus is incomprehensible. …\frac {dy}{dx} is the limit of a ratio of dy to dx… Once we have realized this sufficiently clearly, we may, under certain circumstances, treat \frac {dy}{dx} so as if it were a ratio… and multiply by dx to achieve the separation of variables. We get
{dy} = \frac {\omega^2x}{g}xdx”

Mathematical Methods in Science (1977)

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Hungarian mathematician 1887–1985

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“The discovery of Hippocrates amounted to the discovery of the fact that from the relation
(1)\frac{a}{x} = \frac{x}{y} = \frac{y}{b}it follows that(\frac{a}{x})^3 = [\frac{a}{x} \cdot \frac{x}{y} \cdot \frac{y}{b} =] \frac{a}{b}and if a = 2b, [then (\frac{a}{x})^3 = 2, and]a^3 = 2x^3.The equations (1) are equivalent [by reducing to common denominators or cross multiplication] to the three equations
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the 2 solutions of Menaechmusand the solutions of Menaechmus described by Eutocius amount to the determination of a point as the intersection of the curves represented in a rectangular system of Cartesian coordinates by any two of the equations (2).
Let AO, BO be straight lines placed so as to form a right angle at O, and of length a, b respectively. Produce BO to x and AO to y.
The first solution now consists in drawing a parabola, with vertex O and axis Ox, such that its parameter is equal to BO or b, and a hyperbola with Ox, Oy as asymptotes such that the rectangle under the distances of any point on the curve from Ox, Oy respectively is equal to the rectangle under AO, BO i. e. to ab. If P be the point of intersection of the parabola and hyperbola, and PN, PM be drawn perpendicular to Ox, Oy, i. e. if PN, PM be denoted by y, x, the coordinates of the point P, we shall have

\begin{cases}y^2 = b. ON = b. PM = bx\\ and\\ xy = PM. PN = ab\end{cases}whence\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.
In the second solution of Menaechmus we are to draw the parabola described in the first solution and also the parabola whose vertex is O, axis Oy and parameter equal to a.”

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The point P where the two parabolas intersect is given by<center><math>\begin{cases}y^2 = bx\\x^2 = ay\end{cases}</math></center>whence, as before,<center><math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.</math></center>
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