Page:A Source Book in Mathematics.djvu/661

 (Hence

$d\sqrt[2]{y}=\frac{dy}{2\sqrt[2]{y}}$,|undefined

for in this case $$a$$ is 1, and $$b$$ is 2; therefore $$\frac{a}{b}\sqrt[b]{x^{a-b}}$$ is ½$$\sqrt[2]{y^{-1}}$$; now $$y^{-1}$$ is the same as $$\frac{1}{y}$$, from the nature of the exponents of a geometric progression, and $$\frac{1}{\sqrt[2]{y}}$$ is $$\sqrt[2]{y^{-1}}$$.)

$d\frac{1}{\sqrt[b]{x^a}}=\frac{-adx}{b\sqrt[b]{x^{a+b}}}$.|undefined

Again the rule for an integral power would suffice for determining fractions as well as roots, for the power may be a fraction while the exponent is negative, and it is changed into a root when the exponent is a fraction; but I have preferred to deduce these consequences myself rather than to leave them to be deduced by others, since they are completely general and of frequent occurrence; and in a matter which is itself involved it is preferable to take ease [of operation] into account.

From this rule, known as an algorithm, so to speak, of this calculus, which I call differential, all other differential equations may be found by means of a general calculus, and maxima and minima, as well as tangents [may be] obtained, so that there may be no need of removing fractions, nor irrationals, nor other aggregates, which nevertheless formerly had to be done in accordance with the methods published up to the present. The demonstration of all [these things] will be easy for one versed in these matters, who also takes into consideration this one point which has not received sufficient attention heretofore, that $$dx$$, $$dy$$, $$dv$$, $$dw$$, $$dz$$, can be treated as proportional to the momentaneous differences, whether increments or decrements, of $$x$$, $$y$$, $$v$$, $$w$$, $$z$$ (each in its order).

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