# Meaning of historical fluxion notation

I've noticed that in 18th century books on calculus writers would say that 'the fluxion of $ax$ is $a\dot{x}$' and 'the fluxion of $x^n$ is $n x^{n-1} \dot{x}$'. What does this extra '$\dot{x}$' at the end of the formulas for fluxions (derivatives) signify?

P.S. Apologies if historical questions are not allowed here.

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The notation is equivalent to thinking of the derivative as the linear approximation to the function. You need a new variable because the derivative varies from point to point. –  Ryan Budney Jun 13 '13 at 21:14
@Ryan Budney: I think the concept of "fluxion" just used to indicate derivative w.r.t. time, and the "differential" was a different concept (small change in the variable). So it made sense to write $dx=\dot{x}dt$ –  Qfwfq Jun 13 '13 at 21:18
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## 2 Answers

Every variable is thought of as depending on the parameter "time" (explicitly indicated or just omitted), and the "fluxion" of a variable $x$ stands for the change rate of $x$ with respect to "the time parameter" $t$, i.e. $\dot{x}$ means $\frac{dx(t)}{dt}$, and the formulas you quote are just derivatives of composite functions.

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See Newton's notation: en.wikipedia.org/wiki/Newton%27s_notation –  Joel Reyes Noche Jun 14 '13 at 12:48
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It is surprising that this question has not yet been closed as below research level. But since it exists I may generalize the existing, perfectly correct answers a bit.

Here we see a simple application of the chain rule: $\frac{df(y)}{dx} = \frac{df(y)}{dy} \frac{dy}{dx}$ as it has to be applied, for instance when implicitly differentiating expressions like $x^2 - t^2 = const$ with respect to time $t$ to arrive at $2x\frac{dx}{dt} -2 t = 0$ or, in Newton's notation, $2x\dot{x} -2 t = 0$

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