MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've noticed that in 18th century English 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.

share|cite|improve this question
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

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.

share|cite|improve this answer
See Newton's notation: – Joel Reyes Noche Jun 14 '13 at 12:48

In the physics applications that Newton was interested in, his functions were mostly functions of time. Since he was usually differentiating with respect to time, it was OK that his dot notation, unlike Leibniz's notation, didn't indicate what he was differentiating with respect to. Newton used the symbol $o$ to indicate a fixed infinitesimal interval of time. So in Newton's notation $\dot{x}$ would be Leibniz's $dx/dt$, while $\dot{x}o$ would be $dx$ --- an infinitesimal difference, called a "moment."

Newton also used a notational shortcut that caused confusion. He used a convention that when it was clear from context that he was talking about a moment, then $\dot{x}$ would implicitly mean $\dot{x}o$, i.e., the $o$ could be left out because it was too cumbersome to write it all the time. Because of this, English mathematicians began to muddy the notational waters by not distinguishing the two notions. See Boyer, p. 201, and p. 114 for the definition of "moment."

So when someone using Newton's notation says that the fluxion of $x^n$ is $nx^{n-1}\dot{x}$, what they mean could be two different but equivalent things in Leibniz notation. Either:

(1) $d(x^n)=nx^{n-1}dx$ (where $dx$ is the same as $\dot{x}o$, and the $o$ is implicit); or

(2) $d(x^n)/dt=nx^{n-1}dx/dt$

In 17th- and 18th-century mathematics, the difference between (1) and (2) is purely a matter of dividing both sides of the equation by $dt$. (There was no notion, as in modern NSA, that a derivative is the standard part of the ratio of infinitesimals, which is not quite the same thing as the ratio of infinitesimals.)

Boyer, The History of the Calculus and its Conceptual Development.

share|cite|improve this answer

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, $2x\dot{x} -2 t = 0$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.