Proof of quotient rule using Leibniz differentials

[SamRoss]

I know how to prove the quotient rule by using the definition of a derivative using limits (Newton's style). I just saw a proof of the product rule using Leibniz's concept of differentials on Wikipedia. https://en.wikipedia.org/wiki/Product_rule#Discovery

Does anyone know of a Leibniz-style proof of the quotient rule? I was not able to find it and when I tried doing it myself I got an extra term in the denominator.

 

[jedishrfu]

You might find a proof in here:

https://www.math.wisc.edu/~keisler/calc.html

THis book is based on hyper-reals and how you can use them like real numbers without the need for limit considerations.

 

[fresh_42]

I know how to prove the quotient rule by using the definition of a derivative using limits (Newton's style). I just saw a proof of the product rule using Leibniz's concept of differentials on Wikipedia. https://en.wikipedia.org/wiki/Product_rule#Discovery

Does anyone know of a Leibniz-style proof of the quotient rule? I was not able to find it and when I tried doing it myself I got an extra term in the denominator.

Isn't this the same? What happens if you use the product rule on $u \cdot v^{-1}$ and to find $v^{-1}$ the equation $1=v\cdot v^{-1}$ ?

 

[SamRoss]

You might find a proof in here:

https://www.math.wisc.edu/~keisler/calc.html

THis book is based on hyper-reals and how you can use them like real numbers without the need for limit considerations.

Thank you

 

[SamRoss]

Isn't this the same? What happens if you use the product rule on $u \cdot v^{-1}$ and to find $v^{-1}$ the equation $1=v\cdot v^{-1}$ ?

Okay, I got it to work that way but I still don't see where I went wrong here. I probably should have posted this originally but I was unfamiliar with the Latex notation.

$d(uv)=\frac{u+du}{v+dv}-\frac{u}{v}=\frac{v(u+du)-u(v+dv)}{v(v+dv)}=\frac{vdu-udv}{v^2+vdv}$

<Moderator's note: edited LaTex code for better readability>

 

[fresh_42]

Okay, I got it to work that way but I still don't see where I went wrong here. I probably should have posted this originally but I was unfamiliar with the Latex notation.

$d(uv)=\frac{u+du}{v+dv}-\frac{u}{v}=\frac{v(u+du)-u(v+dv)}{v(v+dv)}=\frac{vdu-udv}{v^2+vdv}$

<Moderator's note: edited LaTex code for better readability>

I do not follow your equations, especially the first one, to be honest.

The Leibniz rule gets us $d(uv) = u \cdot (dv) + (du) \cdot v$. This can be used to calculate $dv^{-1}$ from $0=d(\operatorname{id}) =d(vv^{-1})$. Now with $dv^{-1}$ known, the first equation can be used a second time to get $d(uv^{-1})$. All along the same lines as in the Wikipedia entry you quoted. If you like you can as well add $\frac{1}{dx}$ every time, but it isn't necessary here.

 

[SamRoss]

I do not follow your equations, especially the first one, to be honest.

Oops. I meant for the first part of my equation to be d(u/v), not d(uv). I am subtracting u/v following the same logic in the Wikipedia article. I hope that makes sense now. So where am I going wrong?

 

[fresh_42]

With this approach, you need a similar argument to "Since the term du·dv is "negligible" (compared to du and dv), Leibniz concluded that ..." to neglect $vdv$. This would be: "Since the term v·dv is "negligible" (compared to v·v) ... " which it is. We also "neglect" such small terms in the limit process. Then you get the correct formula $d \frac{u}{v} = \frac{vdu-udv}{v^2}$.

To be honest, I'm not sure whether this can be done this way at all. It's really a bit hand wavy. Certainly it shouldn't. It reduces the formal concept of our modern view with limits and open neighborhoods to an intuitive argument about infinitesimals. From this point of view, I guess @jedishrfu's link in post #2 is a good source.

Me for my part can live well with the concept that derivations (by definition), especially a derivative (by the limit process), obey the product rule. The rest follows from there without any obscure "negligibles". To Leibniz' honor has to be mentioned, that already the term infinitesimal had been revolutionary in the 17th century and our modern language developed a good century later.

 

[SamRoss]

With this approach, you need a similar argument to "Since the term du·dv is "negligible" (compared to du and dv), Leibniz concluded that ..." to neglect $vdv$. This would be: "Since the term v·dv is "negligible" (compared to v·v) ... " which it is. We also "neglect" such small terms in the limit process. Then you get the correct formula $d \frac{u}{v} = \frac{vdu-udv}{v^2}$.

To be honest, I'm not sure whether this can be done this way at all. It's really a bit hand wavy. Certainly it shouldn't. It reduces the formal concept of our modern view with limits and open neighborhoods to an intuitive argument about infinitesimals. From this point of view, I guess @jedishrfu's link in post #2 is a good source.

I had a feeling that extra term in the denominator had to be simply neglected and I also felt it was a bit "hand wavy" as you said. Still, we get so close to the true derivative that it seems worthwhile to take a closer look at what's going on. I think I will take another look at @jedishrfu's link. Thanks again for your help.

 

[Niels Ohlsen]

Gents, I was in the same state of doubt as SamRoss.
Motivation: Whatever information is within the system will remain therein, lest we destroy it by ill advised manipulation. So of course we should be able to demontrate this directly (mathematics does not care in which order we apply logic).

I have made a slight rearrangement that may make this slightly easier to defend: Let's split the denominator into two factors:
\begin{align*}
d(\frac{u}{v})=\frac{(u+du)v-u(v+dv)}{v(v+dv)}\\
=\frac{(u+du)v-u(v+dv)}{v+dv}(\frac{1}{v})\\
=\frac{v(du)-u(dv)}{v+dv}(\frac{1}{v})\\
\end{align*}
Given that situation, considering the leftmost fraction, adding the infinitesimally small element dv to the denominator v should be negligible, given that the numerators consist of multipla of the functions themselves (assumption being that the increment remains much smaller than the current value of the functions themselves, which is not unreasonable, since we can choose to 'cut' however finely we like).

So what follows is, hence, division by v followed by another division by v, or, in effect, by v squared.

In other words: Leibniz knew what he was doing; there is no extra information gained by dividing by some increment along the independent axis. A lot of textbooks could be a heck of a lot shorter.