# B Proof of quotient rule using Leibniz differentials

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1. Jun 10, 2017

### 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.

2. Jun 10, 2017

### Staff: Mentor

3. Jun 10, 2017

### Staff: Mentor

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 $1=v\cdot v^{-1}$ ?

4. Jun 13, 2017

### SamRoss

5. Jun 13, 2017

### SamRoss

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>

6. Jun 13, 2017

### Staff: Mentor

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.

7. Jun 15, 2017

### SamRoss

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?

Last edited by a moderator: Jun 16, 2017
8. Jun 15, 2017

### Staff: Mentor

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.

9. Jun 16, 2017

### SamRoss

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.