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How do we prove the quotient rule for differentiation? The proof in my book from the defintion is very long. Are there some elegant proofs?

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Well, it's really just a rewriting of the product rule. If you can prove the product rule, prove that, then use it with $f \cdot \frac 1g$. –  Mike Miller Apr 6 at 16:24
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The proof I usually see is to prove that $1/g$ is differentiable with derivative $-g'/g^2$ in the points with $g(x) \neq 0$ - that's not hard - and then the general formula directly follows from the product rule. –  Daniel Fischer Apr 6 at 16:28
    
@MichaelHardy Oh I see. Can there be some easy proofs for chain rule for $\log(f(x))$? –  125a8owp Apr 6 at 16:49
    
@user87 : Good question. There's the proof of the chain rule in general, but if one wants to prove only a special case like this, then maybe there are simpler proofs. But I've never thought about whether there might be such a simpler proof for this particular special case. –  Michael Hardy Apr 6 at 16:54
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@jack : The proof works, but the question seemed to ask for a proof that's not so long. This proof is short only because it omits the fairly subtle proof of the chain rule, although it relies on the chain rule. –  Michael Hardy Apr 6 at 16:56

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It is actually quite simple. Observe $\frac{f}{g}=f\cdot g^{-1}.$ So $(\frac{f}{g})'=f'\cdot g^{-1}-(g)'\cdot g^{-2}\cdot f.$ Then just find a common denominator to add the terms. So $f'\cdot g^{-1}-(g)'\cdot g^{-2}\cdot f=\frac{f'\cdot g}{g^2}-\frac{f\cdot g'}{g^2}$. If you want to prove that $\frac{1}{g(x)}$ is differentiable. Let $r(x)=\frac{1}{g(x)}$ and then use the definition of the derivative to show $r(x)'$ exists.

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Thanks, this is more of what I was expecting! –  125a8owp Apr 6 at 16:40
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Using an asterisk for ordinary multiplication in MathJax or TeX or the like is horribly vulgar. One can write $f \cdot g$ of $f\times g$. ${}\qquad{}$ –  Michael Hardy Apr 6 at 16:43
    
I am sorry if its vulgar, but it gets the point across one and I don't have that symbol two. –  Jack Apr 6 at 16:48
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@jack : You do have that symbol: \cdot ${}\qquad{}$ –  Michael Hardy Apr 6 at 17:53
    
Um, ..., $(f \cdot g^{-1})' = f' g^{-1} + f(-1)g^{-2}g'$. That last factor is not $(g^{-1})'$, nor should $(g^{-1})'$ become $g'$ when rewriting as fractions. –  Eric Towers Apr 6 at 19:26

If you know the chain rule and the derivative of log, then let $$y={f(x)\over g(x)}$$ so that $$\ln(y)=\ln{f(x)}-\ln{g(x)}$$ Thus, taking derivatives on both sides, $${y'\over y}={f'(x)\over f(x)}-{g'(x)\over g(x)}$$ so $$y'={f(x)\over g(x)}\left({f'(x)\over f(x)}-{g'(x)\over g(x)}\right)=\frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2}$$

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But the proof of the chain rule is much subtler than the proof of the quotient rule. You can prove the quotient rule without that subtlety. –  Michael Hardy Apr 6 at 16:42
    
A rigorous proof is, but the chain rule is intuitively obvious. I think the bigger problem with this derivation is using logs. –  user138335 Apr 6 at 16:43

A weak version of the quotient rule follows from the product rule. You want $\left(\dfrac f g\right)'$. You know that $$ f=\frac f g \cdot g $$ Differentiate both sides, using the product rule for the right side: $$ f' = \left( \frac f g \right)' g + g'\frac f g $$ Subtract the last term from both sides: $$ f'-g'\frac f g = \left(\frac f g \right)' g $$ Then divide both sides by $g$: $$ \frac{f'}{g} - \frac{g'f}{g^2} = \left( \frac f g \right)' $$ Then recall that you subtract fractions by finding a common denominator, which in this case is $g^2$, and, you've got the quotient rule.

EXCEPT that you don't quite. The product rule works if both functions being multiplied are differentiable. That means all this works if $f/g$ is differentiable. What you would like the quotient rule to say is that if $f$ and $g$ are differentiable, then so is $f/g$ (except at points where $g=0$) and the derivative is given by the usual expression that you see in the quotient rule. The argument above fails to tell you that $f/g$ is differentiable, but tells you that the quotient rule holds if it's differentible. The fact that it doesn't tell you that $f/g$ is differentiable is why I called it a "weak" version.

The full quotient rule, proving not only that the usual formula holds, but also that $f/g$ is indeed differentaible, begins of course like this: $$ \frac{d}{dx} \frac{f(x)}{g(x)} = \lim_{\Delta x\to 0} \frac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}. $$ (If the words "of course" used above seem out of place to you, then that's what you need to learn before working on proofs of things like the quotient and product rules.)

Where the proof goes from there is where all the work is. Maybe I'll continue this answer later by going into that.

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@jack You're correct that the $f \cdot g$ is differentiable. You're incorrect that the product rule applies (because the product rule calls for $g'$, which doesn't exist at all points). –  Mike Miller Apr 6 at 17:17
    
@jack : You're confused. Certainly it's true that in some cases $fg$ is differentiable even though $g$ is not, but that has no relevance here. For one thing, the product rule doesn't apply to those cases, and for another, I was writing about the situation in which the two factors are both differentiable. I wrote "if", not "only if". –  Michael Hardy Apr 6 at 17:52

From my old calculus notes:

We calculate the numerator of the difference quotient: $$\begin{align}\require{cancel} \Delta\left(\dfrac u v\right) &= \dfrac{u+\Delta u}{v+\Delta v}-\dfrac uv \\ &=\dfrac{\cancel{uv}+(\Delta u)v\cancel{-uv}-u\Delta v}{(v+\Delta v)v} \\ &= \dfrac{(\Delta u)v-u\Delta v}{(v+\Delta v)v} \end{align}$$ So the difference quotient is: $$\begin{align}\require{cancel} \dfrac{\Delta\left(\dfrac u v\right)}{\Delta x}&=\dfrac{\dfrac{\Delta u}{\Delta x}v-u\dfrac{\Delta v}{\Delta x}}{(v+\Delta v)v}\,\,\underset{\Delta x\to0}{\xrightarrow[]{}}\,\, \dfrac{\mathrm d}{\mathrm dx}\left(\dfrac u v\right)=\dfrac{\dfrac{\mathrm du}{dx}v-u\dfrac{\mathrm dv}{\mathrm dx}}{v^2} \end{align}$$

I hope this helps.
Best wishes, $\mathcal H$akim.

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