How to prove the quotient rule?

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 $$y'={f(x)\over g(x)}\left({f'(x)\over f(x)}-{g'(x)\over g(x)}\right)=\frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2}$$

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.

Observe,

$$\frac{f(x)}{g(x)}=f(x)\cdot g(x)^{-1}$$

Then using the chain rule. The first part becomes $f'(x)\cdot g(x)^{-1}$. The second part is a little trickier.

The derivative of $f(x)\cdot g(x)^{-1}$, where $f(x)$ is the constant becomes

$$-1 \cdot f(x)\cdot g(x)^{-2}\cdot g'(x)$$

This from using the chain rule, where we let $u=g(x)$.

$$\frac{d}{d(x)}(\frac{1}{g(x)})=\frac{d}{du}\cdot\frac{1}{u}\cdot \frac{du}{dx}$$ $$=\frac{1}{u^2}\cdot u'(x)$$ Substituting back in. $$\frac{1}{g(x)^2}\cdot g'(x)$$

Then find a common denominator to add the terms.

$$f'(x)\cdot g(x)^{-1}-g'(x)\cdot g(x)^{-2}\cdot f(x)$$ $$=\frac{f'(x)}{g(x)}\cdot\frac{g(x)}{g(x)}-\frac{f(x)\cdot g'(x)}{g(x)^2}$$ $$=\frac{f'(x)\cdot g(x)}{g(x)^2}-\frac{f(x)\cdot g'(x)}{g(x)^2}$$

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.