Proving the Quotient Rule Directly I have been trying to prove the quotient rule algebraically. As a reminder,$$(f/g)'(c)=\frac{g(c)f'(c)-f(c)g'(c)}{[g(c)]^2}$$The definition of a derivative that I want to use is the following:$$g'(c)=\lim_{x\to c}\frac{g(x)-g(c)}{x-c}.$$Since I could not get it directly, I tried to prove it the other way around:$$\frac{g(c)f'(c)-f(c)g'(c)}{[g(c)]^2}=\frac{g(c)\left[\frac{f(x)-f(c)}{x-c}\right]-f(c)\left[\frac{g(x)-g(c)}{x-c}\right]}{[g(c)]^2}=$$$$\frac{\frac{g(c)f(x)-g(c)f(c)}{x-c}+\frac{f(c)g(c)-f(c)g(x)}{x-c}}{[g(c)]^2}=\frac{\frac{g(c)f(x)-g(c)f(c)}{[g(c)]^2}+\frac{f(c)g(c)-f(c)g(x)}{[g(c)]^2}}{x-c}=$$$$\frac{\frac{g(c)f(x)}{[g(c)]^2}-\frac{g(c)f(c)}{[g(c)]^2}+\frac{f(c)g(c)}{[g(c)]^2}-\frac{f(c)g(x)}{[g(c)]^2}}{x-c}=\frac{\frac{f(x)}{g(c)}-\frac{f(c)}{g(c)}+\frac{f(c)}{g(c)}-\frac{f(c)g(x)}{[g(c)]^2}}{x-c}.$$However, look at those terms; they make no sense to me. Applying the second equation to the LHS of the first equation, you can expect to obtain$$\frac{\frac{f(x)}{g(x)}-\frac{f(c)}{g(c)}}{x-c},$$which is nowhere to be (implicitely) found in the last expression above. Am I doing something wrong? Thanks in advance.
 A: Work with the limits! You cannot hope to transform $\frac{f'(c)g(c) - f(c)g'(c)}{[g'(c)]^2}$ into $\frac{\frac{f}{g}(x) - \frac{f}{g}(c)}{x-c}$ by simple algebraic manipulations, and if you substitute expressions without using limits, then you aren't respecting equalities. This is a simple case of using the limit directly and following your nose (with the experience of having seen how the product rule is proven, namely by using a "stepping stone" to go from $f(x)g(x)-f(c)g(c)$ to $f(x)g(x) - f(c)g(x) + f(c)g(x) - f(c)g(c)$). 
The reason you run into trouble with your attempt is, as you will see below, that in the derivation of the quotient rule there is a crucial step when a limit of $g(x)$ needs to be evaluated; you don't see it in your attempted manipulation and is the source of that extra $g(c)$ and missing $g(x)$. The problem is that, since you have no limits, there is absolutely no warrant for replacing a $g(c)$ with a $g(x)$. (In fact, there was no warrant for your replacement of the derivatives with the difference quotients!)
$$\begin{align*}
\lim_{x\to c}\frac{(\frac{f}{g})(x) - (\frac{f}{g})(c)}{x-c} &= \lim_{x\to c}\frac{\frac{f(x)}{g(x)} - \frac{f(c)}{g(c)}}{x-c}\\
&= \lim_{x\to c}\frac{\quad\frac{f(x)g(c) - f(c)g(x)}{g(x)g(c)}\quad}{x-c}\\
&= \lim_{x\to c}\frac{f(x)g(c)-f(c)g(x)}{(x-c)g(x)g(c)}\\
&= \lim_{x\to c}\frac{f(x)g(c) - f(c)g(c) + f(c)g(c) - f(c)g(x)}{(x-c)g(x)g(c)}\\
&= \lim_{x\to c}\left(\left(\frac{f(x)-f(c)}{x-c}\right)\frac{g(c)}{g(x)g(c)} - \frac{f(c)}{g(x)g(c)}\left(\frac{g(x)-g(c)}{x-c}\right)\right)\\
&= \left(\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\right)\left(\lim_{x\to c}\frac{1}{g(x)}\right)\\&\qquad\quad - \left(\lim_{x\to c}\frac{f(c)}{g(x)g(c)}\right)\left(\lim_{x\to c}\frac{g(x)-g(c)}{x-c}\right)\\
&= f'(c)\left(\frac{1}{g(c)}\right) - \left(\frac{f(c)}{[g(c)]^2}\right)g'(c)\\
&= \frac{f'(c)g(c) - f(c)g'(c)}{[g(c)]^2}.
\end{align*}$$
We've used that $g(c)\neq 0$, and that $f$ and $g$ are differentiable at $c$ (so in particular that $g$ is continuous at $c$). 
