# 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.

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I don't much like that you are manipulating expressions when you should be trying to manipulate limits. You are taking what is the final result of a limit, and playing with it; it's going to be rather difficult to get from that to your original limit, just like you'll find it difficult to start with $2c$ and arrive at $\frac{x^2-c^2}{x-c}$. – Arturo Magidin Feb 28 '12 at 5:19
You have to be a little more careful since when you substitute $f'(c)$ in the first line of your proof you are using $\frac{f(x)-f(c)}{x-c}$ when you should be using limits. – Daniel Montealegre Feb 28 '12 at 5:24

## 1 Answer

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 $c$).

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You are completely right. For some reason, I totally forgot that there could be some potential limit evaluations amid the entire process; I simply treated the whole thing as an algebraic expression, hoped to "chunk" the limit in right at the end and see if it worked. Your post changed my whole perspective on the entire process and it was clear enough for me to understand it in its entirety. Thank you so much! +1 – EsX_Raptor Feb 28 '12 at 5:38
This argument also uses the fact that differentiable functions are continuous. – Michael Hardy Feb 28 '12 at 5:41
@Michael: As I believe I noted in the parenthetical comment at the end... – Arturo Magidin Feb 28 '12 at 5:45