# Simplifying function notation

For example, in the process of proving that $$\left({\frac{f}{g}}\right)'\left({a}\right)= \frac{f'\left({a}\right)g\left({a}\right)-f\left({a}\right)g'\left({a}\right)}{\left[{g\left({a}\right)}\right]^{2}}$$ I'd like to tidy things up a bit by writing $$\left({\frac{f}{g}}\right)'\left({a}\right)= \left[\frac{f'\cdot g-f\cdot g'}{{g}^{2}}\right]\left({a}\right).$$ I believe that this is true, but I'm not confident why it is true: what assumptions am I making in rewriting this way, and is there a name for this change of notation?

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Since I'm not quite sure how to describe the notational step I'm taking, I could also use some help with the title. –  raxacoricofallapatorius Nov 13 '11 at 1:29
Note that you have already taken the notational step in the left hand side of your first expression. To avoid it, you might start: $\text{If }h(a)=\dfrac{f(a)}{g(a)}\text{ then }h'(a)=\cdots$ –  Henry Nov 13 '11 at 1:40
I apologize to raxa for the false information. –  analysisj Nov 13 '11 at 2:06
@analysisj: No problem! –  raxacoricofallapatorius Nov 13 '11 at 2:07
Wow, none of us noticed the sign error in the original question (or my answer, until I fixed it) :) –  leslie townes Nov 13 '11 at 2:30

It is just a change in perspective, from thinking of the value of the derivative as a numerical expression in the values of the functions $f$ and $g$ at $a$, to the value of the single function $\frac{f' \cdot g - f \cdot g'}{g^2}$ at $a$. So in the second you emphasize the algebra of functions ($\cdot$, $−$, etc. are operations on functions) instead of numbers.
So is it fair to say that the re-expression is a more precise way of proceeding, since the subsequent manipulations on $\left({\frac{f}{g}}\right)'\left({a}\right)$ will rely most directly on the algebra of functions and not of numbers? –  raxacoricofallapatorius Nov 13 '11 at 2:19
They are equally "precise" (when the functional one works at all, i.e. all values are taken at the same $a$), but one can be more convenient than the other. Going back and forth between them at will is a useful skill. –  Henning Makholm Nov 13 '11 at 2:21
I wouldn't say that it's more "precise", it is just a change in perspective. One advantage it has--- not in a proof of the quotient rule, but elsewhere--- is that you can suppress the $a$ entirely: $(\frac{f}{g})' = \frac{f' g - f g'}{g^2}$ makes perfectly good sense in the algebra of functions (and maybe emphasizes the generality of the rule). In a proof of a quotient rule, most of the work is in showing that it makes sense to write $(\frac{f}{g})'(a)$ in the first place--- i.e. that a certain limit exists. You probably shouldn't write $(\frac{f}{g})'(a)$ before then. –  leslie townes Nov 13 '11 at 2:26
And (again, just to confirm my understanding) I need to be very careful when I "suppress $a$". In cases like the one illustrated, for example, there is an additional requirement that both $f$ and $g$ are differentiable at $a$ (and, in fact, that $g(a)\neq 0$), so I'm either going to have to leave the $(a)$, or make a not somewhere that what I'm doing only works when this conditions are satisfied. Correct? –  raxacoricofallapatorius Nov 13 '11 at 16:12