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?
Simplifying function notation
2
$\begingroup$
calculus
real-analysis
functions
notation
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0@leslietownes: Fixed. Missed that (as we all did)—good catch. – 2011-11-13
1 Answers
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(Reposting my comment as an answer)
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.
I don't know a name for this, but it is totally OK (if your audience is familiar with operations on functions). I should emphasize it is a notational difference only; it doesn't "do" anything in a proof.
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0And (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? – 2011-11-13