The proof starts like this:

${d\over dx}(uv) = \lim \limits_{\Delta x \to 0}{\Delta(uv)\over\Delta x} = \lim \limits_{\Delta x \to 0} \left(u{\Delta v \over \Delta x} + v{\Delta u \over \Delta x} +\Delta u{\Delta v \over \Delta x}\right)$

The three terms in the parentheses, how is it that we get these from ${\Delta(uv)\over\Delta x}$? I'm not that familiar with this syntax (from Stewart's Calculus Early Transcendentals, 7th edition) as in class we don't use the Delta symbols (yet anyway). Why doesn't $\Delta(uv)$ distribute to just $\Delta uv$ like $3(ab) => 3ab$?


$\Delta(uv)$ means the change in the product $uv$.

So, writing it out:

$\begin{align*}\Delta(uv) & = (u+\Delta u)(v+\Delta v) - uv \\ & = uv + u\Delta v + v\Delta u + \Delta u \Delta v -uv \\ & = u\Delta v + v\Delta u + \Delta u \Delta v\end{align*}$

Then $\dfrac{\Delta(uv)}{\Delta x} = \dfrac{u\Delta v + v\Delta u + \Delta u \Delta v}{\Delta x} = u\dfrac{\Delta v}{\Delta x} + v\dfrac{\Delta u}{\Delta x} + \Delta u \dfrac{\Delta v}{\Delta x}$

  • $\begingroup$ Now I feel a bit embarrassed. Your answer is stated more or less in the book right above the syntax in question (and quite obviously too - oh my!). I obviously didn't realize they just substituted that result in the proof despite reading it twice, and somehow missed that it was the same thing. Maybe I'm just tired right now. $\endgroup$ – Matt Jun 13 '14 at 0:53

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