In any calculus course, one of the first thing we learn is that $(uv)'=u'v+v'u$ rather than the what I've written in the title. This got me wondering: when is this dream product rule true? There are of course trivial examples, and also many instances where the equality is true at a handful of points. Less obvious though, is the following:

Are there non-constant $u,v$ such that there exists an interval $I$ where $(uv)'=u'v'$ over $I?$

I have a feeling there should be, but I am having trouble constructing such a pair.

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    $\begingroup$ Take $u=x$ and $v=\frac{1}{x-1}$ then $(\frac{x}{x-1})'=\frac{-1}{(x-1)^2}=(x)'(\frac{1}{x-1})'.$ $\endgroup$ Apr 11 '14 at 2:42
  • $\begingroup$ I was just about to post that exact example. Pretty much no matter what you make say u, you can solve an ode for v that will work. $\endgroup$ Apr 11 '14 at 2:47

There are many cases where this will happen. By simple algebra, the relation $(uv)'=u'v'$ can be written as $$\Bigl(\frac{u'}{u}-1\Bigr)\Bigl(\frac{v'}{v}-1\Bigr)=1$$ and this leads to $$\ln u(t)=\int\frac{v'}{v'-v}\,dt\ .$$ You will have to make sure that the integral exists and gives a suitable result, but this should give you a solution for many functions $v$.

  • $\begingroup$ There isn't a way to simplify that integral is there? I've been killing myself trying to figure that out but I don't think it's possible. $\endgroup$
    – user142299
    Apr 11 '14 at 2:48
  • 1
    $\begingroup$ As $v$ could be many different functions, I would be surprised if there were any simplification in general. But let us know if you find anything! $\endgroup$
    – David
    Apr 11 '14 at 2:52

Let $u$ and $v$ be functions of $t$.
then $(uv)'=u'v'$
$\iff u'v+uv'=u'v'$
$\iff u'(v-v')+v'u=0$
$\iff u'+\frac {v'}{v-v'}u=0$
Let u be the unkown function, then multiply both sides by $e^{\int \frac{v'}{v-v'}}dt$: $u'(e^{\int \frac{v'}{v-v'}dt})+(e^{\int \frac{v'}{v-v'}dt}\frac {v'}{v-v'})u=0 (*)$
Notice that $(e^{\int \frac{v'}{v-v'}dt})'=e^{\int \frac{v'}{v-v'}dt}\frac {v'}{v-v'}$,
therefore (*) becomes: $u'(e^{\int \frac{v'}{v-v'}dt})+(e^{\int \frac{v'}{v-v'}dt})'u=0 $
or $(ue^{\int \frac{v'}{v-v'}dt})'=0$
or $ue^{\int \frac{v'}{v-v'}dt}=C$
or $u=Ce^{-\int \frac{v'}{v-v'}dt}$
Note: While evaluating $\int \frac{v'}{v-v'}dt$, please omit the integration constant. $u$ and $v$ can be replaced by each other while the equation still holds.
Replacing $u$ or $v$ by a given function and assign C a given value, we can solve for the other one.
For example, choose $v=t^2, C=1,integral constant=0$, then $u=e^{-\int \frac{2t}{t^2-2t}dt}=e^{-\int \frac 2{t-2}dt}=e^{-2ln|t-2|}=e^{ln \frac 1{(t-2)^2}}=\frac 1{(t-2)^2}$


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