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I have encountered a somewhat odd notation for a derivative, and I can't seem to work it out. I was given that...

$d(u^3v) = 3u^2vdu + u^3dv$

which sure, makes sense, however, If I now integrate both sides...

$\int{d(u^3v)} = 3v\int u^2du + u^3 \int dv $

I get that

$u^3v = 3v\frac{u^3}{3} + u^3v = 2u^3v$

which seems to have doubled the original function. This "effect" seems to happen depending on how many functions I have in the d() notation. As another example...

$d(u^3v^4w^5) = v^4w^5 (3u^2)du + u^3w^5 (4v^3) dv + u^3v^4 (5w^4)dw$

...which then integrates back to

$u^3v^4w^5 = 3u^3v^4w^5$

I am sure I am misunderstanding this notation somehow because I can clearly see for example in the single variable case, that...

$d(cos(\theta)) = -sin(\theta)d\theta$

because I can also write it as:

$\frac{d}{d\theta}\left(\cos(\theta)\right) = - \sin(\theta)$

Can anyone help me understand what I am doing wrong here? This notation seems to be appearing more frequently in my studies and it has seemingly come out of nowhere. Thank you in advance!

BoozyBear
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1 Answers1

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Suppose $u$ and $v$ are functions of $t$. The notation then comes from $$\frac{d}{dt}((u(t))^3 v(t)) = 3(u(t))^2 v(t) \frac{du}{dt}(t) + (u(t))^3 \frac{dv}{dt}(t).$$

If you integrate with respect to $t$, you are basically re-deriving integration by parts.

angryavian
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  • If the variables are not functions of t though, how would I fix the extra factor of 'n' hanging out in front after integrating? When I encountered the notation the variables inside the d(), that is

    $d(u^3v)$ were not functions of any other variable. So

    $\frac{du}{dt} = 0$,

    but

    $\frac{d}{du}(u^3) = 3u^2$

    I am really just confused how the notation is working and why I can not recover the function upon integrating.

     – BoozyBear Feb 03 '22 at 20:34
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    @BoozyBear What I'm trying to say is that your interpretation of the expression $d(u^3 v) = 3u^2 v , du + u^3 , dv$ is not correct; I think it is shorthand for what I've written in my answer, and that $u$ and $v$ are actually functions. – angryavian Feb 05 '22 at 20:25