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Off we go. So let $b:X\rightarrow Y$ be a function from $X$ to $Y$ endowed with as much structure as it needs to make sense of the question :) and $a:Y\rightarrow \mathbb R$ a function into the reals. Then $a\circ b$ is a function from $X$ into the reals. It is an element of the algebra $\mathcal A$ of functions on $X$. Let $D$ be a derivation on $\mathcal A$, i.e. it is a linear map satisfying the Leibniz rule. But how does it act on $a\circ b$? Or is there simply no general rule? The naive guess for the chain rule would be:

\begin{equation} D(a\circ b)=(D'a\circ b)\cdot D''(b) \end{equation} Where $D'a$ should be some sort of derivation on the algebra $A'$ of functions from $Y$ into $\mathbb R$. More worrying even is $D''(b)$ what is that supposed to be? Do those functions even have an algebra structure?

So unless we know what the derivations are on $ A'$ and what $D''$ is, we can't say anything about the way $D$ acts on $(a\circ b)$?

Of course i have diff geo in mind, where $X$ and $Y$ are manifolds. In my case $Y=TM$ is the tangent bundle for some mfld $M$ and $X$ is actually the mfld $\mathbb R$ in my case. So very simple.

Thanks for reading. and looking forward to the moment where i notice the simplicity of the answer :)

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  • $\begingroup$ Leibniz with a B. $\endgroup$ Mar 7, 2015 at 4:04
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    $\begingroup$ Have you considered a special case, like $a(x) = x^n$? Leibnitz' rule should give something. $\endgroup$ Mar 13, 2015 at 23:40

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