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In a part of an expansion for a formula for elasticity, in the expression dB/dp, the B gets replaced by pX (Because B = pX). The expansion then is said to equal (pdX + Xdp)/dp. On what definition does this equality rest? A link to a wiki page will suffice, I just don't seem to find the exact thing I'm looking for. I was thinking total derivative, but it's not exactly the same, is it?

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    It's the differential on smooth functions or exterior derivative, see https://en.wikipedia.org/wiki/Exterior_derivative. – 2017-01-08
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    Thx, but, hmm, I'm not sure how to link it to my problem. Any specific formula I should have in mind? – 2017-01-08
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    Does that help? – 2017-01-09

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You are probably not familiar with the language of differential forms, but to see how this applies to what you have, you just need to know that: (smooth) functions are 0-forms, and the wedge product ($\wedge$) of 0-forms is just the ordinary multiplication of functions (defined pointwise). Therefore the third axiom says that \begin{equation} \begin{split} \mathrm{d}B&=\mathrm{d}(pX) \\ &=\mathrm{d}p\wedge X+(-1)^0p\wedge\mathrm{d}X \\ &=X\mathrm{d}P+p\mathrm{d}X, \end{split} \end{equation} where the last line should be considered as mere notation, i.e. for $\mathrm{d}p\wedge X$ we simply write $X\mathrm{d}P$ (since the space of differential forms is a module over the smooth functions). Note that this "commutativity" ($\mathrm{d}p\wedge X=X\mathrm{d}p$) does not apply to the wedge product of general $p$- and $q$-forms.

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    Basically this is just the chain rule for differential calculus when u consider p and X as separate functions, no? – 2017-01-09
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    No. The chain rule would involve have to involve something like $dp(dX)$, which is not even defined here. This is some sort of product rule/Leibniz identity, analogous to the product rule for ordinary derivatives: $(fg)'=f'g+fg'$. Note the difference with the chain rule, which states that $D_p(f\circ g)=D_{g(p)}(f)\circ D_p(g)$, or in more calculus like notation, $(f\circ g)'(x)=f'(g(x))g'(x)$. In general a linear map F on an algebra A which satisfies $d(f\cdot g)=df\cdot g+f\cdot dg$ is called a derivation, and the exterior derivative is due to the factor $(-1)^{p}$ an antiderivation. – 2017-01-10