differential of a product?

Question

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?

Answer 1

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.