This might be a stupid question, but I can't seem to understand what's going on with a Taylor expansion I did for a physical problem. I expanded around the small parameter $\Delta x$ and got something like: $$ f(x)=c\frac{d}{dx}(g(x)h(x))\Delta x. $$ Now, $\Delta x$ is of course very small, so I can put it inside the derivative. However, in principle, it is also a function of $x$: $$ \Delta x(x)=tp(x). $$ Here, the constant $t$ is very small, but $p(x)$ is not. So the two functions: $$ f(x)=ct\frac{d}{dx}(g(x)h(x))p(x), $$ and $$ f(x)=c\frac{d}{dx}(g(x)h(x)\Delta x(x))=ct\frac{d}{dx}(g(x)h(x)p(x)), $$ behave very differently. In particular, I am interesting in f(x) in the form of: $$ f(x)=cg(x)h(x)\left(\frac{d\ln g}{dx}+\frac{d\ln g}{dx}\right), $$ and the behaviour of the quantity in the parenthesis. So now, if I put $\Delta x$ inside the derivative, I have: $$ f(x)=cg(x)h(x)\Delta x(x)\left(\frac{d\ln g}{dx}+\frac{d\ln h}{dx}+\frac{d\ln\Delta x}{dx}\right)=ctg(x)h(x)p(x)\left(\frac{d\ln g}{dx}+\frac{d\ln h}{dx}+\frac{d\ln p}{dx}\right). $$ The additional derivative inside the parenthesis has a huge impact on the behaviour of $f(x)$, as it can, for example, change its sign. But the $p(x)$ outside of the derivative won't.
What am I missing here? $\Delta x(x)$ is small, so its derivative is also small. It is small because of the $t$ constant, which I suppose is why the logarithmic derivative isn't small. But then, the two forms of $f(x)$ (derivative expanded or not) should be equivalent. What boggles me is the sign change of the whole parenthesis that can occur because of the additional term. The thing is, from a physical point of view, that $\Delta x(x)$ term should be inside, otherwise I have a flux conservation problem.