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I was working on something and I arrived at some inequality of the kind

$$ \frac {\partial Y}{\partial X}\bigg(\frac{A}{B}\bigg) >\frac {\partial Z}{\partial X} \tag{1}$$

Could this be decomposed as follows?

$$ \bigg(\frac{A}{B}\bigg) >\frac {\partial Z}{\partial X}\frac {\partial X}{\partial Y} \tag{2}$$

Because I know that $\frac {\partial X}{\partial Y} = A$. So that equation 2) becomes

$$ \frac {\partial Z}{\partial X} <\frac{1}{B} \tag{3}$$

trying
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pafnuti
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  • What is it the role of $(2)$? From $(1)$ you can go directly to $(3)$ because you know that $\partial X/\partial Y=A$ – trying Aug 07 '17 at 22:23
  • So the answer to my question is yes? – pafnuti Aug 07 '17 at 22:29
  • I thought you are evaluating it at $(A/B)$ – IAmNoOne Aug 07 '17 at 22:34
  • @Nameless, what do you mean? – pafnuti Aug 07 '17 at 22:40
  • @tbes you are asking two different things. From $(1)$ and the fact that $\partial X/\partial Y=A$ you can get $(3)$. But from $(1)$ and nothing else it does not follow $(2)$: when $\partial Y/\partial X>0$, $(2)$ is ok, when $\partial Y/\partial X<0$, in $(2)$ you must change $>$ with $<$, when $\partial Y/\partial X=0$ you can take the reciprocal. – trying Aug 07 '17 at 22:45
  • @trying why does (2) not follow from (1)? That's kind of my question – pafnuti Aug 07 '17 at 22:49
  • @tbes Probably you wnat to read this: https://math.stackexchange.com/q/185004 – trying Aug 08 '17 at 08:16

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