Chain rule / Taylor expansion / functional derivative

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SUMMARY

The discussion centers on proving the inequality ##\rho(p',s)>\rho(p',s') => (\frac{\partial\rho}{\partial s})_p\frac{ds}{dz}<0##, where ##p=p(z)##, ##p'=p(z+dz)##, and ##s'=s(z+dz)##. Participants suggest using Taylor expansions of the function ##\rho(p',s')## in the variable ##s## to approach the problem. A key insight is that the terms in the inequality differ by a differential amount, and the inclusion of "dz" in the product of the derivatives is necessary for the proof. Consensus among readers is sought regarding the necessity of "dz" being non-positive.

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Homework Statement



To show that ##\rho(p',s)>\rho(p',s') => (\frac{\partial\rho}{\partial s})_p\frac{ds}{dz}<0##
where ##p=p(z)##, ##p'=p(z+dz)##, ##s'=s(z+dz)##, ##s=s(z)##

Homework Equations



I have no idea how to approach this. I'm thinking functional derivatives, taylor expansions, but if someone could please give me a clue where to start.

The Attempt at a Solution


as above.

many thanks in advance.
 
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Suggest a Taylor expansion of ## \rho(p',s') ## in just the s variable about s. One comment is since s and s' differ by a differential amount, the the terms of the first inequality differ by only a differential amount. I think the inequality you are trying to prove actually needs a "dz" tacked on to (multiplying) the product of the derivatives. I would be interested in seeing if other readers concur. "dz" does not need to be positive.
 

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