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In the lecture notes (thermodynamics) the following mathematical identity is often used:

$$ \left(\frac{\partial A}{\partial X}\right)_Z = \left(\frac{\partial A}{\partial X}\right)_Y + \left(\frac{\partial A}{\partial Y}\right)_X\left(\frac{\partial Y}{\partial X}\right)_Z = \left(\frac{\partial A}{\partial X}\right)_Y - \left(\frac{\partial A}{\partial Z}\right)_X\left(\frac{\partial Z}{\partial X}\right)_Y $$

As I understand, this is a standard notation in thermodynamics, see e.g. this or this questions. However, I do not understand the mathematical meaning of the above identity. I am trying to figure out the procedure of how to obtain these derivatives but I am stuck. Could someone please clarify this?

Quillo
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Lambda
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    hello! @Lambda can you please try to edit it in form of latex or mathml? it would help'; and also is this homework? I think maybe this would suit the math stack exchange instead (?) – William Martens Nov 17 '22 at 10:02
  • thanks; that made it better; will check it out - by the way is it a homework or? – William Martens Nov 17 '22 at 10:30
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    Yes, I have to prove the chain rule for the Jacobi determinants using this formula, but I don't quite understand how exactly to form these terms in my task – Lambda Nov 17 '22 at 10:43
  • I have to prove the chain rule for the Jacobi determinants using this formula – How did you get that idea? The respective equation is not even mentioned near the task that asks you to prove the chain rule for Jacobi determinants. – Wrzlprmft Nov 17 '22 at 12:32
  • As for your question: Did you understand the thermodynamic notation for partial derivatives? If yes, the mathematical rule and its usefulness should be either clear or you should be able to formulate some more specific doubt. If not, start there – you really need to understand the notation first. Also, what do you mean by form these derivatives? – Wrzlprmft Nov 17 '22 at 12:38
  • @Lambda, I edited your question to try to make it more clear (not sure I actually improved it, for me it was already quite clear). Anyway, the answers to this question may help you: https://physics.stackexchange.com/q/274679/226902 . Alternatively: https://physics.stackexchange.com/a/148735/226902 or https://physics.stackexchange.com/q/623344/226902 – Quillo Nov 17 '22 at 14:16

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