Partial Derivatives Instead of Implicit

  • Thread starter Jameson
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  • #1
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Main Question or Discussion Point

On MathWorld's site, they said that

[tex](\frac{\partial{y}}{\partial{x}}){_f} = -\frac{(\frac{\partial{f}}{\partial{x}})_{y}}{(\frac{\partial{f}}{\partial{y}})_{x}}[/tex]

So can this method be used instead of implicit differentiation? Will I get the same result? This seems kind of like a parametrics process if this is true.

Thanks,
Jameson
 
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Answers and Replies

  • #2
dextercioby
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Yep,that's the formula from the statement of the theorem of implicit functions.

Daniel.
 
  • #3
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Is that for f(x,y)? If so isnt the right hand side = -1?

y(f,x)?
 
  • #4
yes, and there is nothing mysterious about this equation, btw:

for [itex]f(x,y) = 0[/itex]

[tex] df = \frac{\partial{f}}{\partial{x}}dx + \frac{\partial{f}}{\partial{y}}dy = 0[/tex]

[tex]\frac{\partial{f}}{\partial{x}}dx = -\frac{\partial{f}}{\partial{y}}dy[/tex]

[tex]\frac{dy}{dx} = -\frac{\frac{\partial{f}}{\partial{x}}}{\frac{\partial{f}}{\partial{y}}}[/tex]
 
  • #5
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Quetzalcoatl9 I'm not sure if you addressed my question. Isnt there a theorem that says the double partial of a function with respect to alternating variables are equal?

[tex] f(x,y), f_{xy} = f_{yx} [/tex] Right? Isn't that whats going on here?
 
  • #6
whozum said:
Quetzalcoatl9 I'm not sure if you addressed my question. Isnt there a theorem that says the double partial of a function with respect to alternating variables are equal?

[tex] f(x,y), f_{xy} = f_{yx} [/tex] Right? Isn't that whats going on here?
I don't understand what you are asking, whozum...do you mean to ask if partial derivatives commute?
 
  • #7
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I dont know what you mean by commute, but please verify that

[tex]For \ f(x,y), \ f_{xy} = f_{yx} [/tex]

If so, then please explain why the right hand side of the original OP's equation isnt just -1
 
  • #8
Gza
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If so, then please explain why the right hand side of the original OP's equation isnt just -1
where'd you get the idea the right side was -1?
 
  • #9
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[tex] \left(\frac{\delta f}{\delta x}\right)_y = f_{xy} [/tex]

[tex] \left(\frac{\delta f}{\delta y}\right)_x = f_{yx} [/tex]

[tex] f_{yx} = f_{xy} [/tex]

edit: I'm obviously missing something, I just want someone to point out what it is.
 
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  • #10
dextercioby
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Yes,you are.Missing something,that is.We use this notation

[tex] \left (\frac{\partial f}{\partial x}\right)_{y}\equiv \frac{\partial f(x,y)}{\partial x} [/tex]

,that is we explain which variables are kept constant during the partial differentiaition,viz. "y" in this case.

This notation,or convention,if u prefer,is very common in thermodynamics.

Daniel.
 
  • #11
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So that hanging y on the LHS is just to indicate that its constant, then. I had that confused as a derivative.
 
  • #12
arildno
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The notation is excellent in a field like thermo-dynamics, where it is convenient to switch between which quantities are to be regarded as independent quantities and which are to be regarded as dependent quantities.

It is rather redundant in a field where there exisst a "natural" choice of the independent variables.
 
  • #13
dextercioby
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To give u an idea

[tex] \frac{\partial F}{\partial V} [/tex]

,where F is the free energy/Helmholtz potential & V is the volume,doesn't mean anything in thermodynamics.

Daniel.
 
  • #14
arildno
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Are you talking to me??
 
  • #15
dextercioby
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Of course,not.I gave that example to Whozum...

Hmmmmm...:tongue2:

Daniel.
 
  • #16
arildno said:
Are you talking to me??
...<in bronx-accented Al Pacino voice> :rofl:
 
  • #17
arildno
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I haven't bothered to get a driving license yet.
Perhaps I should..:devil:
 

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