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jesuslovesu
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[SOLVED] Partial Derivatives
Whoops, never mind my calculus book explained it.
F(x,y,z) = 0
[tex](\frac{\partial x}{\partial y})\right)_{z} (\frac{\partial y}{\partial x})\right)_{z} [/tex] = 0
Show
[tex](\frac{\partial x}{\partial y})\right)_{z} (\frac{\partial y}{\partial z})\right)_{x} (\frac{\partial z}{\partial x})\right)_{y}[/tex] = -1
Well I drew out a diagram
F
dF/dx dF/dy dF/dz
dx/dy dx/dz dy/dx dy/dz dz/dx dz/dy
So I would assume that the reason the second expression is = -1 because of the chain rule, however, I really don't see why it would be -1...
If it were dx/dy dy/dx dz/dx would it be 1? Just because the middle term is dy/dz the sign will change?
Whoops, never mind my calculus book explained it.
Homework Statement
F(x,y,z) = 0
[tex](\frac{\partial x}{\partial y})\right)_{z} (\frac{\partial y}{\partial x})\right)_{z} [/tex] = 0
Show
[tex](\frac{\partial x}{\partial y})\right)_{z} (\frac{\partial y}{\partial z})\right)_{x} (\frac{\partial z}{\partial x})\right)_{y}[/tex] = -1
The Attempt at a Solution
Well I drew out a diagram
F
dF/dx dF/dy dF/dz
dx/dy dx/dz dy/dx dy/dz dz/dx dz/dy
So I would assume that the reason the second expression is = -1 because of the chain rule, however, I really don't see why it would be -1...
If it were dx/dy dy/dx dz/dx would it be 1? Just because the middle term is dy/dz the sign will change?
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