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Partial Derivatives Proof

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that the equation F(x,y,z) = 0 implicitly defines each of the three variables x, y and z as functions of the other two: z = f(x,y), y = g(x,z), z = h(y,z). If F is differentiable and Fx, Fy and Fz are all nonzero, show that

    [itex]\frac{∂z}{∂x}[/itex] [itex]\frac{∂x}{∂y}[/itex] [itex]\frac{∂y}{∂z}[/itex] = -1


    3. The attempt at a solution

    I have been scribbling away at this for a little while now and can't see it.
    I assume that I need to prove that each partial derivative = -1, thus -1*-1*-1=-1

    Don't want you guys to simply give me the answer but any tips on where to start would be greatly appreciated =D
     
    Last edited: Sep 13, 2011
  2. jcsd
  3. Sep 13, 2011 #2

    Dick

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    Try a simple example. Say, x+2y+3z=0. You can then solve for f, g and h explicitly and show your product is (-1). And you'll also see that not all of the partials need to be (-1). That should be a good hint to start.
     
  4. Sep 13, 2011 #3
    Thanks for the reply.
    That worked...but how can I solve for all cases?
     
  5. Sep 14, 2011 #4
    I tried ax+by+cz=0 and solved it and ended up with -1....but this still doesn't solve it for all cases does it?
     
  6. Sep 14, 2011 #5

    Dick

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    It sort of does. Since F is differentiable and invertible it can be approximated by a linear function well enough. If you want to be more formal then use the chain rule (read all of the following d's as partial derivatives). E.g. let's try finding dx/dy in terms of F. When you are taking a partial derivative you need to specify what variable is being held constant. In this case it must be z. Now the chain rule for partial derivatives tells you (dF/dx)*(dx/dy)+(dF/dy)*(dy/dy)+(dF/dz)*(dz/dy)=0, right? If you hold z constant, can you solve that for dx/dy in terms of partial derivatives of F? Repeat that for dz/dx and dy/dz and multiply the results together.
     
    Last edited: Sep 14, 2011
  7. Sep 14, 2011 #6
    Thank you so much for that.
    I had tried using the chain rule, but I mixed it up =( I only started teaching myself partial derivatives 2 days ago so I'll give my self a break for stuffing up =P

    Thank you again, I'll give that a shot and see how it goes
     
  8. Sep 14, 2011 #7

    Dick

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    That's ok. This one is a little subtle. I had to ponder it a while.
     
  9. Sep 14, 2011 #8
    haha I'm glad you had trouble with it to, makes me feel less dumb =P
     
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