Implicit Differentiation of Multivariable Functions

[KingBigness]

Homework Statement

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

\frac{∂z}{∂x} \frac{∂x}{∂y} \frac{∂y}{∂z} = -1

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

 

[Dick]

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.

 

[KingBigness]

Thanks for the reply.
That worked...but how can I solve for all cases?

 

[KingBigness]

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?

 

[Dick]

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?

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.

 

[KingBigness]

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

 

[Dick]

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

That's ok. This one is a little subtle. I had to ponder it a while.

 

[KingBigness]

haha I'm glad you had trouble with it to, makes me feel less dumb =P