Finding an equation of Partial Derivatives

[sardonic]

Homework Statement

If f(x,y,z) = 0, then you can think of z as a function of x and y, or z(x,y). y can also be thought of as a function of x and z, or y(z,x)
Therefore:

$$dz= \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y} dy$$
and
$$dy= \frac{\partial y}{\partial x}dx + \frac{\partial y}{\partial z} dz$$

Show that
$$1= \frac{\partial z}{\partial y} \frac{\partial y}{\partial z}$$
and then
$$-1= \frac{\partial x}{\partial z} \frac{\partial y}{\partial x}\frac{\partial z}{\partial y}$$

Homework Equations

The Attempt at a Solution

Substituting $dy$ into the $dz$ equation you get
$$dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y} \frac{\partial y}{\partial x}dx + \frac{\partial y}{\partial z}\frac{\partial z}{\partial y}dz$$

This can be rearranged to show
$$dz (1-\frac{\partial z}{\partial y}\frac{\partial y}{\partial z}) = dx(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x})$$
and then
$$(1-\frac{\partial z}{\partial y}\frac{\partial y}{\partial z}) = \frac{dx}{dz}(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x})$$

In order to show that $\frac{\partial z}{\partial y}\frac{\partial y}{\partial z} = 1$, I only need to show that $(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x})=0$, but I'm not sure how to do that. As for the second equation, I'm not sure where to get a $\frac{\partial x}{\partial z}$ into the equation in the first place.

 

[hedipaldi]

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[sardonic]

Thanks for the reply!

What do $f_y$ and $f_z$ refer to?

 

[hedipaldi]

These are the partial derivatives of f with respect to y and z,respectivly.

 

[sardonic]

Oh okay, thing is, that's the way the instructor did it, while we were asked to derive expressions for the total differential for dz and dy, then substitute the latter into the former, sorry if I wasn't clear. Thanks though!

 

[hedipaldi]

Attached