# Homework Help: Finding an equation of Partial Derivatives

1. Oct 27, 2012

### sardonic

1. The problem statement, all variables and given/known data

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
$$-1= \frac{\partial x}{\partial z} \frac{\partial y}{\partial x}\frac{\partial z}{\partial y}$$
2. Relevant equations

3. 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.

2. Oct 27, 2012

### hedipaldi

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3. Oct 27, 2012

### sardonic

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

4. Oct 27, 2012

### hedipaldi

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

5. Oct 27, 2012

### 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!

6. Oct 27, 2012

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