# Partial differentiation with 3 variables

1. Nov 9, 2013

### _Stew_

Given a function: z(x,y) = 2x +2y^2
Determine ∂x/∂y [the partial differentiation of x with respect to y],

Method 1:

x = (z/2) - y^2
∂x/∂y = -2y

Method 2:

∂z/∂x = 2
∂z/∂y = 4y
∂x/∂y = ∂x/∂z X ∂z/∂y = (1/2) X 4y = 2y

One or both of these is wrong. Can someone point out where I went wrong? I started to wonder why this was happening as I was looking over some fluid mechanics problems (above problem is unrelated to my original problem but if I understand the above problem it will help)

Thanks

2. Nov 10, 2013

### DeIdeal

You've got your cyclic rule wrong, it's (-1)× what you have, ie

$$\frac{\partial{x}}{\partial{y}}\frac{\partial{y}}{\partial{z}}\frac{ \partial {z}}{\partial{x}}=-1\implies\frac{\partial{x}}{\partial{y}}=-\frac{\partial{x}}{\partial{z}}\frac{\partial{z}}{\partial{y}}$$

3. Nov 10, 2013

### _Stew_

I hadn't heard of that rule before. Thanks for clearing this up !!!

4. Nov 10, 2013

### DeIdeal

Oh, it looked like you were trying to apply that. In that case, you should be *much more careful* about partial derivatives, you can rarely use chain rule with partial derivatives like that. If you really want to, use the multivariable chain rule, but *don't* just say that "(∂x/∂y)= (∂x/∂z)(∂z/∂y)". It will get you into trouble very quickly.

5. Nov 13, 2013

### Staff: Mentor

Here's a way of getting at it that I think you will like:
$$dz=\left(\frac{\partial z}{\partial x}\right)dx+\left(\frac{\partial z}{\partial y}\right)dy$$
If you are taking the partial of x with respect to y at constant z, then dz must be zero. So,
$$0=\left(\frac{\partial z}{\partial x}\right)dx+\left(\frac{\partial z}{\partial y}\right)dy$$
So, $$\left(\frac{\partial x}{\partial y}\right)=-\left(\frac{\partial z}{\partial y}\right)/\left(\frac{\partial z}{\partial x}\right)$$

6. Nov 14, 2013

### _Stew_

I actually don't understand your first equation.

dz= (∂z/∂x)dx + (∂z/∂y)dy

I tried to look up the proof on Wikipedia but it also doesn't explain it. There must be something intuitive about it but I don't see how you can mix dx,dy,dz with ∂x,∂y,∂z. My fluid mechanics lecturer used the same equation but with Pressure as a function of x,y,z and didn't explain why:

dP= (∂P/∂x)dx + (∂P/∂y)dy + (∂P/∂z)dz

7. Nov 14, 2013

### SteamKing

Staff Emeritus
You must have been out of class in Calc I when they went over the chain rule. Here is a refresher:

http://tutorial.math.lamar.edu/Classes/CalcIII/ChainRule.aspx

http://tutorial.math.lamar.edu/Classes/CalcIII/Differentials.aspx

8. Nov 14, 2013

### _Stew_

I looked over the two links you gave me, thanks. They don't give a proof but bring up some good ideas. I noticed it is similar to implicit differentiation.

9. Nov 14, 2013

### SteamKing

Staff Emeritus
What sort of proof do you require?