# Total differential

1. Apr 17, 2014

### chemphys1

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

If z(x,y) = f(x/y)
show that

x(∂z/∂x)y + y(∂z/∂y)x = 0

2. Relevant equations

so I understand z(x,y)
means I can write
dz = (∂z/∂x)ydx + (∂z/∂y)x dy

I do not understand the = f(x/y) bit though?
does that mean this?
df= (∂f/∂x)y dx+ (∂f/∂y)x dy
and (∂f/∂x)y = -y/x2 (∂f/∂y)x = 1/x

although that seems wrong can't manipulate to get the answer

any help on the method or explaining f(x/y) equalling z(x,y) would be appreciated. maths is not my strongest so if possible go as basic as it comes

2. Apr 17, 2014

### LCKurtz

That isn't true. Try it for z = f(x/y) = x/y. Generally speaking, if everything is continuous, you would have $z_{xy} = z_{yx}$ and you wouldn't expect to multiply one by x and the other by y and have them be equal with opposite signs. Did you copy the problem correctly, parentheses and all?

3. Apr 17, 2014

### chemphys1

complete question is attached, but the information in the original post is correctly copied as far as I can see

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4. Apr 17, 2014

### LCKurtz

Like I said, it is false without more context. Check my example yourself.

5. Apr 17, 2014

### chemphys1

Sorry, I really do not follow
I find it hard to understand mathematical notation, so re: z = f(x/y) = x/y
I can't see how to check the example

6. Apr 17, 2014

### CAF123

You are familiar with the notation f(x) as describing a function of x. Similarly, f(x/y) just means you have some function in the variable x/y. You can instead consider u=x/y and then you are back to the more familiar f(u).

But with LCKurtz's example of f(x/y)=x/y, I get the equation to be satisfied. I also get it to be satisfied in general. I think LCKurtz simply misread the notation, that's all.

7. Apr 17, 2014

### LCKurtz

Are you saying that $\left(\frac {\partial z}{\partial x}\right)_y$ means something other than $\frac \partial {\partial y}\left (\frac{\partial z}{\partial x}\right )$ or, as I wrote $z_{xy}$?

8. Apr 18, 2014

### CAF123

Yes, I took $\left(\frac{\partial z}{\partial x}\right)_y$ to mean, say, differentiate z wrt x, keeping y held fixed. Similarly for the other case. I actually thought that was a standard notation, although I have seen cases where they simply suppress the variable being held constant because in a sense it is obvious from the problem.

9. Apr 18, 2014

### LCKurtz

Well, that's a new one on me. In over 40 years of teaching calculus using many different texts, I never encountered that notation.