# Question about partial derivatives with three unknown

1. Apr 9, 2013

### Outrageous

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

If z=f(x,y) with u= x^2 -y^2 and v=xy , find the expression for (∂x/∂u).
the (∂x/∂u) will be used to calsulate ∂z/∂u.
my question is how to find (∂x/∂u).
I don't know what to keep constant. Maybe the question has some problem.
Can you please guide me .

2. Relevant equations

3. The attempt at a solution
I use u= x^2 -y^2, keep y constant , then get du/dx=2x , then 1/(du/dx)
or keep y constant, then x=√(u+y^2), get dx/du= 1/2x?

Or I should not keep anything constant? or the question is wrong?

really thank

2. Apr 9, 2013

### Simon Bridge

$u=x^2-y^2 \Rightarrow x^2 = u+y^2 \\ \Rightarrow \frac{\partial}{\partial u}(x^2 = u+y^2)=\cdots$ ... that help?

3. Apr 9, 2013

### Outrageous

keep y constant?? if so the answer is 1/2x also.
if not keeping y constant ......
I really dont know how to get this (∂x/∂u)=(x/2)/(x^2+y^2).

4. Apr 9, 2013

### Simon Bridge

$$\frac{\partial}{\partial u}(x^2 = u+y^2)= \frac{\partial}{\partial u}x^2= 1 + \frac{\partial}{\partial u}y^2$$ ... should you keep y constant?
I think that's what you need to decide - is it $x(y,u)=u+y(x,u)$?
Definition of the partial derivative?

5. Apr 9, 2013

### Outrageous

Quote from wolfram math world:
Partial derivatives are defined as derivatives of a function of multiple variables when all but the variable of interest are held fixed during the differentiation.
∂y/∂u , y=f(u,x), since u varies here, so y will vary. Can you please guide how should I calculate ∂y/∂u?
Thank you.

6. Apr 9, 2013

### Simon Bridge

$$u=x^2-y^2$$ ... and you want to show that $$\frac{\partial x}{\partial u}=\frac{x}{2(x^2+y^2)}$$... is that correct?

7. Apr 9, 2013

8. Apr 9, 2013

### Simon Bridge

Normally when you have y=f(x) then dx/dy = 1/(dy/dx)
Does this work for partials?

$$\frac{\partial}{\partial x}u(x,y) \Rightarrow \left . \frac{\partial u}{\partial x} \right |_y$$

If you consider x to be a function of u and y, then the partial - holding y constant - cannot come out the way you want.

9. Apr 9, 2013

### Outrageous

This only applies to partial , is not? I learned that from thermo.

If the question says find (∂x/∂u) keeping y constant , then we can use du/dx = 1/(dx/du) to get the answer without care about y=f(u,x) because u depends on x and y , but x and y are independent. the question no need to give us a equation of y in term of x to calculate dy/dx.
Is that correct?

10. Apr 9, 2013

### Outrageous

Here is the therma's proof

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11. Apr 9, 2013

### Simon Bridge

That's the idea ... $$\left . \frac{\partial x}{\partial u}\right |_y = \left ( \left . \frac{\partial u}{\partial x}\right |_y \right )^{-1}$$ ... works because the same variable is being held constant.

However: $$u=x^2-y^2 \Rightarrow \frac{\partial u}{\partial x} = 2x$$... like you already noticed but this does not give you what you need.

It follows that there is something else going on.

12. Apr 9, 2013

### Outrageous

Is the question not complete, it should be (∂x/∂u) keeping v constant.
Thank for wasting so much time on me. Really sorry.

13. Apr 9, 2013

### Simon Bridge

You said, post #1, you didn't know what to keep constant.
... nothing there about keeping v constant - I wondered why it was relevant.

Ideally you want to express x as a function of u and v then.
Maybe express u as a function of x and v and evaluate du/dx|v and invert it.

14. Apr 10, 2013

### Outrageous

Yup ,can get answer from that. Thank you.

15. Apr 10, 2013

### Simon Bridge

Well done - sometimes you don't see what's there until you have to talk about it to someone else :)